1

### JEE Main 2019 (Online) 10th January Morning Slot

In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is
A
42
B
102
C
1
D
38

## Explanation

Let n(A) = number of students opted mathematic = 70,

n(B) = number of studens opted Physics = 46,

n(C) = number of students opted Chemistry = 28,

n(A $\cap$ B) = 23,

n(B $\cap$ C) = 9,

n(A $\cap$ C) = 14,

n(A $\cap$ B $\cap$ C) = 4,

Now n(A $\cup$ B $\cup$ C)

= n(A) + n(B) + n(C) $-$ n(A $\cap$ B) $-$ n(B $\cap$ C)

$-$ n(A $\cap$ C) + n(A $\cap$ B $\cap$ C)

= 70 + 46 + 28 $-$ 23 $-$ 9 $-$ 14 + 4 = 102

So number of students not opted for any course

Total $-$ n(A $\cup$ B $\cup$ C)

= 140 $-$ 102 = 38
2

### JEE Main 2019 (Online) 10th January Morning Slot

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is -
A
6 : 7
B
10 : 3
C
4 : 9
D
5 : 8

## Explanation

Let two observations are x1 & x2

mean = ${{\sum {{x_i}} } \over 5} = 5$

$\Rightarrow 1 + 3 + 8 + {x_1} + {x_2} = 25$

$\Rightarrow {x_1} + {x_2} = 13$      . . . . (1)

variance $\left( {{\sigma ^2}} \right)$ = ${{\sum {x_i^2} } \over 5} - 25 = 9.20$

$\Rightarrow$  ${\sum {x_i^2 = 171} }$

$\Rightarrow$  $x_1^2 + x_2^2 = 97$      . . . . . (2)

$\Rightarrow$(x1 + x2)2 $-$ 2x1x2 = 97

$\Rightarrow$ 169 - 2x1x2 = 97

or   x1x2 = 36

$\therefore$  x1 : x2 = 4 : 9
3

### JEE Main 2019 (Online) 10th January Evening Slot

If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, ….., x5 and –50 is equal to
A
582.5
B
507.5
C
586.5
D
509.5

## Explanation

$\overline x = 10 \Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 50}$

S.D.  $= \sqrt {{{\sum\limits_{i = 1}^5 {x_i^2} } \over 5} - {{\left( {\overline x } \right)}^2}} = 8$

$\Rightarrow \,\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} = 109$

variance $= \,\,{{\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} + {{\left( { - 50} \right)}^2}} \over 6} - \left( {\sum\limits_{i = 1}^5 {{{{x_i} - 50} \over 6}} } \right)$

$= \,\,507.5$
4

### JEE Main 2019 (Online) 11th January Morning Slot

The outcome of each of 30 items was observed; 10 items gave an outcome ${1 \over 2}$ – d each, 10 items gave outcome ${1 \over 2}$ each and the remaining 10 items gave outcome ${1 \over 2}$+ d each. If the variance of this outcome data is ${4 \over 3}$ then |d| equals :
A
${2 \over 3}$
B
${{\sqrt 5 } \over 2}$
C
${\sqrt 2 }$
D
2

## Explanation

Variance is independent of region. So we shift the given data by ${1 \over 2}$.

so,   ${{10{d^2} + 10 \times {0^2} + 10{d^2}} \over {30}} - {\left( 0 \right)^2} = {4 \over 3}$

$\Rightarrow$  d2 $=$ 2 $\Rightarrow$ $\left| d \right| = \sqrt 2$