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Working alone at their respective constant rates, A can complete a tas
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Bunuel
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Working alone at their respective constant rates, A can complete a tas [#permalink] New post  04 Nov 2019, 02:56
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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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Archit3110
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Re: Working alone at their respective constant rates, A can complete a tas [#permalink] New post  04 Nov 2019, 03:39
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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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lacktutor
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Re: Working alone at their respective constant rates, A can complete a tas [#permalink] New post  04 Nov 2019, 04:01
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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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wishmasterdj
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Re: Working alone at their respective constant rates, A can complete a tas [#permalink] New post  24 Feb 2021, 05:11
lacktutor wrote:
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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I am going to ask a very silly question here. How did you solve the 2 equations. I couldn't get the values of a and b, even after doing everything right upto that point.
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Brian123
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Re: Working alone at their respective constant rates, A can complete a tas [#permalink] New post  26 Feb 2021, 18:27
wishmasterdj wrote:
lacktutor wrote:
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.................. eq (1)
6(1/b)+ (9/2)(1/a)=1................eq (2)

—> 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

Posted from my mobile device


I am going to ask a very silly question here. How did you solve the 2 equations. I couldn't get the values of a and b, even after doing everything right upto that point.


wishmasterdj
eq1 = 6b +4a = ab and eq2 = 6a+ 4b = ab.
Subtract Eq1 from eq2

2a-1.5b = 0. Hence, 2a=1.5b......... Eq3.

Use Eq3 in either Eq1 or Eq2 and you'll get your answer.

Let me know if any step isn't clear!
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Bambi2021
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Re: Working alone at their respective constant rates, A can complete a tas [#permalink] New post  28 Mar 2021, 15:05
One thing that I dont understand here. Why do you choose 6/a and 4/b and not 6*a and 4*b?

I want to take 6 times "the working rate of b" + 4 times "the working rate of a" = 1 unit

(units/day)*days = units

Seems most logical to me...

6a + 4b = 1
4,5a + 6b = 1
-->
1,5a = 2b and a = 4b/3
-->
24b/3 + 12b/3 = 1
-->
36b/3 = 1 --> 36b = 3 --> b = 12 days
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Re: Working alone at their respective constant rates, A can complete a tas [#permalink]
28 Mar 2021, 15:05
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