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A tough Quant QS

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by GMATGuruNY » Sat Jun 01, 2013 6:16 am
harsh.jain wrote:If speed increases by 33.33%, what is the percent reduction in the time taken to travel the same
distance?

It's a challenging question for me, please some one can help me solve this.
Rate and time are RECIPROCALS.
1/3 faster = 4/3 of the regular speed.
4/3 of the regular speed implies 3/4 of the regular time.
Thus, the time decreases by 1/4 = 25%.

An alternate approach is to plug in.
Let the distance = 12 miles.

Let the regular rate = 3 miles per hour.
Regular time = d/r = 12/3 = 4 hours.

1/3 faster = 3 + (1/3)3 = 4 miles per hour.
Time required at the faster rate = d/r = 12/4 = 3 hours.
Time decrease = 4-3 = 1 hour.

Percent decrease in the time = (time difference)/(regular time) * 100 = 1/4 * 100 = 25.
Last edited by GMATGuruNY on Sat Jun 01, 2013 6:20 am, edited 1 time in total.
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by Brent@GMATPrepNow » Sat Jun 01, 2013 6:19 am
harsh.jain wrote:If speed increases by 33.33%, what is the percent reduction in the time taken to travel the same
distance?
If you're not sure how to tackle this question, try plugging in some nice values that meet the given conditions.

Let's say we're traveling at 3 mph (since it's easy to increase this speed by 1/3) and we want to travel a distance of 12 miles (since 3 divides nicely into 12)

time = distance/speed
So, time = 12/3 = 4
It takes 4 hours to travel the 12 miles at 3 mph

Now let's increase the speed by 1/3 (aka 33.3333%)
(1/3) of 3 is 1, so we increase the speed from 3 mph to 4 mph
time = distance/speed
So, time at this new speed = 12/4 = 3
It now takes 3 hours to travel the 12 miles

The travel time decreased from 4 hours to 3 hours.
This represents a 25% decrease in travel time.

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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