GRE 0568 question 11

jijicavalerul · Post by **jijicavalerul** » Thu Oct 30, 2008 8:39 pm

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!

Nameless · Post by **Nameless** » Thu Oct 30, 2008 11:36 pm

You have to use f(x,y)=(x)^(1/2)*(y)^(3/2)
then choose x=1, y=256

amateur · Post by **amateur** » Fri Oct 31, 2008 1:12 am

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

ralphhumacho · Post by **ralphhumacho** » Fri Oct 31, 2008 7:49 pm

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.

nate · Post by **nate** » Wed Nov 05, 2008 7:39 pm

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