Let the mean and variance of 8 numbers $-10,-7,-1, x, y, 9,2,16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively. Then the mean of 4 numbers $x, y, x+y+1,|x-y|$ is:

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Statistics

Let the mean and variance of 8 numbers

- 10, - 7, - 1, x, y, 9, 2, 16

\frac{7}{2}

and

\frac{293}{4}

, respectively. Then the mean of 4 numbers

x, y, x + y + 1, ∣ x - y ∣

is:

Answer: (A)

The mean condition gives

\frac{( x + y ) + 9}{8} = \frac{7}{2}

, so

x + y = 19

. The squares of the known entries sum to

100 + 49 + 1 + 81 + 4 + 256 = 491

, so the variance condition

\frac{491 + x ^{2} + y ^{2}}{8} - (\frac{7}{2})^{2} = \frac{293}{4}

gives

x^{2} + y^{2} = 193

. Then

2 x y = (x + y)^{2} - (x^{2} + y^{2}) = 361 - 193 = 168

, so

x y = 84

and

x, y

are the roots of

t^{2} - 19 t + 84 = 0

, namely

12

and

7

. The four numbers are

12, 7, 20, 5

, whose mean is

\frac{12 + 7 + 20 + 5}{4} = \frac{44}{4} = 11.

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option