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Working alone at their respective constant rates, A can complete a tas

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Working alone at their respective constant rates, A can complete a tas  [#permalink]

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New post 04 Nov 2019, 03:56
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

46% (03:25) correct 54% (03:35) wrong based on 24 sessions

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Working alone at their respective constant rates, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together?

A. 46/9
B. 50/9
C. 50/11
D. 36/7
E. 210/41


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Re: Working alone at their respective constant rates, A can complete a tas  [#permalink]

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New post 04 Nov 2019, 04:39
2
rate of A = 1/a
and rate of B = 1/b
if a starts work it completes in 1,2 , 5,6 , 9,10 = 6 days
and b in 3,4 , 7,8 ; 4 days
6/a + 4/b = 1---(1)
and if b starts work we will have b rate ; 6/b and a ends with half day more ; 4.5/a
6/b+4.5/a=1--(2)
solve for 1 &2
we get
a=9 and b = 12
so combined rate = 1/9+1/12 = 7/36
36 /7 days IMO D

Bunuel wrote:
Working alone at their respective constant rates, A can complete a task in ‘a’ days and B in ‘b’ days. They take turns in doing the task with each working 2 days at a time. If A starts they finish the task in exactly 10 days. If B starts, they take half a day more. How long does it take to complete the task if they both work together?

A. 46/9
B. 50/9
C. 50/11
D. 36/7
E. 210/41


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Re: Working alone at their respective constant rates, A can complete a tas  [#permalink]

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New post 04 Nov 2019, 05:01
If A starts:
—> 2(1/a)+ 2(1/b)+ 2(1/a)+ 2(1/b)+2(1/a)= 1

If b starts:
—> 2(1/b)+ 2(1/a)+ 2(1/b)+2(1/a)+2(1/b)+(1/2)(1/a)= 1

6(1/a)+ 4(1/b)= 1
6(1/b)+ (9/2)(1/a)=1

—> a= 9
—> b= 12

(1/a) + (1/b)= 1/x
1/9+ 1/12= 21/108= 1/x
—> x= 108/21= 36/7

The answer is D

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Re: Working alone at their respective constant rates, A can complete a tas   [#permalink] 04 Nov 2019, 05:01
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