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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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