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Walking at 60% of his usual speed a man takes...

by swerve » Fri Feb 09, 2018 6:48 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

The OA is B.

Please, can any expert explain this PS question for me? I can't get the correct answer. I need your help. Thanks.
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by EconomistGMATTutor » Fri Feb 09, 2018 7:25 am
swerve wrote: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

The OA is B.

Please, can any expert explain this PS question for me? I can't get the correct answer. I need your help. Thanks.
Hi Swerve.
Let's take a look at your question.

Let's consider the following variables: $$V_u\ -\ \text{usual}\ \text{speed}$$ $$T_u\ -\ \text{Usual}\ \text{time}$$ Hence, we have th following $$V_u\cdot T_u\ =\ 60\%\cdot V_u\cdot\left(T_u+24\right)\ \Leftrightarrow\ \ T_u=\frac{3}{5}\left(T_u+24\right)$$ $$5T_u=3T_u+72\ \Leftrightarrow\ \ \ 2T_u=72\ \ \ \Leftrightarrow\ \ \ T_u=36.$$ Therefore, the correct answer is the option [spoiler]B=36[/spoiler].

I hope it helps you.

I'm available if you'd like a follow-up.

Regards.
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by GMATGuruNY » Fri Feb 09, 2018 7:27 am
swerve wrote: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
Time and rate have a RECIPROCAL RELATIONSHIP:
If Amy's rate is 2 times Bob's rate, then Amy's time is 1/2 Bob's time.
If Amy's rate is 3 times Bob's rate, then Amy's time is 1/3 Bob's time.

Since the man is walking at 60% = 3/5 of his normal speed, he will take 5/3 of his normal time.
5/3 of his normal time implies 2/3 MORE than his normal time.
Implication:
24 minutes more represents 2/3 of his normal time:
24 = (2/3)t
72 = 2t
t = 36.

The correct answer is B.
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by ErikaPrepScholar » Fri Feb 09, 2018 7:34 am
We know that distance = rate * time. Let's set the man's initial rate to r and his initial time to t. This gives us:

distance=r*t

If the man decreases his speed (the same thing as rate) by 60%, he is now walking at 0.6r. We know this causes him to walk an additional 24 minutes for the same distance, for a total of t + 24 minutes. This gives us:

distance = 0.6r*(t+24)

Since the distance in both equations is the same, we can set the right sides of the equations equal to each other:

r*t=0.6r*(t+24)

Cancelling r and simplifying gives:

t=0.6t+14.4
0.4t=14.4
t=36

So the man's usual time to cover the distance is 36 minutes.
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by Brent@GMATPrepNow » Fri Feb 09, 2018 7:47 am
swerve wrote: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
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

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by Jeff@TargetTestPrep » Mon Feb 12, 2018 4:43 pm
swerve wrote: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
We can let r = his usual speed, t = his usual time and d = distance. Thus we have:

r x t = d

and

0.6r x (t + 24) = d

Dividing the second equation by the first, we have:

0.6(t + 24)/t = 1

0.6t + 14.4 = t

14.4 = 0.4t

t = 14.4/0.4 = 144/4 = 36

Answer: B

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