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Time, Speed, Distance - Variable Speed Meeting Point

CAT 1998 Slot 1 · Quantitative Ability · Medium · Arithmetic

This is a medium Quantitative Ability question from the CAT 1998 Slot 1 paper. It tests Arithmetic. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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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/hr4\text{ km/hr}. While the other man travelled with varying speed as follows: in the first hour his speed was 2 km/hr2\text{ km/hr}, in the second hour it was 2.5 km/hr2.5\text{ km/hr}, in the third hour it was 3 km/hr3\text{ km/hr}, and so on. When will they meet each other?

  1. A.

    7 hr

  2. B.

    10 hr

  3. C.

    35 km from A

  4. D.

    Mid-way between A and B

Answer

D

Explanation

Relative distance covered in each hour nn is 4+[2+0.5(n−1)]=6+0.5(n−1)4 + [2 + 0.5(n-1)] = 6 + 0.5(n-1). Sum after nn hours =n2[12+0.5(n−1)]=72  ⟹  n(11.5+0.5n)=144  ⟹  n2+23n−288=0  ⟹  (n−9)(n+32)=0  ⟹  n=9 hours= \frac{n}{2}[12 + 0.5(n-1)] = 72 \implies n(11.5 + 0.5n) = 144 \implies n^2 + 23n - 288 = 0 \implies (n-9)(n+32) = 0 \implies n = 9\text{ hours}. In 9 hours, person from A covers 9×4=36 km9 \times 4 = 36\text{ km}, which is exactly half of 72 km72\text{ km}. So they meet mid-way between A and B.

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