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Director  Joined: 05 Jan 2008
Posts: 614
Matt and Peter can do together a piece of work in 20 days  [#permalink]

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3
7 00:00

Difficulty:   5% (low)

Question Stats: 87% (02:01) correct 13% (02:33) wrong based on 283 sessions

### HideShow timer Statistics Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.

A. 26 days
B. 27 days
C. 23 days
D. 25 days
E. 24 days

Originally posted by prasannar on 11 Mar 2008, 03:56.
Last edited by Bunuel on 23 Oct 2012, 05:22, edited 1 time in total.
Renamed the topic and edited the question.
Intern  Joined: 11 Apr 2009
Posts: 1

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5
Another easy way is ..

M&P complete 60% of the work in 12 days (since they complete 100% in 20 days)
P completes the remaining 40% in 10 days .. => to complete 100% he would need 25 days.
##### General Discussion
Director  Joined: 10 Sep 2007
Posts: 831

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1
Suppose rate of work for per day Matt = M & Rate of work per day for Peter = P
Together in a Day they can finish = P + M units
Total Work Done in 20 Days = 20(P+M)
Total Work Done in 12 Days = 12(P+M)
Work Done by Peter in 10 Days = 10P

Since total work is same we can say that 20(P+M) = 12(P+M) + 10P => 8M = 2P => P = 4M
So total work done by them together in 20 Days = 20(4M+M) = 100M
Since Pete does 4M unit of work per day, it will take him 25 Days (100M/4M) to finish up the work.

VP  Joined: 28 Dec 2005
Posts: 1396

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1
2
together, they can do a piece in 20 days, i.e. 1/m + 1/p = 1/20

In 12 days, they can finish 12*(1/20) = 3/5 of the piece. After Matt leaves, 2/5 still needs to be done by Peter, which he does in 10 days.

10/(2/5) = 25
Retired Moderator Joined: 18 Jul 2008
Posts: 809

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Isn't this the same question? time-work-61136.html#p442431 But why do the answers differ...

pmenon wrote:
together, they can do a piece in 20 days, i.e. 1/m + 1/p = 1/20

In 12 days, they can finish 12*(1/20) = 3/5 of the piece. After Matt leaves, 2/5 still needs to be done by Peter, which he does in 10 days.

10/(2/5) = 25
Director  Joined: 14 Aug 2007
Posts: 635

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Isn't this the same question? time-work-61136.html#p442431 But why do the answers differ...

Because Matt and Peter have some issues working together ..Just kidding.

Note that this problem is asking the the time taken by thr guy who did NOT stop working after 12 days while the question in link is asking the time taken by the guy who stopped working after 12 days.
Manager  Joined: 14 Nov 2008
Posts: 164
Schools: Stanford...Wait, I will come!!!

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Best way is this. 1) Get the unit of quantity of work. Make it a number which you find LCM of the given digits.
2) Calculate the rate of work for each person.
And then calculate what is asked.

So as per above,
Lets assume that 20*12=240 unit of work is there.
Assume rate of work per day for Matt is m, and for Peter is p;
so..
Matt and Peter can do together a piece of work in 20 days. implies..
(m+p)*20= 240
m+p=12
Now, peter works for 22 day while, matt works for 12 days.
so, 22p+12m=240
Solving the equations, we find the value of m and p as 2.4, and 9.6 unit of work/day respectively.
So peter will take, 240/9.6=25 days.

prasannar wrote:
Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.

26days

27days

23days

25days

24 days

What is the best way to solve these problems?
Senior Manager  Joined: 28 Aug 2006
Posts: 289

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Work done by M&P in 12 days = 12/20 = 3/5

Remaining 2/5 is done by P alone in 10 days.

So P alone can do the entire work in (5/2) X 10 = 25 days
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Manager  Joined: 13 Aug 2009
Posts: 150
Schools: Sloan '14 (S)

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24.Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.
26days
27days
23days
25days
24 days

Rate Together * # of days working together + Rate of Peter * # of days working alone = 1 completed job

let P = 3 of hours Peter can complete one job alone

(1/20)*12 + (1/P)*10 = 1

P = 25
Intern  Joined: 02 Feb 2010
Posts: 2

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Together they complete the job in 20 days means they complete 12/20 of the job after 12 days.

Peter completes the remaining (8/20) of the job in 10 days which means that the whole job(1) can be completed in X days.

<=> 8/20->10 <=> X=10/(8/20)= 25 Thus the answer is D.
1 -> X
Intern  B
Joined: 18 Jan 2012
Posts: 45
Location: United States
Schools: IIM A '15 (A)
Re: Matt and Peter can do together a piece of work in 20 days.  [#permalink]

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2
To me, the most intuitive approach to solve work /rate problems is to use smart numbers.
Then we need to find the work rate - work done each entity in 1 day. The subsequent steps are then very easy .

If 2 entities A and B work together, then Work done by A in one day + Work Done by B in one day = Total work done by A and B in one day.
Example - If a machine produces 10 widgets per day and another machine produces 20 widgets per day, then working together both machines can produce 30 (10 + 20) widgets per day.

Let's choose a nr that is divisible by all the numbers given in the question stem - 20,12,10
LCM of 20,12,10 = 60
Let's assume that Total work = 60 units.
Matt and Peter work together to complete the work in 20 days. So the work done by both of them together is 3 units per day (60/20)
Now we are almost done

Matt and Peter worked together for 12 days. Hence working together, they completed 12 x 3 = 36 units of work
What remains is 24 units and Peter completed this work all by himself in 10 days

Hence peter's work rate = 24/10 units per day

Therefore, Time taken by peter to complete the 60 units of work = Total Work /Peter's work rate = (60)/(24/10) = 25 days
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Math Expert V
Joined: 02 Sep 2009
Posts: 56266
Re: Matt and Peter can do together a piece of work in 20 days  [#permalink]

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2
1
prasannar wrote:
Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.

A. 26 days
B. 27 days
C. 23 days
D. 25 days
E. 24 days

Matt and Peter together would complete 12/20=3/5th of the work in 12 days, thus the remaining 2/5th is done by Peter alone in 10 days.

Therefore Peter can complete the work alone in 10/(2/5)=25 days.

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Senior Manager  Joined: 13 Aug 2012
Posts: 415
Concentration: Marketing, Finance
GPA: 3.23
Re: Matt and Peter can do together a piece of work in 20 days  [#permalink]

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1
$$\frac{1}{M}+\frac{1}{P}= \frac{1}{20}$$

Calculate work done together in 12 days:
$$\frac{1}{20}x12==>\frac{12}{20}=\frac{3}{5}$$

Remaining work is 1-3/5.
Calculate the days left for P to perform work alone:
$$\frac{1}{P}x10days=1-\frac{3}{5}$$
$$\frac{10}{P}=\frac{2}{5}$$
$$P=25 days$$

A. 26 days
B. 27 days
C. 23 days

D. 25 days
E. 24 days
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Joined: 08 Dec 2013
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Location: India
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GMAT 1: 630 Q47 V30 WE: Operations (Non-Profit and Government)
Matt and Peter can do together a piece of work in 20 days  [#permalink]

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prasannar wrote:
Matt and Peter can do together a piece of work in 20 days. After they have worked together for 12 days Matt stops and Peter completes the remaining work in 10 days. In how many days Peter complete the work separately.

A. 26 days
B. 27 days
C. 23 days
D. 25 days
E. 24 days

let Matt & Peter be machines with efficiency m and p, so

m*p*20 = m*p*12 + p*10
m= (10/8)

Let Peter can complete in 10 days
m*p*20 = p*k
k=25

Bunuel is this approach correct, could you provide me with the repository of such sums.
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I never gave up what I wanted- Matt and Peter can do together a piece of work in 20 days   [#permalink] 14 May 2019, 19:34
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