What is the largest integer n such that 1/2^n> 0.01?
a)5
b)6
c)7
d)10
e)51
The solution to this problem mentioned in the OG is as follows
Since 1/2^n > 0.01 is equivalent to 2n < 100, find
2"
the largest integer n such that 2" < 100. Using
trial and error, 2^6 = 64 and 64 < 100, but
2^7 = 128 and 128 > 100. Therefore, 6 is the largest
integer such that 2^n > 0.01.
The correct answer is B.
My question is
The inequality sign only flips if u multiply or divide by a negative number, so why does the sign flip in the above case.
Thanks
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Inequalities: Reciprocal Rule
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Hi AkiB,
This question asks us to compare fractions. The decimal .01 can be written as 1/100. Since 1/(2^N) and 1/100 both have numerators that equal 1, we can focus on the denominators. We're asked how big we can make N so that 1/(2^N) > 1/100. For the first fraction to be bigger than the second, the denominator must be SMALLER than 100, BUT we still have to make it as BIG as possible.
In answer to your question, the sign doesn't "flip"; you're actually cross-multiplying the inequality, so....
1/(2^N) > 1/100
becomes....
100 > 2^N
GMAT assassins aren't born, they're made,
Rich
This question asks us to compare fractions. The decimal .01 can be written as 1/100. Since 1/(2^N) and 1/100 both have numerators that equal 1, we can focus on the denominators. We're asked how big we can make N so that 1/(2^N) > 1/100. For the first fraction to be bigger than the second, the denominator must be SMALLER than 100, BUT we still have to make it as BIG as possible.
In answer to your question, the sign doesn't "flip"; you're actually cross-multiplying the inequality, so....
1/(2^N) > 1/100
becomes....
100 > 2^N
GMAT assassins aren't born, they're made,
Rich
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Matt@VeritasPrep
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Here's a nice algebraic approach.
(1/2)� > .01 is really
(1/2)� > 1/100, or
1/2� > 1/100
Since 2� and 100 are both positive, when we cross-multiply, the sign WON'T flip. So we have
100 > 2�
We know that 2�=128 > 100 > 2�=64, so n must be 6.
(1/2)� > .01 is really
(1/2)� > 1/100, or
1/2� > 1/100
Since 2� and 100 are both positive, when we cross-multiply, the sign WON'T flip. So we have
100 > 2�
We know that 2�=128 > 100 > 2�=64, so n must be 6.
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GMATinsight
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For better understanding of the rules of Inequality, Please see the rules below:AkiB wrote:What is the largest integer n such that 1/2^n> 0.01?
a)5
b)6
c)7
d)10
e)51
The solution to this problem mentioned in the OG is as follows
Since 1/2^n > 0.01 is equivalent to 2n < 100, find
2"
the largest integer n such that 2" < 100. Using
trial and error, 2^6 = 64 and 64 < 100, but
2^7 = 128 and 128 > 100. Therefore, 6 is the largest
integer such that 2^n > 0.01.
The correct answer is B.
My question is
The inequality sign only flips if u multiply or divide by a negative number, so why does the sign flip in the above case.
Thanks
Inequality Rules
Transitive Property of Inequalities
If a < b and b < c , then a < c.
If a ≤ b and b ≤ c , then a ≤ c.
If a > b and b > c , then a > c.
If a ≥ b and b ≥ c , then a ≥ c.
"GMATinsight"Bhoopendra Singh & Sushma Jha
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