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A and B can do a job together in 7 days. A is 1.75 times as efficient as B. The same job can be done by A alone is :
  • a)
    9.33 days
  • b)
    11 days
  • c)
    12.25 days
  • d)
    16.33 days
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A and B can do a job together in 7 days. A is 1.75 times as efficient ...
Given:
- A and B can do a job together in 7 days.
- A is 1.75 times as efficient as B.

To find:
- The time taken by A to complete the job alone.

Solution:

Let's assume that B takes x days to complete the job alone.

Since A is 1.75 times as efficient as B, it means that A can do 1.75 times the work done by B in the same amount of time.

Therefore, the work done by A in a day = 1.75 * (work done by B in a day)

Calculating the work done by A and B in a day:
- Let's assume the total work to be done is 1 unit.
- B can do 1 unit of work in x days, so the work done by B in a day = 1/x unit.
- A is 1.75 times as efficient as B, so the work done by A in a day = 1.75 * (1/x) = 1.75/x unit.

Calculating the time taken by A and B together:
- The work done by A and B together in a day = (work done by A in a day) + (work done by B in a day)
- = 1.75/x + 1/x = (1.75 + 1)/x = 2.75/x unit

Given that A and B can do the job together in 7 days, it means that the work done by A and B together in a day is equal to 1/7th of the total work.

Therefore, 2.75/x = 1/7

Calculating the value of x:
- Cross-multiplying, we get 2.75 * 7 = x
- x = 19.25

So, B takes 19.25 days to complete the job alone.

Calculating the time taken by A alone:
Since A is 1.75 times as efficient as B, it means that A can do the job in 1.75 times less time than B.

Therefore, the time taken by A alone = 19.25 / 1.75 = 11 days.

Hence, the correct answer is option 'B' - 11 days.
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A and B can do a job together in 7 days. A is 1.75 times as efficient as B. The same job can be done by A alone is :a)9.33 daysb)11 daysc)12.25 daysd)16.33 daysCorrect answer is option 'B'. Can you explain this answer?
Question Description
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