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OG QUANT #86- Operations w/ rational # (exponents)

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by DCS80 » Sat Jan 12, 2013 9:49 am
sup guys,

What is the largest integer "n" such that 1/2^n > 0.01

I'm very poor with exponents, which offsets my strengths unfortunately.

could someone provide a brief step by step? I understand you need to create an equality.

A)5
B) 6
C) 7
D) 10
E) 51

THX
ds
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by vkb001 » Sat Jan 12, 2013 6:22 pm
Answer is B.

1/2^n > 0.01
=> 1/2^n > 1/100
=> 2^n < 100

find a value of n for which 2^n is less than 100. You can either do this by going over each of the choices here, or figure it out. At n = 6, 2^n = 64 < 100. At n = 7, 2^n = 128 > 100.
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by Brian@VeritasPrep » Mon Jan 14, 2013 4:53 pm
Great explanation, vkb - and just to add a little bit of bigger-picture strategy to this: When decimal problems look ugly (particularly with exponents) try looking at them as fraction problems like vkb did here. Without the use of a calculator, you can typically go a lot farther with fractions than with decimals, so often just changing "0.01" to "1/100" is enough to reframe the problem and give yourself a fair shot at it.

The authors of GMAT problems are pretty good at finding inconvenient ways to provide you with information, and often your first job is to look at what they gave you and see if you can display it a different way to make it more convenient. Fractions, a lot of the time, are a great way to take inconvenient decimals and make the problem easier to work with.
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by maxk297 » Sat Dec 06, 2014 7:27 pm
DCS80 wrote:sup guys,

What is the largest integer "n" such that 1/2^n > 0.01

I'm very poor with exponents, which offsets my strengths unfortunately.

could someone provide a brief step by step? I understand you need to create an equality.

A)5
B) 6
C) 7
D) 10
E) 51

THX
ds
Made the same mistake.. so i saw the solution sheet.
There the solution is 1/2^n>0.01 therefore 2^n<100... how did they reach this (i am a little slow with inequality)

Then i Figured it out..
1/2^n>1/100
1>2^n/100 - Multiply both sides by 2^n
100>2^n or 2^n<100 Multiply both sides by 100..

I know this seems trivial but that's the way I understood it. Thereby by understanding of inequalities a little more... A finally my speed..
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by maxk297 » Sat Dec 06, 2014 7:28 pm
DCS80 wrote:sup guys,

What is the largest integer "n" such that 1/2^n > 0.01

I'm very poor with exponents, which offsets my strengths unfortunately.

could someone provide a brief step by step? I understand you need to create an equality.

A)5
B) 6
C) 7
D) 10
E) 51

THX
ds
Made the same mistake.. so i saw the solution sheet.
There the solution is 1/2^n>0.01 therefore 2^n<100... how did they reach this (i am a little slow with inequality) Though the problem itself is simple...

Then i Figured it out..
1/2^n>1/100
1>2^n/100 - Multiply both sides by 2^n
100>2^n or 2^n<100 Multiply both sides by 100..

I know this seems trivial but that's the way I understood it. Thereby by understanding of inequalities a little more... And finally my speed..
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by [email protected] » Sat Dec 06, 2014 10:51 pm
Hi maxk297,

Since the answers to this question are numbers, we can use them to our advantage. Let's TEST THE ANSWERS.

We're told that N has to be an INTEGER and we want to make N as LARGE as possible so that 1/(2^N) > .01

Since this inequality uses a fraction on one side and a decimal on the other, I'm going to convert the decimal to a fraction. This gives us....

1/(2^N) > 1/100

We want to make N as LARGE as possible while still maintaining the inequality. This means that we have to make 2^N as BIG as possible BUT it still has to be less than 100.

One of the 5 answer choices MUST be correct, so let's TEST THE ANSWERS....

If N = 5, 2^5 = 32 1/32 > 1/100
If N = 6, 2^6 = 64 1/64 > 1/100
If N = 7, 2^7 = 128 1/128 is NOT > 1/100

So the BIGGEST that N could be is 6.

Final Answer: B

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