1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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