If the median A and mode B of following distribution satisfy the relation 7 (B – A) = 9C, then find the value of C : Class | 0-30 | 30-60 | 60-90 | 90-120 Frequency | 4 | 5 | 7 | 4

Correct — option (4), 6.

Step 1 — total and cumulative frequencies.
N = 4 + 5 + 7 + 4 = 20
Cumulative frequencies: 4, 9, 16, 20

Step 2 — the median class.
N ÷ 2 = 10; the cumulative frequency first passes 10 in the class 60-90.
L = 60, cf before the class = 9, f = 7, h = 30

Step 3 — median A.
A = L + [(N/2 − cf) ÷ f] × h
A = 60 + [(10 − 9) ÷ 7] × 30 = 60 + 30/7 = 450/7

Step 4 — the modal class.
The highest frequency, 7, makes 60-90 the modal class.
f₁ = 7, f₀ = 5 (class before), f₂ = 4 (class after)

Step 5 — mode B.
B = L + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h
B = 60 + [2 ÷ 5] × 30 = 60 + 12 = 72

Step 6 — the gap B − A.
72 − 450/7 = 504/7 − 450/7 = 54/7

Step 7 — solve for C.
7 (B – A) = 7 × 54/7 = 54
9C = 54
C = 6

The idea to remember: for grouped data, find the class first (N/2 for the median, the highest frequency for the mode), then interpolate inside it.

Why the others are wrong

• (1)
  9 — Option (1) would need 9C = 81, so B − A = 81/7 ≈ 11.6.

  The working gives B = 72 and A = 450/7 ≈ 64.3, a gap of 54/7 ≈ 7.7. Then 7 (B – A) = 54, and C = 54 ÷ 9 = 6.

• (2)
  8 — Option (2) would need 7 (B – A) = 72, a gap of 72/7 ≈ 10.3 between mode and median.

  The figure 72 is the mode B itself, but the relation uses seven times the gap B − A. With A = 450/7, that gap is 54/7, so 7 (B – A) = 54 and C = 6.

• (3)
  7 — Option (3) would need 7 (B – A) = 63, so B − A = 9 exactly.

  The median is not a whole number here: A = 60 + 30/7 ≈ 64.29. So B − A ≈ 72 − 64.29 = 7.71, seven times that is 54, and C = 6.

For grouped data with class width h, the median lies in the class where the cumulative frequency first reaches N/2:
Median = L + [(N/2 − cf) ÷ f] × h, with L the lower limit of that class, cf the cumulative frequency before it and f its frequency.

The mode lies in the class with the highest frequency:
Mode = L + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h, with f₀ and f₂ the frequencies of the classes before and after.

Both give estimates from the grouped table, not exact values from the raw data.

Key facts

• N = 20; cumulative frequencies 4, 9, 16, 20; the median class and the modal class are both 60-90.
• Median A = 60 + (1/7) × 30 = 450/7 ≈ 64.29.
• Mode B = 60 + (2/5) × 30 = 72.
• B − A = 54/7, so 7 (B – A) = 54 = 9C and C = 6.
• Empirical relation for a moderately skewed distribution: Mode ≈ 3 Median − 2 Mean.

Common traps

• Using the empirical relation in place of the formula. Mode ≈ 3 Median − 2 Mean gives about 69.9, which makes 7 (B – A) = 39 and C = 13/3, not a whole number.
• Taking the cumulative frequency of the median class itself, 16, in place of the class before, 9. That gives a median below 60, outside its own class.
• Swapping f₀ and f₂ in the mode formula. f₀ is the class before (5) and f₂ the class after (4); swapped, the numerator becomes 7 − 4 = 3 and the mode 78.

UnlockIAS compared this question with questions from other RAS Prelims papers and found none similar enough to link.

• practice — not a real PYQ

  Find the mode of the following distribution : Class | 0-10 | 10-20 | 20-30 | 30-40 Frequency | 3 | 8 | 5 | 4

  • (a)15
  • (b)16.25
  • (c)18.75
  • (d)13.75
  Answer(2) — Modal class 10-20 (f₁ = 8), with f₀ = 3 and f₂ = 5: Mode = 10 + [(8 − 3) ÷ (16 − 3 − 5)] × 10 = 10 + (5/8) × 10 = 16.25.

  Option (1) is the mid-point of the modal class, option (3) is the median of this distribution, and option (4) swaps f₀ and f₂.

• practice — not a real PYQ

  In a moderately skewed distribution, the mean is 40 and the median is 38. Using the empirical relation between mean, median and mode, what is the mode ?

  • (a)34
  • (b)36
  • (c)42
  • (d)44
  Answer(1) — Mode ≈ 3 × Median − 2 × Mean = 3 × 38 − 2 × 40 = 114 − 80 = 34. Option (4) swaps the mean and median in the relation (3 × 40 − 2 × 38 = 44); options (2) and (3) do not come from the relation.