Let X= x N:1 x 19 and for some a,b R, Y= ax+b:x X. If the mean an… — Solution | JEE Mathematics

MCQMediumJEE 2026Measures of Dispersion

Let $X=\{x\in\mathbb{N}:1\le x\le19\}$ and for some $a,b\in\mathbb{R}$ , $Y=\{ax+b:x\in X\}$ . If the mean and variance of the elements of $Y$ are $30$ and $750$ respectively, then the sum of all possible values of $b$ is

A
$60$
B
$80$
C
$100$
D
$20$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: $X=\{x\in\mathbb{N}:1\le x\le19\}$ , $Y=\{ax+b:x\in X\}$ , mean of $Y$ is $30$ , and variance of $Y$ is $750$ .

Find: The sum of all possible values of $b$ .

Step 1: Mean and variance of set $X$ .

\text{Mean of } X = \frac{1+19}{2}=10

Variance of first $n$ natural numbers is

\frac{n^2-1}{12}

So,

\text{Var}(X)=\frac{19^2-1}{12}=30

Step 2: Use transformation properties. For $Y=ax+b$ ,

\text{Mean}(Y)=a\,\text{Mean}(X)+b

\text{Var}(Y)=a^2\text{Var}(X)

Step 3: Apply variance condition.

a^2(30)=750

a^2=25

a=\pm5

Step 4: Apply mean condition.

a(10)+b=30

For $a=5$ ,

b=30-50=-20

For $a=-5$ ,

b=30+50=80

Step 5: Sum of all possible values of $b$ .

-20+80=60

Therefore, the sum of all possible values of $b$ is $60$ , so the correct option is A.

Common mistakes

Using the variance transformation incorrectly as $\text{Var}(Y)=a\,\text{Var}(X)+b$ . This is wrong because variance is unaffected by translation and scales by the square of the multiplicative factor. Use $\text{Var}(Y)=a^2\text{Var}(X)$ instead.
Taking only $a=5$ from $a^2=25$ and missing $a=-5$ . This is wrong because both signs satisfy the variance condition. Check both values of $a$ before finding all possible values of $b$ .
Computing the variance of $X$ incorrectly by using the formula for the mean or by treating the set as $0$ to $19$ . This is wrong because $X$ contains the integers from $1$ to $19$ . Use the variance of the first $19$ natural numbers: $\frac{19^2-1}{12}=30$ .

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