Question:
Distance between A and B is 72 km. Two men started walking from A and B at the same time towards each other. The person who started from A travelled uniformly with average speed of 4 km/hr. While the other man travelled with varying speed as follows: in the first hour his speed was 2 km/hr, in the second hour it was 2.5 km/hr, in the third hour it was 3 km/hr, and so on. When will they meet each other?
Distance between A and B is 72 km. Two men started walking from A and B at the same time towards each other. The person who started from A travelled uniformly with average speed of 4 km/hr. While the other man travelled with varying speed as follows: in the first hour his speed was 2 km/hr, in the second hour it was 2.5 km/hr, in the third hour it was 3 km/hr, and so on. When will they meet each other?
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The second man covers 2, 2.5, 3, ... km in successive hours, an arithmetic progression. Add his distance to the first man's 4 km per hour until the total reaches 72 km.
Updated On: Nov 4, 2026
- 7 hr
- 10 hr
- Mid-way between A and B
- 35 km from A
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The Correct Option is C
Solution and Explanation
Step 1: Distance covered by the man from A:
He walks at 4 km/hr, so in t hours he covers \( 4t \) km.
Step 2: Distance covered by the man from B:
His speeds in successive hours are 2, 2.5, 3, ... km/hr, so the distances in successive hours are 2, 2.5, 3, ... km. This is an arithmetic progression with first term \( 2 \) and common difference \( 0.5 \). In t hours (t a whole number) he covers \[ S_t = \frac{t}{2}\left[2\times 2 + (t-1)\times 0.5\right] = 2t + \frac{t(t-1)}{4}. \]
Step 3: Total distance equals 72 km when they meet:
\[ 4t + 2t + \frac{t(t-1)}{4} = 72 \]
Multiplying by 4: \( 24t + t^2 - t = 288 \), so \( t^2 + 23t - 288 = 0 \), i.e. \( (t-9)(t+32) = 0 \). Hence \( t = 9 \) hours.
Step 4: Find where they meet:
The man from A has walked \( 4\times 9 = 36 \) km. The man from B has walked \( 2\times 9 + \frac{9\times 8}{4} = 18 + 18 = 36 \) km. Total \( 36 + 36 = 72 \) km, as required. Since 36 km is exactly half of 72 km, they meet mid-way between A and B, after 9 hours.
Step 5: Check the options:
(1) 7 hr: the total distance by then is \( 28 + 24.5 = 52.5 \) km, not 72 km. (2) 10 hr: the total would be \( 40 + 42.5 = 82.5 \) km, more than 72 km, so they have already met. (4) 35 km from A: the man from A has walked 36 km, not 35 km. (3) Mid-way: matches.
Final Answer:
The correct option is (3): Mid-way between A and B (they meet after 9 hours, 36 km from each end).
He walks at 4 km/hr, so in t hours he covers \( 4t \) km.
Step 2: Distance covered by the man from B:
His speeds in successive hours are 2, 2.5, 3, ... km/hr, so the distances in successive hours are 2, 2.5, 3, ... km. This is an arithmetic progression with first term \( 2 \) and common difference \( 0.5 \). In t hours (t a whole number) he covers \[ S_t = \frac{t}{2}\left[2\times 2 + (t-1)\times 0.5\right] = 2t + \frac{t(t-1)}{4}. \]
Step 3: Total distance equals 72 km when they meet:
\[ 4t + 2t + \frac{t(t-1)}{4} = 72 \]
Multiplying by 4: \( 24t + t^2 - t = 288 \), so \( t^2 + 23t - 288 = 0 \), i.e. \( (t-9)(t+32) = 0 \). Hence \( t = 9 \) hours.
Step 4: Find where they meet:
The man from A has walked \( 4\times 9 = 36 \) km. The man from B has walked \( 2\times 9 + \frac{9\times 8}{4} = 18 + 18 = 36 \) km. Total \( 36 + 36 = 72 \) km, as required. Since 36 km is exactly half of 72 km, they meet mid-way between A and B, after 9 hours.
Step 5: Check the options:
(1) 7 hr: the total distance by then is \( 28 + 24.5 = 52.5 \) km, not 72 km. (2) 10 hr: the total would be \( 40 + 42.5 = 82.5 \) km, more than 72 km, so they have already met. (4) 35 km from A: the man from A has walked 36 km, not 35 km. (3) Mid-way: matches.
Final Answer:
The correct option is (3): Mid-way between A and B (they meet after 9 hours, 36 km from each end).
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