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

**Problem**

**Problem**

If a man walks to his office at 3/4^{th} of his usual rate, he reaches office 1/3^{rd} of an hour later than usual. How much time does he usually take to reach his office?

- 1/2 hour
- 1 hour
- 1/4 hour
- 3/2 hours
- 2/3 hour

**The Usual Method**

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Assuming the distance between his home and office as ‘4’, his usual speed as ‘*s*’ and the usual time as ‘*t*’.

Hence,

When *s* becomes 3/4 of *s*, *t* becomes

Hence usual time = 1 hour.

**(Ans: 2)**

*Estimated Time to arrive at the answer = 60 seconds.*

**Using Technique**

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Observe that by a 25% reduction in speed, the person reaches his office later by 33% of an hour. Thus by increasing speed by 33% of the present (not the usual speed), the person will take 25% of an hour less than what he took earlier (not the usual time).

Assuming *t* = hour. A 33% increase in speed on 3/4*s*, will give 1*s*. Thus 25% decrease of time on (*t* + 1/3) hours will give *t* hours which is the usual time. This satisfies the condition of traveling at usual speed *s* and also the arithmetical relationship that an increase of 33% can be brought to usual by a decrease of 25%. Thus *t* = 1 hour is true.

**(Ans: 2)**

*Estimated Time to arrive at the answer = 10 seconds.*

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