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

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Abe and Beth both start from a common point and travel in different [#permalink] New post   20 Apr 2023, 01:30
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Question Stats:

47% (02:16) correct 53% (02:42) wrong based on 70 sessions
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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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A
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Re: Abe and Beth both start from a common point and travel in different [#permalink] New post   20 Apr 2023, 03:21
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
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Re: Abe and Beth both start from a common point and travel in different [#permalink] New post   20 Apr 2023, 10:57
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
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Re: Abe and Beth both start from a common point and travel in different [#permalink] New post   21 Apr 2023, 10:21
RICHA1189 wrote:
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


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
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Re: Abe and Beth both start from a common point and travel in different [#permalink] New post   21 Apr 2023, 10:23
sivakumarm786 wrote:
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


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
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Re: Abe and Beth both start from a common point and travel in different [#permalink] New post   22 Apr 2023, 06:21
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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%
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Re: Abe and Beth both start from a common point and travel in different [#permalink] New post   28 Apr 2023, 15:49
Quote:
Beth took 1 hour & Abe took 1.2 hours. So, Abe took 20% more time as compared to Beth.


Gauravji21 how did you determine that Abe took 1.2 hours from the question stem?
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Re: Abe and Beth both start from a common point and travel in different [#permalink] New post   28 Apr 2023, 22:14
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
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Re: Abe and Beth both start from a common point and travel in different [#permalink]
28 Apr 2023, 22:14
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