Abe and Beth both start from a common point and travel in different directions towards their respective destinations. Abesaverage speed is 25% greater than Beths average speed but Abe needs to cover 50% greater distance than Beth. Which of thefollowing 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%Correct answer is option 'A'. Can you explain this answer?

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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%

Correct answer is option 'A'. Can you explain this answer?

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Verified Answer

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

Given:

Let Abe’s Average Speed, Distance covered and Time taken be S , D and T respectively
Let Beth’s Average Speed, Distance covered and Time taken be S , D and T respectively

To Find: approx. % by which T_B is lesser than

Approach:

% by which T is lesser than
So, to answer the question, we need to find:
- Either the values of T_A and T_B
- Or T_A in terms of T_B
Using the given relations between S and S , and between D_A and D_B , we’ll express T in terms of T and hence, answer the question

Working out:

So, % by which T_A is lesser than T_B =

Looking at the answer choices, we see that Option A is correct

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Most Upvoted Answer

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

Problem: 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?

Solution:

Let's assume that Beth's average speed is 'S' and she covers a distance of 'D' to reach her destination.

According to the problem, Abe's average speed is 25% greater than Beth's average speed. So, Abe's average speed will be:

Abe's average speed = S + 0.25S = 1.25S

Also, Abe needs to cover 50% greater distance than Beth. So, Abe's distance will be:

Abe's distance = D + 0.5D = 1.5D

Now, we can calculate the time taken by Beth and Abe to reach their respective destinations.

Time taken by Beth = Distance / Speed = D / S

Time taken by Abe = Distance / Speed = 1.5D / 1.25S = 6D / 5S

So, the percentage by which the travel time of Beth is lesser than that of Abe can be calculated as:

Percentage = [(Time taken by Abe - Time taken by Beth) / Time taken by Abe] x 100

Percentage = [(6D / 5S - D / S) / (6D / 5S)] x 100

Percentage = [(5D / 5S) / (6D / 5S)] x 100

Percentage = (5/6) x 100

Percentage = 83.33%

Therefore, the closest option to the percentage by which the travel time of Beth is lesser than that of Abe is option 'A' which is 17%.

Answered by Talent Skill Learning · View profile

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Abe and Beth both start from a common point and travel in different directions towards their respective destinations. Abe’saverage speed is 25% greater than Beth’s average speed but Abe needs to cover 50% greater distance than Beth. Which of thefollowing 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%Correct answer is option 'A'. Can you explain this answer?

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