Machine A produces bolts at a uniform

IMO

Machine·····W·······T
A·············120······40
B·············100······20
A+B·········200·······x

Let´s equal the work they do, with the LCM:

Machine·····W·······T
A·············120······40 (*5)
B·············100······20 (*6)

So:

Machine·····W·······T
A·············600······200
B·············600······120

Since now we know the time every machine needs to produce the same work (600), we can apply, in order to obtain the time needed for A+B to produce 600 bolts:

T(A+B) = [T(A) * T(B)] / [T(A) + T(B)]

T(A+B) = (120 * 200) / 120 + 200

T(A+B) = (120 * 200) / 320

T(A+B) = (12Ø * 200) / 32Ø

T(A+B) = (2^2 * 3 * 5^5 * 2^3) / 2^5

T(A+B) = 3 * 25 = 75

So, A+B need 75 seconds to produce 600. To produce 3 time less (200), they will need 3 time less seconds

75 / 3 = 25.

Jan 26, 2015 • Comment

Mike McGarry, Magoosh Tutor

Dec 28, 2013 • Comment