Statistics & Linear ProgrammingmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Grouped-Data Mean = 21 gives k = 10, Root of 2x²-19x-10 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
If the mean of the data with classes 55–10,1010, 10–15,1515, 15–20,2020, 20–25,2525, 25–30,3030, 30–3535 and frequencies 2,k,28,54,k+1,52, k, 28, 54, k+1, 5 is 2121, then kk is one of the roots of the equation
A2x2−23x−10=02x^2-23x-10=0
B4x2−35x+24=04x^2-35x+24=0
C2x2−19x−10=02x^2-19x-10=0correct
D2x2−35x+98=02x^2-35x+98=0
Solution
Step 1: Class midpoints are 7.5,12.5,17.5,22.5,27.5,32.57.5, 12.5, 17.5, 22.5, 27.5, 32.5. Mean formula:
21=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=\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}.
 Step 2: Compute the numerator constants:
15+12.5k+490+1215+27.5k+27.5+162.5=1910+40k,15+12.5k+490+1215+27.5k+27.5+162.5=1910+40k,
 and the total frequency =90+2k=90+2k. So
21=1910+40k90+2k.21=\frac{1910+40k}{90+2k}.
 Step 3: Cross-multiply:
21(90+2k)=1910+40k ⇒ 1890+42k=1910+40k ⇒ 2k=20 ⇒ k=10.21(90+2k)=1910+40k\ \Rightarrow\ 1890+42k=1910+40k\ \Rightarrow\ 2k=20\ \Rightarrow\ k=10.
 Step 4: Test which equation has 1010 as a root:
2(10)2−19(10)−10=200−190−10=0.2(10)^2-19(10)-10=200-190-10=0.
 So k=10k=10 satisfies 2x2−19x−10=02x^2-19x-10=0. Correct answer: (3)
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