A jogger desires to run a certain course in 1/4 less time than she usu

let d = 10 t= 60 mins s=10 kmph

d= 10 t =45 mins

speed = 10x4/3 = 13.33

inc = 100/3

(C) imo

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%

"Theory: \(X*Y=Z ≈ X(1+\frac{1}{3})*Y(1-\frac{1}{4})=z ≈ X*\frac{4}{3}*Y*\frac{3}{4}=z ≈ X*Y=Z\)

So, increasing one element of a product by \(33 \frac{1}{3}\)% and decreasing another by 25% do not affect the result at all. Basically the same thing as doubling one and halving another."

R*T=D ; If T to decrease by 25% then R must increase by \(33 \frac{1}{3}\)%. Answer: C

NB: The same goes for a 25% increase and a 20% decrease.

I solved this by reverse percentage concept.

If time is reduced by 1/4 -> 25% then speed will increase by reverse% increase.

100 -> 75 (time reduced)

reverse % = 25/75 = 1/3 => 33.33%.

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