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ribbons in bag

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by sumasajja » Wed Aug 03, 2011 7:23 am
there are only two kinds of ribbons in a bag of 8 m and 13 m
combined length of all the ribbons is 100
A)no of 8m ribbons
B) no of 13m ribbons

a) A is greater
b) B is greater
c) both A and B are equal
d) data insufficient
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by edge » Wed Aug 03, 2011 7:41 am
Multiples of 13: 13, 26, 39, 52, 65, 78, 91
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96

The only pair of numbers that add up exactly to 100 are 52 (13 * 4) and 48 (8 * 6).

Therefore, A = 6 and B = 4 and A > B.
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by beatthegmat.garry » Wed Aug 03, 2011 7:51 am
Lets assume n1 be the no. of 8m ribbons.
Lets assume n2 be the no. of 13m ribbons.

Since total length=100, we can write:
8*n1 +13*n2=100 .........(1)

Lets guess values of n1 and n2 such that eqn.(1) will be satisfied
You will only find just one combination which is n1=6, n2=4.

So I believe there are 6 ribbons of 8m and 4 ribbon of 13m.

Hence (A) is greater.
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by tpr-becky » Wed Aug 03, 2011 7:58 am
This is a GRE problem but the ideas are the same (question type isn't on the GMAT)

Another way to approach these is to methodically approach the different multiples:

the most 13m ribbons you could have is 7 - 7(13) is 91 100 - 91 = 9 (not divisible by 8)
if you have 6 then 13(6) = 78 100 - 78 = 22 (not divisible by 8)
if you have 5 13(5) = 65; 100 - 65 = 35 (not divisible by 8)
if you have 4 13(4)= 52; 100 - 52 = 48 (that is 6(8)) - so there are more 8's elminate B and C.

you can then try the rest but none are divisible by 8 so the answer is A.
Becky
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The Princeton Review
Irvine, CA
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by gmatboost » Wed Aug 03, 2011 8:29 am
Consider 8a + 13b = 100

8a is EVEN
100 is EVEN

So, 13b must also be EVEN, since E + E = E

That means b must be EVEN, since 13 is not.

So, you should only spend time testing even values of b (6, 4, 2, maybe 0)
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