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Working together, A and B can do a job in 6 days. B and C

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Working together, A and B can do a job in 6 days. B and C [#permalink] New post 29 Oct 2005, 10:21
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Working together, A and B can do a job in 6 days. B and C can do the same job in 10 days, while C and A can do it in 7.5 days. How long will it take if all A, B and C work together to complete the job?

(1) 8 days
(2) 5 days
(3) 3 days
(4) 7 days
(5) 6 days


How long will it take for A alone to complete the job?

(1) 8 days
(2) 6 days
(3) 10 days
(4) 20 days
(5) 5 days
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 [#permalink] New post 29 Oct 2005, 10:38
A B and C together will do the Job in 5 days .

A alone in 10 Days......?

what is OA ..?
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 [#permalink] New post 29 Oct 2005, 11:48
I get 5 for number one and 10 for A alone as well, but the calculations took me quite a while, around 3-5 min! Is there some simple way of doing this one? What is OE
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Re: PS Work Ascent 4/02 [#permalink] New post 29 Oct 2005, 19:30
GMATT73 wrote:
Working together, A and B can do a job in 6 days. B and C can do the same job in 10 days, while C and A can do it in 7.5 days. How long will it take if all A, B and C work together to complete the job?

(1) 8 days
(2) 5 days
(3) 3 days
(4) 7 days
(5) 6 days


How long will it take for A alone to complete the job?

(1) 8 days
(2) 6 days
(3) 10 days
(4) 20 days
(5) 5 days


Q.1: Using the productivity to solve:
Let x,y,z be the work A,B and C can do in one day.
We have:
x+y=1/6
y+z=1/10
x+z=1/7.5
---> 2(x+y+z)= 1/6+1/10+1/7.5= 24/60=2/5 ---> x+y+z=1/5 -----> if they work together, they'll finish the work in 5 days

Q.2 (x+y)+(x+z)-(y+z)= 2x= 1/6+1/7.5 -1/10= 12/60=1/5---> x=1/10
----> A alone can finish the work in 10 days.
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 [#permalink] New post 30 Oct 2005, 17:47
Great explanations laxie...

for the first question, why do u multiply (x+y+z) by 2? Is it because of the double-counting?
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 [#permalink] New post 31 Oct 2005, 00:00
Question 1
A and B can do 1/6 job in 1 day
B and C can do 1/10 job in 1 day
C and A can do 2/15 job in 1 day

1/A + 1/B = 1/6 ----(1)
1/B + 1/C = 1/10 ---(2)
1/C + 1/A = 2/15 --(3)

Adding (1), (2) and (3), we get:

2/A + 2/C + 2/B = 2/5
1/A + 1/C + 2/B = 1/5 job in 1 day
So A + B + C willl finishi 1 job in 5 days

Question 2
(1) - (2) --> 1/A - 1/C = 1/15 --(4)
(4) + (3) --> 2/A = 1/5
So A alone finish 1/10 job in 1 day = 10 days for 1 job
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Re: PS Work Ascent 4/02 [#permalink] New post 01 Nov 2005, 00:35
laxieqv wrote:
GMATT73 wrote:
Working together, A and B can do a job in 6 days. B and C can do the same job in 10 days, while C and A can do it in 7.5 days. How long will it take if all A, B and C work together to complete the job?

(1) 8 days
(2) 5 days
(3) 3 days
(4) 7 days
(5) 6 days


How long will it take for A alone to complete the job?

(1) 8 days
(2) 6 days
(3) 10 days
(4) 20 days
(5) 5 days


Q.1: Using the productivity to solve:
Let x,y,z be the work A,B and C can do in one day.
We have:
x+y=1/6
y+z=1/10
x+z=1/7.5
---> 2(x+y+z)= 1/6+1/10+1/7.5= 24/60=2/5 ---> x+y+z=1/5 -----> if they work together, they'll finish the work in 5 days

Q.2 (x+y)+(x+z)-(y+z)= 2x= 1/6+1/7.5 -1/10= 12/60=1/5---> x=1/10
----> A alone can finish the work in 10 days.


Got it Laxie! Thanx for the explanation. Is there any reason why you are so fond of using the variables x, y, and z?
Re: PS Work Ascent 4/02   [#permalink] 01 Nov 2005, 00:35
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