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 Post subject: II - Arithmetic Problem#89 P:105Posted: Sat Jan 08, 2011 1:30 pm

Joined: Sat Jan 08, 2011 1:22 pm
Posts: 1
I don't understand why is answer choice III: Y2/25 (Y to the power of 2)/25also correct
I understood why Y/15 can be correct as we need a 3 and a 5 but not the above answer choice. Could you plz explain to me if you have figured it out and if there is any easier or quicker way to solve it than the one in the book. Thank you!

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 Post subject: Re: II - Arithmetic Problem#89 P:105Posted: Sun Jan 09, 2011 8:14 pm

Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
The basic way to solve this is to look at the factors of 375:
5*5*5*3
Since we need pairs of everything for this to be a perfect square, we know that y must include, at a minimum, one 5 and one 3. There may be other terms, but these are a minimum. We'll call the other terms a*a (since they must also be in pairs)

Looking at statement III:
y^2 contains two 3's and four 5's at a minimum.
This gives us
3*3*5*5*a*a / 25.
We know that the 25 cancels with two of the 5's, leaving us with an integer value in the numerator and nothing in the denominator. Thus, the answer is an integer.

Veritas Help

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