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16 Nov 2021, 06:15
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Question Stats:
49% (03:12) correct
51%
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A, B and C can independently do a work in 15 days, 20 days and 30 days, respectively. They work together for sometime after which C leaves. A total of $18000 is paid for the work and B gets $6000 more than C. For how many days did A work?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 10
(A) 2
(B) 4
(C) 6
(D) 8
(E) 10
18 Nov 2021, 05:41
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Independently, A does 1/15 of the work in one day, B does 1/20 of the work in one day & C does 1/30 of the work in one day.
1. If they work for an equal number of days, the salary of A & B will be in the ratio of the rate of their work per day i.e. Salary of A:B = (1/15)/(1/20) = 4/3
(Notice that A works the best as he is able to complete the work fastest compared to others or in other words in one day, A does a higher amount of work compared to say, B. Hence, he is also expected to get a better salary compared to B or C).
From pt. 1, let's say, A gets 4x & B gets 3x salary in the entire period they worked till completion (this holds good since both of them worked for an equal number of days). Therefore, C gets, 3x - 6000 salary since it is given in the question that B gets 6000 more salary C.
2. Salary of A+B+C = 4x + 3x + (3x-6000) = 18000 (from the question) => x = 2400
From pt. 2, it follows that C got 3x2400 - 6000 = 1200 salary
Till the time A, B & C worked together, C got 1200 salary & now let's try to figure out how much salary did A & B get in the same time frame. To arrive at the same, note that the ratio of salary of B:C (same logic from pt. 1)= (1/20)/(1/30) = 3/2 & hence, B would have got 3/2 x 1200 = 1800 salary & A would have got 4/3 x 1800 = 2400 salary. So, in the first phase of work when all of them worked together, total salary would amount to: 2400+1800+1200 = 5400
This essentially means in the first phase of all working together, (5400/18000) = 3/10th of the work got completed & 7/10th of the work remains to be completed.
No of days (n) it must have taken to complete 3/10th of the work (1st phase):-
n.[1/15 + 1/20 + 1/30] = 3/10
=> n = (3/10)/[(4+3+2)/60] => n = (3/10)/(9/60) = 2
In the 2nd phase when only A & B worked together & 7/10th of the work got completed, no of days (n') it must have taken to complete this work:-
n'.[1/15+1/20]=7/10
=>n'=(7/10)/[(4+3)/60] =>n'=(7/10)/(7/60) = 6
Since A worked for complete duration across phase 1 & 2, total number of days A worked is n+n' = 2+6 = 8.
So, answer is Option D.
Thanks
Devneet
1. If they work for an equal number of days, the salary of A & B will be in the ratio of the rate of their work per day i.e. Salary of A:B = (1/15)/(1/20) = 4/3
(Notice that A works the best as he is able to complete the work fastest compared to others or in other words in one day, A does a higher amount of work compared to say, B. Hence, he is also expected to get a better salary compared to B or C).
From pt. 1, let's say, A gets 4x & B gets 3x salary in the entire period they worked till completion (this holds good since both of them worked for an equal number of days). Therefore, C gets, 3x - 6000 salary since it is given in the question that B gets 6000 more salary C.
2. Salary of A+B+C = 4x + 3x + (3x-6000) = 18000 (from the question) => x = 2400
From pt. 2, it follows that C got 3x2400 - 6000 = 1200 salary
Till the time A, B & C worked together, C got 1200 salary & now let's try to figure out how much salary did A & B get in the same time frame. To arrive at the same, note that the ratio of salary of B:C (same logic from pt. 1)= (1/20)/(1/30) = 3/2 & hence, B would have got 3/2 x 1200 = 1800 salary & A would have got 4/3 x 1800 = 2400 salary. So, in the first phase of work when all of them worked together, total salary would amount to: 2400+1800+1200 = 5400
This essentially means in the first phase of all working together, (5400/18000) = 3/10th of the work got completed & 7/10th of the work remains to be completed.
No of days (n) it must have taken to complete 3/10th of the work (1st phase):-
n.[1/15 + 1/20 + 1/30] = 3/10
=> n = (3/10)/[(4+3+2)/60] => n = (3/10)/(9/60) = 2
In the 2nd phase when only A & B worked together & 7/10th of the work got completed, no of days (n') it must have taken to complete this work:-
n'.[1/15+1/20]=7/10
=>n'=(7/10)/[(4+3)/60] =>n'=(7/10)/(7/60) = 6
Since A worked for complete duration across phase 1 & 2, total number of days A worked is n+n' = 2+6 = 8.
So, answer is Option D.
Thanks
Devneet
Although it is not explicitly stated, we can assume that each worker is paid according to the portion of the total work that he does. After C's departure, A and B continue working until the job is finished so they work for the same number of days. Since A can do the whole work in 15 days and B in 20 days, A does (20/15)=(4/3) of the work that B does and is thus paid 4/3 of the money that B gets.
Let's say C gets 'x' dollars. Then B gets $(6000+x) and A gets $(4/3)(6000+x).
Since total payment for the work is $18,000, [$(4/3)(6000+x) + $(6000+x) + x] = $18,000.......> x=1200.
So A gets (4/3)*(7200)=$9600 which is 9600/18000=(8/15)th of the total payment which means A does (8/15)th of the total work. Since we are told that A does the whole work in 15 days, he does (8/15)th of the work in 8 days.
ANS: D
Let's say C gets 'x' dollars. Then B gets $(6000+x) and A gets $(4/3)(6000+x).
Since total payment for the work is $18,000, [$(4/3)(6000+x) + $(6000+x) + x] = $18,000.......> x=1200.
So A gets (4/3)*(7200)=$9600 which is 9600/18000=(8/15)th of the total payment which means A does (8/15)th of the total work. Since we are told that A does the whole work in 15 days, he does (8/15)th of the work in 8 days.
ANS: D
