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# A worker can load 1 full truck in 6 hours. A second worker

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A worker can load 1 full truck in 6 hours. A second worker [#permalink]

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30 Mar 2012, 13:29
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A worker can load 1 full truck in 6 hours. A second worker can load the same truck in 7 hours. If both workers load one truck simultaneously while maintaining their constant rates, approximately how long, in hours, will it take them to fill 1 truck?

A. 0.15
B. 0.31
C. 2.47
D. 3.23
E. 3.25

The site where I pulled this question states that the answer is E. 3.25. I'm convinced that 3hr 3/13 minutes is closer to E. 3.23. Who is correct? Their reasoning which I think must contain an error: At this point, we may not be able to decide between (D) or (E). However, the decimal is important. Because the denominator is 13, we know the decimal cannot equal .25. We can also see that 3/12 will yield .25, so 3/13 will be slightly lower. Choice (E).
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30 Mar 2012, 13:39
I concur the site must be wrong
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Re: A worker can load 1 full truck in 6 hours. A second worker [#permalink]

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30 Mar 2012, 14:58
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ralanko wrote:
A worker can load 1 full truck in 6 hours. A second worker can load the same truck in 7 hours. If both workers load one truck simultaneously while maintaining their constant rates, approximately how long, in hours, will it take them to fill 1 truck?

A. 0.15
B. 0.31
C. 2.47
D. 3.23
E. 3.25

The site where I pulled this question states that the answer is E. 3.25. I'm convinced that 3hr 3/13 minutes is closer to E. 3.23. Who is correct? Their reasoning which I think must contain an error: At this point, we may not be able to decide between (D) or (E). However, the decimal is important. Because the denominator is 13, we know the decimal cannot equal .25. We can also see that 3/12 will yield .25, so 3/13 will be slightly lower. Choice (E).

You are right, answer should be D, not E.

Remember we can add the rates of individual entities to get the combined rate.

Generally for multiple entities: $$\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+...+\frac{1}{t_n}=\frac{1}{T}$$, where $$T$$ is time needed for these entities to complete a given job working simultaneously and $$t_1$$, $$t_2$$, ..., $$t_n$$ are individual times needed for them to complete the job alone.

So for two pumps, workers, etc. we'll have $$\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{T}$$ --> $$T=\frac{t_1*t_2}{t_1+t_2}$$ (general formula for 2 workers, pumps, ...).

Back to the original problem: for two outlets the formula becomes: $$\frac{1}{6}+\frac{1}{7}=\frac{1}{T}$$ --> $$\frac{13}{42}=\frac{1}{T}$$ --> $$T=\frac{42}{13}\approx{3.23}$$ (or directly $$T=\frac{t_1*t_2}{t_1+t_2}=\frac{6*7}{6+7}=\frac{42}{13}\approx{3.23}$$).

Check this for more on this subject: two-consultants-can-type-up-a-report-126155.html#p1030079

Hope it helps.

P.S. Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/ and DS questions in the DS subforum: gmat-data-sufficiency-ds-141/

No posting of PS/DS questions is allowed in the main Math forum.

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Re: A worker can load 1 full truck in 6 hours. A second worker [#permalink]

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30 Apr 2016, 23:23
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Re: A worker can load 1 full truck in 6 hours. A second worker [#permalink]

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01 May 2016, 02:10
ralanko wrote:
A worker can load 1 full truck in 6 hours. A second worker can load the same truck in 7 hours. If both workers load one truck simultaneously while maintaining their constant rates, approximately how long, in hours, will it take them to fill 1 truck?

A. 0.15
B. 0.31
C. 2.47
D. 3.23
E. 3.25

Let the capacity of the truck be 42 units (LCM of 6 & 7 )

First worker can fill 7 units / hour
Second worker can fill 6 units / hour

Working together they will fill 13 units/hour

So, to full the entire truck they will need 42/13 ~ 3.23 hours

Re: A worker can load 1 full truck in 6 hours. A second worker   [#permalink] 01 May 2016, 02:10
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