M28-41

Question

Machine A and B working together at their constant rates can complete a certain task in 6 days. In how many days, working alone, can machine A complete the task?

(1) The average (arithmetic mean) of the respective times A and B would each take to complete the task working alone is 12.5 days.
 
(2) It would take machine A 5 more days to complete the task alone than it would take machine B to complete the task.

Bunuel · Accepted Answer

Official Solution:

Machine A and B working together at their constant rates can complete a certain task in 6 days. In how many days, working alone, can machine A complete the task? 
 
Let A and B be the times needed for machines A and B to complete the task working alone, respectively. Thus, we have  \frac{1}{A}+\frac{1}{B}=\frac{1}{6} .
 
(1) The average (arithmetic mean) of the respective times A and B would each take to complete the task working alone is 12.5 days. 
 
This implies that  A+B=2*12.5=25 . However, since we do not know which machine is faster (we cannot differentiate between A and B), even if we substitute B with  25-A  into  \frac{1}{A}+\frac{1}{B}=\frac{1}{6}  to get  \frac{1}{A} + \frac{1}{25-A} = \frac{1}{6}  and solve, we must get two different answers for A and B:  A \lt B  and  A \gt B . Not sufficient.
 
(2) It would take machine A 5 more days to complete the task alone than it would take machine B to complete the task. 
 
This implies that  A=B+5 . Substituting this into the equation, we get  \frac{1}{A}+\frac{1}{A-5}=\frac{1}{6} . We can solve for  A  to get  A=2  or  A=15 . However,  A=2  cannot be true since it would make  B  negative, so  A = 15  is the only valid solution. Sufficient.

Answer: B

Nightmare007 · Answer

Hi  Bunuel , 
 Can you spot my error

(A+B)/AB = 1/6.

Statement 1 : A + B = 25
AB = 150.

A-B = ((A+B)^2 - 4AB)^1/2
A-B = constant.
A+B = 25

Can't we get statement 1 enough.

???

Bunuel · Answer

From (1) we get two solutions: A = 10 and B = 15 OR A = 15 and B = 10.

aliakberza · Answer

chetan2u   Bunuel

Hi. Could someone please explain why  \frac{1}{B}+\frac{1}{B+5}=\frac{1}{6}  as a solution to st-2 is not a valid equation. Solving this yields b=3, but clearly  \frac{1}{3}+\frac{1}{8}  is not equal to  \frac{1}{6} , which led me to believe I had done sth else wrong or was applying the incorrect methodology altogether. Appreciate your help please.

chetan2u · Answer

Hi..
You are going wrong in finding solution..
The solutions will be 10 and -3..
(1/b)+(1/(b+5))=1/6..
6(b+5+b)=b^2+5b...12b+30=b^2+5b...
b^2-7b-30=0....(b-10)(b+3)=0..
So b=10 and -3

Hope it helps

Posted from my mobile device

Oppenheimer1945 · Answer

chetan2u   GMATinsight   Bunuel 
how are we getting this slution for st 1? We are getting A=10 and -5 . Why would we randomly consider other variations

Bunuel · Answer

For (1) we have 1/A + 1/B = 1/6 and A + B = 25. After substituting B = 25 - A into 1/A + 1/B = 1/6 we get 1/A + 1/(25 - A) = 1/6, which gives A^2 -25A + 150 = 0, which gives A = 10 or A = 15. But as explained in the solution, you don't really need to do all that if you recognize that for (1) we have no way of distinguishing machines A and B from one another and thus we have to get two solutions for A and B:  A \lt B  and  A \gt B .

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

joe123x · Answer

while I did answer B.

I couldn't get to solve  \frac{1}{A}+\frac{1}{A-5}=\frac{ 1}{6}

could you please explain how did we get A=2, A= 15? I understand that we will get a quadratic.
but the one I got is:  A^2-5A-6=0

Bunuel · Answer

Sure.

\frac{1}{A}+\frac{1}{A-5}=\frac{1}{6} ;

\frac{A - 5 + A}{A(A-5)}=\frac{1}{6} ;

\frac{2A-5}{A(A-5)}=\frac{1}{6} ;

12A-30 = A(A-5) ;

12A-30 = A^2-5A ;

A^2 - 17A+ 30 = 0 ;

A=2  or  A=15 .

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

einstein801 · Answer

Please explain how you get 2 solutions from (1):

A^2-25A+150 = (A+5)(A-30). --> A=-5 or 30.

Bunuel · Answer

A^2 -25A + 150 = 0

(A - 10)(A - 15) = 0

A = 10 or A = 15.

M28-41

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