whiplash2411 wrote:
rxs0005 wrote:
Last year Harolds 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
So we know that Rate \(= \frac{A}{T}\) where A represents either a distance or a job or whatever. Let's assume that the qualifying event = A, here.
In the first year, we have Rate \(= \frac{A}{3*60} = \frac{A}{180}\) per minute. He increases this speed by 20%, which means that new rate \(= \frac{120}{100}*\frac{A}{180} = \frac{A}{150}\)
From this you can see that the new time is 150 minutes.
Time taken to complete the work = 3hrs
Let the total work be 300 (assumed value. Its always better if u assume good values if the question consists of percentage or percentage incease)
so the speed = 300/3 = 100
Now the speed is increased by 20%. so the new speed = 120 and the total work is still the same = 300.
There for the new time taken to complete the event = 300/120 = 5/2 hrs. = 2hrs 30 mins = 150 mins
Answer A