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Matt and Peter can do together a piece of work in 20 days [#permalink]
11 Mar 2008, 02:56

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88% (02:22) correct
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

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.

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.

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.

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.

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

Re: Matt and Peter can do together a piece of work in 20 days. [#permalink]
22 Oct 2012, 17:57

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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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Re: Matt and Peter can do together a piece of work in 20 days [#permalink]
23 Oct 2012, 04:27

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Expert's post

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.

Re: Matt and Peter can do together a piece of work in 20 days [#permalink]
14 Nov 2012, 05:39

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\(\frac{1}{M}+\frac{1}{P}= \frac{1}{20}\)

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 _________________

Re: Matt and Peter can do together a piece of work in 20 days [#permalink]
27 Jan 2014, 11:57

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Re: Matt and Peter can do together a piece of work in 20 days [#permalink]
13 Feb 2015, 23:32

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