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Re: Last year Harold's average time to finish the qualifying event was [#permalink]
Let the speed of last year be = x; Current year speed = 1.2x
Let time taken this year be = t

Since the distance traveled is the same; 3*x = 1.2*x*t --> t = 3/1.2 = 2.5 hours = 150 mins.

Answer: A
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Re: Last year Harold's average time to finish the qualifying event was [#permalink]
Bunuel wrote:
Last year Harold's average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

A. 150
B. 144
C. 120
D. 90
E. 36



Speed = distance / time

Initially S1=D/T1 -- > T1 = 3 hours --> s1 = D/3

This year S2 = D/T2 -- > S2 = 1.2 S1 (20% increase)

Equate --> 3*s1 = 1.2 * s1 * T2 --> T2 = 3/1.2 * 60 = 150 mins
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Re: Last year Harold's average time to finish the qualifying event was [#permalink]
Bunuel wrote:
Last year Harold's average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

A. 150
B. 144
C. 120
D. 90
E. 36


Let speed be 20 units/hr
Time taken is 3
Total distance covered is 60 units

Now, speed Increases to 24 units/hr
Distance is 60 units
So, Time taken is 60/24*60 = 150 minutes...

Hence, correct answer will be (A) 150
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Re: Last year Harold's average time to finish the qualifying event was [#permalink]
Bunuel wrote:
Last year Harold's average time to finish the qualifying event was three hours. If he knows that he can increase his speed this year by 20%, how many minutes should it take him to complete the event?

A. 150
B. 144
C. 120
D. 90
E. 36



Let the distance be "X"

Initial speed S=X/3*60=X/180;

After increase S1=(X/180) + 0.2 (X/180);
S1= 1.2X/180

1.2X/180= X/T
T=150 min.
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Last year Harold's average time to finish the qualifying event was [#permalink]
Last year time = 3 * 60 = 180 min.

Let the distance = 180 meter...............speed =d/t =180/180= 1 m/min

This year speed =1.2 * 1= 1.2 m/min

time required this year = 180/1.2 = 1800/ 12 = 18 * 100/12= 150 min

Answer: A
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Re: Last year Harold's average time to finish the qualifying event was [#permalink]
Hello from the GMAT Club BumpBot!

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Re: Last year Harold's average time to finish the qualifying event was [#permalink]
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