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If Tom and Huckleberry working at their respective rates can each whit

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Bunuel
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If Tom and Huckleberry working at their respective rates can each whit [#permalink] New post   24 Nov 2017, 00:48
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If Tom and Huckleberry working at their respective rates can each whitewash 600 square feet of fence in x and y hours, respectively, how long will it take both of them working together at their own rates to whitewash 600 square feet of fence?

(1) x - y = 1

(2) (xy)/(x + y) = 6/5
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amanvermagmat
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Re: If Tom and Huckleberry working at their respective rates can each whit [#permalink] New post   24 Nov 2017, 03:57
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Alexey1989x wrote:
we need to find their combined rate which would be 1/x + 1/y = (y+x)/xy
(1) x-y=1
Insufficient, since x and y can take different values
(2) we have the reciprocal of what we exactly are looking for
(x+y)/xy=5/6
Sufficient

Answer B



Hi Alexey

You are correct, answer is coming from second statement. But a small thing: 1/x + 1/y OR (x+y)/xy is the combined per hour work. What the question asks is the time required, which would be reciprocal of this OR xy/(x+y).
So, what is given in second statement = 6/5 is exactly what we are looking for. Together they will take 6/5 hours only.

Not that it matters here because in either case answer is B only, but since its about a concept I thought I will put forth my logic.
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Re: If Tom and Huckleberry working at their respective rates can each whit [#permalink] New post   25 Nov 2017, 13:05
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Bunuel wrote:
If Tom and Huckleberry working at their respective rates can each whitewash 600 square feet of fence in x and y hours, respectively, how long will it take both of them working together at their own rates to whitewash 600 square feet of fence?

(1) x - y = 1

(2) (xy)/(x + y) = 6/5



\(\frac{600}{x} + \frac{600}{y}= \frac{600(x + y)}{xy}\)

A. \(x - y = 1\) => \(y = x- 1\)

\(\frac{600(x + x-1)}{x(x + 1)}\) = \(\frac{600(2x-1)}{x^2 + x}\) A is insufficient because we don't know a value of x

B. \(\frac{xy}{(x+y)} =\frac{6}{5}\) => \(\frac{(x+y)}{xy}\) = \(\frac{5}{6}\) = \(\frac{1}{1.2}\)

\(\frac{600(x + y)}{xy}\) = \(\frac{600*1}{1.2}\) B is sufficient it will take 1.2 hours to finish the job
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Re: If Tom and Huckleberry working at their respective rates can each whit [#permalink] New post   24 Nov 2017, 00:56
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we need to find their combined rate which would be 1/x + 1/y = (y+x)/xy
(1) x-y=1
Insufficient, since x and y can take different values
(2) we have the reciprocal of what we exactly are looking for
(x+y)/xy=5/6
Sufficient

Answer B
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Alexey1989x
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Re: If Tom and Huckleberry working at their respective rates can each whit [#permalink] New post   24 Nov 2017, 04:27
amanvermagmat wrote:
Alexey1989x wrote:
we need to find their combined rate which would be 1/x + 1/y = (y+x)/xy
(1) x-y=1
Insufficient, since x and y can take different values
(2) we have the reciprocal of what we exactly are looking for
(x+y)/xy=5/6
Sufficient

Answer B



Hi Alexey

You are correct, answer is coming from second statement. But a small thing: 1/x + 1/y OR (x+y)/xy is the combined per hour work. What the question asks is the time required, which would be reciprocal of this OR xy/(x+y).
So, what is given in second statement = 6/5 is exactly what we are looking for. Together they will take 6/5 hours only.

Not that it matters here because in either case answer is B only, but since its about a concept I thought I will put forth my logic.


Thanks for careful notice! I rushed through and missed the point! :)
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Re: If Tom and Huckleberry working at their respective rates can each whit [#permalink] New post   25 Nov 2017, 00:47
As mentioned by Aman, we can get the answer from statement 2. The logic is pretty simple, rate of completing one unit of job is 1/time taken.

Now we have been given the total time taken by both of them, we can determine how much time will it take either of them.

Hence Answer is B, hopefully my logic makes sense.

Thanks Bunnel for great questions!
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Re: If Tom and Huckleberry working at their respective rates can each whit [#permalink] New post   21 Feb 2020, 06:33
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Re: If Tom and Huckleberry working at their respective rates can each whit [#permalink]
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