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OG PS #32) sqrt[(16)(20) + (8)(32)]?

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by factor26 » Sun Jul 31, 2011 12:54 pm
sqrt[(16)(20) + (8)(32)]?


A) 4*sqrt20

B) 24

C) 25

D) 4*sqrt20 + 8*sqrt2

E) 32

OG answer is B.

I tried to prime factorize out;

sqrt [2^4*2^2*5 + 2^3*2^5]

now i'm completely lost on what to do next ... can anyone help?
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by factor26 » Sun Jul 31, 2011 1:24 pm
ok i believe i figured this out ... someone correct me if i'm wrong please ...

sqrt[16*20 + 8*16*2]

factor out 16 ...

sqrt[(16) (20 + 8*2)]

sqrt [16*36]

4 * 6

Answer = 24
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by edge » Sun Jul 31, 2011 1:24 pm
sqrt(16x20 + 16x16) = sqrt(16 x (16 + 20)) = sqrt(16 x 36) = sqrt(16) x sqrt(36) = 4 x 6 = 24
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by gmatboost » Tue Aug 02, 2011 4:51 pm
I happened to post about this question on GMAT Boost: The Blog very recently, check it out:

https://blog.gmatboost.com/2011/08/02/al ... on-5-3-29/
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by HeyArnold » Tue Aug 02, 2011 4:59 pm
My initial strategy for this problem was to factor out a 4..

So

sqrt(16 x 20) + (8 x 32)

becomes 2 x Sqrt(4 x 5) + (2x8)

= 2 x sqrt(36)
=2 * 6 = 12

what is the error in my logic ?!
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by Touseef » Wed Aug 03, 2011 9:14 pm
Hey Arnold,

Please understand this logic

16*20=320

can u write 16*20 =4(4*5)=80.That would be wrong.

because a factor of 16 should be taken not 4 to solve this.

Now coming to your question,this is was the error in your logic

sqrt(16 x 20) + (8 x 32)

becomes 2 x Sqrt(4 x 5) + (2x8)

= 2 x sqrt(36)
=2 * 6 = 12

You are actually factoring out 16 but taking it as 4


sqrt[16/4*20/4 + 8/4*32/4] (4*4=16)

=>sqrt[(16) (20 + 8*2)]

sqrt [16*36]

4 * 6

Answer = 24

Hope this helps....
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