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by ketkoag » Sat Mar 28, 2009 2:10 am
Which of the following fractions has the greatest value?
A.1/((3^2 )(5^2 ))

B.
2/((3^2 )(5^2 ))

C.
7/((3^3 )(5^2 ))

D.
45/((3^3 )(5^3 ))

E.75/((3^4 )(5^5 ))

OA is : c but i am getting answer D. please confirm the answer, it might be possible that OA is incorrect.
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Source: — Data Sufficiency |
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by DanaJ » Sat Mar 28, 2009 3:53 am
The secret here is that you can simplify the last two fractions:

D. 45 = 5*9 = 5*3^2. This makes the fraction:

45/((3^3 )(5^3 )) = [(3^2)*5]/[(3^3)*(5^3)] = 1/[3*(5^2)].

E. 75 = 3*25 = 3*5^2

75/((3^4 )(5^5 )) = [3*(5^2)]/[(3^4)*(5^5)] = 1/[(3^3)*(5^2)].

In order to decide whether D or C is greater, just compare the two:
7/((3^3 )(5^2 )) > 1/[3*(5^2)]
7*3*(5^2) > (3^3)*(5^2) - 3*5^2 can be eliminated from each side:
7 > 9 - false.

I too think that D is greater
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by cramya » Sat Mar 28, 2009 9:32 am
The OA is wrong and u r correct in choosing D

One easy way to solve this problem is to get the powers for 3 and 5 in the denominator to match up.

A.1/((3^2 )(5^2 ))

B.
2/((3^2 )(5^2 ))

C.
7/((3^3 )(5^2 )) = 2.33 / ((3^2 )(5^2 ))


D.
45/((3^3 )(5^3 )) = 3/ ((3^2 )(5^2 ))


E.75/((3^4 )(5^5 )) = 1/15 * 1/ (3^2*5^2)



We can clearly see 3 * 1/(3^2*5^2) > 1/15 * 1/ (3^2*5^2)


Many ways to skin a cat and this is one.....


Regards,
CR
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