In a 100 m race, A can give B a start of 10 m and C a start of 28 m. In the same race, how many metres can B give C?

Problem restatement
A's 100 m time equals B's 90 m time and equals C's 72 m time. Find B's head start over C in a 100 m race.

Given data

• Speeds proportional to covered distance in the same time: v_A : v_B : v_C = 100 : 90 : 72.

Concept/Approach
For the time B runs 100 m, C covers (v_C/v_B) × 100.

Step-by-step calculation
v_C/v_B = 72 ÷ 90 = 4 ÷ 5 C's distance when B runs 100 m = (4/5) × 100 = 80 m Hence B can give C = 100 − 80 = 20 m

Verification/Alternative
Ratio check: A:B:C speeds = 100:90:72 ⇒ for any equal time, distances scale the same way; result is consistent.

Common pitfalls

• Using 28 − 10 = 18 m directly; head starts do not combine linearly across different racers.

Final Answer
20 m