Exponents : GMAT Problem Solving (PS)
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# Exponents

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02 May 2010, 10:39
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$$\frac{(8^2)(3^3)(2^4)}{96^2} =$$

Can you please solve? What level question is this?

Last edited by flash2374 on 02 May 2010, 11:53, edited 1 time in total.
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02 May 2010, 10:58
flash2374 wrote:
$$\frac{(8^2)(3^2)(2^4)}{96^2} =$$

Can you please solve? What level question is this?

I guess this must be around 600-650 types level question not sure though .. !

Anyway you can break down this expression to make it simple

$$\frac{(8^2)(3^2)(2^4)}{96^2} = \frac{(8^ 2) (3^2) (2^4)}{(16 * 6 *16 * 6)}$$

And then u can easily solve this one ..!
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02 May 2010, 11:12
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flash2374 wrote:
$$\frac{(8^2)(3^2)(2^4)}{96^2} =$$

Can you please solve? What level question is this?

Take denominator first = (96^2) = (32*3)^2 = (2^5*3)^2 = 2^10*3^2

Numerator = (8^2)(3^2)(2^4) = 2^6*3^2*2^4 = 2^10*3^2

So answer is $$\frac{2^10*3^2}{2^10*3^2} =$$ = 1
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02 May 2010, 11:52
$$\frac{(8^2)(3^3)(2^4)}{96^2} =$$

I'm sorry, I originally mistyped 3^2 instead of 3^3. Can you please solve it again? The answer should come to 3.
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02 May 2010, 11:56
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flash2374 wrote:
$$\frac{(8^2)(3^3)(2^4)}{96^2} =$$

I'm sorry, I originally mistyped 3^2 instead of 3^3. Can you please solve it again? The answer should come to 3.

Yes, the answer should be 3 in that case ..!
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02 May 2010, 12:20
Thanks. I see it now.
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02 May 2010, 12:33
nitishmahajan wrote:
flash2374 wrote:
$$\frac{(8^2)(3^2)(2^4)}{96^2} =$$

Can you please solve? What level question is this?

I guess this must be around 600-650 types level question not sure though .. !

Anyway you can break down this expression to make it simple

$$\frac{(8^2)(3^2)(2^4)}{96^2} = \frac{(8^ 2) (3^2) (2^4)}{(16 * 6 *16 * 6)}$$

And then u can easily solve this one ..!

I solved the answer with Silver's explanation. I'm just curious to understand your reasoning. I see that 16 * 6 is 96 so 96^2 would be 16 * 6 * 16 * 6. Then you can cancel out 2^4 and one of the 16s. You are left with (8^2)(3^3)/(16*6*6). Then you can make it (4 * 27) / 36. Then (4 * 3) / 4 = 3.
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Posts: 306
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02 May 2010, 17:31
flash2374 wrote:
nitishmahajan wrote:
flash2374 wrote:
$$\frac{(8^2)(3^2)(2^4)}{96^2} =$$

Can you please solve? What level question is this?

I guess this must be around 600-650 types level question not sure though .. !

Anyway you can break down this expression to make it simple

$$\frac{(8^2)(3^2)(2^4)}{96^2} = \frac{(8^ 2) (3^2) (2^4)}{(16 * 6 *16 * 6)}$$

And then u can easily solve this one ..!

I solved the answer with Silver's explanation. I'm just curious to understand your reasoning. I see that 16 * 6 is 96 so 96^2 would be 16 * 6 * 16 * 6. Then you can cancel out 2^4 and one of the 16s. You are left with (8^2)(3^3)/(16*6*6). Then you can make it (4 * 27) / 36. Then (4 * 3) / 4 = 3.

you can actually solve this with many ways, generally I try to cancel the big numbers, so that the calculation becomes easier later on .. and it also decreases the number steps involved for me ... Anyway, one should follow the method with which they are comfortable with ..!
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03 May 2010, 10:02
I personally prefer the way silvers solved it!
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03 May 2010, 10:14
flash2374 wrote:
nitishmahajan wrote:
flash2374 wrote:
$$\frac{(8^2)(3^2)(2^4)}{96^2} =$$

Can you please solve? What level question is this?

I guess this must be around 600-650 types level question not sure though .. !

Anyway you can break down this expression to make it simple

$$\frac{(8^2)(3^2)(2^4)}{96^2} = \frac{(8^ 2) (3^2) (2^4)}{(16 * 6 *16 * 6)}$$

And then u can easily solve this one ..!

I solved the answer with Silver's explanation. I'm just curious to understand your reasoning. I see that 16 * 6 is 96 so 96^2 would be 16 * 6 * 16 * 6. Then you can cancel out 2^4 and one of the 16s. You are left with (8^2)(3^3)/(16*6*6). Then you can make it (4 * 27) / 36. Then (4 * 3) / 4 = 3.

Choose the method that suits u the best and will give u the most confident(hopefully the right) answer in the shortest period of time
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Run towards the things that make you uncomfortable daily. The greatest risk is not taking risks
http://gmatclub.com/forum/from-690-to-730-q50-v38-97356.html

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03 May 2010, 14:22
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gxsingh wrote:
I personally prefer the way silvers solved it!

Agreed! Prime factoring is the way to go!
Re: Exponents   [#permalink] 03 May 2010, 14:22
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