If the mean of the data Class 5-10 10-15 15-20 20-25 25-30 30-35 --- --- --- --- --- --- --- Frequency 2 k 28 54 k+1 5 is 21, then k is one of the roots of the equation :

Mathematics · Statistics

JEE Main 2026 — 2 April, Morning Shift — Question 26

If the mean of the data

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

is 21, then k is one of the roots of the equation :

Option A:
$2x^2 - 23x - 10 = 0$
Option B:
$4x^2 - 35x + 24 = 0$
Option C:
$2x^2 - 19x - 10 = 0$
Correct
Option D:
$2x^2 - 35x + 98 = 0$

Answer: C

Step-by-step solution

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

Compute midpoints: 7.5, 12.5, 17.5, 22.5, 27.5, 32.5. Calculate sum of products: $7.5 \times 2 = 15$ , $12.5 \times k = 12.5k$ , $17.5 \times 28 = 490$ , $22.5 \times 54 = 1215$ , $27.5 \times (k+1) = 27.5k + 27.5$ , $32.5 \times 5 = 162.5$ .

Total sum = $15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5 = 1910 + 40k$ . Total frequency = $2 + k + 28 + 54 + (k+1) + 5 = 90 + 2k$ . Set mean = $\frac{1910 + 40k}{90 + 2k} = 21$ . Solve: $1910 + 40k = 21(90 + 2k) = 1890 + 42k$ , so $20 = 2k$ , $k = 10$ . Substitute $k = 10$ into each quadratic option; only option C gives zero: $2(10)^2 - 19(10) - 10 = 200 - 190 - 10 = 0$ .

Hence k is a root of $2x^2 - 19x - 10 = 0$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Statistics
Topic: Measures of Central Tendency