Statistics & Linear ProgrammingmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Mean, Median, Mean Deviation Data Problem: 2a = 131 | JEE 2026

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

Question
Suppose that the mean and median of the non-negative numbers 21,8,17,a,51,103,b,13,6721,8,17,a,51,103,b,13,67 (with a>ba>b) are 4040 and 2121 respectively. If the mean deviation about the median is 2626, then 2a2a is equal to
A109109
B117117
C161161
D131131correct
Solution
Step 1: There are 99 numbers; mean =40=40:
280+a+b9=40 ⇒ a+b=80.\frac{280+a+b}{9}=40\ \Rightarrow\ a+b=80.
 (The seven known numbers sum to 280280.) Step 2: Median =21=21. Mean deviation about the median =26=26:
∣21−21∣+∣8−21∣+∣17−21∣+∣51−21∣+∣103−21∣+∣13−21∣+∣67−21∣+∣a−21∣+∣b−21∣9=26.\frac{|21-21|+|8-21|+|17-21|+|51-21|+|103-21|+|13-21|+|67-21|+|a-21|+|b-21|}{9}=26.
 Step 3: The known absolute deviations sum to 0+13+4+30+82+8+46=1830+13+4+30+82+8+46=183. With a>21>ba>21>b, ∣a−21∣+∣b−21∣=(a−21)+(21−b)=a−b|a-21|+|b-21|=(a-21)+(21-b)=a-b. So
183+(a−b)=234 ⇒ a−b=51.183+(a-b)=234\ \Rightarrow\ a-b=51.
 Step 4: Solve a+b=80a+b=80 and a−b=51a-b=51: 2a=1312a=131. Correct answer: (4)
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