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20 Mar 2013, 16:12
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Are there any tricks or what are the general rules to finding these?
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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20 Mar 2013, 22:03
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exploringm
Hi esploringm, I'm not sure if I'm misunderstanding the question, but I believe you are asking for the fastest way to get a number that multiplies with the original square root to produce 1. So the reciprocal of 3 is 1/3, as 3*1/3 = 3/3 = 1.
the same exact logic can be used to the reciprocal of \(\sqrt{3}\). Multiply it by 1/\(\sqrt{3}\) and you'll get 1. However, the GMAT doesn't allow for square roots in the denominator, so you have to use clever algebra to manipulate the equation into something you can use.
Instead of writing the answer as 1/\(\sqrt{3}\), change the form by multiplying it by the number 1, conveniently written as \(\sqrt{3}/\sqrt{3}\) for our purposes.
This will give you \(\sqrt{3}/\sqrt{3}*\sqrt{3}\). The numerator stays \(\sqrt{3}\) while the denominator simplifies to just 3, the answer is thus \(\sqrt{3}/3\), which is algebraically and numerically identical to 1/\(\sqrt{3}\) but might show up as an answer choice whereas radicals in the denominator are frowned upon.
This strategy can thus be employed for any number. The reciprocal of \(\sqrt{10}\) is thus 1/\(\sqrt{10}\) or \(\sqrt{10}/10\).
Hope this helps!
-Ron
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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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Thank you for understanding, and happy exploring!
