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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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New post 30 Mar 2012, 13:29
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Question Stats:

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

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New post 30 Mar 2012, 14:58
3
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 \(T=\frac{t_1*t_2}{t_1+t_2}=\frac{6*7}{6+7}=\frac{42}{13}\approx{3.23}\)).

Answer: D.

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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New post 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

Hence answer will be (D)
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Re: A worker can load 1 full truck in 6 hours. A second worker  [#permalink]

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New post 20 Feb 2018, 17:17
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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


One approach is to assign a nice value to the entire job (of filling a truck)

We want a number that works well with the given times (6 hours and 7 hours)
42 is such a number.
So, let's say that filling the truck is equivalent to shoveling 42 scoops of dirt into it.

A worker (we'll call worker A) can load 1 full truck in 6 hours
Rate = output/time = 42 scoops/6 hours = 7 scoops/hour
So, worker A's RATE is 7 scoops/hour

Worker B) can load 1 full truck in 7 hours
Rate = output/time = 42 scoops/7 hours = 6 scoops/hour
So, worker B's RATE is 6 scoops/hour

So, their COMBINED rate = 7 scoops/hour + 6 scoops/hour
= 13 scoops/hour


Worker B) Approximately how long, in hours, will it take them to fill 1 truck?
Time = output/rate
= 42 scoops/13 scoops/hour
= 42/13 hours
= 3 3/13 hours

ASIDE: Notice that 3 3/12 hours = 3.25 hours
So, 3 3/13 hours will equal a little less than 3.25 hours

Answer: D

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

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New post 22 Feb 2018, 09:25
Quote:
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 combined rate of the two workers is 1/7 + 1/6 = 6/42 + 7/42 = 13/42.

Since time is inverse of rate, it will take 42/13 = 3 3/13 = 3.23 hours.

Answer: D
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Re: A worker can load 1 full truck in 6 hours. A second worker   [#permalink] 22 Feb 2018, 09:25
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