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A can complete a project in 20 days and B can complete the same projec [#permalink]

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16 Nov 2011, 00:54

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

44% (01:16) correct
56% (01:13) wrong based on 9 sessions

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A can complete a project in 20 days and B can complete the same project in 30 days. If A and B start working on the project together and A quits 10 days before the project is completed, in how many days will the project be completed?

A) 18 days B) 27 days C) 26.67 days D) 16 days E) 12 days

This is my logic - A's rate is 1/20 and B's rate is 1/30. Together they can complete the project in 12 days. # of days the project has been completed is 'x'. Therefore, A and B work together for (x-10) days, and can complete (x-10)/12 of the project. B alone has to complete x - (x-10)/12 of the project --> 12 x - x + 10 / 12 --> 11x + 10 / 12 So if B can complete the project in 30 days, then he can complete [(11x + 10)/12] of the project in [(11x + 10)/12]*30 days Total time taken to complete the project would then be (x-10)/12 + (330x + 300)/12 = x But that equation is obviously incorrect cause it gives a negative value. What have I missed?

Re: A can complete a project in 20 days and B can complete the same projec [#permalink]

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30 Dec 2011, 04:04

two ways to solve it :-

first :-

use formula:- t-10/20 + t/ 30 =1 we will get t = 18 days

second :-

in last 10 days b has done 1/3 rd of job alone so A anb B together has done 2/3rdof job in first X Days , we have to find value of X , in sinlge day both togther will do 1/20th + 1/30th = 1/12th of job

means 1/12th job is done together by A and B in sinlge day 1/12 job ........................ in a day full job in ...................... 12 days 2/3 rd of job ................ 12*2/3 days = 8 days

so all toghether they will take 10 + 8 = 18 days answer is A - 18

Re: A can complete a project in 20 days and B can complete the same projec [#permalink]

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30 Dec 2011, 11:56

Rate together is= 1/20+1/30=1/12 Total T days needed. So, together they work (T-10) days. therefore, 1/12*(T-10)+(1/30)*10=1 [work done together + work done by B= whole work] solving for T = 18

A can complete a project in 20 days and B can complete the same project in 30 days. If A and B start working on the project together and A quits 10 days before the project is completed, in how many days will the project be completed?

A) 18 days B) 27 days C) 26.67 days D) 16 days E) 12 days

This is my logic - A's rate is 1/20 and B's rate is 1/30. Together they can complete the project in 12 days. # of days the project has been completed is 'x'. Therefore, A and B work together for (x-10) days, and can complete (x-10)/12 of the project. B alone has to complete x - (x-10)/12 of the project --> 12 x - x + 10 / 12 --> 11x + 10 / 12 So if B can complete the project in 30 days, then he can complete [(11x + 10)/12] of the project in [(11x + 10)/12]*30 days Total time taken to complete the project would then be (x-10)/12 + (330x + 300)/12 = x But that equation is obviously incorrect cause it gives a negative value. What have I missed?

The rate of A \(\frac{1}{20}\) job per day; The rate of B \(\frac{1}{30}\) job per day.

Say they need \(t\) days to complete the project.

According to the stem we have that B works for all \(t\) days and A works only for \(t-10\) days, thus \(\frac{1}{20}*(t-10)+\frac{1}{30}*t=1\) --> \(t=18\)days.