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A husband and wife started painting their house, but husband [#permalink]
22 Apr 2009, 04:24

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45% (medium)

Question Stats:

63% (03:18) correct
37% (02:30) wrong based on 430 sessions

A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days?

Re: Work & Rate problem [#permalink]
05 Sep 2009, 02:39

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With the answer choices given there's a faster way.

There is no need to solve the whole thing.

Using the rate eqn for combined work... 1/20 + 1/15 = 7/60

So it would take 60/7 days combined. If the Husband quits working a few days earlier... total time is going to be > 60/7. There is only one value > 60/7 so answer is 75/7

Re: Work & Rate problem [#permalink]
27 Jan 2012, 09:44

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A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days? A. 40/7 B. 50/7 C. 75/7 D. 55/7

As pointed out above there are several ways to solve this problem. Below are probably two shortest approaches:

Approach #1: Rate of husband \frac{1}{20} job/day; Rate of wife \frac{1}{15} job/day; Combined rate: \frac{1}{20}+\frac{1}{15}=\frac{7}{60} job/day;

During the last 5 days, when the wife worked alone, she completed \frac{5}{15}=\frac{1}{3}rd of the job; Hence, remaining \frac{2}{3}rd of the job was done by them working together in time=\frac{job'}{rate}=\frac{(\frac{2}{3})}{(\frac{7}{60})}=\frac{40}{7} days;

Total time needed to complete the whole job: 5+\frac{40}{7}=\frac{75}{7} days.

Answer: C.

Approach #2: It's based on observing the answer choices. On the PS section always look at the answer choices before you start to solve a problem. They might often give you a clue on how to approach the question.

Combined rate of the husband and wife is \frac{7}{60} job/day, which means that working together they'll complete the job in \frac{60}{7} days (time is reciprocal of rate). As they worked together only some part of the total time, then actual time would be more than \frac{60}{7} days. Only \frac{75}{7} is more than this value (answer choice C), so it must be correct.

Re: Work & Rate problem [#permalink]
22 Apr 2009, 07:05

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johnnybravo86 wrote:

Please help me solve this problem

A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days?

a) 40/7 b)50/7 c)75/7 d) 55/7

It's a multiple step problem: 1) figure out how many days (D) it would take the husband and wife to complete the job working together, 2) figure out how much of the house would be painted in D-5 days, 3) figure out how long it would take the wife to paint the remaining portion

1. How many days would it take the husband and wife to complete the house working together?

Each day, the husband can paint 1/20 of the house, and the wife can paint 1/15 of the house, so combined they can paint (1/20 + 1/15 = 7/60) of the house.

At that rate it would take them 60/7 days to complete the house. (7/60 house/day * 60/7 days = 1 house)

2. How much of the house can they paint in D-5 days?

(D-5 = 60/7 - 35/7 = 25/7) If working together they paint 7/60 of the house per day, then in D-5 days they could paint:

7/60 house/day * 25/7 days = 175/420 = 5/12 of the house will be painted in D-5 days.

3. How long will it take the wife, working alone, to paint this portion of the house?

The wife can paint 1/15 of the house each day, and there is 5/12 of the house that needs to be painted, so it would take her (5/12) / (1/15) = 25/4 days.

Well, I've checked my math a few times and came up with the same answer, which isn't one of the options presented. Where did the question come from? _________________

Re: Work & Rate problem [#permalink]
27 Oct 2009, 08:02

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Sol1 : First, in 5 days, the wife work 5/15 or 1/3 of the job Then, together, they use (2/3)(60/7) = 40/7 days to finish 2/3 of the job Thus, in total, use 75/7

Sol2: together, they use 60/7 days to finish the job. Thus, if the husband stops working, it has to take more than 60/7 days to finish the job.

Re: A husband and wife started painting their house, but husband [#permalink]
07 Aug 2012, 09:58

1

This post received KUDOS

shivamayam wrote:

A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days? A. 40/7 B. 50/7 C. 75/7 D. 55/7

hi bunuel and friends,

here is my logic, which is incorrect since the answer that i got is incorrect. but am unable to figure out the gap. can you please help. thanks.

1) combined rate is 7/60; hence total time to complete the job when both work together is 60/7. 2) job that would have been done when husband left work = 7/60 * [(60/7)-5] i.e Combined rate * days worked together = 5/14. 3) Hence, unfinished job = 1-(5/14) = 9/14 4) Time taken by wife to finish 9/14 of the job = 9/14*15 = 135/14 (which is not equal to 5 as given in the question stem). 5) So total time taken to complete the work is 135/14+[(60/7)-5] = 92.5/7 !!!

wow. am i a genius in arriving at wrong answers!

You are wrong in the step 2. Your assumption is below:

days worked together = 60/7 -5 => days worked together + 5 days = 60/7 ??? together they finish in 60/7 from start to end ,not when Husband leaves before days. In that case the total days would be more than 60/7. The correct value is 75/7 which is what the question asks. _________________

"Appreciation is a wonderful thing. It makes what is excellent in others belong to us as well." ― Voltaire Press Kudos, if I have helped. Thanks! shit-happens-my-journey-to-172475.html#p1372807

Re: A husband and wife started painting their house, but husband [#permalink]
08 Jul 2013, 05:48

1

This post received KUDOS

johnnybravo86 wrote:

A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days?

A. 40/7 B. 50/7 C. 75/7 D. 55/7

1. In a normal scenario both the husband and wife would have taken 1/ (1/15+ 1/20) days = 60/7 days to complete the task 2. what did not happen was they definitely did not work together for 5 days. So subtract 5 days work together =5*7/60 = 7/12 th of the work 3. what did happen was the wife definitely worked alone for 5 days. So add 5 days of work of the wife = 5/15 = 1/3 rd of the work= 4/12 th of the work 4. From (2) and (3) we see 7/12- 4/12 = 3/12 th or 1/4 th of the work still remains. 5. this additional 1/4 th work would have been done by both because the wife worked alone for only 5 days which we have accounted for. So time taken for this work is 1/4 * 60/7 =15/7 6. Total time taken to complete the whole work in the altered scenario is 60/7 + 15/7 = 75/7 _________________

Re: Work & Rate problem [#permalink]
22 Nov 2013, 00:50

1

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Expert's post

AccipiterQ wrote:

Bunuel wrote:

A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days? A. 40/7 B. 50/7 C. 75/7 D. 55/7

As pointed out above there are several ways to solve this problem. Below are probably two shortest approaches:

Approach #1: Rate of husband \frac{1}{20} job/day; Rate of wife \frac{1}{15} job/day; Combined rate: \frac{1}{20}+\frac{1}{15}=\frac{7}{60} job/day;

During the last 5 days, when the wife worked alone, she completed \frac{5}{15}=\frac{1}{3}rd of the job; Hence, remaining \frac{2}{3}rd of the job was done by them working together in time=\frac{job'}{rate}=\frac{(\frac{2}{3})}{(\frac{7}{60})}=\frac{40}{7} days;

Total time needed to complete the whole job: 5+\frac{40}{7}=\frac{75}{7} days.

Answer: C.

Approach #2: It's based on observing the answer choices. On the PS section always look at the answer choices before you start to solve a problem. They might often give you a clue on how to approach the question.

Combined rate of the husband and wife is \frac{7}{60} job/day, which means that working together they'll complete the job in \frac{60}{7} days (time is reciprocal of rate). As they worked together only some part of the total time, then actual time would be more than \frac{60}{7} days. Only \frac{75}{7} is more than this value (answer choice C), so it must be correct.

Answer: C.

Hope it helps.

Could you possibly help me figure out why my method didn't work for solving:

I combined their two rates, and came up with \frac{7}{60} per day, so the whole job, combined, would have taken them \frac{60}{7}, or 8 4/7 days to complete. The husband left 5 days before this, so they worked together for 3 4/7 days, or \frac{25}{7}. The amount of work they completed was \frac{7}{60}* \frac{25}{7}=25/60 of the job. This leaves \frac{35}{60}of the job for the wife to complete alone. She can work at a rate of 4/60 per day. So that means she completed the remaining 35/60 in 8 3/4 days, not 5. This combined with the 3 4/7 that they worked on it together gives you a number of about 12.32, which is obviously not correct. Where did I go wrong? I'm having trouble reconciling how the wife only needed 5 days working at the rate given, when we can see how much work they did combined, and it leaves more than 5 days worth of work for her.

If they work together they can complete the job in 60/7 days. But if one of them does not work for all that period then the time to complete would increase. Thus you cannot say that when husband left 5 days before, then they worked together for 60/7-5 days.

Re: Work & Rate problem [#permalink]
06 Sep 2009, 04:28

dhushan wrote:

tkarthi4u wrote:

H take 20 days = 1 W take 15 days = 1 H one day work = 1/20 and W one day work = 1/15

H + W one day work = 1/20 + 1/15 = 7/60.

In a total of T day H+W worked for T-5 and W alone for 5

(7/60) (T-5) + 1/15 * 5 = 1 therefore T = 75/7

Ans : C

Hey guys, can someone explain to my why it (1/5*5) since the question is asking how long it will take to complete the work. I would assume that if it is the women alone, it would now take longer than 5 days to complete the paint job?

I agree with you dhushan, the structure of the question is vague. It should have asked "how long did it take to complete the housing painting?" Cause my understanding of the question also narrowed down to how many days will it take the wife to complete the job. I doubt that GMAC will structure the question the same way.

Re: Work & Rate problem [#permalink]
06 Sep 2009, 04:49

consider this:- husband 1 day work=1/20.....wife 1 day work=1/15.. combined 1 day work=1/20+1/15=7/60 last 5 days women worked alone , so fractin of work completed by her in those 5 days=5*1/15=1/3.. remaining work was completed by both .. so time taken=(fraction of work left)/(combined 1 day work)=(2/3)/(7/60).. or (2*60)/(3*7)=40/7... total days=5+40/7=75/7...

Re: Work & Rate problem [#permalink]
07 Feb 2010, 20:36

Husband alone can complete the work in 20 days i.e in one day he does 5% of work

Wife alone can complete the work in 15 days i.e. in one day she does 6.67% of work

During last three days she worked alone i.e. she completed 33.33% (5* 6.67%) of work working alone.

Both of them working together did = 100-33.33 = 66.67% of work

Therefore no. days for which they worked together = 66.67%/(5%+6.67%) = 66.67/11.67 = 5 point something Total days = 5 +5 point something = 10 point something From options = 75/7 = 10 point something (C)

Re: Work & Rate problem [#permalink]
06 Feb 2012, 10:14

Bunuel wrote:

A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days? A. 40/7 B. 50/7 C. 75/7 D. 55/7

As pointed out above there are several ways to solve this problem. Below are probably two shortest approaches:

Approach #1: Rate of husband \frac{1}{20} job/day; Rate of wife \frac{1}{15} job/day; Combined rate: \frac{1}{20}+\frac{1}{15}=\frac{7}{60} job/day;

During the last 5 days, when the wife worked alone, she completed \frac{5}{15}=\frac{1}{3}rd of the job; Hence, remaining \frac{2}{3}rd of the job was done by them working together in time=\frac{job'}{rate}=\frac{(\frac{2}{3})}{(\frac{7}{60})}=\frac{40}{7} days;

Total time needed to complete the whole job: 5+\frac{40}{7}=\frac{75}{7} days.

Answer: C.

Approach #2: It's based on observing the answer choices. On the PS section always look at the answer choices before you start to solve a problem. They might often give you a clue on how to approach the question.

Combined rate of the husband and wife is \frac{7}{60} job/day, which means that working together they'll complete the job in \frac{60}{7} days (time is reciprocal of rate). As they worked together only some part of the total time, then actual time would be more than \frac{60}{7} days. Only \frac{75}{7} is more than this value (answer choice C), so it must be correct.

Answer: C.

Hope it helps.

The question say the husband left the job 5 days back. Then it is never mentioned that he has re-joined. WIth the current wording, doesn't it imply that its only the wife who is going to complete the job . And if thats the case it would take her 15 days as the question itself mentions... May be its just some missing part in the question.

Re: Work & Rate problem [#permalink]
06 Feb 2012, 14:09

Expert's post

docabuzar wrote:

The question say the husband left the job 5 days back. Then it is never mentioned that he has re-joined. WIth the current wording, doesn't it imply that its only the wife who is going to complete the job . And if thats the case it would take her 15 days as the question itself mentions... May be its just some missing part in the question.

It seems that you misinterpreted the question. We are told that: "a husband and wife, started painting, but husband left 5 days before the completion of the work". So the husband worked 5 fewer days than the wife (he didn't rejoin).

Next, part of the question just tells us the individual rates of the husband and the wife: the husband alone can complete the painting in 20 days and wife in 15 days.

The actual time it took for them to complete the painting would obviously be less than 15 as the part of the time they were working together.

Re: Work & Rate problem [#permalink]
17 Feb 2012, 09:44

Bunuel wrote:

docabuzar wrote:

The question say the husband left the job 5 days back. Then it is never mentioned that he has re-joined. WIth the current wording, doesn't it imply that its only the wife who is going to complete the job . And if thats the case it would take her 15 days as the question itself mentions... May be its just some missing part in the question.

It seems that you misinterpreted the question. We are told that: "a husband and wife, started painting, but husband left 5 days before the completion of the work". So the husband worked 5 fewer days than the wife (he didn't rejoined).

Next, part of the question just tells us the individual rates of the husband and the wife: the husband alone can complete the painting in 20 days and wife in 15 days.

The actual time it took for them to complete the painting would obviously be less than 15 as the part of the time they were working together.

Hope it's clear.

Thanks for the explanation.

Actually what I interpreted from the last part of the question is - the times given (20 days & 15 days) would be required to complete the "remaining part of job" and not the complete job.

Do you think what i interpreted can be the case. ?

Re: Work & Rate problem [#permalink]
17 Feb 2012, 12:53

Expert's post

docabuzar wrote:

Thanks for the explanation.

Actually what I interpreted from the last part of the question is - the times given (20 days & 15 days) would be required to complete the "remaining part of job" and not the complete job.

Do you think what i interpreted can be the case. ?

We are told that " husband left painting 5 days before the completion of the work (by wife)", so wife needed 5 days to complete the remaining job, not 15. So, your interpretation can not be correct. _________________

Re: A husband and wife started painting their house, but husband [#permalink]
06 Aug 2012, 18:50

A husband and wife, started painting their house, but husband left painting 5 days before the completion of the work. How many days will it take to complete the work, which the husband alone would have completed in 20 days and wife in 15 days? A. 40/7 B. 50/7 C. 75/7 D. 55/7

hi bunuel and friends,

here is my logic, which is incorrect since the answer that i got is incorrect. but am unable to figure out the gap. can you please help. thanks.

1) combined rate is 7/60; hence total time to complete the job when both work together is 60/7. 2) job that would have been done when husband left work = 7/60 * [(60/7)-5] i.e Combined rate * days worked together = 5/14. 3) Hence, unfinished job = 1-(5/14) = 9/14 4) Time taken by wife to finish 9/14 of the job = 9/14*15 = 135/14 (which is not equal to 5 as given in the question stem). 5) So total time taken to complete the work is 135/14+[(60/7)-5] = 92.5/7 !!!

Re: A husband and wife started painting their house, but husband [#permalink]
06 Aug 2012, 23:40

i am trying to do it in my approach or rectify my approach. let total work of completion of the house be 60 units. rate of husband per day is 3 units and rate of wife is 4 units per day. both they complete 7 units per day. in 5 days wife completes 4*5= 20 units, so they both work for (60-20) or 40 units of the house. they both work for 40/7 days. so to complete the house under the given condition it will take (40/7+ 5) days or 75/5 days. correct me my approach is wrong under time constraint _________________

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Re: A husband and wife started painting their house, but husband
[#permalink]
06 Aug 2012, 23:40