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strategy..hard problem

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by resilient » Tue Feb 05, 2008 10:16 pm
when looking back the whole question is so easy. However, its all about the approach to the question. The solution explains that answer needs be taken with a common denomitor so that the answers will fit. How should one know how to approach these questions that stagger you at first but are actually easy questions?

(1/2)^-3 * (1/4)^-2 * (1/16)^-1 =

a. (1/2)^-48
b. (1/2)^-11
c. (1/2)^-6
d. (1/8)^-11
e.(1/8)^-6

qa is d... the math isnt that hard but the approach is staggering. what to do?
Appetite for 700 and I scraped my plate!
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Re: strategy..hard problem

by Neo2000 » Tue Feb 05, 2008 11:28 pm
Enginpasa1 wrote:when looking back the whole question is so easy. However, its all about the approach to the question. The solution explains that answer needs be taken with a common denomitor so that the answers will fit. How should one know how to approach these questions that stagger you at first but are actually easy questions?

(1/2)^-3 * (1/4)^-2 * (1/16)^-1 =

a. (1/2)^-48
b. (1/2)^-11
c. (1/2)^-6
d. (1/8)^-11
e.(1/8)^-6

qa is d... the math isnt that hard but the approach is staggering. what to do?
Look at the answer options. The numbers are either 2 or 8. So you know straight away that you have to convert to either 2 or 8. Start with 2 because you already have a 2, 4 is the same as 2Square and 16 is the same as 2^4
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by sibbineni » Wed Feb 06, 2008 4:15 am
Given Question...

(1/2)^-3 * (1/4)^-2 * (1/16)^-1

just remember the following formuale for this kind of questions.

1. a^m*a^n=a^m+n
2. (a^m)^n=a^mn
3. a^m/a^n=a^m-n where a!=0
4. (a*b)=a^m*b^m
5. 1/a^-m=a^m

so the question can be written as

2^3*4^2*16^1

2^3*2^4*2^4

2^11

=1/2^-11
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