Smart Math Time Speed Distance

[Smart Math] Time Speed Distance Problem 2

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

Time Speed Distance


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

  1. 1/2 hour
  2. 1 hour
  3. 1/4 hour
  4. 3/2 hours
  5. 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, t=\frac{d}{s}

When s becomes 3/4 of s, t becomes \frac{d}{\frac{3}{4}s}=t+\frac{1}{3}

\therefore t+\frac{1}{3}=\frac{d}{\frac{3}{4}s}

\therefore \frac{d}{s}+\frac{1}{3}=\frac{4d}{3s}

\therefore \frac{1}{3}=\frac{4d}{3s}-\frac{d}{s}

\therefore \frac{1}{3}=\frac{d}{3s}

\therefore \frac{d}{s}=1=t

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 3/4s, will give 1s. 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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