Last year Harold's average time to finish the qualifying event was

Question

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

Kurtosis · Answer

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

SI777 · Answer

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

GMATinsight · Answer

Let the Work of Qualifying event = 60 units
Time taken to finish the work = 3 hours
i.e. Speed per hour = 60/3 = 20 units of work per hour

New speed = 20% more than previous speed = 1.2*20 = 24 units per hour

Time to finish the even at new speed = 60/24 = 2.5 hours = 150 minutes

Answer: OPtion A

Abhishek009 · Answer

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

abhi4212 · Answer

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.

Mo2men · Answer

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

JeffTargetTestPrep · Answer

Recall that rate x time = job (or rate = job/time, or time = job/rate). Since Harold’s time to complete the event was 3 hours, we can say that his rate was 1 job per 3 hours; i.e., his rate is 1/3. Since he’s increasing his speed by 20%, his new rate is:

(1/3) x 1.2

(1/3) x 12/10 = 4/10 = 2/5

Since his new rate is ⅖, his new time is 1/(2/5) = 5/2 = 2.5 hours. Thus he can complete the event in 2.5 x 60 = 150 minutes.

Answer: A

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Last year Harold's average time to finish the qualifying event was

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