Abe and Beth both start from a common point and travel in different

Question

Abe and Beth both start from a common point and travel in different directions towards their respective destinations. Abe’s average speed is 25% greater than Beth’s average speed but Abe needs to cover 50% greater distance than Beth. Which of the following is closest to the percentage by which the travel time of Beth lesser than that of Abe?

A. 17%

B. 20%

C. 25%

D. 40%

E. 60%

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$!vakumar.m · Answer

Abe and Beth both start from a common point and travel in different directions towards their respective destinations. Abe’s average speed is 25% greater than Beth’s average speed but Abe needs to cover 50% greater distance than Beth. Which of the following is closest to the percentage by which the travel time of Beth lesser than that of Abe?

Let speed and distance of Beth be x and y respectively
Beth's travel time is y/x 
Then speed of Abe is 1.25x and distance is 1.5y
Abe's travel time is 1.5y/1.25x = 1.2y/x
so the difference in percentage is (1.2y/x - y/x)/(y/x)*100 = 20%

Answer B

RICHA1189 · Answer

I substituted the values for Abe and Beth

Speed of Beth = 10 m/hr
Distance to be covered by Beth = 10 miles
Time taken = 1 hr

Speed of Abe = 1.25*10 (125% of Beth’s speed) = 12.5 m/hr
Distance to be covered by Abe = 10 + 50% * 10 miles (150% of Beth’s distance) = 15 miles
Time taken = 1.2 hrs

Applying percentage change , (1.2-1)/1 * 100 = 0.2*100 = 20%

B is the answer

Gauravji21 · Answer

Beth took 1 hour & Abe took 1.2 hours. So, Abe took 20% more time as compared to Beth.
In other words, Beth took 0.2 hours less than Abe . This the value which question is asking. How much less time did Beth take as compared to Abe.

So, in my opinion, the in % change, denominator should be Abe' time
0.2/1.2 * 100 = 0.1667 * 100 = 17% - Answer A

Gauravji21 · Answer

Question is asking: How much  less time  did Beth take as compared to Abe.

So, in my opinion, the in % change, denominator should be Abe' time
(1.2y/x - y/x)/(1.2y/x)*100= 17% - Answer A

szcz · Answer

Pretty straightforward question. 
Let Abe's speed = 5x. Hence, Beth's speed = 4x. Further let Distance covered by Beth = D. Hence distance covered by Abe = 1.5D.
Time taken by Abe = 3D/10x. Time taken by Beth = D/4x.
Difference = (3D/10x-D/4x)
 = D/20x
% Difference = (D/20x * 100)/(3D/10x)
 =16.67%

gmattyfatty · Answer

Gauravji21  how did you determine that Abe took 1.2 hours from the question stem?

Kinshook · Answer

Given: Abe and Beth both start from a common point and travel in different directions towards their respective destinations. Abe’s average speed is 25% greater than Beth’s average speed but Abe needs to cover 50% greater distance than Beth.

Asked: Which of the following is closest to the percentage by which the travel time of Beth lesser than that of Abe?

Let

Abe:
Speed = Va
Distance = Da
Time Ta = Da/Va

Beth:
Speed = Vb
Distance = Db
Time Tb = Db/Vb

Va/Vb = 1.25
Da/Db = 1.5

Ta/Tb = Da/Db * Vb/Va = 1.5/1.25 = 1.2
Tb/Ta = 1/1.2 = 83% = 100% - 17%

IMO A

Aarushi225 · Answer

Why are we taking Abe's time in the denominator?

Regor60 · Answer

The question requests that we determine the amount by which Abe's time exceeds Beth's (to keep numbers positive) and to then compare that to Abe's time.

That comparison is the ratio of the two, so Abe's time needs to be in the denominator.

Posted from my mobile device

Aarushi225 · Answer

Oh alright!
I took Beth's time in the denominator. Got it.

Abe and Beth both start from a common point and travel in different

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