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

**Problem**

**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?

- 12 kms
- 16 kms
- 24 kms
- 42 kms
- 60 kms

**The Usual Method**

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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 =

And time taken by B to travel =

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

Hence =

kms

**(Ans: 3)**

*Estimated Time to arrive at the answer = 90 seconds.*

**Using Technique**

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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 = 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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