Two men set out at the same time to walk towards each other from points A and B, 72 km apart. The first man walks at the speed of 4 kmph while the second walks 2 km in the first hour, 2½ km in the second hour, 3 km in the third hour, and so on. The two men will meet
in 8 hours
nearer to A than B
nearer to B than A
midway between A and B
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC CAPF – 2017
Man 1 starts from A at a constant speed of 4 kmph. In T hours, Man 1 covers a distance of 4T km.
Man 2 starts from B. In the first hour, he covers 2 km. In the second hour, 2.5 km. In the third, 3 km, and so on. This is an arithmetic progression of distances covered per hour with first term a = 2 and common difference d = 0.5.
The distance covered by Man 2 in T hours is the sum of the first T terms of this AP:
Sum = (T/2) * [2a + (T-1)d] = (T/2) * [2*2 + (T-1)*0.5] = (T/2) * [4 + 0.5T – 0.5] = (T/2) * [3.5 + 0.5T].
When they meet, the sum of the distances covered by both men equals the total distance:
Distance by Man 1 + Distance by Man 2 = 72
4T + (T/2) * (3.5 + 0.5T) = 72
4T + 1.75T + 0.25T^2 = 72
0.25T^2 + 5.75T – 72 = 0
Multiplying by 4 to clear decimals:
T^2 + 23T – 288 = 0
Using the quadratic formula T = [-b ± sqrt(b^2 – 4ac)] / 2a:
T = [-23 ± sqrt(23^2 – 4*1*(-288))] / 2*1
T = [-23 ± sqrt(529 + 1152)] / 2
T = [-23 ± sqrt(1681)] / 2
Since sqrt(1681) = 41 (as 40^2=1600, 41^2=1681), and time must be positive:
T = (-23 + 41) / 2 = 18 / 2 = 9 hours.
They meet after 9 hours.
Distance covered by Man 1 in 9 hours = 4 kmph * 9 hours = 36 km from A.
Distance covered by Man 2 in 9 hours = Sum of 9 terms of AP (a=2, d=0.5) = (9/2) * [2*2 + (9-1)*0.5] = 4.5 * [4 + 8*0.5] = 4.5 * [4 + 4] = 4.5 * 8 = 36 km from B.
Since they meet after covering 36 km from A and 36 km from B, which adds up to the total distance of 72 km, they meet exactly midway between A and B.