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1

### JEE Main 2021 (Online) 27th August Morning Shift

Numerical
Let n be an odd natural number such that the variance of 1, 2, 3, 4, ......, n is 14. Then n is equal to _____________.

Correct Answer is 13

## Explanation

$${{{n^2} - 1} \over {12}} = 14 \Rightarrow n = 13$$
2

### JEE Main 2021 (Online) 26th August Evening Shift

Numerical
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 ______________.

Correct Answer is 12

## Explanation

$$5 = {{3 + 7 + x + y} \over 4} \Rightarrow x + y = 10$$

Var(x) = $$10 = {{{3^2} + {7^2} + {x^2} + {y^2}} \over 4} - 25$$

$$140 = 49 + 9 + {x^2} + {y^2}$$

$${x^2} + {y^2} = 82$$

x + y = 10

$$\Rightarrow$$ (x, y) = (9, 1)

Four numbers are 21, 9, 10, 8

Mean = $${{48} \over 4}$$ = 12
3

### JEE Main 2021 (Online) 25th July Morning Shift

Numerical
Consider the following frequency distribution :

Class : 10-20 20-30 30-40 40-50 50-60
Frequency : $$\alpha$$ 110 54 30 $$\beta$$

If the sum of all frequencies is 584 and median is 45, then | $$\alpha$$ $$-$$ $$\beta$$ | is equal to _______________.

Correct Answer is 164

## Explanation

$$\because$$ Sum of frequencies = 584

$$\Rightarrow$$ $$\alpha$$ + $$\beta$$ = 390

Now, median is at $${{584} \over 2}$$ = 292th

$$\because$$ Median = 45 (lies in class 40 - 50)

$$\Rightarrow$$ $$\alpha$$ + 110 + 54 + 15 = 292

$$\Rightarrow$$ $$\alpha$$ = 113, $$\beta$$ = 277

$$\Rightarrow$$ | $$\alpha$$ $$-$$ $$\beta$$ | = 164
4

### JEE Main 2021 (Online) 22th July Evening Shift

Numerical
Consider the following frequency distribution :

Class : 0 $$-$$ 6 6 $$-$$ 12 12 $$-$$ 18 18 $$-$$ 24 24 $$-$$ 30
Frequency : a b 12 9 5

If mean = $${{309} \over {22}}$$ and median = 14, then the value (a $$-$$ b)2 is equal to _____________.

Correct Answer is 4

## Explanation

Class Frequency $${x_i}$$ $${f_i}{x_i}$$
0-6 a 3 3a
6-12 b 9 9b
12-18 12 15 180
18-24 9 21 189
24-30 5 27 135
$$N = (26 + a + b)$$ $$(504 + 3a + 9b)$$

Mean = $${{3a + 9b + 180 + 189 + 135} \over {a + b + 26}} = {{309} \over {22}}$$

$$\Rightarrow 66a + 198b + 11088 = 309a + 309b + 8034$$

$$\Rightarrow 243a + 111b = 3054$$

$$\Rightarrow 81a + 37b = 1018$$ $$\to$$ (1)

Now, Median $$= 12 + {{{{a + b + c} \over 2} - (a + b)} \over {12}} \times 6 = 14$$

$$\Rightarrow {{13} \over 2} - \left( {{{a + b} \over 4}} \right) = 2$$

$$\Rightarrow {{a + b} \over 4} = {9 \over 2}$$

$$\Rightarrow a + b = 18$$ $$\to$$ (2)

From equation (1) \$ (2)

a = 8, b = 10

$$\therefore$$ $${(a - b)^2} = {(8 - 10)^2}$$

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