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

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

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New post 22 Feb 2018, 19:39
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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%

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

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New post 22 Feb 2018, 21:25
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%


Formula used: Distance = Time*Speed
Let the speed at which she travels be x and the time she takes be t

If the time the jogger takes to complete the course is 1/4 less than she usually takes,
\(\frac{3}{4}t * A = xt\) -> \(A = \frac{4}{3}*X = (1+\frac{1}{3})X\)

Therefore, the percentage by which the jogger must increase her running speed is \(\frac{1}{3}*100 = 33\frac{1}{3}\) (Option C)
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Re: A jogger desires to run a certain course in 1/4 less time than she usu  [#permalink]

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New post 22 Feb 2018, 22:03
We can find answer quickly by plugging values

Lets say the course is 6 miles and she usually completes the course in 60 Minutes. Now she wants to complete the course in 1/4 less than 60 minutes which is in 45 minutes.

Now if the 6 mile course needs to completed in 1 hour, then the speed needs to be 8 Miles / hour, which is 33.5 % more than original speed.

Ans: C
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Re: A jogger desires to run a certain course in 1/4 less time than she usu  [#permalink]

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New post 23 Feb 2018, 00:58
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%



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

d= 10 t =45 mins

speed = 10x4/3 = 13.33

inc = 100/3

(C) imo
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Re: A jogger desires to run a certain course in 1/4 less time than she usu  [#permalink]

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New post 24 Feb 2018, 15:09
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%
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A jogger desires to run a certain course in 1/4 less time than she usu  [#permalink]

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New post 25 Feb 2018, 17:36
"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.
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A jogger desires to run a certain course in 1/4 less time than she usu &nbs [#permalink] 25 Feb 2018, 17:36
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