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W5XhsmDlO7rM78PkXxjgy2xukfal3yve | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the variance of the following frequency
distribution :<br/><br/>
Class : 10–20 20–30 30–40<br/><br/>
Frequency : 2 x 2<br/><br/>
is 50, then x is equal to____
| [] | null | 4 | x<sub>i</sub> = midpoint of class interval
<br><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266140/exam_images/bkvpquedltlzbkr2djxo.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265188/exam_images/m72otaydfa13nmamaqlp.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264660/exam_images/rmwouh2oixkjbf9ac6cj.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264798/exam_images/cehho8pjccdpteeui8ch.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 4th September Evening Slot Mathematics - Statistics Question 75 English Explanation"></picture>
<br><br>Variance($$\sigma^{2})$$$$ =\frac{\sum f_{i}\left( x_{i}-\bar{x} \right)^{2} }{\sum f_{i}} $$
<br><br>Also, $$\bar{x} =\frac{\sum f_{i}x_{i}}{\sum f_{i}} $$
<br><br>= $$\frac{30+25x+70}{2+2+x} $$ = 25
<br><br>Given, Variance = 50
<br><br>$$ \therefore $$ 50 = $$\frac{200+0+200}{2+2+x} $$
<br><br>$$ \Rightarrow $$ x = 4 | integer | jee-main-2020-online-4th-september-evening-slot |
9eXN0XuwQ7MGk8tOJAjgy2xukf8zg1o2 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and variance of 8 observations are 10 and 13.5, respectively. If 6 of these observations
are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "9"}] | ["C"] | null | Let the two remaining observations be x and y.<br><br>$$ \because $$ $$\bar x = 10 = {{5 + 7 + 10 + 12 + 14 + 15 + x + y} \over 8}$$<br><br>$$ \Rightarrow x + y = 17$$ ....(1)<br><br>$$ \because $$ $${\mathop {\rm var}} (x) = 13.5 = {{25 + 49 + 100 + 144 + 196 + 225 + {x^2} + {y^2}} \over 8} - {(10)^2}$$<br><br>$$ \Rightarrow {x^2} + {y^2} = 169$$ ....(2)<br><br>From (1) and (2)<br><br>(x, y) = (12, 5) or (5, 12)<br><br>So $$\left| {x - y} \right| = 7$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
zf49X1ZTnEJU37kGDbjgy2xukf461jt2 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let x<sub>i</sub>
(1 $$ \le $$ i $$ \le $$ 10) be ten observations of a
random variable X. If <br/>$$\sum\limits_{i = 1}^{10} {\left( {{x_i} - p} \right)} = 3$$ and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - p} \right)}^2}} = 9$$ <br/>where 0 $$ \ne $$ p $$ \in $$ R, then the
standard deviation of these observations is : | [{"identifier": "A", "content": "$${7 \\over {10}}$$"}, {"identifier": "B", "content": "$${9 \\over {10}}$$"}, {"identifier": "C", "content": "$${4 \\over 5}$$"}, {"identifier": "D", "content": "$$\\sqrt {{3 \\over 5}} $$"}] | ["B"] | null | Standard deviation = $$\sqrt {Variance} $$<br><br>$$ = \sqrt {{{\sum {x_1^2} } \over n} - {{(\overline x )}^2}} $$<br><br>$$ = \sqrt {{{\sum\limits_{i = 1}^{10} {{{({x_i} - p)}^2}} } \over {10}} - {{\left( {{{\sum\limits_{i = 1}^{10} {({x_i} - p)} } \over {10}}} \right)}^2}} $$
<br><br>[ Standard deviation
is free from shifting
of origin.]
<br><br>$$ = \sqrt {{9 \over {10}} - {{\left( {{3 \over {10}}} \right)}^2}} $$<br><br>$$ = \sqrt {{9 \over {10}} - {9 \over {100}}} $$<br><br>$$ = \sqrt {{{90 - 9} \over {100}}} $$<br><br>$$ = \sqrt {{{81} \over {100}}} $$<br><br>$$ = {9 \over {10}}$$ | mcq | jee-main-2020-online-3rd-september-evening-slot |
HOGv0aaEIEk875NhPrjgy2xukezfbwzn | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the variance of the terms in an increasing A.P.,<br/> b<sub>1</sub>
, b<sub>2</sub>
, b<sub>3</sub>
,....,b<sub>11</sub> is 90, then the common
difference of this A.P. is_______. | [] | null | 3 | Let the common difference = d<br><br>
and $${b_1} = a$$<br>
$${b_2} = a + d$$<br>
$${b_3} = a + 2d$$ <br>
... $${b_{11}} = a + 10d$$<br><br>
Variance = $${{\sum {a_i^2} } \over {11}} - {\left( {{{\sum {{a_i}} } \over {11}}} \right)^2} = 90$$<br><br>
$$ \Rightarrow {{{a^2} + {{\left( {a + d} \right)}^2} + ... + {{\left( {a + 10d} \right)}^2}} \over {11}} - {\left( {{{a + \left( {a + d} \right) + ... + \left( {a + 10d} \right)} \over {11}}} \right)^2} = 90$$<br><br>
$$ \Rightarrow 11\left[ {11{a^2} + 385{d^2} + 110ad} \right] - {\left[ {11a + 55d} \right]^2} = 10890$$<br><br>
$$ \Rightarrow 1210{d^2} = 10890$$<br><br>
$$ \Rightarrow {d^2} = 9$$<br><br>
$$ \Rightarrow d = \pm 3$$<br><br>
As A.P is increasing so d should be positive<br><br>
$$ \therefore $$ d = 3 | integer | jee-main-2020-online-2nd-september-evening-slot |
hKNc0Jm125H51MVWYj7k9k2k5irxzoy | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let the observations x<sub>i</sub> (1 $$ \le $$ i $$ \le $$ 10) satisfy the
<br/>equations, $$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $$ = 10 and $$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40.
<br/>If $$\mu $$ and $$\lambda $$ are the mean and the variance of the
<br/>observations, x<sub>1</sub> – 3, x<sub>2</sub> – 3, ...., x<sub>10</sub> – 3, then
<br/>the ordered pair ($$\mu $$, $$\lambda $$) is equal to : | [{"identifier": "A", "content": "(6, 6)"}, {"identifier": "B", "content": "(3, 3)"}, {"identifier": "C", "content": "(3, 6)"}, {"identifier": "D", "content": "(6, 3)"}] | ["B"] | null | $$\sum\limits_{i = 1}^{10} {\left( {{x_1} - 5} \right)} $$ = 10
<br><br>$$ \Rightarrow $$ x<sub>1</sub> + x<sub>2</sub> + .... + x<sub>10</sub> = 60 ....(1)
<br><br>$$\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - 5} \right)}^2}} $$ = 40
<br><br>$$ \Rightarrow $$ ($$x_1^2 + x_2^2 + ... + x_{10}^2$$) + 25 $$ \times $$ 10 - <br>10( x<sub>1</sub> + x<sub>2</sub> + .... + x<sub>10</sub>) = 40
<br><br>$$ \Rightarrow $$ $$x_1^2 + x_2^2 + ... + x_{10}^2$$ = 390 .....(2)
<br><br>From question,
<br> $$\mu $$ = $${{\left( {{x_1} - 3} \right) + \left( {{x_2} - 3} \right) + ... + \left( {{x_{10}} - 3} \right)} \over {10}}$$
<br><br>= $${{60 - 3 \times 10} \over {10}}$$ = 3
<br><br>And $$\lambda $$ = variance = $${{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - 3} \right)}^2}} } \over {10}}$$ - $$\mu $$<sup>2</sup>
<br><br>= $${{\left( {x_1^2 + x_2^2 + ... + x_{10}^2} \right) + 90 - 6\left( {\sum {{x_i}} } \right)} \over {10}}$$ - 9
<br><br>= $${{390 + 90 - 360} \over {10}}$$ - 9
<br><br>= 12 - 9 = 3 | mcq | jee-main-2020-online-9th-january-morning-slot |
vMYmelKA42j1sVXgr17k9k2k5hjwvqi | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and variance of 20 observations are
found to be 10 and 4, respectively. On
rechecking, it was found that an observation 9
was incorrect and the correct observation was
11. Then the correct variance is | [{"identifier": "A", "content": "3.98"}, {"identifier": "B", "content": "3.99"}, {"identifier": "C", "content": "4.01"}, {"identifier": "D", "content": "4.02"}] | ["B"] | null | Let 20 observation be x<sub>1</sub>
, x<sub>2</sub>
,....., x<sub>20</sub>
<br><br>Mean = <span style="display: inline-block;vertical-align: middle;">
<div style="text-align: center;border-bottom: 1px solid black;">x<sub>1</sub>
+ x<sub>2</sub>
+, .....+ x<sub>20</sub></div>
<div style="text-align: center;">20</div>
</span> = 10
<br><br>$$ \Rightarrow $$ x<sub>1</sub>
+ x<sub>2</sub>
+, .....+ x<sub>20</sub> = 200
<br><br>Variance = $${{\sum\limits_{i = 1}^{i = n} {x_i^2} } \over n} - {\left( {\overline x } \right)^2}$$
<br><br>$$ \Rightarrow $$ 4 = $${{x_1^2 + x_2^2 + ... + x_{20}^2} \over {20}}$$ - 10<sup>2</sup>
<br><br>$$ \Rightarrow $$ $${x_1^2 + x_2^2 + ... + x_{20}^2}$$ = 2080
<br><br>Also x<sub>1</sub>
+ x<sub>2</sub>
+, .....+ x<sub>20</sub> - 9 + 11 = 202
<br><br>new variance will be
<br><br>= $${{x_1^2 + x_2^2 + ... + x_{20}^2 - 81 + 121} \over {20}}$$ - $${\left( {{{202} \over {20}}} \right)^2}$$
<br><br>= 3.99
| mcq | jee-main-2020-online-8th-january-evening-slot |
Uz0o2Nru9rFJhwY0TQ7k9k2k5gr41hh | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and the standard deviation (s.d.) of
10 observations are 20 and 2 resepectively.
Each of these 10 observations is multiplied by
p and then reduced by q, where p $$ \ne $$ 0 and
q $$ \ne $$ 0. If the new mean and new s.d. become
half of their original values, then q is equal to | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "-20"}, {"identifier": "C", "content": "-10"}, {"identifier": "D", "content": "-5"}] | ["B"] | null | Let observations are
x<sub>1</sub>, x<sub>2</sub>, ...., x<sub>10</sub>
<br><br>Here mean = 20 and standard deviation(S.D) = 2
<br><br>When each of these 10
observations is multiplied by p then new observations are
px<sub>1</sub>, px<sub>2</sub>, ....., px<sub>10</sub>
<br>and new mean = 20p and new standard deviation(S.D) = 2|p|
<br><br>Now when Reduced by q then new observations are
<br>px<sub>1</sub> - q, px<sub>2</sub> - q, ....., px<sub>10</sub> - q
<br><br>and new mean = 20p - q and new standard deviation(S.D) = 2|p|
<br><br>Given 20p - q = $${{20} \over 2}$$ = 10
<br>and 2|p| = $${2 \over 2}$$ = 1
<br><br>$$ \Rightarrow $$ p = $$ \pm $$ $${1 \over 2}$$
<br><br>If p = $${1 \over 2}$$ then q = 0 (not possible as given q $$ \ne $$ 0)
<br><br>If p = - $${1 \over 2}$$ then q = -20 | mcq | jee-main-2020-online-8th-january-morning-slot |
Vv28WUTSQhpfdxt06C7k9k2k5fosp90 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the mean and variance of eight numbers 3, 7, 9, 12, 13, 20, x and y be 10 and 25 respectively,
then x.y is equal to _______. | [] | null | 54 | Mean = $${{3 + 7 + 9 + 12 + 13 + 20 + x + y} \over 8}$$ = 10
<br><br>16 = x + y ....(1)
<br><br>Variance ($${\sigma ^2}$$) = 25
<br><br>$$ \Rightarrow $$ $${{{3^2} + {7^2} + {9^2} + {{12}^2} + {{13}^2} + {{20}^2} + {x^2} + {y^2}} \over 8}$$ - 100 = 25
<br><br>$$ \Rightarrow $$ 125 × 8 = 9 + 49 + 81 + 144 + 169 + 400 + x<sup>2</sup>
+ y<sup>2</sup> - 800
<br><br>$$ \Rightarrow $$ x<sup>2</sup>
+ y<sup>2</sup> = 148
<br><br>We know, (x + y)<sup>2</sup>
= x<sup>2</sup>
+ y<sup>2</sup>
+ 2xy
<br><br>$$ \Rightarrow $$ 256 = 148 + 2xy
<br><br>$$ \Rightarrow $$ x.y = 54 | integer | jee-main-2020-online-7th-january-evening-slot |
GFKNto6vHF2bHElsUH7k9k2k5e4p7gr | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the variance of the first n natural numbers is 10 and the variance of the first m even natural
numbers is 16, then m + n is equal to_____.
| [] | null | 18 | Variance $${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}$$
<br><br>variance of (1, 2, ….. n)
<br><br>10 = $${{{1^2} + {2^2} + .... + {n^2}} \over n} - {\left( {{{1 + 2 + 3 + .... + n} \over n}} \right)^2}$$
<br><br>on solving we get n = 11
<br><br>variance of 2, 4, 6…….2m = 16
<br><br>$$ \Rightarrow $$ $${{{2^2} + {4^2} + .... + {{\left( {2m} \right)}^2}} \over m} - {\left( {m + 1} \right)^2}$$ = 16
<br><br>$$ \Rightarrow $$ m<sup>2</sup> = 49
<br><br>$$ \Rightarrow $$ m = 7
<br><br>$$ \therefore $$ m + n = 18 | integer | jee-main-2020-online-7th-january-morning-slot |
CFSOgEQFEAJx9rVOX1jgy2xukf0q5sme | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | For the frequency distribution :
<br/>Variate (x) : x<sub>1</sub> x<sub>2</sub> x<sub>3</sub>
.... x<sub>15</sub>
<br/>Frequency (f) : f<sub>1</sub>
f<sub>2</sub>
f<sub>3</sub>
...... f<sub>15</sub>
<br/>where 0 < x<sub>1</sub>
< x<sub>2</sub>
< x<sub>3</sub>
< ... < x<sub>15</sub> = 10 and
<br/>$$\sum\limits_{i = 1}^{15} {{f_i}} $$ > 0, the standard deviation cannot be : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2"}] | ["A"] | null | If variate varries from m to M then variance
<br><br>$${\sigma ^2} \le {1 \over 4}{\left( {M - m} \right)^2}$$
<br><br>(M = upper bound of value of any random variable,
<br><br> m = Lower bound of value of any random variable)
<br><br>Here M = 10 and m = 0
<br><br>$$ \therefore $$ $${\sigma ^2} \le {1 \over 4}{\left( {10 - 0} \right)^2}$$
<br><br>$$ \Rightarrow $$ $${\sigma ^2} \le 25$$
<br><br>$$ \Rightarrow $$ $$ - 5 \le \sigma \le 5$$
<br><br>$$ \therefore $$ $$\sigma $$ $$ \ne $$ 6 | mcq | jee-main-2020-online-3rd-september-morning-slot |
7ShyNpbxVy15rDBXBC1klrmgial | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the variance of 10 natural numbers 1, 1, 1, ....., 1, k is less than 10, then the maximum possible value of k is ________. | [] | null | 11 | $${\sigma ^2} = {{\sum {{x^2}} } \over n} - {\left( {{{\sum x } \over n}} \right)^2}$$<br><br>$${\sigma ^2} = {{(9 + {k^2})} \over {10}} - {\left( {{{9 + k} \over {10}}} \right)^2} < 10$$<br><br>$$(90 + {k^2})10 - (81 + {k^2} + 8k) < 1000$$<br><br>$$90 + 10{k^2} - {k^2} - 18k - 81 < 1000$$<br><br>$$9{k^2} - 18k + 9 < 1000$$<br><br>$${(k - 1)^2} < {{1000} \over 9} \Rightarrow k - 1 < {{10\sqrt {10} } \over 3}$$<br><br>$$k < {{10\sqrt {10} } \over 3} + 1$$
<br><br>k $$ \le $$ 11
<br><br>Maximum integral value of k = 11. | integer | jee-main-2021-online-24th-february-evening-slot |
UDM8kSpbt9yLePlNNi1kluz00j2 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let X<sub>1</sub>, X<sub>2</sub>, ......., X<sub>18</sub> be eighteen observations such <br/>that $$\sum\limits_{i = 1}^{18} {({X_i} - } \alpha ) = 36$$ and $$\sum\limits_{i = 1}^{18} {({X_i} - } \beta {)^2} = 90$$, where $$\alpha$$ and $$\beta$$ are distinct real numbers. If the standard deviation of these observations is 1, then the value of | $$\alpha$$ $$-$$ $$\beta$$ | is ____________. | [] | null | 4 | Given, $$\sum\limits_{i = 1}^{18} {({x_1} - \alpha ) = 36} $$<br><br>$$ \Rightarrow \sum {{x_i} - 18\alpha = 36} $$<br><br>$$ \Rightarrow \sum {{x_i} = 18(\alpha + 2)} $$ .... (1)<br><br>Also, $$\sum\limits_{i = 1}^{18} {{{({x_1} - \beta )}^2} = 90} $$<br><br>$$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} - 2\beta \sum {{x_i} = 90} } $$<br><br>$$ \Rightarrow \sum {x_i^2 + 18{\beta ^2} + 2\beta \times 18(\alpha + 2) = 90} $$ (using equation (1))<br><br>$$ \Rightarrow \sum {x_i^2 = 90} - 18{\beta ^2} + 36\beta (\alpha + 2)$$<br><br>Given, $${\sigma ^2} = 1 \Rightarrow {1 \over {18}}{\sum {x_i^2 - \left( {{{\sum {{x_i}} } \over {18}}} \right)} ^2} = 1$$<br><br>$$ = {1 \over {18}}(90 - 18{\beta ^2} + 36\alpha \beta + 72\beta ) - {\left( {{{18(\alpha + 2)} \over {18}}} \right)^2} = 1$$<br><br>$$ \Rightarrow 90 - 18{\beta ^2} + 36\alpha \beta + 72\beta - 18{(\alpha + 2)^2} = 18$$<br><br>$$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {(\alpha + 2)^2} = 1$$<br><br>$$ \Rightarrow 5 - {\beta ^2} + 2\alpha \beta + 4\beta - {\alpha ^2} - 4 - 4\alpha = 1$$<br><br>$$ \Rightarrow {\alpha ^2} - {\beta ^2} + 2\alpha \beta + 4\beta - 4\alpha = 0$$<br><br>$$ \Rightarrow (\alpha - \beta )(\alpha - \beta + 4) = 0$$<br><br>$$ \Rightarrow \alpha - \beta = - 4$$<br><br>$$ \therefore $$ $$|\alpha - \beta |\, = 4$$ $$(\alpha \ne \beta )$$ | integer | jee-main-2021-online-26th-february-evening-slot |
0QmqJySyTWsKx4mmiX1kmhxjwk1 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Consider three observations a, b, and c such that b = a + c. If the standard deviation of a + 2, b + 2, c + 2 is d, then which of the following is true? | [{"identifier": "A", "content": "b<sup>2</sup> = 3(a<sup>2</sup> + c<sup>2</sup>) + 9d<sup>2</sup>"}, {"identifier": "B", "content": "b<sup>2</sup> = 3(a<sup>2</sup> + c<sup>2</sup>) $$-$$ 9d<sup>2</sup>"}, {"identifier": "C", "content": "b<sup>2</sup> = 3(a<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>)"}, {"identifier": "D", "content": "b<sup>2</sup> = a<sup>2</sup> + c<sup>2</sup> + 3d<sup>2</sup>"}] | ["B"] | null | For a, b, c<br><br>mean = $$\overline x = {{a + b + c} \over 3}$$<br><br>$$\overline x = {{2b} \over 3}$$<br><br>We know, S.D. of a + 2, b + 2, c + 2 = S.D. of a, b, c = d<br><br>$${d^2} = {{{a^2} + {b^2} + {c^2}} \over 3} - {{4{b^2}} \over 9}$$<br><br>$${b^2} = 3{a^2} + 3{c^2} - 9{d^2}$$ | mcq | jee-main-2021-online-16th-march-morning-shift |
M8JH1kUryCFdUajSNM1kmiztfmo | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Consider the statistics of two sets of observations as follows :<br/><br/><table>
<thead>
<tr>
<th></th>
<th>Size</th>
<th>Mean</th>
<th>Variance</th>
</tr>
</thead>
<tbody>
<tr>
<td>Observation I</td>
<td>10</td>
<td>2</td>
<td>2</td>
</tr>
<tr>
<td>Observation II</td>
<td>n</td>
<td>3</td>
<td>1</td>
</tr>
</tbody>
</table><br/><br/>If the variance of the combined set of these two observations is $${{17} \over 9}$$, then the value of n is equal to ___________. | [] | null | 5 | For group - 1 : $${{\sum {{x_i}} } \over {10}} = 2 \Rightarrow \sum {{x_i}} = 20$$<br><br>$${{\sum {{x_i^2}} } \over {10}} - {(2)^2} = 2 \Rightarrow \sum {x_i^2} = 60$$<br><br>For group - 2 : $${{\sum {{y_i}} } \over n} = 3 \Rightarrow \sum {{y_i}} = 3n$$<br><br>$${{\sum {y_i^2} } \over n} - {3^2} = 1 \Rightarrow \sum {y_i^2} = 10n$$<br><br>Now, combined variance<br><br>$${\sigma ^2} = {{\sum {\left( {x_i^2 + y_i^2} \right)} } \over {10 + n}} - {\left( {{{\sum {\left( {{x_i} + {y_i}} \right)} } \over {10 + n}}} \right)^2}$$<br><br>$$ \Rightarrow {{17} \over 9} = {{60 + 10n} \over {10 + n}} - {{{{(20 + 3n)}^2}} \over {{{(10 + n)}^2}}}$$<br><br>$$ \Rightarrow $$ 17 (n<sup>2</sup> + 20n + 100) = 9(n<sup>2</sup> + 40n + 200)<br><br>$$ \Rightarrow $$ 8n<sup>2</sup> $$-$$ 20n $$-$$ 100 = 0<br><br>$$ \Rightarrow $$ 2n<sup>2</sup> $$-$$ 5n $$-$$ 25 = 0 $$ \Rightarrow $$ n = 5 | integer | jee-main-2021-online-16th-march-evening-shift |
IlkocwVJmPZsS9fqjq1kmko2650 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal to __________. | [] | null | 68 | Let first 2n observations are x<sub>1</sub>, x<sub>2</sub> ...................., x<sub>2n</sub><br><br>and last n observations are y<sub>1</sub>, y<sub>2</sub> ....................., y<sub>n</sub><br><br>Now, $${{\sum {{x_i}} } \over {2n}} = 6$$, $${{\sum {{y_i}} } \over n} = 3$$<br><br>$$ \Rightarrow \sum {{x_i}} = 12n,\sum {{y_i}} = 3n$$
<br><br>$$ \therefore $$ $${{\sum {{x_i}} + \sum {{y_i}} } \over {3n}} = {{15n} \over {3n}} = 5$$<br><br>Now, $${{\sum {x_i^2} + \sum {y_i^2} } \over {3n}} - {5^2} = 4$$<br><br>$$ \Rightarrow \sum {x_i^2} + \sum {y_i^2} = 29 \times 3n = 87n$$<br><br>Now, mean is $${{\sum {({x_i} + 1) + \sum {({y_i} - 1)} } } \over {3n}} = {{15n + 2n - n} \over {3n}} = {{16} \over 3}$$<br><br>Now, variance is $${{{{\sum {{{({x_i} + 1)}^2} + \sum {({y_i} - 1)} } }^2}} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$<br><br>$$ = {{\sum {x_i^2 + \sum {y_i^2} + 2\left( {\sum {{x_i}} - \sum {{y_i}} } \right) + 3n} } \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$<br><br>$$ = {{87n + 2(9n) + 3n} \over {3n}} - {\left( {{{16} \over 3}} \right)^2}$$<br><br>= $$29 + 6 + 1 - {\left( {{{16} \over 3}} \right)^2}$$<br><br>$$ = {{324 - 256} \over 9} = {{68} \over 9} = k$$<br><br>$$ \Rightarrow $$ 9k = 68<br><br>Therefore, the correct answer is 68. | integer | jee-main-2021-online-17th-march-evening-shift |
hSAG7R7BQD0BoMC2MU1kmm3ajdr | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let in a series of 2n observations, half of them are equal to a and remaining half are equal to $$-$$a. Also by adding a constant b in each of these observations, the mean and standard deviation of new set become 5 and 20, respectively. Then the value of a<sup>2</sup> + b<sup>2</sup> is equal to : | [{"identifier": "A", "content": "425"}, {"identifier": "B", "content": "250"}, {"identifier": "C", "content": "925"}, {"identifier": "D", "content": "650"}] | ["A"] | null | Given series<br><br>(a, a, a, ........ n times), ($$-$$a, $$-$$a, $$-$$a, ...... n times)<br><br>Now $$\overline x $$ = $${{\sum {{x_i}} } \over {2n}} = 0$$<br><br>as, x<sub>i</sub> $$ \to $$ x<sub>i</sub> + b<br><br>then $$\overline x $$ $$ \to $$ $$\overline x $$ + b<br><br>So, $$\overline x $$ + b = 5 $$ \Rightarrow $$ b = 5<br><br>No change in S.D. due to change in origin<br><br>Standard deviation ($$\sigma$$) = $$\sqrt {{{\sum\limits_{i = 1}^{2n} {{{({x_i} - \overline x )}^2}} } \over {2n}}} $$<br><br>$$= \sqrt {{{\sum\limits_{i = 1}^{2n} {x_i^2} } \over {2n}}} = \sqrt {{{2n{a^2}} \over {2n}}} = \sqrt {{a^2}} $$<br><br>$$ \therefore $$ $$20 = \sqrt {{a^2}} \Rightarrow a = 20$$<br><br>$$ \therefore $$ a<sup>2</sup> + b<sup>2</sup> = 425 | mcq | jee-main-2021-online-18th-march-evening-shift |
1krpt023e | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are : | [{"identifier": "A", "content": "10, 11"}, {"identifier": "B", "content": "3, 18"}, {"identifier": "C", "content": "8, 13"}, {"identifier": "D", "content": "1, 20"}] | ["A"] | null | Let other two numbers be a, (21 $$-$$ a)<br><br>Now,<br><br>$$10.25 = {{(4 + 16 + 25 + 49 + {a^2} + {{(21 - a)}^2}} \over 6} - {(6.5)^2}$$<br><br>(Using formula for variance)<br><br>$$ \Rightarrow 6(10.25) + 6{(6.5)^2} = 94 + {a^2} + {(21 - a)^2}$$<br><br>$$ \Rightarrow {a^2} + {(21 - a)^2} = 221$$<br><br>$$\therefore$$ a = 10 and (21 $$-$$ a) = 21 $$-$$ 10 = 11<br><br>So, remaining two observations are 10, 11.<br><br>$$\Rightarrow$$ Option (1) is correct. | mcq | jee-main-2021-online-20th-july-morning-shift |
1krrtye6a | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the mean and variance of six observations 7, 10, 11, 15, a, b are 10 and $${{20} \over 3}$$, respectively, then the value of | a $$-$$ b | is equal to : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "1"}] | ["D"] | null | $$10 = {{7 + 10 + 11 + 15 + a + b} \over 6}$$<br><br>$$\Rightarrow$$ a + b = 17 ..... (i)<br><br>$${{20} \over 3} = {{{7^2} + {{10}^2} + {{11}^2} + {{15}^2} + {a^2} + {b^2}} \over 6} - {10^2}$$<br><br>a<sup>2</sup> + b<sup>2</sup> = 145 ...... (ii)<br><br>Solve (i) and (ii) a = 9, b = 8 or a = 8, b = 9<br><br>| a $$-$$ b | = 1 | mcq | jee-main-2021-online-20th-july-evening-shift |
1krxjeee4 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let the mean and variance of the frequency distribution<br/><br/>$$\matrix{
{x:} & {{x_1} = 2} & {{x_2} = 6} & {{x_3} = 8} & {{x_4} = 9} \cr
{f:} & 4 & 4 & \alpha & \beta \cr
} $$<br/><br/>be 6 and 6.8 respectively. If x<sub>3</sub> is changed from 8 to 7, then the mean for the new data will be : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "$${{17} \\over 3}$$"}, {"identifier": "D", "content": "$${{16} \\over 3}$$"}] | ["C"] | null | Given 32 + 8$$\alpha$$ + 9$$\beta$$ = (8 + $$\alpha$$ + $$\beta$$) $$\times$$ 6<br><br>$$\Rightarrow$$ 2$$\alpha$$ + 3$$\beta$$ = 16 ..... (i)<br><br>Also, 4 $$\times$$ 16 + 4 $$\times$$ $$\alpha$$ + 9$$\beta$$ = (8 + $$\alpha$$ + $$\beta$$) $$\times$$ 6.8<br><br>$$\Rightarrow$$ 640 + 40$$\alpha$$ + 90$$\beta$$ = 544 + 68$$\alpha$$ + 68$$\beta$$<br><br>$$\Rightarrow$$ 28$$\alpha$$ $$-$$ 22$$\beta$$ = 96<br><br>$$\Rightarrow$$ 14$$\alpha$$ $$-$$ 11$$\beta$$ = 48 ..... (ii)<br><br>from (i) & (ii)<br><br>$$\alpha$$ = 5 & $$\beta$$ = 2<br><br>So, new mean = $${{32 + 35 + 18} \over {15}} = {{85} \over {15}} = {{17} \over 3}$$ | mcq | jee-main-2021-online-27th-july-evening-shift |
1krz5gy7q | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The first of the two samples in a group has 100 items with mean 15 and standard deviation 3. If the whole group has 250 items with mean 15.6 and standard deviation $$\sqrt {13.44} $$, then the standard deviation of the second sample is : | [{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "5"}] | ["C"] | null | n<sub>1</sub> = 100<br><br>m = 250<br><br>$$\overline X $$<sub>1</sub> = 15<br><br>$$\overline X $$ = 15.6<br><br>V<sub>1</sub>(x) = 9<br><br>Var(x) = 13.44<br><br>$${\sigma ^2} = {{{n_1}\sigma _1^2 + {n_2}\sigma _2^2} \over {{n_1} + {n_2}}} + {{{n_1}{n_2}} \over {{{({n_1} + {n_2})}^2}}}{({\overline x _1} - {\overline x _2})^2}$$<br><br>n<sub>2</sub> = 150, $${\overline x _2}$$ = 16, V<sub>2</sub>(x) = $$\sigma$$<sub>2</sub><br><br>$$13.44 = {{100 \times 9 + 150 \times \sigma _2^2} \over {250}} + {{100 \times 150} \over {{{(250)}^2}}} \times 1$$<br><br>$$\Rightarrow$$ $${\sigma _2}^2$$ = 16 $$\Rightarrow$$ $$\sigma$$<sub>2</sub> = 4 | mcq | jee-main-2021-online-25th-july-evening-shift |
1ks01t341 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and $${{37} \over 4}$$ <br/><br/>respectively, then (a $$-$$ b)<sup>2</sup> is equal to : | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "32"}, {"identifier": "D", "content": "16"}] | ["D"] | null | Mean = $${{6 + 10 + 7 + 13 + a + 12 + b + 12} \over 8} = 9$$<br><br>60 + a + b = 72<br><br>a + b = 12 .....(1)<br><br>variance $$ = {{\sum {x_i^2} } \over n} - {\left( {{{\sum {x_i^{}} } \over n}} \right)^2} = {{37} \over 4}$$<br><br>$$\sum {x_i^2} = {6^2} + {10^2} + {7^2} + {13^2} + {a^2} + {b^2} + {12^2} + {12^2} = {a^2} + {b^2} + 642$$<br><br>$${{{a^2} + {b^2} + 642} \over 8} - {(9)^2} = {{37} \over 4}$$<br><br>$${{{a^2} + {b^2}} \over 8} + {{321} \over 4} - 81 = {{37} \over 4}$$<br><br>$${{{a^2} + {b^2}} \over 8} = 81 + {{37} \over 4} - {{321} \over 4}$$<br><br>$${{{a^2} + {b^2}} \over 8} = 81 - 71$$<br><br>$$\therefore$$ a<sup>2</sup> + b<sup>2</sup> + 2ab = 144<br><br>80 + 2ab = 144<br><br>$$\therefore$$ 2ab = 64<br><br>$$ \therefore $$ $${(a - b)^2} = {a^2} + {b^2} - 2ab = 80 - 64 = 16$$ | mcq | jee-main-2021-online-27th-july-morning-shift |
1ktbavc5a | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. if $$\alpha$$ and $$\sqrt \beta $$ are the mean and standard deviation respectively for correct data, then ($$\alpha$$, $$\beta$$) is : | [{"identifier": "A", "content": "(11, 26)"}, {"identifier": "B", "content": "(10.5, 25)"}, {"identifier": "C", "content": "(11, 25)"}, {"identifier": "D", "content": "(10.5, 26)"}] | ["D"] | null | Given :<br><br>Mean $$(\overline x ) = {{\sum {{x_i}} } \over {20}} = 10$$<br><br>or $$\Sigma$$x<sub>i</sub> = 200 (incorrect)<br><br>or 200 $$-$$ 25 + 35 = 210 = $$\Sigma$$x<sub>i</sub> (Correct)<br><br>Now correct $$\overline x = {{210} \over {20}} = 10.5$$<br><br>again given S.D = 2.5 ($$\sigma$$)<br><br>$${\sigma ^2} = {{\sum {{x_i}^2} } \over {20}} - {(10)^2} = {(2.5)^2}$$<br><br>or $$\sum {{x_i}^2} = 2125$$ (incorrect)<br><br>or $$\sum {{x_i}^2} = 2125 - {25^2} + {35^2}$$<br><br>= 2725 (correct)<br><br>$$\therefore$$ correct $${\sigma ^2} = {{2725} \over {20}} - {(10.5)^2}$$<br><br>$${{\sigma } ^2}$$ = 26<br><br>or $$\sigma$$ = 26<br><br>$$\therefore$$ $$\alpha $$ = 10.5, $$\beta$$ = 26 | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktd4qfua | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let the mean and variance of four numbers 3, 7, x and y(x > y) be 5 and 10 respectively. Then the mean of four numbers 3 + 2x, 7 + 2y, x + y and x $$-$$ y is ______________. | [] | null | 12 | $$5 = {{3 + 7 + x + y} \over 4} \Rightarrow x + y = 10$$<br><br>Var(x) = $$10 = {{{3^2} + {7^2} + {x^2} + {y^2}} \over 4} - 25$$<br><br>$$140 = 49 + 9 + {x^2} + {y^2}$$<br><br>$${x^2} + {y^2} = 82$$<br><br>x + y = 10<br><br>$$\Rightarrow$$ (x, y) = (9, 1)<br><br>Four numbers are 21, 9, 10, 8<br><br>Mean = $${{48} \over 4}$$ = 12 | integer | jee-main-2021-online-26th-august-evening-shift |
1kteph95b | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let n be an odd natural number such that the variance of 1, 2, 3, 4, ......, n is 14. Then n is equal to _____________. | [] | null | 13 | $${{{n^2} - 1} \over {12}} = 14 \Rightarrow n = 13$$ | integer | jee-main-2021-online-27th-august-morning-shift |
1ktgqbd07 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | An online exam is attempted by 50 candidates out of which 20 are boys. The average marks obtained by boys is 12 with a variance 2. The variance of marks obtained by 30 girls is also 2. The average marks of all 50 candidates is 15. If $$\mu$$ is the average marks of girls and $$\sigma$$<sup>2</sup> is the variance of marks of 50 candidates, then $$\mu$$ + $$\sigma$$<sup>2</sup> is equal to ________________. | [] | null | 25 | $$\sigma _b^2$$ = 2 (variance of boys)<br><br>n<sub>1</sub> = no. of boys<br><br>$${\overline x _b}$$ = 12<br><br>n<sub>2</sub> = no. of girls<br><br>$$\sigma _g^2$$ = 2<br><br>$${\overline x _g}$$ = $${{50 \times 15 - 12 \times {\sigma _b}} \over {30}} = {{750 - 12 \times 20} \over {30}} = 17 = \mu $$<br><br>variance of combined series<br><br>$$\sigma _{}^2 = {{{n_1}\sigma _b^2 + {n_2}\sigma _g^2} \over {{n_1} + {n_2}}} + {{{n_1}.\,{n_2}} \over {{{({n_1} + {n_2})}^2}}}{\left( {{{\overline x }_b} - {{\overline x }_g}} \right)^2}$$<br><br>$$\sigma _{}^2 = {{20 \times 2 + 30 \times 2} \over {20 + 30}} + {{20 \times 30} \over {{{(20 + 30)}^2}}}{(12 - 17)^2}$$<br><br>$$\sigma$$<sup>2</sup> = 8<br><br>$$\Rightarrow$$ $$\mu$$ + $$\sigma$$<sup>2</sup> = 17 + 8 = 25 | integer | jee-main-2021-online-27th-august-evening-shift |
1ktkac2ui | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8, then the variance of the remaining 5 observations is : | [{"identifier": "A", "content": "$${{92} \\over 5}$$"}, {"identifier": "B", "content": "$${{134} \\over 5}$$"}, {"identifier": "C", "content": "$${{536} \\over {25}}$$"}, {"identifier": "D", "content": "$${{112} \\over 5}$$"}] | ["C"] | null | Let 8, 16, x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, x<sub>5</sub> be the observations.<br><br>Now, $${{{x_1} + {x_2} + .... + {x_5} + 14} \over 7} = 8$$<br><br>$$ \Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 42} $$ .... (1)<br><br>Also, $${{x_1^2 + x_2^2 + ...x_5^2 + {8^2} + {6^2}} \over 7} - 64 = 16$$<br><br>$$ \Rightarrow \sum\limits_{i = 1}^5 {x_i^2 = 560 - 100 = 460} $$ ..... (2)<br><br>So variance of x<sub>1</sub>, x<sub>2</sub>, ......., x<sub>5</sub><br><br>$$ = {{460} \over 5} - {\left( {{{42} \over 5}} \right)^2} = {{2300 - 1764} \over {25}} = {{536} \over {25}}$$ | mcq | jee-main-2021-online-31st-august-evening-shift |
1l545sujb | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean and the variance of 5 observations x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub>, x<sub>5</sub> be $${24 \over 5}$$ and $${194 \over 25}$$ respectively. If the mean and variance of the first 4 observation are $${7 \over 2}$$ and a respectively, then (4a + x<sub>5</sub>) is equal to:</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "15"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "18"}] | ["B"] | null | <p>Mean $$(\overline x ) = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$$</p>
<p>Given, $${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = {{24} \over 5}$$</p>
<p>$$ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 24$$ ...... (1)</p>
<p>Now, Mean of first 4 observation</p>
<p>$$ = {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4}$$</p>
<p>Given, $$ = {{{x_1} + {x_2} + {x_3} + {x_4}} \over 4} = {7 \over 2}$$</p>
<p>$$ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} = 14$$ ...... (2)</p>
<p>From equation (1) and (2), we get</p>
<p>$$14 + {x_5} = 24$$</p>
<p>$$ \Rightarrow {x_5} = 10$$</p>
<p>Now, variance of first 5 observation</p>
<p>$$ = {{\sum {x_i^2} } \over n} - {\left( {\overline x } \right)^2}$$</p>
<p>$$ = {{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2}$$</p>
<p>Given,</p>
<p>$${{x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2} \over 5} - {\left( {{{24} \over 5}} \right)^2} = {{194} \over {24}}$$</p>
<p>$$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 5\left( {{{194} \over {25}} + {{576} \over {25}}} \right)$$</p>
<p>$$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 154$$</p>
<p>$$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 + {(10)^2} = 154$$</p>
<p>$$ \Rightarrow x_1^2 + x_2^2 + x_3^2 + x_4^2 = 54$$</p>
<p>Now, variance of first 4 observation</p>
<p>$$ = {{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2}$$</p>
<p>Given,</p>
<p>$${{x_1^2 + x_2^2 + x_3^2 + x_4^2} \over 4} - {\left( {{7 \over 2}} \right)^2} = a$$</p>
<p>$$ \Rightarrow {{54} \over 4} - {{49} \over 4} = a$$</p>
<p>$$ \Rightarrow a = {5 \over 4}$$</p>
<p>$$\therefore$$ $$4a + {x_5}$$</p>
<p>$$ = 4 \times {5 \over 4} + 10 = 15$$</p> | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54tcmsg | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The number of values of a $$\in$$ N such that the variance of 3, 7, 12, a, 43 $$-$$ a is a natural number is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "infinite"}] | ["A"] | null | <p>Given,</p>
<p>5 numbers are 3, 7, 12, a, 43 $$-$$ a</p>
<p>$$\therefore$$ Mean $$(\overline x ) = {{3 + 7 + 12 + a + 43 - a} \over 5}$$</p>
<p>$$ = {{65} \over 5}$$</p>
<p>$$ = 13$$</p>
<p>We know,</p>
<p>Variance $$({\sigma ^2}) = {{\sum {x_1^2} } \over n} - {(\overline x )^2}$$</p>
<p>$$ = {{{3^2} + {7^2} + {{12}^2} + {a^2} + {{(43 - a)}^2}} \over 5} - {(13)^2}$$</p>
<p>$$ = {{9 + 49 + 144 + {a^2} + 1849 + {a^2} - 86a} \over 5} - 169$$</p>
<p>$$ = {{2{a^2} - 86a + 2051} \over 5} - 169$$</p>
<p>$$ = {{2{a^2} - 86a + 2051 - 845} \over 5}$$</p>
<p>$$ = {{2{a^2} - 86a + 1206} \over 5}$$</p>
<p>$$ = {2 \over 5}({a^2} - 43a + 603)$$</p>
<p>Variance will be natural number if $${a^2} - 43a + 603$$ in multiple of 5.</p>
<p>$$\therefore$$ $${a^2} - 43a + 603 = 5n$$</p>
<p>$$ \Rightarrow {a^2} - 43a + 603 - 5n = 0$$ ..... (1)</p>
<p>$$\therefore$$ $$a = {{43\, \pm \,\sqrt D } \over 2}$$</p>
<p>Now "a" will be natural number if</p>
<p>(1) D is a perfect square and</p>
<p>(2) $$43\, \pm \,\sqrt D $$ is multiple of 2</p>
<p>From equation (1),</p>
<p>$$D = {( - 43)^2} - 4\,.\,1\,.\,(603 - 5n)$$</p>
<p>$$ = 1849 - 2412 + 20n$$</p>
<p>$$ = 20n - 563$$</p>
<p>For any value of n, unit digit of 20n is always 0. Then 20n $$-$$ 563 will give a number whose unit digit is 7.</p>
<p>For perfect square numbers ex : 1<sup>2</sup> = 1, 2<sup>2</sup> = 4, 3<sup>2</sup> = 9, 4<sup>2</sup> = 16, 5<sup>2</sup> = 25, 6<sup>2</sup> = 36, 7<sup>2</sup> = 49, 8<sup>2</sup> = 64, 9<sup>2</sup> = 81, 10<sup>2</sup> = 100</p>
<p>So, unit digit is either 1, 4, 6, 5, 6, 9, 0 it can't be 7. So D can't be perfect square.</p> | mcq | jee-main-2022-online-29th-june-evening-shift |
1l55iz4s8 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Suppose a class has 7 students. The average marks of these students in the mathematics examination is 62, and their variance is 20. A student fails in the examination if he/she gets less than 50 marks, then in worst case, the number of students can fail is _________.</p> | [] | null | 0 | <p>According to given data</p>
<p>$${{\sum\limits_{i = 1}^7 {{{({x_i} - 62)}^2}} } \over 7} = 20$$</p>
<p>$$ \Rightarrow \sum\limits_{i = 1}^7 {{{({x_i} - 62)}^2} = 140} $$</p>
<p>So for any x<sub>i</sub>, $${({x_i} - 62)^2} \le 140$$</p>
<p>$$ \Rightarrow {x_i} > 50\,\forall i = 1,2,3,\,\,.....\,\,7$$</p>
<p>So no student is going to score less than 50.</p> | integer | jee-main-2022-online-28th-june-evening-shift |
1l567lkyk | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5. Then, the correct variance is equal to _____________.</p> | [] | null | 17 | <p>$${{\sum {x_i^2} } \over {15}} - {8^2} = 9 \Rightarrow \sum {x_i^2 = 15 \times 73 = 1095} $$</p>
<p>Let $${\overline x _c}$$ be corrected mean $${\overline x _c}$$ = 9</p>
<p>$$\sum {x_c^2 = 1095 - 25 + 400 = 1470} $$</p>
<p>Correct variance $$ = {{1470} \over {15}} - {(9)^2} = 98 - 81 = 17$$</p> | integer | jee-main-2022-online-28th-june-morning-shift |
1l56rfywl | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x < y, are 6 and $${9 \over 4}$$ respectively. Then $${x^4} + {y^2}$$ is equal to :</p> | [{"identifier": "A", "content": "162"}, {"identifier": "B", "content": "320"}, {"identifier": "C", "content": "674"}, {"identifier": "D", "content": "420"}] | ["B"] | null | <p>Mean $$ = {{4 + 5 + 6 + 6 + 7 + 8 + x + y} \over 8} = 6$$</p>
<p>$$\therefore$$ $$x + y = 12$$ ..... (i)</p>
<p>And variance</p>
<p>$$ = {{{2^2} + {1^2} + {0^2} + {0^2} + {1^2} + {2^2} + {{(x - 6)}^2} + {{(y - 6)}^2}} \over 8}$$</p>
<p>$$ = {9 \over 4}$$</p>
<p>$$\therefore$$ $${(x - 6)^2} + {(y - 6)^2} = 8$$ ..... (ii)</p>
<p>From (i) and (ii)</p>
<p>x = 4 and y = 8</p>
<p>$$\therefore$$ $${x^4} + {y^2} = 320$$</p> | mcq | jee-main-2022-online-27th-june-evening-shift |
1l58a5qpa | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean of the numbers a, b, 8, 5, 10 is 6 and their variance is 6.8. If M is the mean deviation of the numbers about the mean, then 25 M is equal to :</p> | [{"identifier": "A", "content": "60"}, {"identifier": "B", "content": "55"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "45"}] | ["A"] | null | <p>$$\because$$ $$\overline x = 6 = {{a + b + 8 + 5 + 10} \over 5} \Rightarrow a + b = 7$$ ...... (i)</p>
<p>And $${\sigma ^2} = {{{a^2} + {b^2} + {8^2} + {5^2} + {{10}^2}} \over 5} - {6^2} = 6.8$$</p>
<p>$$ \Rightarrow {a^2} + {b^2} = 25$$ ..... (ii)</p>
<p>From (i) and (ii) (a, b) = (3, 4) or (4, 3)</p>
<p>Now mean deviation about mean</p>
<p>$$M = {1 \over 5}(3 + 2 + 2 + 1 + 4) = {{12} \over 5}$$</p>
<p>$$ \Rightarrow 25M = 60$$</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l58gj21w | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to :</p> | [{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "36"}, {"identifier": "C", "content": "43"}, {"identifier": "D", "content": "60"}] | ["C"] | null | <p>Given $$\overline x = 15,\,\sigma = 2 \Rightarrow {\sigma ^2} = 4$$</p>
<p>$$\therefore$$ $${x_2} + {x_2} + \,\,.....\,\, + \,\,{x_{50}} = 15 \times 50 = 750$$</p>
<p>$$4 = {{x_1^2 + x_2^2 + \,\,.....\,\, + \,\,x_{50}^2} \over {50}} - 225$$</p>
<p>$$\therefore$$ $$x_1^2 + x_2^2 + \,\,.....\,\, + \,\,x_{50}^2 = 50 \times 229$$</p>
<p>Let a be the correct observation and b is the incorrect observation</p>
<p>then $$a + b = 70$$</p>
<p>and $$16 = {{750 - b + a} \over {50}}$$</p>
<p>$$\therefore$$ $$a - b = 50 \Rightarrow a = 60,\,b = 10$$</p>
<p>$$\therefore$$ Correct variance $$ = {{50 \times 229 + {{60}^2} - {{10}^2}} \over {50}} - 256 = 43$$</p> | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59lchg3 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean deviation about the mean of the numbers 1, 2, 3, .........., n, where n is odd, is $${{5(n + 1)} \over n}$$, then n is equal to ______________.</p> | [] | null | 21 | <p>Mean $$ = {{n{{(n + 1)} \over 2}} \over n} = {{n + 1} \over 2}$$</p>
<p>M.D. $$ = {{2\left( {{{n - 1} \over 2} + {{n - 3} \over 2} + {{n - 5} \over 2} + \,\,\,...\,\,\,0} \right)} \over n} = {{5(n + 1)} \over n}$$</p>
<p>$$ \Rightarrow ((n - 1) + (n - 3) + (n - 5) + \,\,...\,\,0) = 5(n + 1)$$</p>
<p>$$ \Rightarrow \left( {{{n + 1} \over 4}} \right)\,.\,(n - 1) = 5(n + 1)$$</p>
<p>So, $$n = 21$$</p> | integer | jee-main-2022-online-25th-june-evening-shift |
1l6f2xzmg | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean deviation about median for the numbers 3, 5, 7, 2k, 12, 16, 21, 24, arranged in the ascending order, is 6 then the median is :</p> | [{"identifier": "A", "content": "11.5"}, {"identifier": "B", "content": "10.5"}, {"identifier": "C", "content": "12"}, {"identifier": "D", "content": "11"}] | ["D"] | null | <p>Median $$ = {{2k + 12} \over 2} = k + 6$$</p>
<p>Mean deviation $$ = \sum {{{|{x_i} - M|} \over n} = 6} $$</p>
<p>$$ \Rightarrow {{(k + 3) + (k + 1) + (k - 1) + (6 - k) + (6 - k) + (10 - k) + (15 - k) + (18 - k)} \over 8}$$</p>
<p>$$\therefore$$ $${{58 - 2k} \over 8} = 6$$</p>
<p>$$k = 5$$</p>
<p>Median $$ = {{2 \times 5 + 12} \over 2} = 11$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6hzvkr1 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $$\sigma$$ is the standard deviation of the data after omitting the two wrong observations from the data, then $$38 \sigma^{2}$$ is equal to ___________.</p> | [] | null | 238 | <p>$$\mu = {{\sum {{x_i}} } \over {40}} = 30 \Rightarrow \sum {{x_i} = 1200} $$</p>
<p>$${\sigma ^2} = {{\sum {x_i^2} } \over {40}} - {(30)^2} = 25 \Rightarrow \sum {x_i^2 = 37000} $$</p>
<p>After omitting two wrong observations</p>
<p>$$\sum {{y_i} = 1200 - 12 - 10 = 1178} $$</p>
<p>$$\sum {y_i^2 = 37000 - 144 - 100 = 36756} $$</p>
<p>Now $${\sigma ^2} = {{\sum {y_i^2} } \over {38}} - {\left( {{{\sum {{y_i}} } \over {38}}} \right)^2}$$</p>
<p>$$ = {{36756} \over {38}} - {\left( {{{1178} \over {38}}} \right)^2} = - {31^2}$$</p>
<p>$$ = 38{\sigma ^2} = 36756 - 36518 = 238$$</p> | integer | jee-main-2022-online-26th-july-evening-shift |
1l6jdvjzw | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is _____________.</p> | [] | null | 2 | <p>Given $${{\sum\limits_{i = 1}^{10} {{x_i}} } \over {10}} = 15$$ ..... (1)</p>
<p>$$ \Rightarrow \sum\limits_{i = 1}^{10} {{x_i} = 150} $$</p>
<p>and $${{\sum\limits_{i = 1}^{10} {x_i^2} } \over {10}} - {15^2} = 15$$</p>
<p>$$ \Rightarrow \sum\limits_{i = 1}^{10} {x_i^2 = 2400} $$</p>
<p>Replacing 25 by 15 we get</p>
<p>$$\sum\limits_{i = 1}^9 {{x_i} + 25 = 150} $$</p>
<p>$$ \Rightarrow \sum\limits_{i = 1}^9 {{x_i} = 125} $$</p>
<p>$$\therefore$$ Correct mean $$ = {{\sum\limits_{i = 1}^9 {{x_i} + 15} } \over {10}} = {{125 + 15} \over {10}} = 14$$</p>
<p>Similarly, $$\sum\limits_{i = 1}^2 {x_i^2 = 2400 - {{25}^2} = 1775} $$</p>
<p>$$\therefore$$ Correct variance $$ = {{\sum\limits_{i = 1}^9 {x_i^2 + {{15}^2}} } \over {10}} - {14^2}$$</p>
<p>$$ = {{1775 + 225} \over {10}} - {14^2} = 4$$</p>
<p>$$\therefore$$ Correct $$S.D = \sqrt 4 = 2$$.</p> | integer | jee-main-2022-online-27th-july-morning-shift |
1l6p36ojw | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean and the variance of 20 observations $$x_{1}, x_{2}, \ldots, x_{20}$$ be 15 and 9 , respectively. For $$\alpha \in \mathbf{R}$$, if the mean of $$\left(x_{1}+\alpha\right)^{2},\left(x_{2}+\alpha\right)^{2}, \ldots,\left(x_{20}+\alpha\right)^{2}$$ is 178 , then the square of the maximum value of $$\alpha$$ is equal to ________.</p> | [] | null | 4 | <p>Given $$\sum\limits_{{{i = 1} \over {20}}}^{20} {{x_i} = 15 \Rightarrow \sum\limits_{i = 1}^{20} {{x_i} = 300} } $$</p>
<p>and $$\sum\limits_{{{i = 1} \over {20}}}^{20} {x_i^2 - {{\left( {\overline x } \right)}^2} = 9 \Rightarrow \sum\limits_{i = 1}^{20} {x_i^2 = 4680} } $$</p>
<p>Mean $$ = {{{{({x_i} + \alpha )}^2} + {{({x_2} + \alpha )}^2}\, + \,.....\, + \,{{({x_{20}} + \alpha )}^2}} \over {20}} = 178$$</p>
<p>$$ \Rightarrow {{\sum\limits_{i = 1}^{20} {x_i^2 + 2\alpha \sum\limits_{i = 1}^{20} {{x_i} + 20{\alpha ^2}} } } \over {20}} = 178$$</p>
<p>$$ \Rightarrow 4680 + 600\alpha + 20{\alpha ^2} = 3560$$</p>
<p>$$ \Rightarrow {\alpha ^2} + 30\alpha + 56 = 0$$</p>
<p>$$ \Rightarrow {\alpha ^2} + 28\alpha + 2\alpha + 56 = 0$$</p>
<p>$$ \Rightarrow (\alpha + 28)(\alpha + 2) = 0$$</p>
<p>$${\alpha _{\max }} = - 2 \Rightarrow \alpha _{\max }^2 = 4.$$</p> | integer | jee-main-2022-online-29th-july-morning-shift |
1ldo4m5zi | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let $$9=x_{1} < x_{2} < \ldots < x_{7}$$ be in an A.P. with common difference d. If the standard deviation of $$x_{1}, x_{2}..., x_{7}$$ is 4 and the mean is $$\bar{x}$$, then $$\bar{x}+x_{6}$$ is equal to :</p> | [{"identifier": "A", "content": "$$2\\left(9+\\frac{8}{\\sqrt{7}}\\right)$$"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "$$18\\left(1+\\frac{1}{\\sqrt{3}}\\right)$$"}, {"identifier": "D", "content": "34"}] | ["D"] | null | $\begin{aligned} & \text { Mean } \Rightarrow \bar{x}=\frac{\sum\limits_{i=1}^7 x_i}{7}=\frac{\frac{7}{2}[2 a+6 d]}{7}=a+3 d=x_4 \\\\ & \text { Variance }=\frac{\sum\limits_{i=1}^7\left(x_i-\bar{x}\right)^2}{7}=(4)^2 \Rightarrow \frac{\sum\limits_{i=1}^7\left(x_i-x_4\right)^2}{7}=16 \\\\ & \Rightarrow \frac{(3 d)^2+(2 d)^2+d^2+0+d^2+(2 d)^2+(3 d)^2}{7}=16 \\\\ & =4 d^2=16 \Rightarrow d=2 \\\\ & \Rightarrow \bar{x}=9+3(2)=15 \\\\ & x_6=a+5 d=9+5(2)=19 \Rightarrow \bar{x}+x_6=34\end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift |
ldo85ulg | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and $\alpha(>$
0 ), and the mean and standard deviation of marks of class $B$ of $n$ students be respectively 55 and 30
$-\alpha$. If the mean and variance of the marks of the combined class of $100+\mathrm{n}$ studants are
respectively 50 and 350 , then the sum of variances of classes $A$ and $B$ is : | [{"identifier": "A", "content": "450"}, {"identifier": "B", "content": "900"}, {"identifier": "C", "content": "650"}, {"identifier": "D", "content": "500"}] | ["D"] | null | $$
\begin{array}{cll}
\quad \mathbf{A} & \mathbf{B} & \mathbf{A}+\mathbf{B} \\
\overline{\mathrm{x}}_{1}=40 & \overline{\mathrm{x}}_{2}=55 & \overline{\mathrm{x}}=50 \\
\sigma_{1}=\alpha & \sigma_{2}=30-\alpha & \sigma^{2}=350 \\
\mathrm{n}_{1}=100 & \mathrm{n}_{2}=\mathrm{n} & 100+\mathrm{n} \\\\
\overline{\mathrm{x}}=\frac{100 \times 40+55 \mathrm{n}}{100+\mathrm{n}} &
\end{array}
$$
<br/><br/>$$
\begin{aligned}
& 5000+50 \mathrm{n}=4000+55 \mathrm{n} \\\\
& 1000=5 \mathrm{n} \\\\
& \mathrm{n}=200 \\\\
& \sigma_{1}^{2}=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{100}-40^{2} \\\\
& \sigma_{2}^{2}=\frac{\sum \mathrm{x}_{\mathrm{j}}^{2}}{100}-55^{2} \\\\
& 350=\sigma^{2}=\frac{\sum \mathrm{x}_{\mathrm{i}}{ }^{2}+\sum \mathrm{x}_{\mathrm{j}}^{2}}{300}-(\overline{\mathrm{x}})^{2} \\\\
& 350=\frac{\left(1600+\alpha^{2}\right) \times 100+\left[(30-\alpha)^{2}+3025\right] \times 200}{300}-(50)^{2} \\\\
& 2850 \times 3=\alpha^{2}+2(30-\alpha)^{2}+1600+6050 \\\\
& 8550=\alpha^{2}+2(30-\alpha)^{2}+7650 \\\\
& \alpha^{2}+2(30-\alpha)^{2}=900 \\\\
& \alpha^{2}-40 \alpha+300=0 \\\\
& \alpha=10,30 \\\\
& \sigma_{1}^{2}+\sigma_{2}^{2}=10^{2}+20^{2}=500
\end{aligned}
$$ | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldomwdbe | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :</p> | [{"identifier": "A", "content": "1792"}, {"identifier": "B", "content": "1216"}, {"identifier": "C", "content": "1456"}, {"identifier": "D", "content": "1072"}] | ["D"] | null | Let observation $1,3,5, a, b$
<br/><br/>$$
\begin{aligned}
& \text { Mean } = \frac{9+a+b}{5}=5 \\\\
& \text { Variance } = \frac{a^{2}+b^{2}+35}{5}-25=8 \\\\
& \Rightarrow a+b=16 \text { and } a^{2}+b^{2}=130 \\\\
& \therefore a \text { and } b \text { are } 7 \text { and } 9 \\\\
& \therefore a^{3}+b^{3}=7^{3}+9^{3}=1072
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldptxgvi | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the variance of the frequency distribution</p>
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<col style="width: 54px"/>
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<th class="tg-0lax">$$x_i$$</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">3</th>
<th class="tg-baqh">4</th>
<th class="tg-baqh">5</th>
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<th class="tg-baqh">8</th>
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<tr>
<td class="tg-0lax">Frequency $$f_i$$</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">6</td>
<td class="tg-baqh">16</td>
<td class="tg-baqh">$$\alpha$$</td>
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<p>is 3, then $$\alpha$$ is equal to _____________.</p> | [] | null | 5 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lek5dmcv/3fd2ee94-309d-415f-8593-3aa8daeaf873/5aaf9ff0-b525-11ed-ba8b-d3643e26f082/file-1lek5dmcw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lek5dmcv/3fd2ee94-309d-415f-8593-3aa8daeaf873/5aaf9ff0-b525-11ed-ba8b-d3643e26f082/file-1lek5dmcw.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Morning Shift Mathematics - Statistics Question 34 English Explanation">
<br><br>$\sigma_{\mathrm{x}}^{2}=\sigma_{\mathrm{d}}^{2}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}^{2}}{\sum \mathrm{f}_{\mathrm{i}}}-\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\right)^{2}$
<br><br>$=\frac{150}{45+\alpha}-0=3$
<br><br>$\Rightarrow 150=135+3 \alpha$
<br><br>$\Rightarrow 3 \alpha=15 \Rightarrow \alpha=5$ | integer | jee-main-2023-online-31st-january-morning-shift |
ldqvg79v | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of 100 consecutive positive integers $a_1, a_2, a_3, \ldots ., a_{100}$ is 25 . Then $S$ is : | [{"identifier": "A", "content": "$\\{9\\}$"}, {"identifier": "B", "content": "$\\phi$"}, {"identifier": "C", "content": "$\\{99\\}$"}, {"identifier": "D", "content": "N"}] | ["D"] | null | <p>Let $${a_1} = a \Rightarrow {a_2} = a + 1,\,.....\,{a_{100}} = a + 90$$</p>
<p>$$\mu = {{{{100a + (99 \times 100)} \over 2}} \over {100}} = a + {{99} \over 2}$$</p>
<p>M.D $$ = {{\sum {|{a_1} - \mu |} } \over {100}} = {{{{\left( {{{99} \over 2} + {{97} \over 2}\, + \,...\, + \,{1 \over 2}} \right)}^2}} \over {100}} = {{2500} \over {100}} = 25$$</p>
<p>$$\therefore$$ $$a \to z$$</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldr873oc | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and a and b are respectively mean and variance of remaining 6 observation, then $$\mathrm{a+3 b-5}$$ is equal to ___________.</p> | [] | null | 37 | <p>$$\sum {{x_i} = 7 \times 8 = 56} $$</p>
<p>$${{\sum {x_i^2} } \over n} - {\left( {{{\sum {{x_i}} } \over n}} \right)^2} = 16$$</p>
<p>$${{\sum {x_i^2} } \over 7} - 64 = 16$$</p>
<p>$$\sum {x_i^2 = 560} $$</p>
<p>when 14 is omitted</p>
<p>$$\sum {{x_i} = 56 - 14 = 42} $$</p>
<p>New mean $$ = a = {{\sum {{x_i}} } \over 6} = 7$$</p>
<p>$$\sum {x_i^2 = 560 - 196 = 364} $$</p>
<p>new variance, $$b = {{\sum {x_i^2} } \over 6} - {\left( {{{\sum {{x_i}} } \over 6}} \right)^2}$$</p>
<p>$$ = {{364} \over 6} - 49 = {{35} \over 3}$$</p>
<p>$$3b = 35$$</p>
<p>$$a + 3b - 5 = 7 + 35 - 5 = 37$$</p> | integer | jee-main-2023-online-30th-january-morning-shift |
1ldsg3712 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let $$X=\{11,12,13,....,40,41\}$$ and $$Y=\{61,62,63,....,90,91\}$$ be the two sets of observations. If $$\overline x $$ and $$\overline y $$ are their respective means and $$\sigma^2$$ is the variance of all the observations in $$\mathrm{X\cup Y}$$, then $$\left| {\overline x + \overline y - {\sigma ^2}} \right|$$ is equal to ____________.</p> | [] | null | 603 | <p>$$x = \{ 11,12,13\,....,40,41\} $$</p>
<p>$$y = \{ 61,62,63\,....,90,91\} $$</p>
<p>$$\overline x = {{{{31} \over 2}(11 + 41)} \over {31}} = {1 \over 2} \times 52 = 26$$</p>
<p>$$\overline y = {{{{31} \over 2}(61 + 91)} \over {31}} = {1 \over 2} \times 152 = 76$$</p>
<p>$${\sigma ^2} = {{\sum {x_i^2 + \sum {y_i^2} } } \over {62}} - {\left( {{{\sum {x + \sum y } } \over {62}}} \right)^2}$$</p>
<p>$$ = 705$$</p>
<p>Now,</p>
<p>$$\left| {\overline x + \overline y - {\sigma ^2}} \right|$$</p>
<p>$$ = |26 + 76 - 705| = 603$$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
1ldsu9i15 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Three rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable X denote the number of rotten apples. If $$\mu$$ and $$\sigma^2$$ represent mean and variance of X, respectively, then $$10(\mu^2+\sigma^2)$$ is equal to :</p> | [{"identifier": "A", "content": "20"}, {"identifier": "B", "content": "30"}, {"identifier": "C", "content": "250"}, {"identifier": "D", "content": "25"}] | ["A"] | null | <p>3 rotten apples are mixed with 7 good apples.</p>
<p>$$\therefore$$ Total apples = 10</p>
<p>Among those 10 apples 4 are chosen randomly.</p>
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<col style="width: 93px">
<col style="width: 234px">
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<thead>
<tr>
<th class="tg-7btt">$${x_i}$$</th>
<th class="tg-7btt">$${p_i}$$<br></th>
<th class="tg-7btt">$${p_i}{x_i}$$</th>
<th class="tg-7btt">$${p_i}{({x_i})^2}$$<br></th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-c3ow">0</td>
<td class="tg-c3ow">$${{{}^7{C_4}} \over {{}^{10}{C_4}}} = {{35} \over {210}}$$</td>
<td class="tg-c3ow">0</td>
<td class="tg-c3ow">0</td>
</tr>
<tr>
<td class="tg-c3ow">1</td>
<td class="tg-c3ow">$${{{}^3{C_1} \times {}^7{C_3}} \over {{}^{10}{C_4}}} = {{105} \over {210}}$$</td>
<td class="tg-c3ow">$${{105} \over {210}}$$</td>
<td class="tg-c3ow">$${{105} \over {210}}$$</td>
</tr>
<tr>
<td class="tg-c3ow">2</td>
<td class="tg-c3ow">$${{{}^3{C_2} \times {}^7{C_2}} \over {{}^{10}{C_4}}} = {{63} \over {210}}$$</td>
<td class="tg-c3ow">$${{126} \over {210}}$$</td>
<td class="tg-c3ow">$${{252} \over {210}}$$</td>
</tr>
<tr>
<td class="tg-c3ow">3</td>
<td class="tg-c3ow">$${{{}^3{C_3} \times {}^7{C_1}} \over {{}^{10}{C_4}}} = {7 \over {210}}$$</td>
<td class="tg-c3ow">$${{21} \over {210}}$$</td>
<td class="tg-c3ow">$${{63} \over {210}}$$</td>
</tr>
</tbody>
</table></p>
<p>$${x_i}$$ = Number of rotten apples drawn.</p>
<p>$${p_i}$$ = Probability of rotten apple.</p>
<p>We know,</p>
<p>Mean $$(\mu ) = \sum {{p_i}{x_i}} $$</p>
<p>$$ = 0 + {{105} \over {210}} + {{126} \over {210}} + {{21} \over {210}}$$</p>
<p>$$ = {{252} \over {210}} = {6 \over 5}$$</p>
<p>Also,</p>
<p>Variance $$({\sigma ^2}) = \left( {\sum {{p_i}{{({x_i})}^2}} } \right) - {\mu ^2}$$</p>
<p>$$ = {{105} \over {210}} + {{252} \over {210}} + {{63} \over {210}} - {{36} \over {25}}$$</p>
<p>$$ = {1 \over 2} + {{12} \over {10}} + {3 \over {10}} - {{36} \over {25}} = {{14} \over {25}}$$</p>
<p>$$\therefore$$ $$10(\mu^2 + {\sigma ^2})$$</p>
<p>$$ = 10\left( {({6 \over 5})^2 + {{14} \over {25}}} \right)$$</p>
<p>$$ = 10\left( {{{36 + 14} \over {25}}} \right)$$</p>
<p>$$ = 10 \times {{50} \over {25}} = 20$$</p> | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldv1ctei | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to :</p> | [{"identifier": "A", "content": "3.92"}, {"identifier": "B", "content": "4.08"}, {"identifier": "C", "content": "3.96"}, {"identifier": "D", "content": "4.04"}] | ["C"] | null | $\bar{x}=10 ~\&~ \sigma^{2}=4$, No. of students $=N$ (let)
<br/><br/>
$$
\therefore \quad \frac{\sum x_{i}}{N}=10 ~\&~ \frac{\sum x_{i}^{2}}{N}-(10)^{2}=4
$$
<br/><br/>
Now if one of $x_{i}$ is changed from 8 to 12 we have
<br/><br/>
New mean $\frac{\sum x_{i}+4}{N}=10+\frac{4}{N}=10.2$
<br/><br/>
$\Rightarrow N=20$
<br/><br/>
and $\sigma_{\text {new }}^{2}=\frac{\sum x_{i}^{2}-(8)^{2}+(12)^{2}}{20}-(10 \cdot 2)^{2}$
<br/><br/>
$$
\begin{aligned}
& =\frac{\sum x_{i}^{2}}{20}+\frac{144-64}{20}-(10 \cdot 2)^{2} \\\\
& =104+4-(10 \cdot 2)^{2} \\\\
& =108-104.04=3.96
\end{aligned}
$$ | mcq | jee-main-2023-online-25th-january-morning-shift |
1ldwx058i | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the six numbers $$\mathrm{a_1,a_2,a_3,a_4,a_5,a_6}$$, be in A.P. and $$\mathrm{a_1+a_3=10}$$. If the mean of these six numbers is $$\frac{19}{2}$$ and their variance is $$\sigma^2$$, then 8$$\sigma^2$$ is equal to :</p> | [{"identifier": "A", "content": "220"}, {"identifier": "B", "content": "210"}, {"identifier": "C", "content": "105"}, {"identifier": "D", "content": "200"}] | ["B"] | null | <p>$${a_1},{a_2},{a_3},{a_4},{a_5},{a_6}$$ are in AP.</p>
<p>Let</p>
<p>$${a_1} = a$$</p>
<p>$${a_2} = a + d$$</p>
<p>$${a_3} = a + 2d$$</p>
<p>$${a_4} = a + 3d$$</p>
<p>$${a_5} = a + 4d$$</p>
<p>$${a_6} = a + 5d$$</p>
<p>Now Mean of $${a_1},{a_2},{a_3},{a_4},{a_5}$$ and $${a_6}$$ is</p>
<p>$$ = {{{a_1} + {a_2} + {a_3} + {a_4} + {a_5} + {a_6}} \over 6} = {{19} \over 2}$$</p>
<p>$$ \Rightarrow {{6a + 15d} \over 6} = {{19} \over 2}$$</p>
<p>$$ \Rightarrow 2a + 5d = 19$$ ...... (1)</p>
<p>Also, given,</p>
<p>$${a_1} + {a_3} = 10$$</p>
<p>$$ \Rightarrow a + a + 2d = 10$$</p>
<p>$$ \Rightarrow 2a + 2d = 10$$</p>
<p>$$ \Rightarrow a + d = 5$$ ..... (2)</p>
<p>From equation (1) and (2), we get $$a = 2$$ and $$d = 3$$</p>
<p>$$\therefore$$ AP is 2, 5, 8, 11, 14, 17</p>
<p>Now, Variance $$({\sigma ^2}) = {{\sum {x_i^2} } \over 6} - {\left( {\overline x } \right)^2}$$</p>
<p>$$ = {{{2^2} + {5^2} + {8^2} + {{11}^2} + {{14}^2} + {{17}^2}} \over 6} - {\left( {{{19} \over 2}} \right)^2}$$</p>
<p>$$ = {{105} \over 4}$$</p>
<p>$$\therefore$$ $$8{\sigma ^2} = 8 \times {{105} \over 4} = 210$$</p> | mcq | jee-main-2023-online-24th-january-evening-shift |
lgnxgk71 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is : | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "14"}] | ["C"] | null | 1. Calculate the sum of the original observations:
<br/><br/>$$\frac{x_1+x_2+\ldots+x_9+50}{10}=20$$
<br/><br/>$$x_1+x_2+\ldots+x_9=150$$
<br/><br/>2. Calculate the sum of the squares of the original observations using the original variance:
<br/><br/>$$\sigma^2 = 8^2 = 64$$
<br/><br/>$$64 = \frac{x_1^2+x_2^2+\ldots+x_9^2+2500}{10} - 400$$
<br/><br/>$$x_1^2+x_2^2+\ldots+x_9^2 = 2140$$
<br/><br/>3. Calculate the new mean after correcting the error:
<br/><br/>$$\text{New mean} = \frac{150+40}{10} = 19$$
<br/><br/>4. Calculate the new variance using the corrected sum of observations and the corrected sum of squares of observations:
<br/><br/>$$\text{New } \sigma^2 = \frac{2140+1600}{10} - (19)^2$$
<br/><br/>$$ \Rightarrow $$ $$\sigma^2 = 13$$
<br/><br/>The correct variance after correcting the error is 13 (Option C). | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgoxwkg1 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is _________</p> | [] | null | 269 | <ol>
<li>The initial mean is given by:</li>
</ol>
<p>$$\bar{x}=50$$</p>
<p>So, the total sum of the marks initially was:</p>
<p>$$\sum x_i = \bar{x} \times n = 50 \times 10 = 500$$</p>
<ol>
<li>We later realize that two marks were incorrectly read as 45 and 50, when they should have been 20 and 25. Therefore, the corrected sum of the marks is:</li>
</ol>
<p>$$\sum x_{i{\text{correct}}} = \sum x_i - 45 - 50 + 20 + 25 = 500 - 45 - 50 + 20 + 25 = 450$$</p>
<ol>
<li>The initial variance is given as:</li>
</ol>
<p>$$\sigma^2 = 144$$</p>
<p>We know that variance is calculated as the mean of the squares minus the square of the mean. Therefore, rearranging gives:</p>
<p>$$\frac{\sum x_i^2}{n} = \sigma^2 + \bar{x}^2 = 144 + 50^2 = 2644$$</p>
<p>Then, the sum of the squares of the initial marks is:</p>
<p>$$\sum x_i^2 = n \times \frac{\sum x_i^2}{n} = 10 \times 2594 = 26440$$</p>
<ol>
<li>The corrected sum of the squares of the marks is calculated by subtracting the squares of the incorrect marks and adding the squares of the correct marks:</li>
</ol>
<p>$$\sum x_{i{\text{correct}}}^2 = \sum x_i^2 - 45^2 - 50^2 + 20^2 + 25^2 = 26400 - 45^2 - 50^2 + 20^2 + 25^2 = 22940$$</p>
<ol>
<li>Now we can calculate the corrected variance. The variance is the mean of the squares minus the square of the mean. Using the corrected values gives:</li>
</ol>
<p>$$\sigma_{\text{correct}}^2 = \frac{\sum x_{i{\text{correct}}}^2}{n} - \left(\frac{\sum x</em>{i_{\text{correct}}}}{n}\right)^2 = \frac{22940}{10} - \left(\frac{450}{10}\right)^2 = 2294 - 45^2 = 2294 - 2025 = 269$$</p>
<p>Therefore, the correct variance is 269.</p>
| integer | jee-main-2023-online-13th-april-evening-shift |
1lgq0rwyw | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean of the data</p>
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<th class="tg-0lax">$$x$$</th>
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<th class="tg-baqh">3</th>
<th class="tg-baqh">5</th>
<th class="tg-baqh">7</th>
<th class="tg-baqh">9</th>
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<tr>
<td class="tg-0lax">Frequency ($$f$$)</td>
<td class="tg-baqh">4</td>
<td class="tg-baqh">24</td>
<td class="tg-baqh">28</td>
<td class="tg-baqh">$$\alpha$$</td>
<td class="tg-baqh">8</td>
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<p>be 5. If $$m$$ and $$\sigma^{2}$$ are respectively the mean deviation about the mean and the variance of the data, then $$\frac{3 \alpha}{m+\sigma^{2}}$$ is equal to __________</p> | [] | null | 8 | $$
\begin{aligned}
& 5=\bar{x}=\frac{\sum x_i f_i}{\sum f_i}=\frac{4+72+140+7 \alpha+72}{64+\alpha} \\\\
& \Rightarrow 320+5 \alpha=288+7 \alpha \Rightarrow 2 \alpha=32 \Rightarrow \alpha=16
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\sum f_i & =80 \\\\
\text { M.D } & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\
& =\frac{4+4+24 \times 2+0+16 \times 2+8 \times 4}{80} \\\\
& =\frac{8}{5}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\sigma^2 & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\\\
& =\frac{4+16+24 \times 4+0+16 \times 4+8 \times 16}{80}=\frac{22}{5}
\end{aligned}
$$
<br/><br/>So, $$
\frac{3 \alpha}{m+\sigma^2}=\frac{3 \times 16}{\frac{8}{5}+\frac{22}{5}}=8
$$ | integer | jee-main-2023-online-13th-april-morning-shift |
1lgrgr0fx | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the positive numbers $$a_{1}, a_{2}, a_{3}, a_{4}$$ and $$a_{5}$$ be in a G.P. Let their mean and variance be $$\frac{31}{10}$$ and $$\frac{m}{n}$$ respectively, where $$m$$ and $$n$$ are co-prime. If the mean of their reciprocals is $$\frac{31}{40}$$ and $$a_{3}+a_{4}+a_{5}=14$$, then $$m+n$$ is equal to ___________.</p> | [] | null | 211 | Since $a_1, a_2, a_3, a_4, a_5$ are in geometric progression, we can write :
<br/><br/>$a_2 = r a$,
<br/><br/>$a_3 = r^2 a$,
<br/><br/>$a_4 = r^3 a$,
<br/><br/>$a_5 = r^4 a$.
<br/><br/>where $r$ is the common ratio and $a_1$ = $$a$$ is the first term.
<br/><br/>Given that the mean of the series is $\frac{31}{10}$, we get
<br/><br/>$$\frac{1}{5}(a_1 + a_2 + a_3 + a_4 + a_5) = \frac{31}{10}$$
<br/><br/>Substituting the terms with the values of $a_1$ and $r$ gives
<br/><br/>$$\frac{1}{5}(a + r a + r^2 a + r^3 a + r^4 a) = \frac{31}{10}$$
<br/><br/>Simplifying this gives
<br/><br/>$$a (1 + r + r^2 + r^3 + r^4) = \frac{31}{2}$$
<br/><br/>$$
\begin{aligned}
& \frac{a\left(r^5-1\right)}{r-1}=\frac{31}{2} ......(1) \\\\
& \frac{1}{a}\left(1+\frac{1}{r}+\frac{1}{r^2}+\frac{1}{r^3}+\frac{1}{r^4}\right)=\frac{31}{40} \cdot 5=\frac{31}{8} \\\\
& \frac{1}{a}\left(\frac{1-\left(\frac{1}{r}\right)^5}{1-\frac{1}{r}}\right)=\frac{31}{8} \\\\
& \text { or } \frac{1}{a}\left(\frac{r^5-1}{r-1}\right) \frac{1}{r^4}=\frac{31}{8} .........(2)
\end{aligned}
$$
<br/><br/>From (1) and (2)
<br/><br/>$$
\frac{1}{a} \cdot \frac{31}{2 a} \cdot \frac{1}{r^4}=\frac{31}{8}
$$
<br/><br/>$$
a r^2=2
$$
<br/><br/>From (1)
<br/><br/>$$
\begin{aligned}
& \frac{2}{r^2}\left(\frac{r^5-1}{r-1}\right)=\frac{31}{2} \\\\
& \frac{1+r+r^2+r^3+r^4}{r^2}=\frac{31}{4} \\\\
& \left(r^2+\frac{1}{r^2}\right)+\left(r+\frac{1}{r}\right)=\frac{27}{4} \\\\
& t^2-2+t=\frac{27}{4}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
&4 t^2+4 t-35=0\\\\
&4 t^2+14 t-10 t-35=0\\\\
&(2 t-5)(2 t+7)=0\\\\
&t=\frac{5}{2}, \frac{-7}{2} \Rightarrow r=2\\\\
& \therefore r=2, a=\frac{1}{2}
\end{aligned}
$$
<br/><br/>$$
\text { Variance of data set }\left\{\frac{1}{2}, 1,2,4,8\right\}
$$
<br/><br/>$$
\therefore \sigma^2=\frac{\sum{\mathrm{X}^2}}{\mathrm{~N}}-\left(\frac{\sum{\mathrm{X}}}{\mathrm{N}}\right)^2
$$
<br/><br/>$$
\begin{aligned}
& =\frac{\left(\frac{341}{4}\right)}{5}-\left(\frac{31}{10}\right)^2 \\\\
& =\frac{341}{20}-\frac{961}{100}=\frac{1705-961}{100} \\\\
& =\frac{744}{100}
\end{aligned}
$$
<br/><br/>$$
=\frac{186}{25}=\frac{\mathrm{m}}{\mathrm{n}} \Rightarrow 211=\mathrm{m}+\mathrm{n}
$$ | integer | jee-main-2023-online-12th-april-morning-shift |
1lgsvg819 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean of 6 observations $$1,2,4,5, \mathrm{x}$$ and $$\mathrm{y}$$ be 5 and their variance be 10 .
Then their mean deviation about the mean is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{10}{3}$$"}, {"identifier": "B", "content": "$$\\frac{8}{3}$$"}, {"identifier": "C", "content": "$$\\frac{7}{3}$$"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <p>Given that the mean of the observations ${x, y, 1, 2, 4, 5}$ is 5, we get the equation:</p>
<p>$x + y + 1 + 2 + 4 + 5 = 6 \cdot 5 \Rightarrow x + y = 18$ ..........$(1)$</p>
<p>We are also given that the variance of the observations is 10. Using the formula for variance, we have:</p>
<p>$V = \frac{\Sigma x_i^2}{n} - \bar{x}^2$, where $n$ is the number of observations, and $\bar{x}$ is the mean.</p>
<p>Substituting the given values in the variance formula, we get:</p>
<p>$10 = \frac{\Sigma x_i^2}{6} - 5^2 = \frac{\Sigma x_i^2}{6} - 25 \Rightarrow \Sigma x_i^2 = 210$.</p>
<p>Here, $\Sigma x_i^2$ is the sum of the squares of all the observations, thus:</p>
<p>$x^2 + y^2 + 1 + 4 + 16 + 25 = 210 \Rightarrow x^2 + y^2 = 164$ .........$(2)$</p>
<p>By solving the system of equations $(1)$ and $(2)$, we get $x = 8$ and $y = 10$.</p>
<p>Now, the mean deviation about the mean ($\bar{x} = 5$) is calculated by taking the average of the absolute differences of each observation from the mean:</p>
<p>$MD(5) = \frac{1}{6} [|1-5| + |2-5| + |4-5| + |5-5| + |8-5| + |10-5|] = \frac{16}{6} = \frac{8}{3}$.</p>
<p>So, the mean deviation about the mean is $\frac{8}{3}$, and the correct answer is Option B, $\frac{8}{3}$.</p>
| mcq | jee-main-2023-online-11th-april-evening-shift |
1lguu6kre | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of $$\mathrm{A}$$ and adding 2 to each element of $$\mathrm{B}$$. Then the sum of the mean and variance of the elements of $$\mathrm{C}$$ is ___________.</p> | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "40"}, {"identifier": "C", "content": "38"}, {"identifier": "D", "content": "32"}] | ["C"] | null | <p>To solve this problem, let's break it down step by step :</p>
<p><strong>Step 1 :</strong> Determine the mean of set C</p>
<p>The mean of set A = $5$
<br/><br/>The mean of set B = $8$</p>
<p>After subtracting 3 from each element in set A, the new mean becomes $5 - 3 = 2$.
After adding 2 to each element in set B, the new mean becomes $8 + 2 = 10$.</p>
<p>When we combine both modified sets to form set C, the mean of set C is the weighted average of the means of these modified sets:</p>
<p>Mean of C = $\frac{5 \times 2 + 5 \times 10}{10} = \frac{10 + 50}{10} = 6$</p>
<p><strong>Step 2 :</strong> Determine the variance of set C</p>
<p>Variance is defined as the expectation of the squared deviation of a random variable from its mean. One property of variance is that, if you add (or subtract) a constant from each data point in a set, the variance of the set does not change.</p>
<p>Thus, the variance of the elements in set A remains $12$ even after subtracting 3 from each element, and the variance of the elements in set B remains $20$ even after adding 2 to each element.</p>
<p>Now, when combining variances from two datasets into one :
<br/><br/>Variance of C
<br/><br/>= $\frac{n_1 \times \text{Variance of A} + n_1 \times (\text{Mean of modified A} - \text{Mean of C})^2 + n_2 \times \text{Variance of B} + n_2 \times (\text{Mean of modified B} - \text{Mean of C})^2}{n_1 + n_2}$</p>
<p>Given :
<br/><br/>$n_1 = n_2 = 5$
<br/><br/>Variance of A = $12$, Variance of B = $20$
<br/><br/>Mean of modified A = $2$, Mean of modified B = $10$, Mean of C = $6$</p>
<p>Plugging in the values, we get :
<br/><br/>Variance of C = $\frac{5 \times 12 + 5 \times (2 - 6)^2 + 5 \times 20 + 5 \times (10 - 6)^2}{10}$
<br/><br/>Variance of C = $\frac{60 + 80 + 100 + 80}{10} = \frac{320}{10} = 32$</p>
<p><strong>Step 3 :</strong> Sum of the mean and variance of set C
<br/><br/>Sum = Mean of C + Variance of C = $6 + 32 = 38$</p>
<p>So, the correct answer is :
<br/><br/>Option C : 38.</p>
| mcq | jee-main-2023-online-11th-april-morning-shift |
1lgvpxns4 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let $$\mu$$ be the mean and $$\sigma$$ be the standard deviation of the distribution</p>
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<col style="width: 72px"/>
<col style="width: 70px"/>
<col style="width: 71px"/>
<col style="width: 70px"/>
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$${x_i}$$</th>
<th class="tg-baqh">0</th>
<th class="tg-baqh">1</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">3</th>
<th class="tg-baqh">4</th>
<th class="tg-baqh">5</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$${f_i}$$</td>
<td class="tg-baqh">$$k + 2$$</td>
<td class="tg-baqh">$$2k$$</td>
<td class="tg-baqh">$${k^2} - 1$$</td>
<td class="tg-baqh">$${k^2} - 1$$</td>
<td class="tg-baqh">$${k^2} + 1$$</td>
<td class="tg-baqh">$$k - 3$$</td>
</tr>
</tbody>
</table></p>
<p>where $$\sum f_{i}=62$$. If $$[x]$$ denotes the greatest integer $$\leq x$$, then $$\left[\mu^{2}+\sigma^{2}\right]$$ is equal to :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "7"}] | ["B"] | null | We have, $\Sigma f_i=62$
<br/><br/>$$
\begin{aligned}
& \left.(K+2)+2 K+\left(K^2-1\right)\right)+\left(K^2-1\right)+\left(K^2+1\right)+(K-3)=62 \\\\
& \Rightarrow 3 K^2+4 K-64=0 \\\\
& \Rightarrow (3 K+16)(K-4)=0 \\\\
& \Rightarrow K=4 \quad
\end{aligned}
$$
<br/><br/>$$
\left(\because k=\frac{-16}{3} \text { is not possible }\right)
$$
<br/><br/>$$
\begin{array}{|r|c|c|c|}
\hline x_i & f_i & f_i x_i & f_i x_i^2 \\
\hline 0 & 6 & 0 & 0 \\
1 & 8 & 8 & 8 \\
2 & 15 & 30 & 60 \\
3 & 15 & 45 & 135 \\
4 & 17 & 68 & 272 \\
5 & 1 & 5 & 25 \\
\hline \text { Total } & 62 & 156 & 500 \\
\hline
\end{array}
$$
<br/><br/>$$
\mu=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{0+8+30+45+68+5}{62}=\frac{156}{62}
$$
<br/><br/>$$
\begin{aligned}
\sigma^2= & \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2 \\\\
= & \frac{1 \times 8+4 \times 15+9 \times 15+16 \times 17+25 \times 1}{62}-\left(\frac{156}{62}\right)^2 \\\\
& =\frac{500}{62}-\left(\frac{156}{62}\right)^2=\frac{500}{62}-\mu^2
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \therefore \sigma^2+\mu^2=\frac{500}{62} \\\\
& \text { Hence, }\left[\sigma^2+\mu^2\right]=8
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxwcpil | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean of the frequency distribution</p>
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<col style="width: 67px"/>
<col style="width: 71px"/>
<col style="width: 76px"/>
<col style="width: 72px"/>
</colgroup>
<thead>
<tr>
<th class="tg-0lax">Class :</th>
<th class="tg-baqh">0-10</th>
<th class="tg-baqh">10-20</th>
<th class="tg-baqh">20-30</th>
<th class="tg-baqh">30-40</th>
<th class="tg-baqh">40-50</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">Frequency :</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">$$x$$</td>
<td class="tg-baqh">5</td>
<td class="tg-baqh">4</td>
</tr>
</tbody>
</table></p>
<p>is 28, then its variance is __________.</p> | [] | null | 151 | Given mean is 28
<br/><br/>$$
\begin{array}{ll}
\text { So, } \frac{2 \times 5+3 \times 15+x \times 25+5 \times 35+4 \times 45}{14+x}=28 \\\\
\Rightarrow \frac{10+45+25 x+175+180}{14+x}=28 \\\\
\Rightarrow 310+25 x=392+28 x \\\\
\Rightarrow 3 x=18 \Rightarrow x=6
\end{array}
$$
<br/><br/>$$
\begin{aligned}
& \therefore \text { Variance }=\left(\frac{\sum x_i^2 f_i}{\sum f_i}\right)-(\text { mean })^2 \\\\
& =\left(\frac{2 \times 5^2+3 \times 15^2+6 \times 25^2+5 \times 35^2+4 \times 45^2}{20}\right)-(28)^2 \\\\
& =\left(\frac{50+675+3750+6125+8100}{20}\right)-(28)^2 \\\\
& =\left(\frac{18700}{20}\right)-(28)^2 \\\\
& =935-784=151
\end{aligned}
$$ | integer | jee-main-2023-online-10th-april-morning-shift |
1lgylem09 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean and variance of 12 observations be $$\frac{9}{2}$$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $$\frac{m}{n}$$, where $$\mathrm{m}$$ and $$\mathrm{n}$$ are coprime, then $$\mathrm{m}+\mathrm{n}$$ is equal to :</p> | [{"identifier": "A", "content": "317"}, {"identifier": "B", "content": "316"}, {"identifier": "C", "content": "314"}, {"identifier": "D", "content": "315"}] | ["A"] | null | $$
\begin{aligned}
& \text { Since, Mean }=\frac{9}{2} \\\\
& \Rightarrow \Sigma x=\frac{9}{2} \times 12=54
\end{aligned}
$$
<br/><br/>Also, variance $=4$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{\sum x^2}{12}=\left[\frac{\sum x_i}{12}\right]^2=4 \\\\
& \Rightarrow \frac{\sum x^2}{12}=4+\frac{81}{4}=\frac{97}{4} \\\\
& \Rightarrow \sum x^2=291
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \sum x^{\prime}=54-(9+10)+7+14 \\\\
& =54-19+21=56 \\\\
& \text { and } \sum x^2=291-(81+100)+49+196 \\\\
& =291-181+49+196=355
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { So, } \sigma_{\text {new }}^2=\frac{\sum x_{\text {new }}^2}{12}-\left(\frac{\sum x_{\text {new }}}{12}\right)^2 \\\\
& =\frac{355}{12}-\left(\frac{56}{12}\right)^2 \\\\
& =\frac{4260-3136}{144}=\frac{1124}{144}=\frac{281}{36} \\\\
& =\frac{m}{n} \\\\
& \Rightarrow m=281, n=36 \\\\
& \Rightarrow m+n=281+36=317
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh00fjuy | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean and variance of 8 numbers $$x, y, 10,12,6,12,4,8$$ be $$9$$ and $$9.25$$ respectively. If $$x > y$$, then $$3 x-2 y$$ is equal to _____________.</p> | [] | null | 25 | $$
\begin{array}{|c|c|c|}
\hline x_i & (x_i-\bar{x}) & (x_i-\bar{x})^2 \\
\hline x & x-9 & (x-9)^2 \\
\hline y & y-9 & (y-9)^2 \\
\hline 10 & 1 & 1 \\
\hline 12 & 3 & 9 \\
\hline 6 & -3 & 9 \\
\hline 12 & 3 & 9 \\
\hline 4 & -5 & 25 \\
\hline 8 & -1 & 1 \\
\hline x+y+92 & & (x-9)^2+(y-9)^2+54 \\
\hline
\end{array}
$$
<br/><br/>Now, mean $(\bar{x})=9$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{x+y+52}{8}=9 \\\\
& \Rightarrow x+y=20
\end{aligned}
$$
<br/><br/>Also, variance $=9.25$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{(x-9)^2+(y-9)^2+54}{8}=9.25 \\\\
& \Rightarrow x^2+y^2+81+81-2 \times 9(x+y)=20 \\\\
& \Rightarrow x^2+y^2-18 \times 20=-142 \\\\
& \Rightarrow x^2+y^2=218 \\\\
& \Rightarrow x^2+(20-x)^2=218 \\\\
& \Rightarrow x^2+400+x^2-40 x=218 \\\\
& \Rightarrow 2 x^2-40 x+182=0 \\\\
& \Rightarrow x=\frac{40 \pm 12}{4} \\\\
& \Rightarrow x=13 \text { or } x=7 \Rightarrow y=7 \text { or } y=13 \\\\
& \text { But } x>y \\\\
& \therefore x=13 \text { and } y=7 \\\\
& \text { So, } 3 x-2 y=39-14=25
\end{aligned}
$$
<br/><br/><b>Concept :</b>
<br/><br/>(a) Mean $=\frac{\Sigma x_i}{n}$
<br/><br/>(b) Variance $=\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n}$ | integer | jee-main-2023-online-8th-april-morning-shift |
1lh21a7of | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and variance of a set of 15 numbers are 12 and 14 respectively. The mean and variance of another set of 15 numbers are 14 and $$\sigma^{2}$$ respectively. If the variance of all the 30 numbers in the two sets is 13 , then $$\sigma^{2}$$ is equal to :</p> | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "11"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "9"}] | ["C"] | null | We know that if $n_1, n_2$ are the sizes, $\bar{X}_1, \bar{X}_2$ are the means and $\sigma_1, \sigma_2$ are the standard deviation of the series, then the combine variance of the series.
<br/><br/>$$
\begin{array}{ll}
& \sigma^2=\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 \cdot n_2}{\left(n_1+n_2\right)^2}\left(\bar{X}_1-\bar{X}_2\right)^2 \\\\
&\Rightarrow 13=\frac{15 \times 14+15 \times \sigma^2}{15+15}+\frac{15 \times 15}{(15+15)^2}(12-14)^2 \\\\
&\Rightarrow 13=\frac{14+\sigma^2}{2}+\frac{1}{4} \times 4 \\\\
&\Rightarrow 14+\sigma^2=2 \times 12 \\\\
&\Rightarrow \sigma^2=10
\end{array}
$$ | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2ypm8n | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean and variance of the frequency distribution</p>
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<col style="width: 53px"/>
<col style="width: 55px"/>
<col style="width: 55px"/>
<col style="width: 57px"/>
<col style="width: 57px"/>
<col style="width: 55px"/>
<col style="width: 52px"/>
<col style="width: 53px"/>
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$x_i$$</th>
<th class="tg-baqh">2</th>
<th class="tg-baqh">4</th>
<th class="tg-baqh">6</th>
<th class="tg-baqh">8</th>
<th class="tg-baqh">10</th>
<th class="tg-baqh">12</th>
<th class="tg-baqh">14</th>
<th class="tg-baqh">16</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$f_i$$</td>
<td class="tg-baqh">4</td>
<td class="tg-baqh">4</td>
<td class="tg-baqh">$$\alpha$$</td>
<td class="tg-baqh">15</td>
<td class="tg-baqh">8</td>
<td class="tg-baqh">$$\beta$$</td>
<td class="tg-baqh">4</td>
<td class="tg-baqh">5</td>
</tr>
</tbody>
</table></p>
<p>are 9 and 15.08 respectively, then the value of $$\alpha^2+\beta^2-\alpha\beta$$ is ___________.</p> | [] | null | 25 | $$
\begin{array}{lllll}
\hline x_i & f_i & x_i^2 & f_i x_i & f_i x_i^2 \\
\hline 2 & 4 & 4 & 8 & 16 \\
\hline 4 & 4 & 16 & 16 & 64 \\
\hline 6 & \alpha & 36 & 6 \alpha & 36 \alpha \\
\hline 8 & 15 & 64 & 120 & 960 \\
\hline 10 & 8 & 100 & 80 & 800 \\
\hline 12 & \beta & 144 & 12 \beta & 144 \beta \\
\hline 14 & 4 & 196 & 56 & 784 \\
\hline 16 & 5 & 256 & 80 & 1280 \\
\hline
\end{array}
$$
<br/><br/>$$ \therefore $$ $\begin{aligned} & \Sigma f_i=40 +\alpha+\beta\end{aligned}$
<br/><br/>$\begin{aligned} & \Sigma f_i x_i=360+ 6 \alpha+12 \beta\end{aligned}$
<br/><br/>$\begin{aligned} & \Sigma f_i x_i^2=3904 +36 \alpha+144 \beta\end{aligned}$
<br/><br/>Given, mean $=9$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \frac{\Sigma f_i x_i}{\Sigma f_i}=9 \\\\
& \Rightarrow \frac{360+6 \alpha+12 \beta}{40+\alpha+\beta}=9 \\\\
& \Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta) \\\\
& \Rightarrow 3 \beta=3 \alpha \\\\
& \Rightarrow \alpha=\beta .........(i)
\end{aligned}
$$
<br/><br/>$\begin{aligned} & \text { Variance }=15.08 \\\\ & \Rightarrow \frac{\Sigma f_i x_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i x_i}{\Sigma f_i}\right)^2=15.08 \\\\ & \Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(9)^2=15.08 \\\\ & \Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}=81+15.08 \quad[\because \alpha=\beta] \\\\ & \Rightarrow 3904+180 \alpha=96.08(40+2 \alpha) \\\\ & \Rightarrow 3904+180 \alpha=3843.2+192.16 \alpha \\\\ & \Rightarrow 60.8=12.16 \alpha \\\\ & \Rightarrow \alpha=5=\beta \\\\ & \therefore \alpha^2+\beta^2-\alpha \beta=25+25-25=25\end{aligned}$ | integer | jee-main-2023-online-6th-april-evening-shift |
lsamgfx7 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Consider 10 observations $x_1, x_2, \ldots, x_{10}$ such that $\sum\limits_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum\limits_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. Then $\frac{\beta}{\alpha}$ is equal to : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "$\\frac{5}{2}$"}, {"identifier": "D", "content": "$\\frac{3}{2}$"}] | ["A"] | null | We have given $\bar{x}($ mean $)=\frac{6}{5}$
<br/><br/>$$
\begin{aligned}
& \text { Variance }=\frac{84}{25} \\\\
& \sum_{i=1}^{10}\left(x_i-\alpha\right)=2 \\\\
& \Rightarrow x_1+x_2+\ldots+x_{10}-10 \alpha=2 \\\\
& \Rightarrow \frac{x_1+x_2+\ldots+x_{10}}{10}-\alpha=\frac{2}{10} \\\\
& \Rightarrow \frac{6}{5}-\alpha=\frac{2}{10} \\\\
& \Rightarrow \alpha=1
\end{aligned}
$$
<br/><br/>$\begin{aligned} & \text { and } \sum_{i=1}^{10}\left(x_i-\beta\right)^2=40 \\\\ & \left(x_1-\beta\right)^2+\left(x_2-\beta\right)^2+\ldots+\left(x_{10}-\beta\right)^2=40 \\\\ & x_1^2+x_2^2+\ldots+x_{10}^2+10 \beta^2-2 \beta\left(x_1+x_2+\ldots+x_{10}\right)=40 \\\\ & \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}+\beta^2-\frac{2 \beta\left(x_1+x_2+\ldots+x_{10}\right)}{10}=4 \\\\ & \Rightarrow \frac{x_1^2+x_2^2+\ldots+x_{10}^2}{10}-\frac{36}{25}+\frac{36}{25}+\beta^2-2 \beta \times \frac{6}{5}=4\end{aligned}$
<br/><br/>$\left[\right.$ Variance $\left.=\frac{\sum_{i=1}^n x_i^2}{n}-(\bar{x})^2\right]$
<br/><br/>$\begin{aligned} & \Rightarrow \frac{84}{25}+\frac{36}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\\\ & \Rightarrow \frac{120}{25}+\beta^2-\frac{12 \beta}{5}-4=0 \\\\ & \Rightarrow 25 \beta^2-60 \beta+20=0 \\\\ & \Rightarrow 5 \beta^2-12 \beta+4=0 \\\\ & \Rightarrow \beta=2, \frac{2}{5}\end{aligned}$
<br/><br/>Take $\beta=2$
<br/><br/>$$
\frac{\beta}{\alpha}=\frac{2}{1}=2
$$ | mcq | jee-main-2024-online-1st-february-evening-shift |
lsaohwm3 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let the median and the mean deviation about the median of 7 observation $170,125,230,190,210$, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is : | [{"identifier": "A", "content": "31"}, {"identifier": "B", "content": "28"}, {"identifier": "C", "content": "30"}, {"identifier": "D", "content": "32"}] | ["C"] | null | $$
\text { Median }=170 \Rightarrow 125, \mathrm{a}, \mathrm{b}, 170,190,210,230
$$
<br/><br/>Mean deviation about
Median $=$
<br/><br/>$$
\begin{aligned}
& \frac{0+45+60+20+40+170-a+170-b}{7}=\frac{205}{7} \\\\
& \Rightarrow \mathrm{a}+\mathrm{b}=300 \\\\
& \text { Mean }=\frac{170+125+230+190+210+a+b}{7}=175
\end{aligned}
$$
<br/><br/>Mean deviation
About mean $=$
<br/><br/>$$
\frac{50+175-a+175-b+5+15+35+55}{7}=30
$$ | mcq | jee-main-2024-online-1st-february-morning-shift |
lsbl4onq | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let $\mathrm{a}_1, \mathrm{a}_2, \ldots \mathrm{a}_{10}$ be 10 observations such that $\sum\limits_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50$ and $\sum\limits_{\forall \mathrm{k} < \mathrm{j}} \mathrm{a}_{\mathrm{k}} \cdot \mathrm{a}_{\mathrm{j}}=1100$. Then the standard deviation of $\mathrm{a}_1, \mathrm{a}_2, \ldots, \mathrm{a}_{10}$ is equal to : | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "$\\sqrt{115}$"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "$\\sqrt{5}$"}] | ["D"] | null | <p>$$\begin{aligned}
& \sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50 \\
& \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50 \quad \text{.... (i)}\\
& \sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=1100 \quad \text{.... (ii)}\\
& \text { If } \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50 . \\
& \left(\mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}\right)^2=2500 \\
& \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2+2 \sum_{\mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=2500 \\
& \Rightarrow \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=2500-2(1100) \\
& \sum_{\mathrm{i}=1}^{10} \mathrm{a}_{\mathrm{i}}^2=300 \text {, Standard deviation ‘ } \sigma \text { ’ }
\end{aligned}$$</p>
<p>$$\begin{aligned}
& =\sqrt{\frac{\sum \mathrm{a}_{\mathrm{i}}^2}{10}-\left(\frac{\sum \mathrm{a}_{\mathrm{i}}}{10}\right)^2}=\sqrt{\frac{300}{10}-\left(\frac{50}{10}\right)^2} \\
& =\sqrt{30-25}=\sqrt{5}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscoct80 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If $$\mu$$ and $$\sigma^2$$ denote the mean and variance of the correct observations respectively, then $$15\left(\mu+\mu^2+\sigma^2\right)$$ is equal to __________.</p> | [] | null | 2521 | <p>Let the incorrect mean be $$\mu^{\prime}$$ and standard deviation be $$\sigma^{\prime}$$</p>
<p>We have</p>
<p>$$\mu^{\prime}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}}{15}=12 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}=180$$</p>
<p>As per given information correct $$\Sigma \mathrm{x}_{\mathrm{i}}=180-10+12$$</p>
<p>$$\Rightarrow \mu(\text { correct mean})=\frac{182}{15}$$</p>
<p>Also</p>
<p>$$\sigma^{\prime}=\sqrt{\frac{\Sigma \mathrm{x}_{\mathrm{i}}^2}{15}-144}=3 \Rightarrow \Sigma \mathrm{x}_{\mathrm{i}}^2=2295$$</p>
<p>Correct $$\Sigma \mathrm{x}_{\mathrm{i}}{ }^2=2295-100+144=2339$$</p>
<p>$$\sigma^2(\text { correct variance })=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}
$$</p>
<p>Required value</p>
<p>$$\begin{aligned}
& =15\left(\mu+\mu^2+\sigma^2\right) \\
& =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) \\
& =15\left(\frac{182}{15}+\frac{2339}{15}\right) \\
& =2521
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd4v4of | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean and the variance of 6 observations $$a, b, 68,44,48,60$$ be $$55$$ and $$194$$, respectively. If $$a>b$$, then $$a+3 b$$ is</p> | [{"identifier": "A", "content": "180"}, {"identifier": "B", "content": "210"}, {"identifier": "C", "content": "190"}, {"identifier": "D", "content": "200"}] | ["A"] | null | <p>$$\begin{aligned}
& a, b, 68,44,48,60 \\
& \text { Mean }=55 \quad a>b \\
& \text { Variance }=194 \quad a+3 b \\
& \frac{a+b+68+44+48+60}{6}=55 \\
& \Rightarrow 220+a+b=330 \\
& \therefore a+b=110 \ldots . .(1)
\end{aligned}$$</p>
<p>Also,</p>
<p>$$\begin{aligned}
& \sum \frac{\left(x_i-\bar{x}\right)^2}{n}=194 \\
& \Rightarrow(a-55)^2+(b-55)^2+(68-55)^2+(44-55)^2 \\
& +(48-55)^2+(60-55)^2=194 \times 6 \\
& \Rightarrow(a-55)^2+(b-55)^2+169+121+49+25=1164 \\
& \Rightarrow(a-55)^2+(b-55)^2=1164-364=800 \\
& a^2+3025-110 a+b^2+3025-110 b=800 \\
& \Rightarrow a^2+b^2=800-6050+12100 \\
& a^2+b^2=6850 \ldots \ldots . .(2)
\end{aligned}$$</p>
<p>Solve (1) & (2);</p>
<p>$$\begin{aligned}
& a=75, b=35 \\
& \therefore a+3 b=75+3(35)=75+105=180
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsf0j0pi | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean and variance of the data $$65,68,58,44,48,45,60, \alpha, \beta, 60$$ where $$\alpha> \beta$$, are 56 and 66.2 respectively, then $$\alpha^2+\beta^2$$ is equal to _________.</p> | [] | null | 6344 | <p>$$\begin{aligned}
& \overline{\mathrm{x}}=56 \\
& \sigma^2=66.2 \\
& \Rightarrow \frac{\alpha^2+\beta^2+25678}{10}-(56)^2=66.2 \\
& \therefore \alpha^2+\beta^2=6344
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfkircs | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean and variance of five observations are $$\frac{24}{5}$$ and $$\frac{194}{25}$$ respectively and the mean of the first four observations is $$\frac{7}{2}$$, then the variance of the first four observations in equal to</p> | [{"identifier": "A", "content": "$$\\frac{5}{4}$$\n"}, {"identifier": "B", "content": "$$\\frac{4}{5}$$\n"}, {"identifier": "C", "content": "$$\\frac{105}{4}$$\n"}, {"identifier": "D", "content": "$$\\frac{77}{12}$$"}] | ["A"] | null | <p>$$\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}$$</p>
<p>Let first four observation be $$\mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4$$</p>
<p>Here, $$\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$$. ..... (1)</p>
<p>Also, $$\frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4}{4}=\frac{7}{2}$$</p>
<p>$$\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4=14$$</p>
<p>Now from eqn -1</p>
<p>$$\mathrm{x}_5=10$$</p>
<p>Now, $$\sigma^2=\frac{194}{25}$$</p>
<p>$$\begin{aligned}
& \frac{\mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2+\mathrm{x}_5^2}{5}-\frac{576}{25}=\frac{194}{25} \\
& \Rightarrow \mathrm{x}_1^2+\mathrm{x}_2^2+\mathrm{x}_3^2+\mathrm{x}_4^2=54
\end{aligned}$$</p>
<p>Now, variance of first 4 observations</p>
<p>$$\begin{aligned}
\operatorname{Var} & =\frac{\sum_\limits{i=1}^4 x_i^2}{4}-\left(\frac{\sum_\limits{i=1}^4 x_i}{4}\right)^2 \\
& =\frac{54}{4}-\frac{49}{4}=\frac{5}{4}
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsg56yjt | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The variance $$\sigma^2$$ of the data</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 611px">
<colgroup>
<col style="width: 81px"/>
<col style="width: 79px"/>
<col style="width: 74px"/>
<col style="width: 76px"/>
<col style="width: 74px"/>
<col style="width: 74px"/>
<col style="width: 78px"/>
<col style="width: 75px"/>
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$x_i$$</th>
<th class="tg-baqh">0</th>
<th class="tg-baqh">1</th>
<th class="tg-baqh">5</th>
<th class="tg-baqh">6</th>
<th class="tg-baqh">10</th>
<th class="tg-baqh">12</th>
<th class="tg-baqh">17</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$f_i$$</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">6</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">3</td>
</tr>
</tbody>
</table></p>
<p>is _________.</p> | [] | null | 29 | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 519px">
<colgroup>
<col style="width: 94px">
<col style="width: 123px">
<col style="width: 99px">
<col style="width: 203px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$\mathrm{x_i}$$</th>
<th class="tg-baqh">$$\mathrm{f_i}$$</th>
<th class="tg-baqh">$$\mathrm{f_i x_i}$$</th>
<th class="tg-baqh">$$\mathrm{f_i x^2_i}$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">0</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">0</td>
<td class="tg-baqh">0</td>
</tr>
<tr>
<td class="tg-baqh">1</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">2</td>
</tr>
<tr>
<td class="tg-baqh">5</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">15</td>
<td class="tg-baqh">75</td>
</tr>
<tr>
<td class="tg-baqh">6</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">12</td>
<td class="tg-baqh">72</td>
</tr>
<tr>
<td class="tg-baqh">10</td>
<td class="tg-baqh">6</td>
<td class="tg-baqh">60</td>
<td class="tg-baqh">600</td>
</tr>
<tr>
<td class="tg-baqh">12</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">36</td>
<td class="tg-baqh">432</td>
</tr>
<tr>
<td class="tg-baqh">17</td>
<td class="tg-baqh">3</td>
<td class="tg-baqh">51</td>
<td class="tg-baqh">867</td>
</tr>
<tr>
<td class="tg-baqh"></td>
<td class="tg-baqh">$$\Sigma \mathrm{f}_{\mathrm{i}}=22$$</td>
<td class="tg-baqh"></td>
<td class="tg-baqh">$$\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=2048$$</td>
</tr>
</tbody>
</table></p>
<p>$$\begin{aligned}
& \therefore \quad \Sigma \mathrm{f}_{\mathrm{i}} \mathrm{X}_{\mathrm{i}}=176 \\
& \text { So } \overline{\mathrm{x}}=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{176}{22}=8 \\
& \text { for } \sigma^2=\frac{1}{\mathrm{~N}} \sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}{ }^2-(\overline{\mathrm{x}})^2 \\
& =\frac{1}{22} \times 2048-(8)^2 \\
& =93.090964 \\
& =29.0909 \\
&
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-evening-shift |
luxwcxru | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the variance of the frequency distribution</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 492px">
<colgroup>
<col style="width: 70px"/>
<col style="width: 68px"/>
<col style="width: 70px"/>
<col style="width: 70px"/>
<col style="width: 69px"/>
<col style="width: 72px"/>
<col style="width: 73px"/>
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$x$$</th>
<th class="tg-baqh">$$c$$</th>
<th class="tg-baqh">$$2c$$</th>
<th class="tg-baqh">$$3c$$</th>
<th class="tg-baqh">$$4c$$</th>
<th class="tg-baqh">$$5c$$</th>
<th class="tg-baqh">$$6c$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$f$$</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">1</td>
</tr>
</tbody>
</table></p>
<p>is 160, then the value of $$c\in N$$ is</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "7"}] | ["D"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 339px">
<colgroup>
<col style="width: 82px">
<col style="width: 83px">
<col style="width: 86px">
<col style="width: 88px">
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$x_i$$</th>
<th class="tg-baqh">$$f(x_i)$$</th>
<th class="tg-baqh">$$x(f(x)$$</th>
<th class="tg-baqh">$$x^2f(x)$$</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">C</td>
<td class="tg-baqh">2</td>
<td class="tg-baqh">2C</td>
<td class="tg-baqh">2C$$^2$$</td>
</tr>
<tr>
<td class="tg-baqh">2C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">2C</td>
<td class="tg-baqh">4C$$^2$$</td>
</tr>
<tr>
<td class="tg-baqh">3C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">3C</td>
<td class="tg-baqh">9C$$^2$$</td>
</tr>
<tr>
<td class="tg-baqh">4C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">4C</td>
<td class="tg-baqh">16C$$^2$$</td>
</tr>
<tr>
<td class="tg-baqh">5C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">5C</td>
<td class="tg-baqh">25C$$^2$$</td>
</tr>
<tr>
<td class="tg-baqh">6C</td>
<td class="tg-baqh">1</td>
<td class="tg-baqh">6C</td>
<td class="tg-baqh">36C$$^2$$</td>
</tr>
</tbody>
</table></p>
<p>$$\begin{aligned}
& \sigma^2=E\left(x^2\right)-[E(x)], \sum f\left(x_i\right)=7 \\
& E(x)=\sum x f(x)=22 C \\
& E\left(x^2\right)=\sum x^2 f(x)=92 C^2
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \sigma^2=160=\frac{92 C^2}{7}-\left(\frac{22 C}{7}\right)^2 \\
& \Rightarrow C= \pm 7 \text { but } C \in N \\
& \Rightarrow C=7
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
luy6z4to | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The frequency distribution of the age of students in a class of 40 students is given below.</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-1wig{font-weight:bold;text-align:left;vertical-align:top}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 543px">
<colgroup>
<col style="width: 156px"/>
<col style="width: 65px"/>
<col style="width: 65px"/>
<col style="width: 64px"/>
<col style="width: 63px"/>
<col style="width: 65px"/>
<col style="width: 65px"/>
</colgroup>
<thead>
<tr>
<th class="tg-1wig">Age</th>
<th class="tg-baqh">15</th>
<th class="tg-baqh">16</th>
<th class="tg-baqh">17</th>
<th class="tg-baqh">18</th>
<th class="tg-baqh">19</th>
<th class="tg-baqh">20</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-1wig">No of Students</td>
<td class="tg-baqh">5</td>
<td class="tg-baqh">8</td>
<td class="tg-baqh">5</td>
<td class="tg-baqh">12</td>
<td class="tg-baqh">$$x$$</td>
<td class="tg-baqh">$$y$$</td>
</tr>
</tbody>
</table></p>
<p>If the mean deviation about the median is 1.25, then $$4x+5y$$ is equal to :</p> | [{"identifier": "A", "content": "43"}, {"identifier": "B", "content": "46"}, {"identifier": "C", "content": "44"}, {"identifier": "D", "content": "47"}] | ["C"] | null | <p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 448px">
<colgroup>
<col style="width: 95px">
<col style="width: 200px">
<col style="width: 153px">
</colgroup>
<thead>
<tr>
<th class="tg-amwm">Age</th>
<th class="tg-amwm">No. of Students</th>
<th class="tg-amwm">CF</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">15</td>
<td class="tg-baqh">5</td>
<td class="tg-baqh">5</td>
</tr>
<tr>
<td class="tg-baqh">16</td>
<td class="tg-baqh">8</td>
<td class="tg-baqh">13</td>
</tr>
<tr>
<td class="tg-baqh">17</td>
<td class="tg-baqh">5</td>
<td class="tg-baqh">18</td>
</tr>
<tr>
<td class="tg-baqh">18</td>
<td class="tg-baqh">12</td>
<td class="tg-baqh">30</td>
</tr>
<tr>
<td class="tg-baqh">19</td>
<td class="tg-baqh">$$x$$</td>
<td class="tg-baqh">$$30+x$$</td>
</tr>
<tr>
<td class="tg-baqh">20</td>
<td class="tg-baqh">$$y$$</td>
<td class="tg-baqh">$$30+x+y$$</td>
</tr>
</tbody>
</table></p>
<p>$$\begin{aligned}
& 30+x+y=40 \\
& x+y=10
\end{aligned}$$</p>
<p>Median $$=\left(\frac{n+1}{2}\right)^{\text {th }}$$ observation</p>
<p>$$=\frac{40+1}{2}=\frac{41}{2}$$</p>
<p>Median $$=18$$</p>
<p>Mean deviation about median</p>
<p>$$\begin{aligned}
& 5.3+8.2+5.1+12.0+x \cdot 1+y \cdot 2=1.25 \times 40 \\
& 15+16+5+x+2 y=50 \\
& x+2 y=14 \\
& \quad x+y=10 \\
& \Rightarrow \quad x=4 \\
& \Rightarrow \quad y=6 \\
& \begin{aligned}
4 x+5 y & =24+20 \\
& =44
\end{aligned}
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lv0vxcac | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let $$\alpha, \beta \in \mathbf{R}$$. Let the mean and the variance of 6 observations $$-3,4,7,-6, \alpha, \beta$$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :</p> | [{"identifier": "A", "content": "$$\\frac{16}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{11}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{14}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{13}{3}$$"}] | ["D"] | null | <p>$$\begin{aligned}
& \text { Mean }=\frac{-3+4+7+(-6)+\alpha+\beta}{6}=2 \\
& \Rightarrow \alpha+\beta=10 \\
& \text { Variance }=\frac{\sum x_i^2}{n}-\left(\frac{\bar{x}}{n}\right)^2=23 \\
& \Rightarrow \sum x_i^2=27 \times 6 \\
& \Rightarrow 9+16+49+36+\alpha^2+\beta^2=162 \\
& \Rightarrow \alpha^2+\beta^2=52
\end{aligned}$$</p>
<p>We get $$\alpha$$ and $$\beta$$ as 4 and 6</p>
<p>So, mean deviation about mean</p>
<p>$$\begin{aligned}
& =\frac{|-3-2|+|4-2|+|7-2|+|-6-2|+|4-2|+|6-2|}{6} \\
& =\frac{5+2+5+8+2+4}{6} \\
& =\frac{13}{3}
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv3vefev | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let $$\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{N}$$ and $$\mathrm{a}< \mathrm{b}< \mathrm{c}$$. Let the mean, the mean deviation about the mean and the variance of the 5 observations $$9,25, a, b, c$$ be 18, 4 and $$\frac{136}{5}$$, respectively. Then $$2 a+b-c$$ is equal to ________</p> | [] | null | 33 | <p>$$\begin{aligned}
& a, b, c \in N \\
& a< b < c \\
& \text { Mean }=18 \\
& \frac{9+25+a+b+c}{5}=18 \\
& 34+a+b+c=90 \\
& a+b+c=56
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{|9-18|+|25-18|+|a-18|+|b-18|+|c-18|}{5}=4 \\
& 9+7+|a-18|+|b-18|+|c-18|=20 \\
& |a-18|+|b-18|+|c-18|=4 \\
& \frac{136}{5}=\frac{706+a^2+b^2+c^2}{5}-(18)^2 \\
& \Rightarrow 136=706+a^2+b^2+c^2-1620 \\
& \Rightarrow a^2+b^2+c^2=1050 \\
& \text { Consider } a<19 < b< c \\
& \text { Solving } a=17, b=19, c=20 \\
& 2 a+b-c \\
& 34+19-20 \\
& =33
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-evening-shift |
lv9s20ji | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean and the standard deviation of the probability distribution</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 404px">
<colgroup>
<col style="width: 85px"/>
<col style="width: 81px"/>
<col style="width: 79px"/>
<col style="width: 77px"/>
<col style="width: 82px"/>
</colgroup>
<thead>
<tr>
<th class="tg-baqh">$$\mathrm{X}$$</th>
<th class="tg-baqh">$$\alpha$$</th>
<th class="tg-baqh">1</th>
<th class="tg-baqh">0</th>
<th class="tg-baqh">$$-$$3</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$\mathrm{P(X)}$$</td>
<td class="tg-baqh">$$\frac{1}{3}$$</td>
<td class="tg-baqh">$$\mathrm{K}$$</td>
<td class="tg-baqh">$$\frac{1}{6}$$</td>
<td class="tg-baqh">$$\frac{1}{4}$$</td>
</tr>
</tbody>
</table></p>
<p>be $$\mu$$ and $$\sigma$$, respectively. If $$\sigma-\mu=2$$, then $$\sigma+\mu$$ is equal to ________.</p> | [] | null | 5 | <p>Mean $$(\mu)=\Sigma x_i P\left(x_i\right)$$</p>
<p>Standard deviation $$(\sigma)=\sqrt{\left(\Sigma x_i^2 P\left(x_i\right)\right)-\mu^2}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad \mu=\frac{1}{3} \alpha+K-\frac{3}{4} \\
& \sigma=\sqrt{\left(\frac{1}{3} \alpha^2+K+0+\frac{9}{4}\right)-\left(\frac{1}{3} \alpha+K-\frac{3}{4}\right)^2} \\
& \because \Sigma P_i=1 \Rightarrow \frac{1}{3}+K+\frac{1}{6}+\frac{1}{4}=1 \\
& \Rightarrow \quad K=\frac{1}{4} \Rightarrow \mu=\frac{1}{3} \alpha-\frac{1}{2} \\
& \because \sigma-\mu=2 \\
& \sigma^2=(\mu+2)^2
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{1}{3} \alpha^2+\frac{5}{2}-\mu^2=(\mu+2)^2 \\
& \frac{1}{3} \alpha^2+\frac{5}{2}=\left(\frac{1}{3} \alpha-\frac{1}{2}\right)^2+\left(\frac{1}{3} \alpha+\frac{3}{2}\right)^2 \\
& \Rightarrow \alpha=0,6
\end{aligned}$$</p>
<p>$$\begin{array}{ll}
\text { If } \alpha=0, K=\frac{1}{4} & \text { If } \alpha=6, K=\frac{1}{4} \\
\mu=-\frac{1}{2}, \sigma=\frac{3}{2} & \mu=\frac{3}{2}, \sigma=\frac{7}{2} \\
\sigma+\mu=1 & \sigma+\mu=5
\end{array}$$</p>
<p>Both (1) and (5) are correct but according to NTA
(5) is correct</p> | integer | jee-main-2024-online-5th-april-evening-shift |
lvc583dm | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and standard deviation of 20 observations are found to be 10 and 2 , respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is</p> | [{"identifier": "A", "content": "1.94\n"}, {"identifier": "B", "content": "$$\\sqrt{3.96}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{3.86}$$\n"}, {"identifier": "D", "content": "1.8"}] | ["B"] | null | <p>To find the correct standard deviation, we first need to adjust the mean and sum of squares of the observations based on the correction of the erroneous entry. The original erroneous observation was 8, and the correct observation is 12. The mean and standard deviation of the 20 observations before correction were 10 and 2, respectively.</p>
<p>The sum of all 20 original observations ($S_{\text{original}}$) can be calculated from the mean formula:
<p>$ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Number of observations}} $</p>
<p>$ 10 = \frac{S_{\text{original}}}{20} $</p>
<p>$ S_{\text{original}} = 10 \times 20 = 200 $</p></p>
<p>The corrected sum of observations ($S_{\text{correct}}$) will replace the incorrect observation (8) with the correct one (12):
<p>$ S_{\text{correct}} = S_{\text{original}} - 8 + 12 = 200 - 8 + 12 = 204 $</p></p>
<p>The corrected mean ($\mu_{\text{correct}}$) is:
<p>$ \mu_{\text{correct}} = \frac{S_{\text{correct}}}{20} = \frac{204}{20} = 10.2 $</p></p>
<p>To find the corrected standard deviation, we need the sum of squares of the deviations from the mean for both the original and corrected data. The original sum of squares ($SS_{\text{original}}$) is calculated from the original standard deviation formula, where $\sigma = 2$:
<p>$ \sigma^2 = \frac{SS}{n} $</p>
<p>$ 4 = \frac{SS_{\text{original}}}{20} $</p>
<p>$ SS_{\text{original}} = 4 \times 20 = 80 $</p></p>
<p>To calculate the corrected sum of squares ($SS_{\text{correct}}$), we need to adjust $SS_{\text{original}}$ by removing the square of the deviation of the incorrect observation and adding the square of the deviation of the correct observation:
<p>$ SS_{\text{correct}} = SS_{\text{original}} - (8 - 10)^2 + (12 - 10.2)^2 $</p>
<p>$ SS_{\text{correct}} = 80 - (-2)^2 + (1.8)^2 $</p>
<p>$ SS_{\text{correct}} = 80 - 4 + 3.24 = 79.24 $</p></p>
<p>Finally, the corrected standard deviation ($\sigma_{\text{correct}}$) is:
<p>$ \sigma_{\text{correct}} = \sqrt{\frac{SS_{\text{correct}}}{20}} $</p>
<p>$ \sigma_{\text{correct}} = \sqrt{\frac{79.24}{20}} $</p>
<p>$ \sigma_{\text{correct}} = \sqrt{3.962} $</p></p>
<p>The corrected standard deviation is closest to the value given in Option B, which is
<p>$ \sqrt{3.96} $</p></p> | mcq | jee-main-2024-online-6th-april-morning-shift |
2c3WXCVRZ7i0gI95 | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | The lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$ and $$\left( {{p^2} + 1} \right){}^2x + \left( {{p^2} + 1} \right)y + 2q$$ $$=0$$ are perpendicular to a common line for : | [{"identifier": "A", "content": "exactly one values of $$p$$"}, {"identifier": "B", "content": "exactly two values of $$p$$ "}, {"identifier": "C", "content": "more than two values of $$p$$ "}, {"identifier": "D", "content": "no value of $$p$$ "}] | ["A"] | null | If the lines $$p\left( {{p^2} + 1} \right)x - y + q = 0$$
<br><br>and $${\left( {{p^2} + 1} \right)^2}x + \left( {{p^2} + 1} \right)y + 2q = 0$$
<br><br>are perpendicular to a common line then these lines -
<br><br>must be parallel to each other,
<br><br>$$\therefore$$ $${m_1} = {m_2} \Rightarrow - {{p\left( {{p^2} + 1} \right)} \over { - 1}} = - {{{{\left( {{p^2} + 1} \right)}^2}} \over {{p^2} + 1}}$$
<br><br>$$ \Rightarrow \left( {{p^2} + 1} \right)\left( {p + 1} \right) = 0$$
<br><br>$$ \Rightarrow p = - 1$$
<br><br>$$\therefore$$ $$p$$ can have exactly one value. | mcq | aieee-2009 |
XGLyrM09lMV4Qd9v | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | A ray of light along $$x + \sqrt 3 y = \sqrt 3 $$ gets reflected upon reaching $$X$$-axis, the equation of the reflected ray is : | [{"identifier": "A", "content": "$$y = x + \\sqrt 3 $$ "}, {"identifier": "B", "content": "$$\\sqrt 3 y = x - \\sqrt 3 $$ "}, {"identifier": "C", "content": "$$y = \\sqrt 3 x - \\sqrt 3 $$ "}, {"identifier": "D", "content": "$$\\sqrt 3 y = x - 1$$ "}] | ["B"] | null | <p>$$x + \sqrt 3 y = \sqrt 3 $$ or $$y = - {1 \over {\sqrt 3 }}x + 1$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxw9ix7m/ee1f9082-9d8b-46f7-b6e1-9aa6d676fa80/4cfc2f30-6b3f-11ec-8608-9b519146e1b7/file-1kxw9ix7n.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxw9ix7m/ee1f9082-9d8b-46f7-b6e1-9aa6d676fa80/4cfc2f30-6b3f-11ec-8608-9b519146e1b7/file-1kxw9ix7n.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2013 (Offline) Mathematics - Straight Lines and Pair of Straight Lines Question 126 English Explanation"> </p>
<p>Let $$\theta$$ be the angle which the line makes with the positive x-axis.</p>
<p>$$\therefore$$ $$\tan \theta = - {1 \over {\sqrt 3 }} = \tan \left( {\pi - {\pi \over 6}} \right)$$ or $$\theta = \pi - {\pi \over 6}$$</p>
<p>$$\therefore$$ $$\angle ABC = {\pi \over 6}$$; $$\therefore$$ $$\angle DBE = {\pi \over 6}$$</p>
<p>$$\therefore$$ the equation of the line BD is,</p>
<p>$$y = \tan {\pi \over 6}x + c$$ or $$y = {x \over {\sqrt 3 }} + c$$ ..... (1)</p>
<p>The line $$x + \sqrt 3 y = \sqrt 3 $$ intersects the x-axis at $$B(\sqrt 3 ,0)$$ and, the line (1) passes through $$B(\sqrt 3 ,0)$$.</p>
<p>$$\therefore$$ $$0 = {{\sqrt 3 } \over {\sqrt 3 }} + c$$ or, c = $$-$$1</p>
<p>Hence, the equation of the reflected ray is,</p>
<p>$$y = {x \over {\sqrt 3 }} - 1$$ or $$y\sqrt 3 = x - \sqrt 3 $$</p> | mcq | jee-main-2013-offline |
dOqIs62IhCgSeWpccJum5 | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | A ray of light is incident along a line which meets another line, 7x − y + 1 = 0, at the point (0, 1). The ray is then reflected from this point along the line, y + 2x = 1. Then the equation of the line of incidence of the ray of light is : | [{"identifier": "A", "content": "41x \u2212 38y + 38 = 0"}, {"identifier": "B", "content": "41x + 25y \u2212 25 = 0"}, {"identifier": "C", "content": "41x + 38y \u2212 38 = 0"}, {"identifier": "D", "content": "41x \u2212 25y + 25 = 0"}] | ["A"] | null | Let slope of incident ray be m.
<br><br>$$ \therefore $$ angle of incidence = angle of reflection
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266113/exam_images/yoallvxjvujp2wtehbma.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2016 (Online) 10th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 107 English Explanation">
<br><br>$$ \therefore $$ $$\left| {{{m - 7} \over {1 + 7m}}} \right| = \left| {{{ - 2 - 7} \over {1 - 14}}} \right| = {9 \over {13}}$$
<br><br>$$ \Rightarrow $$ $${{m - 7} \over {1 + 7m}} = {9 \over {13}}$$ <br><br> or $${{m - 7} \over {1 + 7m}} = - {9 \over {13}}$$
<br><br>$$ \Rightarrow $$ 13m $$-$$ 91 $$=$$ 9 + 63m
<br><br> or 13m $$-$$ 91 $$=$$ $$-$$ 9 $$-$$ 63m
<br><br>$$ \Rightarrow $$ 50m $$=$$ $$-$$ 100 or 76m $$=$$ 82
<br><br>$$ \Rightarrow $$ m $$=$$ $$ - {1 \over 2}$$
<br><br> or m $$=$$ $${{41} \over {38}}$$
<br><br>$$ \Rightarrow $$ y $$-$$ 1 $$=$$ $$-$$ $${1 \over 2}$$ (x $$-$$ 0) <br><br> or y $$-$$ 1 $$=$$ $${{41} \over {38}}$$ (x $$-$$ 0)
<br><br>i.e x + 2y $$-$$ 2 $$=$$ 0
<br><br> or 38y $$-$$ 38 $$-$$ 41x $$=$$ 0
<br><br>$$ \Rightarrow $$ 41x $$-$$ 38y + 38 $$=$$ 0 | mcq | jee-main-2016-online-10th-april-morning-slot |
zqvZ0QAtIaIYngHVRVOzH | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, $$\beta $$), then $$\beta $$ equals : | [{"identifier": "A", "content": "$${{35} \\over 3}$$"}, {"identifier": "B", "content": "$$-$$ 5"}, {"identifier": "C", "content": "$$-$$ $${{35} \\over 3}$$"}, {"identifier": "D", "content": "5"}] | ["D"] | null | $${{17 - \beta } \over { - 8}} \times {2 \over 3} = - 1$$
<br><br>$$\beta $$ = 5 | mcq | jee-main-2019-online-12th-january-morning-slot |
tqcd69HbzTLtYQQmnxrZH | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | Suppose that the points (h,k), (1,2) and (–3,4) lie
on the line L<sub>1</sub>
. If a line L<sub>2</sub>
passing through the points
(h,k) and (4,3) is perpendicular to L<sub>1</sub>
, then
$$k \over h$$
equals :
| [{"identifier": "A", "content": "$${1 \\over 3}$$"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "-$${1 \\over 7}$$"}] | ["A"] | null | Equation of line L<sub>1</sub> passing through points (1, 2) and (–3, 4) is :
<br><br>y - 2 = $$\left( {{{4 - 2} \over { - 3 - 1}}} \right)$$(x - 1)
<br><br>$$ \Rightarrow $$ y - 2 = $$ - {1 \over 2}$$(x - 1)
<br><br>$$ \Rightarrow $$ 2y – 4 = –x + 1
<br><br>$$ \Rightarrow $$ x + 2y = 5 .......(L<sub>1</sub>)
<br><br>Line L<sub>1</sub> passes through point (h, k).
<br><br>$$ \therefore $$ h + 2k = 5 .......(1)
<br><br>Line L<sub>2</sub> is perpendicular to line L<sub>1</sub>.
<br><br>$$ \therefore $$ Equation of line L<sub>2</sub> :
<br><br>2x - y = $$\lambda $$
<br><br>This line L<sub>2</sub> passes through point (4, 3).
<br><br>$$ \therefore $$ 2(4) - 3 = $$\lambda $$
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 5
<br><br>$$ \therefore $$ Line L<sub>2</sub> :
<br><br>2x - y = 5
<br><br>L<sub>2</sub> also passes through (h, k)
<br><br>$$ \therefore $$ 2h - k = 5 ............(2)
<br><br>Solving (1) and (2) we get,
<br><br>h = 3 and k = 1
<br><br>$$ \therefore $$ $${k \over h} = {1 \over 3}$$ | mcq | jee-main-2019-online-8th-april-evening-slot |
qWfD2NN5mDxLIypv4d18hoxe66ijvwusm9u | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | If the two lines x + (a – 1) y = 1 and
2x + a<sup>2</sup>y = 1 (a$$ \in $$R – {0, 1}) are perpendicular, then
the distance of their point of intersection from the
origin is : | [{"identifier": "A", "content": "$${2 \\over \\sqrt5}$$"}, {"identifier": "B", "content": "$${\\sqrt2 \\over 5}$$"}, {"identifier": "C", "content": "$${2 \\over 5}$$"}, {"identifier": "D", "content": "$$\\sqrt{2 \\over 5}$$"}] | ["D"] | null | Line 1 : x + (a – 1) y = 1
<br><br>Slope of this line (m<sub>1</sub>) = $${{ - 1} \over {a - 1}}$$
<br><br>Line 2 : 2x + a<sup>2</sup>y = 1
<br><br>Slope of this line (m<sub>2</sub>) = $$ - {2 \over {{a^2}}}$$
<br><br>Line 1 and Line 2 are perpendicular to each other.
<br><br>$$ \therefore $$ m<sub>1</sub> m<sub>2</sub> = -1
<br><br>$$ \Rightarrow $$ $$\left( {{{ - 1} \over {a - 1}}} \right)\left( { - {2 \over {{a^2}}}} \right) = - 1$$
<br><br>$$ \Rightarrow $$ $${a^3} - {a^2} + 2 = 0$$
<br><br>$$ \Rightarrow $$ $$\left( {a + 1} \right)$$$$\left( {{a^2} - 2a + 2} \right)$$ = 0
<br><br>$$ \therefore $$ $${a = - 1}$$
<br><br>So lines are
<br><br>Line 1 : x - 2y + 1 = 0
<br><br>Line 2 : 2x + y - 1 = 0
<br><br>Solving these equation we get point of intersection
<br><br>P$$\left( {{3 \over 5}, - {1 \over 5}} \right)$$
<br><br>Now distance of P from origin
<br><br>OP = $$\sqrt {{{\left( {{3 \over 5}} \right)}^2} + {{\left( { - {1 \over 5}} \right)}^2}} $$
<br><br>= $$\sqrt {{{10} \over {25}}} $$
<br><br>= $$\sqrt {{2 \over 5}} $$ | mcq | jee-main-2019-online-9th-april-evening-slot |
S0THmEuUj76nRnKbL11kmlig5f5 | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | The equation of one of the straight lines which passes through the point (1, 3) and makes an angles $${\tan ^{ - 1}}\left( {\sqrt 2 } \right)$$ with the straight line, y + 1 = 3$${\sqrt 2 }$$ x is : | [{"identifier": "A", "content": "$$4\\sqrt 2 x + 5y - \\left( {15 + 4\\sqrt 2 } \\right) = 0$$"}, {"identifier": "B", "content": "$$5\\sqrt 2 x + 4y - \\left( {15 + 4\\sqrt 2 } \\right) = 0$$"}, {"identifier": "C", "content": "$$4\\sqrt 2 x + 5y - 4\\sqrt 2 = 0$$"}, {"identifier": "D", "content": "$$4\\sqrt 2 x - 5y - \\left( {5 + 4\\sqrt 2 } \\right) = 0$$"}] | ["A"] | null | Let slope of line be m<br><br>$$ \therefore $$ $$\left| {{{m - 3\sqrt 2 } \over {1 + 3\sqrt 2 m}}} \right| = \sqrt 2 $$<br><br>$$ \Rightarrow m - 3\sqrt 2 = \pm \,\sqrt 2 \pm 6m$$<br><br>$$ \Rightarrow m \mp 6m = \pm \sqrt 2 + 3\sqrt 2 $$<br><br>$$ \Rightarrow m = - {{4\sqrt 2 } \over 5}$$ or $${{2\sqrt 2 } \over 7}$$<br><br>Hence line can be<br><br>$$y - 3 = {{ - 4\sqrt 2 } \over 5}(x - 1)$$<br><br>$$ \Rightarrow 5y - 15 = - 4\sqrt 2 x + 4\sqrt 2 $$<br><br>$$ \Rightarrow 4\sqrt 2 x + 5y - (15 + 4\sqrt 2 ) = 0$$ | mcq | jee-main-2021-online-18th-march-morning-shift |
lsan6rft | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \angle A=\frac{2 \pi}{3}, A B=A C$ and $B$ is on the positve $x$-axis. If $\mathrm{BC}=4 \sqrt{3}$ and the line $\mathrm{BC}$ intersects the line $y=x+3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is __________. | [] | null | 36 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoggtre/0457091e-f79d-43cd-92f6-312124caeaea/e227d0a0-ccae-11ee-aa98-13f456b8f7af/file-6y3zli1lsoggtrf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoggtre/0457091e-f79d-43cd-92f6-312124caeaea/e227d0a0-ccae-11ee-aa98-13f456b8f7af/file-6y3zli1lsoggtrf.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 24 English Explanation">
<br><br>$\frac{\mathrm{c}}{\sin 30^{\circ}}=\frac{4 \sqrt{3}}{\sin 120^{\circ}}[$ By sine rule $]$
<br><br>$$
2 c=8 \Rightarrow c=4
$$
<br><br>$\begin{gathered}\mathrm{AB}=|(\mathrm{b}+1)|=4 \\\\ \mathrm{~b}=3, \mathrm{~m}_{\mathrm{AB}}=0 \\\\ \mathrm{~m}_{\mathrm{BC}}=\frac{-1}{\sqrt{3}} \\\\ \mathrm{BC}:-\mathrm{y}=\frac{-1}{\sqrt{3}}(\mathrm{x}-3) \\\\ \sqrt{3} \mathrm{y}+\mathrm{x}=3\end{gathered}$
<br><br>Point of intersection : $y=x+3, \sqrt{3} y+x=3$
<br><br>$\begin{aligned} & ({\sqrt{3}+1}) y=6 \\\\ & y=\frac{6}{\sqrt{3}+1} \\\\ & x=\frac{6}{\sqrt{3}+1}-3 \\\\ & =\frac{6-3 \sqrt{3}-3}{\sqrt{3}+1} \\\\ & =3 \frac{(1-\sqrt{3})}{(1+\sqrt{3})}=\frac{-6}{(1+\sqrt{3})^2}\end{aligned}$
<br><br>$\frac{\beta^4}{\alpha^2}=36$ | integer | jee-main-2024-online-1st-february-evening-shift |
lsbkz3t1 | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | The portion of the line $4 x+5 y=20$ in the first quadrant is trisected by the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ passing through the origin. The tangent of an angle between the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ is : | [{"identifier": "A", "content": "$\\frac{30}{41}$"}, {"identifier": "B", "content": "$\\frac{8}{5}$"}, {"identifier": "C", "content": "$\\frac{2}{5}$"}, {"identifier": "D", "content": "$\\frac{25}{41}$"}] | ["A"] | null | <p>Co-ordinates of $$\mathrm{A}=\left(\frac{5}{3}, \frac{8}{3}\right)$$</p>
<p>Co-ordinates of $$\mathrm{B}=\left(\frac{10}{3}, \frac{4}{3}\right)$$</p>
<p>Slope of $$\mathrm{OA}=\mathrm{m}_1=\frac{8}{5}$$</p>
<p>Slope of $$\mathrm{OB}=\mathrm{m}_2=\frac{2}{5}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1e6sji/400a79ad-eefc-44fe-9456-362749aed957/294e71e0-d3cc-11ee-a50b-bb659a2e1d74/file-1lt1e6sjj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1e6sji/400a79ad-eefc-44fe-9456-362749aed957/294e71e0-d3cc-11ee-a50b-bb659a2e1d74/file-1lt1e6sjj.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 22 English Explanation"></p>
<p>$$\begin{aligned}
& \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\
& \tan \theta=\frac{\frac{6}{5}}{1+\frac{16}{25}}=\frac{30}{41} \\
& \tan \theta=\frac{30}{41}
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift |
lv3ve5wl | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | <p>If the line segment joining the points $$(5,2)$$ and $$(2, a)$$ subtends an angle $$\frac{\pi}{4}$$ at the origin, then the absolute value of the product of all possible values of $a$ is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}] | ["A"] | null | <p>To solve this problem, we will use the concept of the slope and tangent of the angle subtended by the line segments at the origin.</p>
<p>Given points: $$(5,2)$$ and $$(2,a)$$</p>
<p>The slope of the line joining the origin and $$(5,2)$$ is:</p>
$$m_1 = \frac{2-0}{5-0} = \frac{2}{5}$$
<p>The slope of the line joining the origin and $$(2,a)$$ is:</p>
$$m_2 = \frac{a-0}{2-0} = \frac{a}{2}$$
<p>The line segment joining these points subtends an angle $$\theta = \frac{\pi}{4}$$ at the origin.</p>
<p>According to the tangent formula for the angle between two lines:</p>
$$ \tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| $$
<p>Here, $$\theta = \frac{\pi}{4}$$ so, $$\tan \left( \frac{\pi}{4} \right) = 1$$. Therefore,</p>
$$ 1 = \left| \frac{\frac{a}{2} - \frac{2}{5}}{1 + \left(\frac{2}{5} \cdot \frac{a}{2}\right)} \right| $$
<p>This simplifies to:</p>
$$ 1 = \left| \frac{\frac{5a - 4}{10}}{1 + \frac{2a}{10}} \right| $$
<p>Multiplying both the numerator and denominator by 10 to simplify:</p>
$$ 1 = \left| \frac{5a - 4}{10 + 2a} \right| $$
<p>This leads to two equations due to the absolute value:</p>
<p>1. $$ 5a - 4 = 10 + 2a $$</p>
$$ 3a = 14 $$
$$ a = \frac{14}{3} $$
<p>2. $$ 5a - 4 = -(10 + 2a) $$</p>
$$ 5a - 4 = -10 - 2a $$
$$ 7a = -6 $$
$$ a = -\frac{6}{7} $$
<p>Thus, the possible values of $$a$$ are $$\frac{14}{3}$$ and $$-\frac{6}{7}$$.</p>
<p>The product of these values is:</p>
$$ \left| \frac{14}{3} \times -\frac{6}{7} \right| $$
$$ = \left| -\frac{84}{21} \right| $$
$$ = \left| -4 \right| $$
$$ = 4 $$
<p>Therefore, the absolute value of the product of all possible values of $$a$$ is 4.</p>
<p><strong>Answer: Option A</strong></p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv3ve74h | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | <p>Let a ray of light passing through the point $$(3,10)$$ reflects on the line $$2 x+y=6$$ and the reflected ray passes through the point $$(7,2)$$. If the equation of the incident ray is $$a x+b y+1=0$$, then $$a^2+b^2+3 a b$$ is equal to _________.</p> | [] | null | 1 | <p>Equation of incident ray : $$a x+b y+1=0$$</p>
<p>Using mirror image,</p>
<p>$$\frac{m-7}{2}=\frac{n-2}{1}=\frac{-2(14+2-6)}{5}$$</p>
$$\begin{array}{l|l}
\frac{m-7}{2}=-4 & n-2=-4 \\
m=-8+7 & n=-2 \\
m=-1 &
\end{array}$$<p></p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4q8kfi/88bdaca2-4aca-49af-a105-f44c4a414e43/484b38e0-1106-11ef-9d25-fd88fb6fcd30/file-1lw4q8kfj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4q8kfi/88bdaca2-4aca-49af-a105-f44c4a414e43/484b38e0-1106-11ef-9d25-fd88fb6fcd30/file-1lw4q8kfj.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 7 English Explanation"></p>
<p>$${ }^*$$ Note: It can be observed from diagram $$A, P, B$$' are collinear.</p>
<p>Equation of Incident Ray,</p>
<p>Using two-point form,</p>
<p>$$\begin{aligned}
& (y-10)=\frac{10+2}{3+1}(x-3) \\
& (y-10)=\frac{12}{4}(x-3) \\
& y-10=+3(x-3) \\
& y-10=+3 x-9 \\
& 3 x-y+1=0
\end{aligned}$$</p>
<p>On comparing,</p>
<p>$$\begin{aligned}
& a=3 \\
& b=-1
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-evening-shift |
lv7v3k2j | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | <p>Let two straight lines drawn from the origin $$\mathrm{O}$$ intersect the line $$3 x+4 y=12$$ at the points $$\mathrm{P}$$ and $$\mathrm{Q}$$ such that $$\triangle \mathrm{OPQ}$$ is an isosceles triangle and $$\angle \mathrm{POQ}=90^{\circ}$$. If $$l=\mathrm{OP}^2+\mathrm{PQ}^2+\mathrm{QO}^2$$, then the greatest integer less than or equal to $$l$$ is :</p> | [{"identifier": "A", "content": "42"}, {"identifier": "B", "content": "46"}, {"identifier": "C", "content": "48"}, {"identifier": "D", "content": "44"}] | ["B"] | null | <p>$$O P=O Q$$ ($$\triangle P Q R$$ is isosceles triangle)</p>
<p>Let slope of line $$O P \rightarrow m_1$$</p>
<p>So, equation $$\rightarrow y=m_1 x$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwgf0viw/7dffe883-cd31-494e-b982-28a57ce172bd/d5c35180-1773-11ef-bded-616e4abb66b9/file-1lwgf0vix.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwgf0viw/7dffe883-cd31-494e-b982-28a57ce172bd/d5c35180-1773-11ef-bded-616e4abb66b9/file-1lwgf0vix.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 4 English Explanation"></p><p>
</p><p>$$\begin{aligned}
&\begin{aligned}
& \tan 45^{\circ}=\left|\frac{m_1-m_2}{m_1 m_2}\right| \\
& 1=\left|\frac{m_1+\frac{3}{4}}{1-\frac{3}{4} m_1}\right| \\
& \Rightarrow 1-\frac{3}{4} m_1=m_1+\frac{3}{4} \\
& \frac{1}{4}=\frac{7}{4} m_1 \Rightarrow m_1=\frac{1}{7}
\end{aligned}\\
&\text { Equation } O P \rightarrow y=\frac{1}{7} x
\end{aligned}$$</p>
<p>Point of intersection of $$O P$$ & line $$3 x+4 y=12$$ is $$P\left(\frac{84}{25}, \frac{12}{25}\right)$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad O P^2=a^2=\left(\frac{84}{25}\right)^2+\left(\frac{12}{25}\right)^2=\frac{288}{25} \\
& \therefore \quad I=O P^2+P Q^2+Q O^2 \\
&=a^2+a^2+2 a^2 \\
&=4 a^2 \\
&=4 \times \frac{288}{25} \\
& I=46.08 \\
& {[I] }=46
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
rp3i2cj8qy6duBh0 | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | Let $$P = \left( { - 1,0} \right),\,Q = \left( {0,0} \right)$$ and $$R = \left( {3,3\sqrt 3 } \right)$$ be three point. The equation of the bisector of the angle $$PQR$$ is : | [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}x + y = 0$$ "}, {"identifier": "B", "content": "$$x + \\sqrt {3y} = 0$$ "}, {"identifier": "C", "content": "$$\\sqrt 3 x + y = 0$$ "}, {"identifier": "D", "content": "$$x + {{\\sqrt 3 } \\over 2}y = 0$$ "}] | ["C"] | null | <b>Given :</b> The coordinates of points $$P,Q,R$$ are $$(-1,0),$$
<br><br>$$\left( {0,0} \right),\,\left( {3,3\sqrt 3 } \right)$$ respectively
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266475/exam_images/xlctrwy5tbqvjebu61yj.webp" loading="lazy" alt="AIEEE 2007 Mathematics - Straight Lines and Pair of Straight Lines Question 134 English Explanation">
<br><br>Slope of $$QR$$ $$ = {{{y_2} - {y_1}} \over {{x_2} - {x_1}}} = {{3\sqrt 3 } \over 3}$$
<br><br>$$ \Rightarrow \tan \theta = \sqrt 3 \Rightarrow \theta = {\pi \over 3}$$
<br><br>$$ \Rightarrow \angle RQX = {\pi \over 3}$$
<br><br>$$\therefore$$ $$\angle RQP = \pi - {\pi \over 3} = {{2\pi } \over 3}$$
<br><br>Let $$QM$$ bisects the $$\angle PQR,$$
<br><br>$$\therefore$$ Slope of the line $$QM=tan$$ $${{2\pi } \over 3} = - \sqrt 3 $$
<br><br>$$\therefore$$ Equation of line $$QM$$ is $$\left( {y - 0} \right) = - \sqrt 3 \left( {x - 0} \right)$$
<br><br>$$ \Rightarrow y = - \sqrt 3 \,x \Rightarrow \sqrt 3 x + y = 0$$ | mcq | aieee-2007 |
0M6VvF2gupICRk8L | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | The lines $${L_1}:y - x = 0$$ and $${L_2}:2x + y = 0$$ intersect the line $${L_3}:y + 2 = 0$$ at $$P$$ and $$Q$$ respectively. The bisector of the acute angle between $${L_1}$$ and $${L_2}$$ intersects $${L_3}$$ at $$R$$.
<p><b>Statement-1:</b> The ratio $$PR$$ : $$RQ$$ equals $$2\sqrt 2 :\sqrt 5 $$
<br/><b>Statement-2:</b> In any triangle, bisector of an angle divide the triangle into two similar triangles.</p> | [{"identifier": "A", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is false."}, {"identifier": "C", "content": "Statement-1 is false, Statement-2 is true."}, {"identifier": "D", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1."}] | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265867/exam_images/lvhph7vjigkfkhimhf1y.webp" loading="lazy" alt="AIEEE 2011 Mathematics - Straight Lines and Pair of Straight Lines Question 128 English Explanation">
<br><br>$${L_1}:y - x = 0$$
<br><br>$${L_2}:2x + y = 0$$
<br><br>$${L_3}:y + 2 = 0$$
<br><br>On solving the equation of line $${L_1}$$ and $${L_2}$$ we get their point of
<br><br>intersection $$(0, 0)$$ i.e., origin $$O.$$
<br><br>On solving the equation of line $${L_1}$$ and $${L_3},$$
<br><br>we get $$P=(-2, -2).$$
<br><br>Similarly, we get $$Q = \left( { - 1, - 2} \right)$$
<br><br>We know that bisector of an angle of a triangle, divide the opposite side the triangle in the ratio of the sides including the angle [ Angle Bisector Theorem of a Triangle ]
<br><br>$$\therefore$$ $${{PR} \over {RQ}} = {{OP} \over {OQ}} = {{\sqrt {{{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}} } \over {\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}} }}$$
<br><br>$$ = {{2\sqrt 2 } \over {\sqrt 5 }}$$ | mcq | aieee-2011 |
1ktbg9g4w | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | Let ABC be a triangle with A($$-$$3, 1) and $$\angle$$ACB = $$\theta$$, 0 < $$\theta$$ < $${\pi \over 2}$$. If the equation of the median through B is 2x + y $$-$$ 3 = 0 and the equation of angle bisector of C is 7x $$-$$ 4y $$-$$ 1 = 0, then tan$$\theta$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264104/exam_images/x4hc1zf1v4mjcoywea4i.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266949/exam_images/fvxlje6drmq1itzrm6n3.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265458/exam_images/cy1htpps8nbvy7zqa0x1.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265739/exam_images/vsdoxcyq1slggciks1wh.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 60 English Explanation"></picture> <br><br>$$\therefore$$ $$M\left( {{{a - 3} \over 2},{{b + 1} \over 2}} \right)$$ lies on 2x + y $$-$$ 3 = 0<br><br>$$\Rightarrow$$ 2a + b = 11 ...........(i)<br><br>$$\because$$ C lies on 7x $$-$$ 4y = 1<br><br>$$\Rightarrow$$ 7a $$-$$ 4b = 1 ......... (ii)<br><br>$$\therefore$$ by (i) and (ii) : a = 3, b = 5<br><br>$$\Rightarrow$$ C(3, 5)<br><br>$$\therefore$$ m<sub>AC</sub> = 2/3<br><br>Also, m<sub>CD</sub> = 7/4<br><br>$$ \Rightarrow \tan {\theta \over 2} = \left| {{{{2 \over 3} - {4 \over 4}} \over {1 + {{14} \over {12}}}}} \right| \Rightarrow \tan {\theta \over 2} = {1 \over 2}$$<br><br>$$ \Rightarrow \tan \theta = {{2.{1 \over 2}} \over {1 - {1 \over 4}}} = {4 \over 3}$$ | mcq | jee-main-2021-online-26th-august-morning-shift |
1ldomnas6 | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | <p>The combined equation of the two lines $$ax+by+c=0$$ and $$a'x+b'y+c'=0$$ can be written as
<br/><br/>$$(ax+by+c)(a'x+b'y+c')=0$$.</p>
<p>The equation of the angle bisectors of the lines represented by the equation $$2x^2+xy-3y^2=0$$ is :</p> | [{"identifier": "A", "content": "$$3{x^2} + xy - 2{y^2} = 0$$"}, {"identifier": "B", "content": "$${x^2} - {y^2} - 10xy = 0$$"}, {"identifier": "C", "content": "$${x^2} - {y^2} + 10xy = 0$$"}, {"identifier": "D", "content": "$$3{x^2} + 5xy + 2{y^2} = 0$$"}] | ["B"] | null | For pair of straight lines in this form
<br/><br/>$a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$
<br/><br/>Equation of angle bisector is
<br/><br/>$$
\frac{x^2-y^2}{a-b}=\frac{x y}{h}
$$
<br/><br/>for $2 x^2+x y-3 y^2=0$
<br/><br/>$$
a=2, b=-3, h=\frac{1}{2}
$$
<br/><br/>Equation of angle bisector is
<br/><br/>$$
\begin{aligned}
& \frac{x^2-y^2}{5}=\frac{x y}{1 / 2} \\\\
& \Rightarrow \mathrm{x}^2-\mathrm{y}^2-10 \mathrm{xy}=0
\end{aligned}
$$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1lgswc9j4 | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | <p>If the line $$l_{1}: 3 y-2 x=3$$ is the angular bisector of the lines $$l_{2}: x-y+1=0$$ and $$l_{3}: \alpha x+\beta y+17=0$$, then $$\alpha^{2}+\beta^{2}-\alpha-\beta$$ is equal to _________.</p> | [] | null | 348 | $$
\begin{aligned}
& L_1: 3 y-2 x=3 \\\\
& L_2: x-y+1=0 \\\\
& L_3: \alpha x+\beta y+17=0
\end{aligned}
$$
<br/><br/>Point of intersection of $L_1 $ and $ L_2$ is $(0,1)$, should lie on $L_3 \Rightarrow \beta=-17$
<br/><br/>Any point, say $\left(\frac{-3}{2}, 0\right)$ on $L_1$ should be equidistant from the lines $L_2 $ and $ L_3$
<br/><br/>$$
\begin{aligned}
& \Rightarrow\left|\frac{\frac{-3}{2}-0+1}{\sqrt{1^2+1^2}}\right|=\left|\frac{\frac{-3 \alpha}{2}+0+17}{\sqrt{\alpha^2+(-17)^2}}\right| \\\\
& \Rightarrow (\alpha-7)(\alpha-17)=0
\end{aligned}
$$
<br/><br/>For $\alpha=17, L_2 $ and $ L_3$ coincides $\Rightarrow \alpha=7$
<br/><br/>$$
\begin{aligned}
\alpha^2+\beta^2-\alpha-\beta & =(7)^2+(-17)^2-7+17 \\\\
& =348
\end{aligned}
$$ | integer | jee-main-2023-online-11th-april-evening-shift |
jaoe38c1lseyt6uy | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | <p>In a $$\triangle A B C$$, suppose $$y=x$$ is the equation of the bisector of the angle $$B$$ and the equation of the side $$A C$$ is $$2 x-y=2$$. If $$2 A B=B C$$ and the points $$A$$ and $$B$$ are respectively $$(4,6)$$ and $$(\alpha, \beta)$$, then $$\alpha+2 \beta$$ is equal to</p> | [{"identifier": "A", "content": "42"}, {"identifier": "B", "content": "39"}, {"identifier": "C", "content": "48"}, {"identifier": "D", "content": "45"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt2yegk8/aaaae12e-3775-4328-9765-0804dbc280e4/fdc1fd80-d4a7-11ee-bdd1-01c80c3e2d9a/file-1lt2yegk9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt2yegk8/aaaae12e-3775-4328-9765-0804dbc280e4/fdc1fd80-d4a7-11ee-bdd1-01c80c3e2d9a/file-1lt2yegk9.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Morning Shift Mathematics - Straight Lines and Pair of Straight Lines Question 16 English Explanation"></p>
<p>$$\begin{aligned}
& A D: D C=1: 2 \\
& \frac{4-\alpha}{6-\alpha}=\frac{10}{8} \\
& \alpha=\beta \\
& \alpha=14 \text { and } \beta=14
\end{aligned}$$</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
UPCzVVhJscdJ5RBH | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | Let A $$\left( {h,k} \right)$$, B$$\left( {1,1} \right)$$ and C $$(2, 1)$$ be the vertices of a right angled triangle with AC as its hypotenuse. If the area of the triangle is $$1$$ square unit, then the set of values which $$'k'$$ can take is given by : | [{"identifier": "A", "content": "$$\\left\\{ { - 1,3} \\right\\}$$ "}, {"identifier": "B", "content": "$$\\left\\{ { - 3, - 2} \\right\\}$$ "}, {"identifier": "C", "content": "$$\\left\\{ { 1,3} \\right\\}$$"}, {"identifier": "D", "content": "$$\\left\\{ {0,2} \\right\\}$$ "}] | ["A"] | null | <b>Given :</b> The vertices of a right angled triangle $$A\left( {1,k} \right),$$
<br><br>$$B\left( {1,1} \right)$$ and $$C\left( {2,1} \right)$$ and area of $$\Delta ABC = 1$$ square unit
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264479/exam_images/nm6k8ccujpr0qdcb1imw.webp" loading="lazy" alt="AIEEE 2007 Mathematics - Straight Lines and Pair of Straight Lines Question 135 English Explanation">
<br><br>We know that, area of night angled triangle
<br><br>$$ = {1 \over 2} \times BC \times AB = 1 = {1 \over 2}\left( 1 \right)\left| {\left( {k - 1} \right)} \right|$$
<br><br>$$ \Rightarrow \pm \left( {k - 1} \right) = 2 \Rightarrow k = - 1,3$$ | mcq | aieee-2007 |
Cr16jGUNappIEabn | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices $$(0, 0)$$ $$(0, 41)$$ and $$(41, 0)$$ is : | [{"identifier": "A", "content": "820 "}, {"identifier": "B", "content": "780 "}, {"identifier": "C", "content": "901 "}, {"identifier": "D", "content": "861"}] | ["B"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxw9pbpa/316e0bc4-d33f-4ef5-89c3-8879b661f4ca/ff0879e0-6b3f-11ec-8608-9b519146e1b7/file-1kxw9pbpb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxw9pbpa/316e0bc4-d33f-4ef5-89c3-8879b661f4ca/ff0879e0-6b3f-11ec-8608-9b519146e1b7/file-1kxw9pbpb.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2015 (Offline) Mathematics - Straight Lines and Pair of Straight Lines Question 122 English Explanation"> </p>
<p>The number of integral points lie inside the triangle are</p>
<p>1. If x = 1, then y may be 1, 2, 3, ....., 39</p>
<p>2. If x = 2, then y may be 1, 2, 3, ....., 38</p>
<p>3. If x = 3, then y may be 1, 2, 3, ....., 37</p>
<p>$$ \vdots $$</p>
<p>39. If x = 39, then the value of y is 1.</p>
<p>Hence, the number of interior points are</p>
$$1 + 2 + 3 + .... + 39 = {{39 \times 40} \over 2} = 780$$<p></p> | mcq | jee-main-2015-offline |
FB22zgLrjeNln8qFXdE78 | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x + y = 5 and x = 4. Then area of $$\Delta $$ ABC (in sq. units) is : | [{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "9"}] | ["D"] | null | Median through C is x = 4
<br><br>So the coordinate of C is 4. Let C = (4, y), then the midpoint of A(1, 2) and
<br><br>C(4, y) is D which lies on the median through B.
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264030/exam_images/c4whvpojuhs4abrlpy4l.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2018 (Online) 15th April Morning Slot Mathematics - Straight Lines and Pair of Straight Lines Question 113 English Explanation">
<br><br>$$ \therefore $$$$\,\,\,$$ D = $$\left( {{{1 + 4} \over 2},{{2 + y} \over 2}} \right)$$
<br><br>Now, $${{1 + 4 + 2 + y} \over 2}$$ = 5 $$ \Rightarrow $$ y = 3.
<br><br>So, C $$ \equiv $$ (4, 3).
<br><br>The centroid of the triangle is the intersection of the mesians, Here the medians x = 4 and x + 4 and x + y = 5 intersect at G (4, 1).
<br><br>The area of triangle $$\Delta $$ABC = 3 $$ \times $$ $$\Delta $$AGC
<br><br>= 3 $$ \times $$ $${1 \over 2}$$ [1(1 $$-$$ 3) + 4(3 $$-$$ 2) + 4(2 $$-$$ 1)] = 9. | mcq | jee-main-2018-online-15th-april-morning-slot |
XxNZxBTU8Au457SbdoyVr | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | If 5, 5r, 5r<sup>2</sup> are the lengths of the sides of a triangle, then r cannot be equal to : | [{"identifier": "A", "content": "$${7 \\over 4}$$"}, {"identifier": "B", "content": "$${5 \\over 4}$$"}, {"identifier": "C", "content": "$${3 \\over 4}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}] | ["A"] | null | <p>For three lengths to form a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This is known as the triangle inequality theorem.</p>
<p>Let's apply this to our lengths: $5$, $5r$, and $5r^2$. We have three inequalities to check:</p>
<ol>
<li>$5 + 5r > 5r^2$</li>
<li>$5 + 5r^2 > 5r$</li>
<li>$5r + 5r^2 > 5$</li>
</ol>
<p>Solving these inequalities will give us a range of values for $r$. Any value outside of this range will mean that $r$ cannot form a triangle with the given sides.</p>
<ol>
<li>1. $1 + r > r^2$</li>
</ol>
<p>We can rewrite this as:
$r^2 - r - 1 < 0$
<br/><br/>$\Rightarrow\left[r-\left(\frac{1-\sqrt{5}}{2}\right)\right]\left[r-\left(\frac{1+\sqrt{5}}{2}\right)\right]<0$</p>
<p>The roots of this quadratic equation are $\frac{1-\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$, so the inequality holds for $r$ in the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$.</p>
<ol>
<li>2. $1 + r^2 > r$</li>
</ol>
<p>We can rewrite this as:
$r^2 - r + 1 > 0$</p>
$$
\begin{array}{lc}
\Rightarrow r^2-2 \cdot \frac{1}{2} r+\left(\frac{1}{2}\right)^2+1-\left(\frac{1}{2}\right)^2>0 \\\\
\Rightarrow \left(r-\frac{1}{2}\right)^2+\frac{3}{4}>0 \\\\
\Rightarrow r \in R
\end{array}
$$
<p>This inequality is true for all real numbers since the left-hand side is always positive, which means the inequality holds for all real $R$.</p>
<ol>
<li>3. $r + r^2 > 1$</li>
</ol>
<p>We can rewrite this as:
$r^2 + r - 1 > 0$</p>
<br/>$$
\begin{aligned}
& \Rightarrow {\left[r-\left(\frac{-1-\sqrt{5}}{2}\right)\right]\left[r-\left(\frac{-1+\sqrt{5}}{2}\right)\right]>0} \\\\
& {\left[\because r^2+r-1=0 \Rightarrow r=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2}\right]} \\\\
& \Rightarrow r \in\left(-\infty, \frac{-1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)
\end{aligned}
$$
<p>Taking the intersection of the intervals from the three inequalities, we get $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ as the valid range for $r$.</p>
<p>Given the solution of the inequalities $r \in \left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$, we need to determine which of the options are outside this interval. </p>
<ol>
<li>Option A: $\frac{7}{4}$</li>
</ol>
<p>The decimal representation of $\frac{7}{4}$ is 1.75, which is outside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$ or approximately (-0.618, 1.618). Hence, r cannot be equal to $\frac{7}{4}$.</p>
<ol>
<li>Option B: $\frac{5}{4}$</li>
</ol>
<p>The decimal representation of $\frac{5}{4}$ is 1.25, which is inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{5}{4}$.</p>
<ol>
<li>Option C: $\frac{3}{4}$</li>
</ol>
<p>The decimal representation of $\frac{3}{4}$ is 0.75, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{4}$.</p>
<ol>
<li>Option D: $\frac{3}{2}$</li>
</ol>
<p>The decimal representation of $\frac{3}{2}$ is 1.5, which is also inside the interval $\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$. Hence, r can be equal to $\frac{3}{2}$.</p>
<p>Therefore, the only value that r cannot be equal to, among the given options, is $\frac{7}{4}$. So, the correct answer is:</p>
<p>Option A: $\frac{7}{4}$</p>
| mcq | jee-main-2019-online-10th-january-morning-slot |
0Mvxsbf6feqHxYDR1S7k9k2k5gr1i1o | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | Let two points be A(1, –1) and B(0, 2). If a point
P(x', y') be such that the area of $$\Delta $$PAB = 5 sq.
units and it lies on the line, 3x + y – 4$$\lambda $$ = 0,
then a value of $$\lambda $$ is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "-3"}, {"identifier": "D", "content": "3"}] | ["D"] | null | Point P(x', y') lies on the line 3x + y – 4$$\lambda $$ = 0
<br><br>$$ \therefore $$ 3x' + y' – 4$$\lambda $$ = 0
<br><br>Area of $$\Delta $$PAB = 5
<br><br>$$ \Rightarrow $$ $${1 \over 2}\left| {\matrix{
1 & 1 & { - 1} \cr
1 & 0 & 2 \cr
1 & {x'} & {y'} \cr
} } \right|$$ = $$ \pm $$ 5
<br><br>$$ \Rightarrow $$ [(–2x') – (y' – 2) – (x')] = $$ \pm $$ 10
<br><br>$$ \Rightarrow $$ – 3x' – y' + 2 = $$ \pm $$ 10
<br><br>$$ \Rightarrow $$ 2 - 4$$\lambda $$ = $$ \pm $$ 10
<br><br>$$ \Rightarrow $$ $$\lambda $$ = -2 or $$\lambda $$ = 3 | mcq | jee-main-2020-online-8th-january-morning-slot |
xo2beH5Tsuz7Cfu7TO1kmiywcgk | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | Let A($$-$$1, 1), B(3, 4) and C(2, 0) be given three points. <br/>A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A<sub>1</sub> and A<sub>2</sub> be the areas of $$\Delta$$ABC and $$\Delta$$PQC respectively, such that A<sub>1</sub> = 3A<sub>2</sub>, then the value of m is equal to : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$${4 \\over {15}}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265203/exam_images/f7qd8qi2cxp1fjerej1f.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 72 English Explanation"><br><br>$${A_1} = \Delta ABC = {1 \over 2}\left| {\left| {\matrix{
{ - 1} & 1 & 1 \cr
2 & 0 & 1 \cr
3 & 4 & 1 \cr
} } \right|} \right|$$<br><br>$${A_1} = {{13} \over 2}$$<br><br>Equation of line AC is $$y - 1 = - {1 \over 3}(x + 1)$$<br><br>Line AC intersect with line y = mx at P,
<br><br>Solving we get $$P\left( {{2 \over {3m + 1}},{{2m} \over {3m + 1}}} \right)$$<br><br>Equation of line BC is y $$-$$ 0 = 4(x $$-$$ 2)<br><br>Line BC intersect with line y = mx at Q,
<br><br>Solving we get $$Q\left( {{{ - 8} \over {m - 4}},{{ - 8m} \over {m - 4}}} \right)$$<br><br>A<sub>2</sub> = Area of $$\Delta PQC = {1 \over 2}\left| {\left| {\matrix{
2 & 0 & 1 \cr
{{2 \over {3m + 1}}} & {{{2m} \over {3m + 1}}} & 1 \cr
{{{ - 8} \over {m - 4}}} & {{{ - 8m} \over {m - 4}}} & 1 \cr
} } \right|} \right|\, = {{{A_1}} \over 3} = {{13} \over 6}$$<br><br>$$ \Rightarrow $$ $$ {1 \over 2}\left( {2\left( {{{2m} \over {3m + 1}} + {{8m} \over {m - 4}}} \right) - 1\left( {{{ - 16m} \over {(3m + 1)(m - 4)}} + {{16m} \over {(3m + 1)(m - 4)}}} \right)} \right)$$$$ = \pm {{13} \over 6}$$<br><br>$$ \Rightarrow $$ $${{26{m^2}} \over {3{m^2} - 11m - 4}} = \pm {{13} \over 6}$$<br><br>$$ \Rightarrow 12{m^2} = \pm (3{m^2} - 11m - 4)$$<br><br>taking +ve sign<br><br>9m<sup>2</sup> + 11m + 4 = 0 (Rejected $$ \because $$ m is imaginary)<br><br>taking $$-$$ve sign<br><br>15m<sup>2</sup> $$-$$ 11m $$-$$ 4 = 0<br><br>$$m = 1, - {4 \over {15}}$$
<br><br>$$ \therefore $$ m = 1 (As given m > 0) | mcq | jee-main-2021-online-16th-march-evening-shift |
1ktk9nxgh | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | Let A be the set of all points ($$\alpha$$, $$\beta$$) such that the area of triangle formed by the points (5, 6), (3, 2) and ($$\alpha$$, $$\beta$$) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is : | [{"identifier": "A", "content": "$${4 \\over {\\sqrt 5 }}$$"}, {"identifier": "B", "content": "$${16 \\over {\\sqrt 5 }}$$"}, {"identifier": "C", "content": "$${8 \\over {\\sqrt 5 }}$$"}, {"identifier": "D", "content": "$${12 \\over {\\sqrt 5 }}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265668/exam_images/sxvhou3saadfcy0c0w3l.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 55 English Explanation"> <br><br>$$\left| {{1 \over 2}\left| {\matrix{
5 & 6 & 1 \cr
3 & 2 & 1 \cr
\alpha & \beta & 1 \cr
} } \right|} \right| = 12$$<br><br>4$$\alpha$$ $$-$$ 2$$\beta$$ = $$\pm$$ 24 + 8<br><br>$$\Rightarrow$$ 4$$\alpha$$ $$-$$ 2$$\beta$$ = + 24 + 8 $$\Rightarrow$$ 2$$\alpha$$ $$-$$ $$\beta$$ = 16<br><br>2x $$-$$ y $$-$$ 16 = 0 ..... (1)<br><br>$$\Rightarrow$$ 4$$\alpha$$ $$-$$ 2$$\beta$$ = $$-$$ 24 + 8 $$\Rightarrow$$ 2$$\alpha$$ $$-$$ $$\beta$$ = $$-$$8<br><br>2x $$-$$ y + 8 = 0 ...... (2)<br><br>Perpendicular distance of (1) from (0, 0)<br><br>$$\left| {{{0 - 0 - 16} \over {\sqrt 5 }}} \right| = {{16} \over {\sqrt 5 }}$$<br><br>Perpendicular distance of (2) from (0, 0)<br><br>$$\left| {{{0 - 0 + 8} \over {\sqrt 5 }}} \right| = {8 \over {\sqrt 5 }}$$ | mcq | jee-main-2021-online-31st-august-evening-shift |
1ktob4v4k | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | Let the points of intersections of the lines x $$-$$ y + 1 = 0, x $$-$$ 2y + 3 = 0 and 2x $$-$$ 5y + 11 = 0 are the mid points of the sides of a triangle $$\Delta $$ABC. Then, the area of the $$\Delta $$ABC is _____________. | [] | null | 6 | Intersection point of given lines are (1, 2), (7, 5), (2, 3)<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwjjr9ui/956a8127-b9f9-4bb1-93a5-f77f19c4361a/e8b04fa0-5074-11ec-8f7b-45f8e5a35226/file-1kwjjr9uj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kwjjr9ui/956a8127-b9f9-4bb1-93a5-f77f19c4361a/e8b04fa0-5074-11ec-8f7b-45f8e5a35226/file-1kwjjr9uj.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 1st September Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 57 English Explanation"><br>$$\Delta = {1 \over 2}\left| {\matrix{
1 & 2 & 1 \cr
7 & 5 & 1 \cr
2 & 3 & 1 \cr
} } \right|$$<br><br>$$ = {1 \over 2}[1(5 - 3) - 2(7 - 2) + 1(21 - 10)]$$<br><br>$$ = {1 \over 2}[2 - 10 + 11]$$<br><br>$$\Delta$$DEF $$ = {1 \over 2}(3) = {3 \over 2}$$<br><br>$$\Delta$$ABC = 4$$\Delta$$DEF $$ = 4\left( {{3 \over 2}} \right) = 6$$ | integer | jee-main-2021-online-1st-september-evening-shift |
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