Yogesh S.
asked • 03/18/16Rate, time and work
Andy and Ben can finish painting a wall in x hours working together. Andy takes y hours more than Ben to finish the same work when both of them work alone. What is the number of hours that Ben takes to finish the same work alone?
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1 Expert Answer
John K. answered • 03/19/16
Tutor
4.9
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Math and Engineering Tutor, Professional Engineer
Well shucks let's use numbers for a start. Let A = hrs for A and B = hrs for B to complete the job. Suppose they can finish the job in 10 hrs working at the same time but independently then A+B=10 and A=10-B, or B=A-10. Suppose y=2. Then A=B+2 (Andy takes two hours more than Ben) or substituting for A, (B+2)+B=10 -> 2B=8 -> A=6, B=4. We used x=10, y=2 then symbolically 2B=x-y, -> B=(x-y)/2, and A=(x+y)/2. Note that x must be larger than y to avoid negative time.
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Imad A.
Time for Andy = y + Time for Ben
A = y + B
B= A - y
Rate at which Andy works = Work / A = work / (B + y)
Rate at which Ben Works = Work / B
Rate at which they work together = Work / (B + y) + Work / B
= Work ( 1/(B+y) + 1/B)
= Work (B + B +y )/ (B^2 + By)
x hours = Work / Rate at which they work together
x = Work / [Work (B + B +y )/ (B^2 + By)]
x = (b^2 + by) /(2b + y)
b^2 + by = 2bx+xy
b^2 + by - 2bx - xy = 0
b^2 + B(Y-2X) -xy = 0
Solve the quadratic equation for B and you have the answer.
Answer options would help.
03/18/16