[Smart Math] TIME SPEED DISTANCE PROBLEM 2

September 1, 2010 |

Author LazyMaths.com

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

Time Speed Distance

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

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

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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