when working alone A takes 24 more days than time taken by A and B together to finish certain work. If A is 50% more efficient than B, then find the time taken by B to finish the work alone.
Question
when working alone A takes 24 more days than time taken by A and B together to finish certain work. If A is 50% more efficient than B, then find the time taken by B to finish the work alone.
Solution
Sorry, there seems to be a misunderstanding. I can't see any specific language in your text. However, I can help you solve the problem you've given. Here's how:
Let's denote the time taken by A and B together to finish the work as 'x' days. Therefore, the time taken by A alone to finish the work is 'x + 24' days.
Given that A is 50% more efficient than B, we can say that the amount of work done by A in one day is 1.5 times the amount of work done by B in one day.
So, if A and B work together in one day, they can finish 1/x of the work. If A works alone in one day, he can finish 1/(x + 24) of the work. And if B works alone in one day, he can finish 1/(1.5*(x + 24)) of the work.
According to the problem, the amount of work done by A and B together in one day is equal to the sum of the amount of work done by A alone in one day and the amount of work done by B alone in one day. So, we can write the equation as:
1/x = 1/(x + 24) + 1/(1.5*(x + 24))
Solving this equation, we get x = 48 days.
So, the time taken by B to finish the work alone is 1.5*(x + 24) = 1.5*(48 + 24) = 108 days.
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