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Source: — Problem Solving |

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by Rahul@gurome » Fri Dec 03, 2010 5:14 am
N:Dure wrote:Walking at 4/5 of his usual speed, a man is 10 minutes too late. find his usual time.
A. 81 minutes
B. 64 minutes
C. 52 minutes
D. 40 minutes
E. none
By walking at 4/5 of usual speed he will take 5/4 of his usual time.

Extra time = (5/4) of usual time - usual time = 10 minutes
=> (1/4) of usual time = 10 minute
=> usual time = 40 minutes

The correct answer is D.
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by N:Dure » Fri Dec 03, 2010 5:20 am
Thanks Rahul! But I don't get this part: "By walking at 4/5 of usual speed he will take 5/4 of his usual time"
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by gmatusa2010 » Fri Dec 03, 2010 5:26 am
RT=D

4/5R(T)=1, so T has to be 5/4.
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by N:Dure » Fri Dec 03, 2010 5:30 am
Why D=1?
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by gmatusa2010 » Fri Dec 03, 2010 6:16 am
1 is 100%, as in completing the job. IF at R rate and T time you can complete 1 job then RT=1; 4/5R(T)=1? if T is the the same then you can only complet 4/5 the job or 4/5R(T)=1(4/5) right?
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by GMATGuruNY » Fri Dec 03, 2010 9:00 am
N:Dure wrote:Walking at 4/5 of his usual speed, a man is 10 minutes too late. find his usual time.



A. 81 minutes

B. 64 minutes

C. 52 minutes

D. 40 miuutes

E. none
We can plug in the answer choices, which represent the normal time. Since the problem includes 4/5, the correct answer is likely to be a multiple of 4 and 5. Let's try answer choice D.

Answer choice D: Normal time = 40 minutes.
Let distance = 200 miles.
Rate = 200/40 = 5 miles/minute.
4/5 rate = 4/5*5 = 4 miles/minute.
Time at slower rate = 200/4 = 50 minutes.
50-40 = 10 minute increase. Success!

The correct answer is D.
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