JEE Main 2026 — Mathematics Probability & Statistics

The class marks (mid-points) $x_i$xi​ for the given classes are $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$7.5,12.5,17.5,22.5,27.5,32.5.

The sum of the frequencies is:

$$\sum f_i = 2 + k + 28 + 54 + (k + 1) + 5 = 2k + 90$$∑fi​=2+k+28+54+(k+1)+5=2k+90

The sum of the products $f_i x_i$fi​xi​ is:

$$\sum f_i x_i = 2(7.5) + k(12.5) + 28(17.5) + 54(22.5) + (k + 1)(27.5) + 5(32.5)$$∑fi​xi​=2(7.5)+k(12.5)+28(17.5)+54(22.5)+(k+1)(27.5)+5(32.5)

$$\sum f_i x_i = 15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5$$∑fi​xi​=15+12.5k+490+1215+27.5k+27.5+162.5

$$\sum f_i x_i = 40k + 1910$$∑fi​xi​=40k+1910

The mean is given as $21$21. Using the formula for the mean $\displaystyle \bar{x} = \dfrac{\sum f_i x_i}{\sum f_i}$xˉ=∑fi​∑fi​xi​​:

$$\dfrac{40k + 1910}{2k + 90} = 21$$2k+9040k+1910​=21

$$40k + 1910 = 21(2k + 90)$$40k+1910=21(2k+90)

$$40k + 1910 = 42k + 1890$$40k+1910=42k+1890

$$2k = 20 \Rightarrow k = 10$$2k=20⇒k=10

Now, substituting $x = 10$x=10 into the given options to check which equation has $10$10 as a root:

For $2x^2 - 19x - 10 = 0$2x2−19x−10=0:

$$2(10)^2 - 19(10) - 10 = 200 - 190 - 10 = 0$$2(10)2−19(10)−10=200−190−10=0

Thus, $k = 10$k=10 is a root of $2x^2 - 19x - 10 = 0$2x2−19x−10=0.

Answer: $2x^2 - 19x - 10 = 0$2x2−19x−10=0