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Can you solve this in 2 mins?

Expert replies
by netigen » Tue May 20, 2008 10:12 pm
This is a 600-700 difficulty level question. Do you think you can solve it in less than 2 mins?
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Source: — Problem Solving |

by Magellan » Tue May 20, 2008 11:25 pm
Could you give the potential solutions?
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by akshatsingh » Tue May 20, 2008 11:58 pm
I wont exactly sove this, but the answer should be something greater than 5 and less than 6.

Depending on the answer choices..it can be lot easier.
Aks
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by netigen » Wed May 21, 2008 12:22 am
here are the answers:

1. 2xsqrt(3x(sqrt(5))
2. 3
3. 3sqrt(3)
4. 9+4sqrt(5)
5. 3+2sqrt(5)
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Re: Can you solve this in 2 mins?

by simplyjat » Wed May 21, 2008 1:02 am
netigen wrote:This is a 600-700 difficulty level question. Do you think you can solve it in less than 2 mins?
The question can very well be solved under 60 seconds. You need to know that while evaluating radicals you have to remove radicals from the denominator. Also, you need to know that (a+b)(a-b)=a^2 - b^2.

In the given case start out by multiplying both numerator and denominator of the second term by 9 - 4 sqrt(5)...
simplyjat
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by vishalsongra » Wed May 21, 2008 2:05 am
Thanks ... simplyjat

Very much solve in 60 sec ..
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by vishalsongra » Wed May 21, 2008 2:10 am
Thanks ... simplyjat

Very much solve in 60 sec ..
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by m51v50 » Wed May 21, 2008 10:01 am
Solution:
√[(3√80 +3/(9+4√5)]
=√[12√5 +27-12√5]
=√27
=3√3
Answer is 3√3.
Have 790 or die
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by AleksandrM » Wed May 21, 2008 10:08 am
I couldn't do this under 60 secs, but I did this under 120 secs. Glad I took my college algebra class to know to flip the sign when mulitplying the denominator in the first step.
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by VP_Jim » Wed May 21, 2008 10:35 am
Allow me to chime in - I'd do this by estimating:

The square root of 80 is something very close to 9, so let's just call it 9. 3*9=27.

Now, we have 27+some fraction. Let's just drop the fraction.

Now, we have the square root of 27, which is 3 root 3. Easy! That took less than 30 seconds.

My hunch is that this is how the GMAC would like you to solve such a question. Very rarely do they test strict arithmetic rules - always look for a trick.
Jim S. | GMAT Instructor | Veritas Prep
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