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# A husband and wife started painting their house, but husband

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Re: A husband and wife started painting their house, but husband [#permalink]  14 Nov 2012, 06:36
Setup the Rate Equation:

$$\frac{1}{20}(d-10) + \frac{1}{15}(d)=1$$
$$3d-15+4d=60$$
$$d=\frac{75}{7}$$

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Re: A husband and wife started painting their house, but husband [#permalink]  08 Jul 2013, 00:09
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Re: A husband and wife started painting their house, but husband [#permalink]  08 Jul 2013, 04:34
Where am i going wrong?

Combined work rate= 7/60

Time H and W should have taken= 60/7

Time H & W worked together = 60/7 - 5= 25/7

In 25/7 days they did= 25/7 * 7/60= 5/12

Work left for wife to do= 7/12

Time taken by wife to do 7/12= 7/12 * 15= 35/4

Total Time- 35/4+25/7 ?????
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Re: A husband and wife started painting their house, but husband [#permalink]  08 Jul 2013, 04:49
Expert's post
rahulmathur9818 wrote:
Where am i going wrong?

Combined work rate= 7/60

Time H and W should have taken= 60/7

Time H & W worked together = 60/7 - 5= 25/7

In 25/7 days they did= 25/7 * 7/60= 5/12

Work left for wife to do= 7/12

Time taken by wife to do 7/12= 7/12 * 15= 35/4

Total Time- 35/4+25/7 ?????

Actual time would be more than 60/7 days. Check here: a-husband-and-wife-started-painting-their-house-but-husband-77973.html#p1035303

Hope it helps.
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Re: A husband and wife started painting their house, but husband [#permalink]  08 Jul 2013, 05:48
1
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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
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Re: A husband and wife started painting their house, but husband [#permalink]  07 Nov 2013, 11:41
Husband(working alone) completes the job in 20 days, implying a per day completion of 5% of total job
Wife(working alone) completes the job in 15 days, implying a per day completion of 20/3% of total job

If we let the total no of days for work completion to be equal to 'T' days, then,

Wife works for all T days; husband works for T-5 days (leaves work 5 days before completion of work)

Setting up an equation for 100% completion of work -

(T-5)*5% + T*20/3% = 100%

Solving for T, T = 75/7 days
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Re: A husband and wife started painting their house, but husband [#permalink]  16 Nov 2013, 12:41
Hey johnnybravo86, whenever you come across such problems where you people work for different number of days or some days more or less, the best approach is to add up their individual work and not as a whole,
In this problem, lets consider the work to be done as a single unit i.e. 1.
now going according to the words of the question, let the wife work for x days. Thus, the husband works for x-5 days.
Now their respective efficiencies(fraction of work done per day) are given as,
H W
1/20 1/15
The final equation is

x/15 + (x-5)/20 = 1
solve for x, it comes out to be 75/7, which is the OA.
Kudos me!
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Re: A husband and wife started painting their house, but husband [#permalink]  17 Nov 2013, 05:16
t/15 + (1/20)(t-5) = 1
4t/60 + 3(t-5)/60 = 1
7t -15 = 60
7t = 75
t = 75/7
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Re: Work & Rate problem [#permalink]  21 Nov 2013, 09:32
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.

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.

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.
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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.

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.

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.

Hope it's clear.
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Re: Work & Rate problem [#permalink]  22 Nov 2013, 16:30
Bunuel wrote:
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.

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.

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.

Hope it's clear.

absolutely, thank you!
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Re: A husband and wife started painting their house, but husband [#permalink]  27 Apr 2014, 09:46
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

Here's my simple solution.

Let total time taken for the job together = T
Husband worked for days = T-5
Wife worked for days = T

Then, according to individual rates,

(T-5)/20 + T/15 = 1
=> 35T = 375
or, T = 75/7

Ans. C. 75/7
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Re: Work & Rate problem [#permalink]  28 Apr 2014, 02:39
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.

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.

Hope it helps.

Approach # 2 definitely the smart way to think and work over here !!
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Re: A husband and wife started painting their house, but husband [#permalink]  09 Mar 2015, 05:41
In how many days it will finish the work . together they can finish in 60/7 days, but someone left the works with 5 days to spare..Now it means it will take more than 60/7 only option giving is 75/7 days hence this is the answer.Hope this helps.
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Re: A husband and wife started painting their house, but husband [#permalink]  11 Mar 2015, 09:20
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

If we have a look at the problem then we come to know that the work was completed in T days when both worked together. But when Husband moves out then the remaining work must have taken more than 5 days to complete as earlier it was both husband and wife and now it is only wife.The work would have got completed in 5 days provide both of them have worked.Also it is not mentioned nywhere in the problem that the work gets completed on time.Please correct me if I am wrong
Re: A husband and wife started painting their house, but husband   [#permalink] 11 Mar 2015, 09:20

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