Math JEE Adv. 2025 paper 1

 

To solve the problem, we'll analyze the given frequency distribution step-by-step and match the entries in List-I to List-II.

Given Data:

  • Values (x): 4, 5, 8, 9, 6, 12, 11

  • Frequencies (f): 5, f1f2, 2, 1, 1, 3

  • Total frequency: 19

  • Median: 6

Step 1: Determine f1 and f2

The total frequency is:

5+f1+f2+2+1+1+3=19    f1+f2=7.

Since the median is 6, the cumulative frequency up to the median must cover the 10th observation (as 19+12=10).

Cumulative Frequencies:

  • Up to 4: 5

  • Up to 5: 5+f1

  • Up to 6: 5+f1+1=6+f1

For the median to be 6, the 10th observation must fall in the 6-value category:

5+f1106+f1    f1=4.

Thus, f2=7f1=3.

Verification of 7f1+9f2:

7(4)+9(3)=28+27=55.

So, (P)55.

Step 2: Compute the Mean (μ)

First, calculate the sum of x×f:

4×5+5×4+8×3+9×2+6×1+12×1+11×3=20+20+24+18+6+12+33=133.

The mean is:

μ=13319=7.

Step 3: Compute Mean Deviation about the Mean (α)

Calculate xμ×f for each value:

47×5=15,57×4=8,87×3=3,97×2=4,67×1=1,127×1=5,117×3=12.

Total deviation:

15+8+3+4+1+5+12=48.

Thus, 19α=48, so (Q)48.

Step 4: Compute Mean Deviation about the Median (β)

The median is 6. Calculate x6×f:

46×5=10,56×4=4,86×3=6,96×2=6,66×1=0,126×1=6,116×3=15.

Total deviation:

10+4+6+6+0+6+15=47.

Thus, 19β=47, so (R)47.

Step 5: Compute Variance (σ2)

Calculate (xμ)2×f:

(47)2×5=45,(57)2×4=16,(87)2×3=3,(97)2×2=8,(67)2×1=1,(127)2×1=25,(117)2×3=48.

Total squared deviation:

45+16+3+8+1+25+48=146.

Thus, 19σ2=146, so (S)146.

Final Matching:

  • (P)55

  • (Q)48 (3)

  • (R)47 (2)

  • (S)146 (1)

Correct Option: C

C

Comments