JEE Main 2026 — Apr 2, Shift 1 — Mathematics solution

JEE Main2026Apr 2, Shift 1Mathematics

Class	5-10	10-15	15-20	20-25	25-30	30-35
Frequency	$2$	$k$	$28$	$54$	$k + 1$	$5$

If the mean of the data above is $21$ , then $k$ is one of the roots of the equation :

Solution

Use class midpoints $7.5, 12.5, 17.5, 22.5, 27.5, 32.5$ . The mean is

$\overset{x}{ˉ} = \frac{2 ( 7.5 ) + k ( 12.5 ) + 28 ( 17.5 ) + 54 ( 22.5 ) + ( k + 1 ) ( 27.5 ) + 5 ( 32.5 )}{2 + k + 28 + 54 + ( k + 1 ) + 5} = 21$

Numerator $= 15 + 12.5 k + 490 + 1215 + 27.5 k + 27.5 + 162.5 = 1910 + 40 k$ . Denominator $= 90 + 2 k$ .

$21 (90 + 2 k) = 1910 + 40 k \Rightarrow 1890 + 42 k = 1910 + 40 k \Rightarrow 2 k = 20 \Rightarrow k = 10$ .

Test $k = 10$ in the options: $2 (100) - 19 (10) - 10 = 200 - 190 - 10 = 0$ . Hence $k$ satisfies $2 x^{2} - 19 x - 10 = 0$ .

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.