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# [Smart Math] Time Speed Distance Problem 3

Here’s and example of a SMART MATH problem for TIME SPEED DISTANCE.

### Problem

A and B have to go from X to Y which are at a distance of 230 kms from each other. A drives at a constant speed of 100 kms/hr for the first 115 kms and then at an average speed of 50 kms/hr for the remaining distance. B starts with an initial speed of 50 kms/hr at X and drives with constant acceleration such that when he reaches Y, his speed is 100 kms/hr. Which of the following statement is true?

1. A reaches Y first
2. B reaches Y first
3. A and B reach Y together
4. Cannot calculate from given data
5. None of these

### The Usual Method

A covers half of the distance at 100 kms/hr and the remaining half at 50 kms/hr. Hence average speed of A = $\frac{2\times 50\times 100}{50+100}=\frac{10000}{150}=$ 66.67 kms/hr

Average speed of B for the whole journey = $\frac{100+50}{2}=\frac{150}{2}=$ 75 kms/hr

$\because$ Average speed of B > Average speed of A,

B will reach Y before A does.

(Ans: 2)

Estimated Time to arrive at the answer = 45 seconds.

### Using Technique

Note that speed of B follows Arithmetic mean and that of A follows Harmonic mean.

We know that for a given set of numbers, Harmonic Mean is always < Arithmetic Mean

Hence speed of A < speed of B, so B reaches Y before A.

(Ans: 2)

Estimated Time to arrive at the answer = 10 seconds.

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