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GRE 0568 question 11

Posted: Thu Oct 30, 2008 8:39 pm
by jijicavalerul
Hey guys,

Can anyone help me with this problem please:

What is the best approximation of 1.5^{1/2}*266^{3/2}?

I've tried using a Taylor approximation on f(x)=x*266^x, with f(1)=266, but don't get the right answer 5,300. Does anyone have any ideas?

Thanks!

Posted: Thu Oct 30, 2008 11:36 pm
by Nameless
You have to use f(x,y)=(x)^(1/2)*(y)^(3/2)
then choose x=1, y=256

Posted: Fri Oct 31, 2008 1:12 am
by amateur
Try 1.5^(1/2) * 266 ^ (3/2) = [ (1.5 * 266) ^ (3/2) ] / 1.5 = (3 / 2 * 266) ^ 3/2 / (3/2) = (399)^(3/2) / (3/2)

Your answer will be slightly higher if you use 400 instead of 399, so try (400)^(3/2) / 1.5 = 8000 / 1.5 = 16000 / 3 = 5333.3333

Posted: Fri Oct 31, 2008 7:49 pm
by ralphhumacho
Here's how I did it:

266^(3/2) = sqrt(266)sqrt(266)sqrt(266) = 16 * 16 * 16 = 4096 = 4100, since 16^2 = 256, which is definitely close enough in this case considering the range of answer choices.

To find sqrt(1.5) = sqrt(3)/sqrt(2) = 1.7/1.4 (rough approx). Thus, the final answer will be significantly higher than 4100, so 5300 makes the most sense.


Note that since the answers are each like 1000 units apart, you have HUGE leeway in approximating.

Posted: Wed Nov 05, 2008 7:39 pm
by nate
This is what I did:
266^(3/2) is 266*266^(1/2)
so we can rewrite as 266*(1.5*266)^(1/2)
= 266*sqrt(399)
which is approximately 266*sqrt(400) or 266*20 = 5320