quants problems

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quants problems

by nagar.sandeep » Fri Sep 10, 2010 10:24 pm
Hi,

It will be nice if anyone could help me with couple of quants problems, which are listed down:

Q1 What is the median number of employees assigned per project for the projects at Company Z?
(1) 25 percent of the projects at Company Z have 4 or more employees assigned to each
project.
(2) 35 percent of the projects at Company Z have 2 or fewer employees assigned to each
project.


Q2 For any positive integer x, the 2-height of x is defined to be the greatest nonnegative
integer n such that 2n is a factor of x. If k and m are positive integers, is the 2-height of k
greater than the 2-height of m ?
(1) k > m
(2)k/m is an even integer



Q3 In the decimal representation of x, where 0 < x < 1, is the tenths digit of x nonzero?
(1) 16x is an integer.
(2) 8x is an integer.

Any help will be appreciated.

Thanks,
Sandeep
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by Rahul@gurome » Fri Sep 10, 2010 11:53 pm
Q1. What is the median number of employees assigned per project for the projects at Company Z?
(1) 25 percent of the projects at Company Z have 4 or more employees assigned to each project.
(2) 35 percent of the projects at Company Z have 2 or fewer employees assigned to each project.

Explanation:

(1) Median is the value that separates the upper and lower half of the sample. (1) gives that 25% of projects have 4 or more employees to each project . But the % of projects that have less than 4 employees is not given. So, (1) is NOT SUFFICIENT.

(2) From this statement, the % of projects that have more than 2 employees is not given. As above, (2) is also NOT SUFFICIENT.

Combining (1) and (2), 35% projects have 2 or fewer employees, 25% of the projects have 4 or more employees, so the remaining % of projects = 100 - (35 + 25) = 40% projects should have exactly 3 employees (between 2 and 4).
Therefore, median number of employees assigned per project = 3

The correct answer is [spoiler](C)[/spoiler].
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by Rahul@gurome » Sat Sep 11, 2010 12:18 am
Q2. For any positive integer x, the 2-height of x is defined to be the greatest nonnegative
integer n such that 2n is a factor of x. If k and m are positive integers, is the 2-height of k
greater than the 2-height of m?
(1) k > m
(2)k/m is an even integer

Explanation:

(1) k > m does not imply that the 2-height (that is the number of 2's in the prime factorization of an integer) of k is greater than the 2-height of m. It may or may not be greater than the 2-height of m.
If k = 22 and m = 12, then 2-height of k is less than 2-height of m.
If k = 12 and m = 10, then 2-height of k is greater than 2-height of m.
No unique answer.
So, (1) is NOT SUFFICIENT.

(2) k/m can be an even integer if k's 2-height is greater than the m's 2-height.
If k = 14 and m = 7, then k's 2-height is greater than the m's 2-height.
If k = 12 and m = 6, then k's 2-height is greater than the m's 2-height.
So, (2) is SUFFICIENT.

The correct answer is [spoiler](B)[/spoiler].
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by Rahul@gurome » Sat Sep 11, 2010 12:25 am
Q3. In the decimal representation of x, where 0 < x < 1, is the tenths digit of x nonzero?
(1) 16x is an integer.
(2) 8x is an integer.

Explanation:

(1) If x = 1/2 = 0.5, then tenths digit of x = 5
If x = 1/16 = 0.0625, then tenths digit of x = 0
So, the tenths digit of x may or may not be zero. No unique answer.
So, (1) is NOT SUFFICIENT.

(2) If x = 1/2 = 0.5, then tenths digit of x = 5, which is non-zero.
If x = 1/8 = 0.125, then tenths digit of x = 1, which is non-zero. This implies that all numbers in the tenth place between 0 < x < 1 will always have a non-zero integer.
So, (2) is SUFFICIENT.

The correct answer is [spoiler](B)[/spoiler].
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by Crystal W » Thu Apr 14, 2016 6:01 am
Rahul@gurome wrote:Q2. For any positive integer x, the 2-height of x is defined to be the greatest nonnegative
integer n such that 2n is a factor of x. If k and m are positive integers, is the 2-height of k
greater than the 2-height of m?
(1) k > m
(2)k/m is an even integer

Explanation:

(1) k > m does not imply that the 2-height (that is the number of 2's in the prime factorization of an integer) of k is greater than the 2-height of m. It may or may not be greater than the 2-height of m.
If k = 22 and m = 12, then 2-height of k is less than 2-height of m.
If k = 12 and m = 10, then 2-height of k is greater than 2-height of m.
No unique answer.
So, (1) is NOT SUFFICIENT.

(2) k/m can be an even integer if k's 2-height is greater than the m's 2-height.
If k = 14 and m = 7, then k's 2-height is greater than the m's 2-height.
If k = 12 and m = 6, then k's 2-height is greater than the m's 2-height.
So, (2) is SUFFICIENT.

The correct answer is [spoiler](B)[/spoiler].
Thank you for your explanation! Can you explain more about what is the meaning of the question?

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by [email protected] » Thu Apr 14, 2016 9:43 am
Hi Crystal W,

The prior posts are all over 5 years old, so it's likely that the original posters are long gone. What specifically would you like to know about this prompt?

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by Crystal W » Thu Apr 14, 2016 1:58 pm
[email protected] wrote:Hi Crystal W,

The prior posts are all over 5 years old, so it's likely that the original posters are long gone. What specifically would you like to know about this prompt?

GMAT assassins aren't born, they're made,
Rich
I cannot understand the question meaning. The description makes me so confused. Can you explain it?

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by OptimusPrep » Thu Apr 14, 2016 6:44 pm
Crystal W wrote: Thank you for your explanation! Can you explain more about what is the meaning of the question?
Hi Crystal W,

First things first, i think there is a problem with the question. The question should read "For any positive integer x, the 2-height of x is defined to be the greatest non-negative integer n such that 2^n is a factor of x"

Now let us talk about 2-height of a number.
Assume x = 20 = 2^2*5
Here 2-height of 20 = 2, since the highest power of 2 is 2
Assume x = 10 = 2^1*5
Here 2-height of 10 = 1, since the highest power of 2 is 1

You simply need to find out the powers of two in the prime factorization.
Does this help?

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by Matt@VeritasPrep » Fri Apr 15, 2016 12:49 pm
Crystal W wrote:
[email protected] wrote:Hi Crystal W,

The prior posts are all over 5 years old, so it's likely that the original posters are long gone. What specifically would you like to know about this prompt?

GMAT assassins aren't born, they're made,
Rich
I cannot understand the question meaning. The description makes me so confused. Can you explain it?
It's essentially asking how many times you can divide x by 2 and still have an integer.

For example, take the number 24.

24/2 = 12
12/2 = 6
6/2 = 3

So we can divide 24 by 2 THREE TIMES and still have an integer. (If we kept going, 3/2 wouldn't be, so we have to stop at this point.) So the 2-height of 24 is 3.

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by Crystal W » Fri Apr 15, 2016 2:16 pm
OptimusPrep wrote:
Crystal W wrote: Thank you for your explanation! Can you explain more about what is the meaning of the question?
Hi Crystal W,

First things first, i think there is a problem with the question. The question should read "For any positive integer x, the 2-height of x is defined to be the greatest non-negative integer n such that 2^n is a factor of x"

Now let us talk about 2-height of a number.
Assume x = 20 = 2^2*5
Here 2-height of 20 = 2, since the highest power of 2 is 2
Assume x = 10 = 2^1*5
Here 2-height of 10 = 1, since the highest power of 2 is 1

You simply need to find out the powers of two in the prime factorization.
Does this help?
It is really helpful! Thank you so much!

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by Crystal W » Fri Apr 15, 2016 2:18 pm
Matt@VeritasPrep wrote:
Crystal W wrote:
[email protected] wrote:Hi Crystal W,

The prior posts are all over 5 years old, so it's likely that the original posters are long gone. What specifically would you like to know about this prompt?

GMAT assassins aren't born, they're made,
Rich
I cannot understand the question meaning. The description makes me so confused. Can you explain it?
It's essentially asking how many times you can divide x by 2 and still have an integer.

For example, take the number 24.

24/2 = 12
12/2 = 6
6/2 = 3

So we can divide 24 by 2 THREE TIMES and still have an integer. (If we kept going, 3/2 wouldn't be, so we have to stop at this point.) So the 2-height of 24 is 3.
Thank you for your help! Hoewever, I believe in OG 2016, it's 2^n instead 2n