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pump a, b, & c operate at constant rates

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pump a, b, & c operate at constant rates

by lunarpower » Fri Jul 11, 2008 8:49 pm
courtesy of user 'karenmeow'

1. pump a, b, & c operate at constant rates. pump a & b, simultaneously, fill a tank in 6/5 hr; pump a & c, simultaneously, fill a tank in 3/2 hr; pump b & c, simultaneously, fill a tank in 2 hrs. how many hrs does it take a, b, & c, simultaneously, to fill a tank?
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Re: pump a, b, & c operate at constant rates

by lunarpower » Fri Jul 11, 2008 8:55 pm
lunarpower wrote:courtesy of user 'karenmeow'

1. pump a, b, & c operate at constant rates. pump a & b, simultaneously, fill a tank in 6/5 hr; pump a & c, simultaneously, fill a tank in 3/2 hr; pump b & c, simultaneously, fill a tank in 2 hrs. how many hrs does it take a, b, & c, simultaneously, to fill a tank?
like many rate problems, this one presents information about rates not as actual rates, but as TIMES taken to complete jobs. these times are the reciprocals of the rates, and must be transformed accordingly:

rate for A and B together = (1 tank) / (6/5 hr) = 5/6 tank/hr (or you could just take the reciprocal)
rate for A and C together = (1 tank) / (3/2 hr) = 2/3 tank/hr (or you could just take the reciprocal)
rate for B and C together = (1 tank) / (2 hr) = 1/2 tank/hr (or you could just take the reciprocal)

note that these are rates for machines working together, and are thus additive. in other words, the 'rate for A and B together' is actually just the sum of A's rate and B's rate, and likewise for the other two compound rates. (no such relationship holds for times, a fact that's the principal reason why we have to bother with all these reciprocals in the first place.)
we're interested in finding the sum of ALL THREE rates. as in many other gmat problems, symmetry comes to the rescue: if we simply add together all the given data, we arrive at a symmetric expression in terms of the three rates.

adding everything up:
(rate for A and B together) + (rate for A and C together) + (rate for B and C together)
= 2(rate A) + 2(rate B) + 2(rate C)
= 5/6 + 2/3 + 1/2 = 5/6 + 4/6 + 3/6 = 12/6 = 2

dividing by two gives
rate A + rate B + rate C = 1 tank/hr

the time taken is, once again, the reciprocal of this quantity (or you could solve the equation rate x time = total output, or, since the number is so nice, you could just solve the problem by inspection).
it takes 1 hour.
Ron has been teaching various standardized tests for 20 years.

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