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Andy and Ben can finish painting a wall in x hours working together. [#permalink]
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Carcass wrote:
Let the speed at which Andy and Ben do the work is A units/hour \& B units/hour respectively.

So, the time taken by Andy and Bob to finish the work would be $\(\frac{1}{\mathrm{~A\)$ hours \& $\(\frac{1}{\mathrm{~B\)$ hours respectively. ( $\(\because\)$ Speed \& Time are inversely proportional when Work is constant)

As we know that Andy and Ben together finish the work in ' $\(x\)$ ' houirs, so their combined speeds is $\(\frac{1}{x}\)$, we get $\(A+B=\frac{1}{x} \ldots\)$ (1) Also it is given that the time taken by Andy to finish the work alone is ' $\(y\)$ ' hours more than the time taken by Ben alone to finish the work, so we get $\(\frac{1}{A}-\frac{1}{B}=y \ldots . .(2)\)$

Subtracting equations (1) \& (2) we get $\(\frac{2}{B}=x-y \Rightarrow \frac{1}{B}=\frac{x-y}{2}\)$ which is equal to the time that Ben would take to finish the work alone.

Hence column (A) has same quantity as column (B), so the answer is (A).

Thank you, but when A + B = 1/x, shouldn't it be x = 1/(A+B)?
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Re: Andy and Ben can finish painting a wall in x hours working together. [#permalink]
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:thumbsup:
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Re: Andy and Ben can finish painting a wall in x hours working together. [#permalink]
in that case, the subtraction of the two equations would be quite impossible to calculate, don't you think?
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Re: Andy and Ben can finish painting a wall in x hours working together. [#permalink]
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shinobi07 wrote:
in that case, the subtraction of the two equations would be quite impossible to calculate, don't you think?


did not get the point sir ?
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Re: Andy and Ben can finish painting a wall in x hours working together. [#permalink]
Considering the fact that A + B = 1/x (1) and 1/A - 1/B = y (2),

then, subtracting the equations (1) and (2) will get us x - y = 1/(A+B) - 1/A + 1/B, which is not as simple as quoted in your solution.

How should we proceed from here, or is it just not solvable?
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Re: Andy and Ben can finish painting a wall in x hours working together. [#permalink]
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\(A+B=\frac{1}{x}\)

1) \(\frac{1}{A}-\frac{1}{B}=y\)

\(\frac{1}{A}+\frac{1}{B}=x\)

2) \(\frac{1}{A}=x-\frac{1}{B}\)


The 2) into the 1)

\(x-\frac{1}{B}-\frac{1}{B}=y\)

\(-\frac{1}{B}-\frac{1}{B}=y-x\)

\(-\frac{2}{B}=y-x\)

\(\frac{2}{B}=x-y\)

\(\frac{1}{B}=\frac{x-y}{2}\)
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Re: Andy and Ben can finish painting a wall in x hours working together. [#permalink]
I apologize for not making myself clear, but I guess you are missing the point.

Earlier, we all agreed that from A + B = 1/x, we will have x = 1/(A+B).

From 1/(A+B), how can we have 1/A + 1/B like you mentioned in your reply, or am I missing something in this concept of GRE quant?
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Re: Andy and Ben can finish painting a wall in x hours working together. [#permalink]
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1. If a man can do a piece of work in N days (or hours or any other unit of time), then the work done by him in one day will be \(\frac{1}{N}\) of the total work.



Now Earlier, we all agreed that from \(A + B = \frac{1}{x}\), we will have x = 1/(A+B).

the red part is wrong

\(A+B=\frac{1}{x}\)

Then we do have

\(\frac{1}{A}+\frac{1}{B}=x\) NOT \(\frac{1}{A+B}\) this is not the same.

We take into account always that a job is = units/hour

So A+B become 1/A+1/B because speed and time are inversionally proportional when they are equal to a constant which is x in this case

From the math standing point \(\frac{1}{A+B} \neq \frac{1}{A}+\frac{1}{B}\)

See more here for theory https://gre.myprepclub.com/forum/gre-wo ... 24921.html
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Re: Andy and Ben can finish painting a wall in x hours working together. [#permalink]
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