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Mira's telescopes! Can we use LCM to solve this problem?

Expert replies
by wazzawayne » Mon May 20, 2013 7:50 am
Hi,

I got this problem in a Manhattan GMAT CAT:
Mira is making telescopes, each consisting of 2 lenses, 1 tube, and 1 eyepiece. Lenses can be purchased only in packs of 50, tubes only in packs of 10, and eyepieces only in packs of 30. However, half of the lenses in each pack are not usable for telescopes. If all parts are used only for the telescopes, what is the minimum number of lenses Mira must purchase to make a set of telescopes with no leftover components other than the unusable lenses?
a)75
b)150
c)300
d)600
e)7,500


The official explanation solves this using Ratios / Back-solving.
I just wanted if my approach using LCM is correct here:
We buy the elements in packs:
1 pack of Lenses - can contribute to (25/2) telescopes .. final number will be a multiple of (25/2)
1 pack of Tubes - can contribute to 10 Telescopes .. final number will be a multiple of 10
1 pack of Eyepieces - contribute to 30 telescopes .. final number will be a multiple of 30.

So, LEAST number of telescopes that will need to be built to use all usable elements = LCM (10,25/2,30) = {1/2 * LCM (20,25, 60)} = 150 telescopes
150 telescopes use 300 usable LENSES.
but to get 300 USABLE lenses I will need to buy 600 Lenses in total (since half are unusable in each pack).

So answer is 600.

Please let me know if this approach is correct, or it gives the correct answer by fluke :)
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Source: — Problem Solving |

by Matt@VeritasPrep » Mon May 20, 2013 1:21 pm
Wayne, your logic seems sound to me. Here's how I'd think of it:

The LCM of 50, 10, and 30 is the minimum # of telescopes that you can make. (So 150 telescopes.)

To make 150 telescopes, you need 150 * 2 = 300 compatible lenses, and to get 300 compatible lenses, you need 300 * 2 = 600 total lenses.

Good work!
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