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# [Smart Math] Time Speed Distance Problem 9

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

A and B cycle from Mumbai to Pune, a distance of 192 kms at 18 km/hr and 14 km/hr respectively. A reaches Pune and starts back for Mumbai. How far from Pune will he meet B?

1. 12 kms
2. 16 kms
3. 24 kms
4. 42 kms
5. 60 kms

### The Usual Method As can be seen from the figure above that A will have to travel a distance of (192 + x) kms and B travels a distance of (192 – x) kms, so that they both meet at point M.

Thus time taken by A to travel = $\frac{192+x}{18}$

And time taken by B to travel = $\frac{192-x}{14}$

Since, both A and B start together, time taken by both to meet will be equal.

Hence $\frac{192+x}{18}$ = $\frac{192-x}{14}$ $\therefore \frac{192+x}{9}=\frac{192-x}{7}$ $\therefore 1344+7x=1728-9x$ $\therefore 16x=1728+1344=384$ $\therefore x=\frac{384}{16}=24$kms

(Ans: 3)

Estimated Time to arrive at the answer = 90 seconds.

### Using Technique

Distance traveled by A and B will be in the ratio of their speeds since time taken to meet is same for both. Thus ratio of distance traveled by A and B = ratio of speeds of A and B = 18 : 14 = 9 : 7.

Total distance traveled by both A and B together = twice the distance between Mumbai and Pune = 2 x 192 = 384 kms. Dividing 384 kms between A and B in the ratio of 9 : 7, we get distance traveled by A = $\frac{384}{(9+7)}\times 9=$ 216 kms.

Hence the meeting point is 192 – 216 = 24 kms from Pune.

(Ans: 3)

Estimated Time to arrive at the answer = 15 seconds.

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