question_id stringlengths 8 35 | subject stringclasses 1
value | chapter stringclasses 32
values | topic stringclasses 178
values | question stringlengths 26 9.64k | options stringlengths 2 1.63k | correct_option stringclasses 5
values | answer stringclasses 293
values | explanation stringlengths 13 9.38k | question_type stringclasses 3
values | paper_id stringclasses 149
values |
|---|---|---|---|---|---|---|---|---|---|---|
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/m72... | 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>$$ \Rig... | mcq | jee-main-2020-online-4th-september-morning-slot |
zf49X1ZTnEJU37kGDbjgy2xukf461jt2 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let x<sub>i</sub>
(1 $$ \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 $$ R, then the
standard deviation of these observati... | [{"identifier": "A", "content": "$${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}} $$
<br><br>[ Standard deviation
is free from... | mcq | jee-main-2020-online-3rd-september-evening-slot |
HOGv0aaEIEk875NhPrjgy2xukezfbwzn | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the variance of the terms in an increasing A.P.,<br/> b<sub>1</sub>
, b<sub>2</sub>
, b<sub>3</sub>
,....,b<sub>11</sub> is 90, then the common
difference of this A.P. is_______. | [] | null | 3 | Let the common difference = d<br><br>
and $${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(... | 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,... | [{"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 -... | 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... | 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(... | 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 + ... | 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>$$ \Rightarrow $$ $${{{2^2} +... | integer | jee-main-2020-online-7th-january-morning-slot |
CFSOgEQFEAJx9rVOX1jgy2xukf0q5sme | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | For the frequency distribution :
<br/>Variate (x) : x<sub>1</sub> x<sub>2</sub> x<sub>3</sub>
.... x<sub>15</sub>
<br/>Frequency (f) : f<sub>1</sub>
f<sub>2</sub>
f<sub>3</sub>
...... f<sub>15</sub>
<br/>where 0 < x<sub>1</sub>
< x<sub>2</sub>
< x<sub>3</sub>
< ... < x<sub>15</sub> = 1... | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2"}] | ["A"] | null | If variate varries from m to M then variance
<br><br>$${\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( {... | 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>$$9{k^2} - 18... | integer | jee-main-2021-online-24th-february-evening-slot |
UDM8kSpbt9yLePlNNi1kluz00j2 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let X<sub>1</sub>, X<sub>2</sub>, ......., X<sub>18</sub> be eighteen observations such <br/>that $$\sum\limits_{i = 1}^{18} {({X_i} - } \alpha ) = 36$$ and $$\sum\limits_{i = 1}^{18} {({X_i} - } \beta {)^2} = 90$$, where $$\alpha$$ and $$\beta$$ are distinct real numbers. If the standard deviation of these observation... | [] | null | 4 | Given, $$\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 {{... | 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"... | ["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>
<... | [] | 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>$${{\sum {y_i^2} } \over n} - {3^2} = 1 \Rightarrow \s... | integer | jee-main-2021-online-16th-march-evening-shift |
IlkocwVJmPZsS9fqjq1kmko2650 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new set is constructed by adding 1 into each of first 2n numbers, and subtracting 1 from each of the remaining n numbers. If the variance of the new set is k, then 9k is equal t... | [] | null | 68 | Let first 2n observations are x<sub>1</sub>, x<sub>2</sub> ...................., x<sub>2n</sub><br><br>and last n observations are y<sub>1</sub>, y<sub>2</sub> ....................., y<sub>n</sub><br><br>Now, $${{\sum {{x_i}} } \over {2n}} = 6$$, $${{\sum {{y_i}} } \over n} = 3$$<br><br>$$ \Rightarrow \sum {{x_i}} = 1... | 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 $$ \Rightarrow... | mcq | jee-main-2021-online-18th-march-evening-shift |
1krpt023e | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are : | [{"identifier": "A", "content": "10, 11"}, {"identifier": "B", "content": "3, 18"}, {"identifier": "C", "content": "8, 13"}, {"identifier": "D", "content": "1, 20"}] | ["A"] | null | Let other two numbers be a, (21 $$-$$ a)<br><br>Now,<br><br>$$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$... | 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 $$-$$... | 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... | [{"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 +... | 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>$${\sigma ^2} = {{{n_1}\sigma _1^2 + {n_2}\sigma _2^2} \over {{n_1} + {n_2}}} + {{{n_1}{n_2}} \over {{{({n_1} + {n_2})}^2}}}{({\overline x _1} - {\overline ... | mcq | jee-main-2021-online-25th-july-evening-shift |
1ks01t341 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and $${{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} = ... | 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>$${\sigma ... | mcq | jee-main-2021-online-26th-august-morning-shift |
1ktd4qfua | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let the mean and variance of four numbers 3, 7, x and y(x > y) be 5 and 10 respectively. Then the mean of four numbers 3 + 2x, 7 + 2y, x + y and x $$-$$ y is ______________. | [] | null | 12 | $$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... | 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 ma... | [] | 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 ... | 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, $${{x_1^2 + x_2^2 + ...x_5^2 + {8^2} + {6^2}} \over 7} - 64 = 16$$... | 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>$$ = {{{x_1} + {x_2} + {x_3} + ... | mcq | jee-main-2022-online-29th-june-morning-shift |
1l54tcmsg | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The number of values of a $$\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}^... | 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 stu... | 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>$$\therefore$$ $${(x - 6)^2} + {(y - 6)^2} = 8$$ ..... ... | 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 ... | 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>$$\therefore$$ $$x_1^2 + x_2^2 + \,\,.....\,\, + \,\,x_{50}^2 = 50 ... | mcq | jee-main-2022-online-26th-june-evening-shift |
1l59lchg3 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean deviation about the mean of the numbers 1, 2, 3, .........., n, where n is odd, is $${{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 \... | 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 $$ = {{2 \ti... | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6hzvkr1 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and standard deviation of 40 observations are 30 and 5 respectively. It was noticed that two of these observations 12 and 10 were wrongly recorded. If $$\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 ... | 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>... | 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 $$\alpha$... | [] | null | 4 | <p>Given $$\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}... | 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 ... | 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
... | [{"identifier": "A", "content": "450"}, {"identifier": "B", "content": "900"}, {"identifier": "C", "content": "650"}, {"identifier": "D", "content": "500"}] | ["D"] | null | $$
\begin{array}{cll}
\quad \mathbf{A} & \mathbf{B} & \mathbf{A}+\mathbf{B} \\
\overline{\mathrm{x}}_{1}=40 & \overline{\mathrm{x}}_{2}=55 & \overline{\mathrm{x}}=50 \\
\sigma_{1}=\alpha & \sigma_{2}=30-\alpha & \sigma^{2}=350 \\
\mathrm{n}_{1}=100 & \mathrm{n}_{2}=\mathrm{n} & 100+\mathrm{n} \\\\
\overline{\mathrm{x}}... | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldomwdbe | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :</p> | [{"identifier": "A", "content": "1792"}, {"identifier": "B", "content": "1216"}, {"identifier": "C", "content": "1456"}, {"identifier": "D", "content": "1072"}] | ["D"] | null | Let observation $1,3,5, a, b$
<br/><br/>$$
\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}... | 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>
<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;b... | [] | 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/fi... | 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 $$ = {{\sum {|{a_1} - \mu |} } \over {100}} = {{{{\left( {{{99} \over 2} + {{97} \over 2}\, + \,...\, + \,{1 \over 2}} \right)}^2}} \over {100}}... | mcq | jee-main-2023-online-30th-january-evening-shift |
1ldr873oc | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and variance of 7 observations are 8 and 16 respectively. If one observation 14 is omitted and a and b are respectively mean and variance of remaining 6 observation, then $$\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}} } \ove... | 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 $$\left| {\overline x + \overline y - {\sigma ^2}} \right|$... | [] | null | 603 | <p>$$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}... | 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>
<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-s... | 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/>
$\Rightarrow N=20$
<b... | 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>$$ = {{{a_1} + {a_2} + {a_3}... | mcq | jee-main-2023-online-24th-january-evening-shift |
lgnxgk71 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is : | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "12"}, {"identifier": "C", "content": "13"}, {"identifier": "D", "content": "14"}] | ["C"] | null | 1. Calculate the sum of the original observations:
<br/><br/>$$\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... | 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 corre... | 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>
<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;bor... | [] | 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/>$$
\begin{aligned}
\sum f_i & =80 \\\\
\text { M.D } & =\frac{\sum f_i\left|x_i-5\right|}{\sum f_i} \\... | integer | jee-main-2023-online-13th-april-morning-shift |
1lgrgr0fx | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the positive numbers $$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$$ is equal to ______... | [] | null | 211 | Since $a_1, a_2, a_3, a_4, a_5$ are in geometric progression, we can write :
<br/><br/>$a_2 = r a$,
<br/><br/>$a_3 = r^2 a$,
<br/><br/>$a_4 = r^3 a$,
<br/><br/>$a_5 = r^4 a$.
<br/><br/>where $r$ is the common ratio and $a_1$ = $$a$$ is the first term.
<br/><br/>Given that the mean of the series is $\frac{31}{10}... | integer | jee-main-2023-online-12th-april-morning-shift |
1lgsvg819 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean of 6 observations $$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}$$"}, {"identifier": "D", "content": "3"}] | ["B"] | null | <p>Given that the mean of the observations ${x, y, 1, 2, 4, 5}$ is 5, we get the equation:</p>
<p>$x + y + 1 + 2 + 4 + 5 = 6 \cdot 5 \Rightarrow x + y = 18$ ..........$(1)$</p>
<p>We are also given that the variance of the observations is 10. Using the formula for variance, we have:</p>
<p>$V = \frac{\Sigma x_i^2}{n} -... | mcq | jee-main-2023-online-11th-april-evening-shift |
1lguu6kre | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let sets A and B have 5 elements each. Let the mean of the elements in sets A and B be 5 and 8 respectively and the variance of the elements in sets A and B be 12 and 20 respectively. A new set C of 10 elements is formed by subtracting 3 from each element of $$\mathrm{A}$$ and adding 2 to each element of $$\mathrm{B... | [{"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,... | 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>
<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-bre... | [{"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/>$$
\left(\because k=\frac{-16}{3} \text { is not p... | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgxwcpil | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean of the frequency distribution</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;borde... | [] | null | 151 | Given mean is 28
<br/><br/>$$
\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}
& \therefor... | 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 $$\mathrm{m}+\mathrm{... | [{"identifier": "A", "content": "317"}, {"identifier": "B", "content": "316"}, {"identifier": "C", "content": "314"}, {"identifier": "D", "content": "315"}] | ["A"] | null | $$
\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... | 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 | $$
\begin{array}{|c|c|c|}
\hline x_i & (x_i-\bar{x}) & (x_i-\bar{x})^2 \\
\hline x & x-9 & (x-9)^2 \\
\hline y & y-9 & (y-9)^2 \\
\hline 10 & 1 & 1 \\
\hline 12 & 3 & 9 \\
\hline 6 & -3 & 9 \\
\hline 12 & 3 & 9 \\
\hline 4 & -5 & 25 \\
\hline 8 & -1 & 1 \\
\hline x+y+92 & & (x-9)^2+(y-9)^2+54 \\
\hline
\end{array}
$$
<... | 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/>$$
\begin{array}{ll}
& \sigma^2=\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 \cdot n_2}{\left(n_1+n_2\right)^2}... | mcq | jee-main-2023-online-6th-april-morning-shift |
1lh2ypm8n | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean and variance of the frequency distribution</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-colo... | [] | null | 25 | $$
\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... | 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 ... | [{"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/>$$
\begin{aligned}
& \text { Variance }=\frac{84}{25} \\\\
& \sum_{i=1}^{10}\left(x_i-\alpha\right)=2 \\\\
& \Rightarrow x_1+x_2+\ldots+x_{10}-10 \alpha=2 \\\\
& \Rightarrow \frac{x_1+x_2+\ldots+x_{10}}{10}-\alpha=\frac{2}{10} \\\\
& \Rightarrow \frac{6}{5}-\alpha... | mcq | jee-main-2024-online-1st-february-evening-shift |
lsaohwm3 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | Let the median and the mean deviation about the median of 7 observation $170,125,230,190,210$, a, b be 170 and $\frac{205}{7}$ respectively. Then the mean deviation about the mean of these 7 observations is : | [{"identifier": "A", "content": "31"}, {"identifier": "B", "content": "28"}, {"identifier": "C", "content": "30"}, {"identifier": "D", "content": "32"}] | ["C"] | null | $$
\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{al... | 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 $\mathrm{a}_1, \mathrm{a}_2... | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "$\\sqrt{115}$"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "$\\sqrt{5}$"}] | ["D"] | null | <p>$$\begin{aligned}
& \sum_{\mathrm{k}=1}^{10} \mathrm{a}_{\mathrm{k}}=50 \\
& \mathrm{a}_1+\mathrm{a}_2+\ldots+\mathrm{a}_{10}=50 \quad \text{.... (i)}\\
& \sum_{\forall \mathrm{k}<\mathrm{j}} \mathrm{a}_{\mathrm{k}} \mathrm{a}_{\mathrm{j}}=1100 \quad \text{.... (ii)}\\
& \text { If } \mathrm{a}_1+\mathrm{a}_2+\ldots... | mcq | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscoct80 | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If $$\mu$$ and $$\sigma^2$$ denote the mean and variance of the correct observations respectively, then $$15\left(\mu+\mu^2+\sigma^2\right)$$ is eq... | [] | null | 2521 | <p>Let the incorrect mean be $$\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$$</p>
<p>$$\Ri... | 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>$$\begin{aligned}
& \sum \frac{\left(x_i-\bar{x}\right)^2}{n}=194 \\
... | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsf0j0pi | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>If the mean and variance of the data $$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}$$</p>
<p>$... | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsg56yjt | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>The variance $$\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-st... | [] | 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,... | 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;b... | [{"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,... | 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;w... | [{"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,... | 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{... | 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>$$\begin{aligned}
& \frac{|9-18|+|25-18|+|a-18|+|b-18|+|c-18|}{5}=4 \\
& 9+7+|a-18|+|b-18|+|c-18|=20 \\
& |a-18|+|b-18|+|c-18|=4 \\
& \frac{136}{5}=\frac{706+a^2+b... | integer | jee-main-2024-online-8th-april-evening-shift |
lv9s20ji | maths | statistics | calculation-of-standard-deviation,-variance-and-mean-deviation-of-grouped-and-ungrouped-data | <p>Let the mean and the standard deviation of the probability distribution</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.... | [] | null | 5 | <p>Mean $$(\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-... | 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... | 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>$$\therefore$$ $${m_1} = {m_2} \Rightarrow - {{p\left( {{p^2} +... | mcq | aieee-2009 |
XGLyrM09lMV4Qd9v | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | A ray of light along $$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$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxw9ix7m/ee1f9082-9d8b-46f7-b6e1-9aa6d676fa80/4cfc2f30-6b3f-11ec-8608-9b519146e1b7/file-1kxw9ix7n.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxw9i... | mcq | jee-main-2013-offline |
dOqIs62IhCgSeWpccJum5 | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | A ray of light is incident along a line which meets another line, 7x − y + 1 = 0, at the point (0, 1). The ray is then reflected from this point along the line, y + 2x = 1. Then the equation of the line of incidence of the ray of light is : | [{"identifier": "A", "content": "41x \u2212 38y + 38 = 0"}, {"identifier": "B", "content": "41x + 25y \u2212 25 = 0"}, {"identifier": "C", "content": "41x + 38y \u2212 38 = 0"}, {"identifier": "D", "content": "41x \u2212 25y + 25 = 0"}] | ["A"] | null | Let slope of incident ray be m.
<br><br>$$ \therefore $$ angle of incidence = angle of reflection
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266113/exam_images/yoallvxjvujp2wtehbma.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" ... | mcq | jee-main-2016-online-10th-april-morning-slot |
zqvZ0QAtIaIYngHVRVOzH | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, $$\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 $$ x + 2y = 5 .......(L<sub>1</sub>)
<br><br>L... | mcq | jee-main-2019-online-8th-april-evening-slot |
qWfD2NN5mDxLIypv4d18hoxe66ijvwusm9u | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | If the two lines x + (a – 1) y = 1 and
2x + a<sup>2</sup>y = 1 (a$$ \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 $$ m<sub>1</sub> m<sub>2</sub... | mcq | jee-main-2019-online-9th-april-evening-slot |
S0THmEuUj76nRnKbL11kmlig5f5 | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | The equation of one of the straight lines which passes through the point (1, 3) and makes an angles $${\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 -... | ["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 $${{2\sqrt 2 } \over... | mcq | jee-main-2021-online-18th-march-morning-shift |
lsan6rft | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | Let $A B C$ be an isosceles triangle in which $A$ is at $(-1,0), \angle A=\frac{2 \pi}{3}, A B=A C$ and $B$ is on the positve $x$-axis. If $\mathrm{BC}=4 \sqrt{3}$ and the line $\mathrm{BC}$ intersects the line $y=x+3$ at $(\alpha, \beta)$, then $\frac{\beta^4}{\alpha^2}$ is __________. | [] | null | 36 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoggtre/0457091e-f79d-43cd-92f6-312124caeaea/e227d0a0-ccae-11ee-aa98-13f456b8f7af/file-6y3zli1lsoggtrf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoggtre/0457091e-f79d-43cd-92f6-312124caeaea/e227d0a0-ccae-11ee-aa... | integer | jee-main-2024-online-1st-february-evening-shift |
lsbkz3t1 | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | The portion of the line $4 x+5 y=20$ in the first quadrant is trisected by the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ passing through the origin. The tangent of an angle between the lines $\mathrm{L}_1$ and $\mathrm{L}_2$ is : | [{"identifier": "A", "content": "$\\frac{30}{41}$"}, {"identifier": "B", "content": "$\\frac{8}{5}$"}, {"identifier": "C", "content": "$\\frac{2}{5}$"}, {"identifier": "D", "content": "$\\frac{25}{41}$"}] | ["A"] | null | <p>Co-ordinates of $$\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}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly... | mcq | jee-main-2024-online-27th-january-morning-shift |
lv3ve5wl | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | <p>If the line segment joining the points $$(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}$$
<p>The slope of the line jo... | mcq | jee-main-2024-online-8th-april-evening-shift |
lv3ve74h | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | <p>Let a ray of light passing through the point $$(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}$$<p></p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4q8kfi... | integer | jee-main-2024-online-8th-april-evening-shift |
lv7v3k2j | maths | straight-lines-and-pair-of-straight-lines | angle-between-two-lines | <p>Let two straight lines drawn from the origin $$\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$$, then the greatest int... | [{"identifier": "A", "content": "42"}, {"identifier": "B", "content": "46"}, {"identifier": "C", "content": "48"}, {"identifier": "D", "content": "44"}] | ["B"] | null | <p>$$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$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwgf0viw/7dffe883-cd31-494e-b982-28a57ce172bd/d5c35180-1773-11ef-bded-616e4abb66b9/file-1lwgf0vi... | mcq | jee-main-2024-online-5th-april-morning-shift |
rp3i2cj8qy6duBh0 | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | Let $$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)$$ respectively
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266475/exam_images/xlctrwy5tbqvjebu61yj.webp" loading="lazy" alt="AIEEE 2007 Ma... | mcq | aieee-2007 |
0M6VvF2gupICRk8L | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | The lines $${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 $$
<br/><b>Statement-... | [{"identifier": "A", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is false."}, {"identifier": "C", "content": "Statement-1 is false, Statement-2 is true."}, {"identifier": "D", "content... | ["B"] | null | <img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265867/exam_images/lvhph7vjigkfkhimhf1y.webp" loading="lazy" alt="AIEEE 2011 Mathematics - Straight Lines and Pair of Straight Lines Question 128 English Explanation">
<br><br>$${L_1}:y - x = 0$$
<br><br>$${L_2}:2x + y = 0$$
<br><br... | mcq | aieee-2011 |
1ktbg9g4w | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | Let ABC be a triangle with A($$-$$3, 1) and $$\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}$$"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264104/exam_images/x4hc1zf1v4mjcoywea4i.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266949/exam_images/fvxlje6drmq1itzrm6n3.webp"><source media="(max-wid... | mcq | jee-main-2021-online-26th-august-morning-shift |
1ldomnas6 | maths | straight-lines-and-pair-of-straight-lines | angle-bisector | <p>The combined equation of the two lines $$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{align... | 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... | 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... | 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/nm6k8ccujpr0qdcb... | 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-8608-9b519146e1b... | 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... | 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>$5 + 5r > 5r^2$</l... | 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>$$ \Rightarrow $$ $${1 \over 2}\left| {\matrix{
1 & 1 & { - 1} \cr
1 & 0 & 2 \cr
1 & {x'} & {y'} \cr
} } \right|$$ = $$ \p... | mcq | jee-main-2020-online-8th-january-morning-slot |
xo2beH5Tsuz7Cfu7TO1kmiywcgk | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | Let A($$-$$1, 1), B(3, 4) and C(2, 0) be given three points. <br/>A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A<sub>1</sub> and A<sub>2</sub> be the areas of $$\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}}$$"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265203/exam_images/f7qd8qi2cxp1fjerej1f.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 72 Engli... | mcq | jee-main-2021-online-16th-march-evening-shift |
1ktk9nxgh | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | Let A be the set of all points ($$\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 }}$$"}] | ["C"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265668/exam_images/sxvhou3saadfcy0c0w3l.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Evening Shift Mathematics - Straight Lines and Pair of Straight Lines Question 55 Engl... | mcq | jee-main-2021-online-31st-august-evening-shift |
1ktob4v4k | maths | straight-lines-and-pair-of-straight-lines | area-of-triangle-and-condition-of-collinearity | Let the points of intersections of the lines x $$-$$ y + 1 = 0, x $$-$$ 2y + 3 = 0 and 2x $$-$$ 5y + 11 = 0 are the mid points of the sides of a triangle $$\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/956a812... | integer | jee-main-2021-online-1st-september-evening-shift |
Subsets and Splits
Chapter Question Count Chart
Displays the number of questions in each chapter and a graphical representation, revealing which chapters have the most questions for focused study.
SQL Console for archit11/jee_math
Counts the number of occurrences of each paper ID, which could help identify duplicates but lacks deeper analytical insight.