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A company wants to spend equal amounts of money for

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Guest2
 Post subject: A company wants to spend equal amounts of money for
 Post Posted: Sat Sep 06, 2008 8:42 pm 
A company wants to spend equal amounts of money for the purchase of two types of computer printers costing $600 and $375 per unit, respectively. What is the fewest number of computer printers that the company can purchase?
(A) 13
(B) 12
(C) 10
(D) 8
(E) 5
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divya
 Post subject:
 Post Posted: Sat Sep 06, 2008 10:46 pm 
A company wants to spend equal amounts of money for the purchase of two types of computer printers costing $600 and $375 per unit, respectively. What is the fewest number of computer printers that the company can purchase?
(A) 13
(B) 12
(C) 10
(D) 8
(E) 5

Is the answer A ?

This is how I approached it :
since equal amounts of money needs to be spent on the 2 type of computer ( $ 600, $ 375)
let x be the number of computer printers for type --A --> $ 600
Let y be the number of computer printers for type --b --> $ 375

600x = 375y

x/y = 375/600
x/y = 5/8

Thus minimally for 5 Type X, for the costs to be equal there has to be 8 type Y
you can also check:
5 * 600 = 3000
8 * 375 = 3000

Thus, total number of computers: 5 + 8 = 13
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Guest2
 Post subject:
 Post Posted: Mon Sep 08, 2008 8:21 am 
The correct answer is A,
Thanks
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RonPurewal
 Post subject:
 Post Posted: Mon Sep 29, 2008 5:58 am 
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ManhattanGMAT Staff


Posts: 6765
the above method is great.

you can also realize that this problem is a thinly veiled way of asking you for the least common multiple of the two prices. in fact, that shouldn't be that hard to see here: you're looking to spend equal amounts of money on the two types of items, so, those equal amounts must be multiples of both prices.
since you're looking for the smallest such multiple, you're looking for the smallest number that's a multiple of both prices - which is the exact definition of the least common multiple.

to find the least common multiple, you could just experiment with numbers, but, if you're looking for a more systematic way, you could use prime factorizations:
600 = 6 x 100 = (2 x 3)(2^2 x 5^2) = 2^3 x 3 x 5^2
375 = 3 x 125 = 3 x 5^3
the least common multiple takes the highest power of every prime number appearing anywhere in these factorizations. therefore, the least common multiple is 2^3 x 3 x 5^3, or 3000.
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