Bunuel wrote:
A jogger desires to run a certain course in 1/4 less time than she usually takes. By what percent must she increase her average running speed to accomplish this goal?
(A) 20%
(B) 25%
(C) 33 1/3%
(D) 50%
(E) 75%
Hello Bunuel,
I did it with a different approach but not really sure if it is the proper way to resolve this type of PS. The answer was very close to the one in the answer options. Please let me know what I did wrong.
Let's give some values to the initial time, speed, and distance that she usually takes.
D=S x T 10 =10mph x 60 min. finding S. we have S = D/T therefore S = 10/60min S1= 1/6
Now let use the information given in the argument, however, S and D are the same values. Here the time is 1/4 less than she usually takes. So T = 60 - (60 * 1/4) = 45
D = S X T 10 =10mph x 45 min. we need to find The average speed. So S = D/T therefore S = 10/45min S2= 2/9
Now we have S1=1/6 and S2=2/9.
S1---100%
S2--- X
let's find X: X = (2/9 1/6) X 100 Solution: X = 133.33% Then 133.33% - 100 % = 33.33%