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Walking at \(60\%\) of his usual speed a man takes \(24\) minutes more to cover a distance. What is his usual time to

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by VJesus12 » Wed Jun 02, 2021 10:50 am

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Walking at \(60\%\) of his usual speed a man takes \(24\) minutes more to cover a distance. What is his usual time to cover this distance?

A. 30
B. 36
C. 42
D. 48
E. 54

Answer: B

Source: GMAT Club Tests
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Source: — Problem Solving |

VJesus12 wrote:
Wed Jun 02, 2021 10:50 am
Walking at \(60\%\) of his usual speed a man takes \(24\) minutes more to cover a distance. What is his usual time to cover this distance?

A. 30
B. 36
C. 42
D. 48
E. 54

Answer: B

Source: GMAT Club Tests
Let's start with a WORD EQUATION:
distance traveled at REGULAR speed = distance traveled at REDUCED speed

distance = (rate)(time)
Let R = REGULAR speed
So 0.6R = REDUCED speed
Let t = travel time (in minutes) at REGULAR speed
So t + 24 = travel time (in minutes) at REDUCED speed

Now plug this information into our WORD EQUATION to get:
Rt = (0.6R)(t + 24)
Divide both sides by R to get: t = (0.6)(t + 24)
Rewrite 0.6 as 3/5 to get: t = (3/5)(t + 24)
Expand: t = 3t/5 + 72/5
Multiply both sides by 5 to get: 5t = 3t + 72
Solve to get: t = 36

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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VJesus12 wrote:
Wed Jun 02, 2021 10:50 am
Walking at \(60\%\) of his usual speed a man takes \(24\) minutes more to cover a distance. What is his usual time to cover this distance?

A. 30
B. 36
C. 42
D. 48
E. 54

Answer: B

Source: GMAT Club Tests
Speed is inversely proportional to time

Walking at 60% of speed means \(\dfrac{3}{5}s\) takes \(\dfrac{5}{3}t.\)
It takes \(24\) minutes extra to cover the distance. Then

\(\dfrac{5}{3}t=t+24\)
\(5t=3t+72\)
\(2t=72\)
\(t=36.\)
Join the discussion