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# [Smart Math] Time Speed Distance Problem 10

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

### Problem

In a kilometer race B can give C a 100 meters and A 150 meters start. How many meters start can C give A?

1. 50
2. 49
3. 500/9
4. 8500/9
5. 50/9

### The Usual Method

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When B gives C a 100 meters start in kilometer race, B travels 1000 meters and C 900 meters. Also when B travels 1000 meters, A will travel 850 meters. Thus we can write as follows:

A                     B                     C

850                  1000                900

Thus when C travels 1000 meters, A will travel ‘x’ meters.

A                     C

x 1000

$\therefore x=\frac{850\times 1000}{900}=\frac{8500}{9}$

Hence C can give A an start of 1000 – $\frac{8500}{9}$ = $\frac{500}{9}$meters

(Ans: 3)

Estimated Time to arrive at the answer = 45 seconds.

### Using Technique

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Note that when B travels 1000 meters, C travels 900 meters and A 850 meters. Thus when C travels 900 meters, A travels 850 meters. Thus in a 900 meters race C can give a lead of a little over 50 meters to A. Hence answer has to be closer to 50 meters but more than that. From the options, only option ‘3’ is closest to 50 and more than that (500/9 > 50).

(Ans: 3)

Estimated Time to arrive at the answer = 5 seconds.

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