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JEE Mathematics 2025 Question with Solution

Let the Mean and Variance of five observations xix_i, i=1,2,3,4,5i = 1, 2, 3, 4, 5 be 55 and 1010 respectively. If three observations are x1=1,x2=3,x3=ax_1 = 1, x_2 = 3, x_3 = a and x4=7,x5=bx_4 = 7, x_5 = b with a>ba > b, then the Variance of the observations n+xnn + x_n for n=1,2,3,4,5n = 1, 2, 3, 4, 5 is

  • A

    1717

  • B

    16.416.4

  • C

    17.417.4

  • D

    1616

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Mean of the five observations is 55 and variance is 1010. The observations are 1,3,a,7,b1, 3, a, 7, b with a>ba>b.

Find: The variance of the transformed observations n+xnn+x_n for n=1,2,3,4,5n=1,2,3,4,5.

Using the mean,

1+3+a+7+b5=5\frac{1+3+a+7+b}{5}=5

so,

11+a+b=2511+a+b=25 a+b=14a+b=14

Using the variance formula,

12+32+a2+72+b25−(5)2=10\frac{1^2+3^2+a^2+7^2+b^2}{5}-(5)^2=10 1+9+a2+49+b25−25=10\frac{1+9+a^2+49+b^2}{5}-25=10 59+a2+b2=17559+a^2+b^2=175 a2+b2=116a^2+b^2=116

Now use

(a+b)2=a2+b2+2ab(a+b)^2=a^2+b^2+2ab

Substituting the known values,

142=116+2ab14^2=116+2ab 196=116+2ab196=116+2ab 2ab=80⇒ab=402ab=80 \Rightarrow ab=40

So aa and bb are roots of

t2−(a+b)t+ab=0t^2-(a+b)t+ab=0

that is,

t2−14t+40=0t^2-14t+40=0 (t−10)(t−4)=0(t-10)(t-4)=0

Hence,

t=10 or t=4t=10 \text{ or } t=4

Since a>ba>b,

a=10,b=4a=10, \quad b=4

Therefore the original observations are

1,3,10,7,41, 3, 10, 7, 4

The new observations n+xnn+x_n are

1+x1=2,2+x2=5,3+x3=13,4+x4=11,5+x5=91+x_1=2, \quad 2+x_2=5, \quad 3+x_3=13, \quad 4+x_4=11, \quad 5+x_5=9

So the new set is

2,5,13,11,92, 5, 13, 11, 9

Its mean is

2+5+13+11+95=405=8\frac{2+5+13+11+9}{5}=\frac{40}{5}=8

Now compute the variance:

22+52+132+112+925−(8)2\frac{2^2+5^2+13^2+11^2+9^2}{5}-(8)^2 =4+25+169+121+815−64=\frac{4+25+169+121+81}{5}-64 =4005−64=80−64=16=\frac{400}{5}-64=80-64=16

Therefore, the variance of the transformed observations is 1616. The correct option is D.

Use centered values directly

Given: The original mean is 55 and variance is 1010.

Find: The variance of n+xnn+x_n.

From the mean condition,

a+b=14a+b=14

So the mean of the transformed observations is

(1+1)+(2+3)+(3+a)+(4+7)+(5+b)5=26+a+b5=8\frac{(1+1)+(2+3)+(3+a)+(4+7)+(5+b)}{5}=\frac{26+a+b}{5}=8

Now measure each transformed value from the new mean 88:

2−8=−6,5−8=−3,(3+a)−8=a−5,11−8=3,(5+b)−8=b−32-8=-6, \quad 5-8=-3, \quad (3+a)-8=a-5, \quad 11-8=3, \quad (5+b)-8=b-3

Using the working from the extracted solution,

(a−5)2+(b−5)2=26(a-5)^2+(b-5)^2=26

Then the variance is evaluated there as

36+9+26+95=16\frac{36+9+26+9}{5}=16

Therefore, the variance of the new observations is 1616. The correct option is D.

Common mistakes

  • Using the variance formula incorrectly by omitting the square on the mean. Variance here is ∑xi25−(xˉ)2\frac{\sum x_i^2}{5}-(\bar{x})^2, not ∑xi25−xˉ\frac{\sum x_i^2}{5}-\bar{x}. Always square the mean before subtracting.

  • Ignoring the condition a>ba>b after solving the quadratic. The roots are 1010 and 44, and the larger value must be assigned to aa. Otherwise the observations are mislabeled.

  • Computing the new observations as xn+n−1x_n+n-1 or changing every term by the same amount. The transformation is position-wise: first term is 1+x11+x_1, second is 2+x22+x_2, and so on.

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