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What is probability that u/v/w and x/y/z are reciprocal [#permalink]
29 Sep 2008, 23:02

2

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What is probability that u/v/w and x/y/z are reciprocal fractions? 1: v,w,y,z are each randomly chosen from the first 100 positive integers. 2: The product of U X is the median of 100 consecutive integers.

Re: reciprocal probability [#permalink]
30 Sep 2008, 11:48

yes, it should be C.

I - v w y z could be any of the first 100 positive integers. no information about u and x thus insufficient

II - ux could be any number of the form x.5 where x is any integer. for example, 0.5, 1.5, 2.5, 3.5 etc.. like if we take "first" 100 consecutive integers, ux is 50.5. doesn't tell anything about v w y z

I and II - we have to find out probability of us being equal to vwyz; product of v w y z (all integers) can never be a fraction of the form x.5. probability is 0.

Re: reciprocal probability [#permalink]
30 Sep 2008, 19:05

aim2010 wrote:

yes, it should be C.

I - v w y z could be any of the first 100 positive integers. no information about u and x thus insufficient

II - ux could be any number of the form x.5 where x is any integer. for example, 0.5, 1.5, 2.5, 3.5 etc.. like if we take "first" 100 consecutive integers, ux is 50.5. doesn't tell anything about v w y z

I and II - we have to find out probability of us being equal to vwyz; product of v w y z (all integers) can never be a fraction of the form x.5. probability is 0.

thus C.

The product of U X is the median of 100 consecutive integers.

How can the median of 100 consecutive integers take the form of x.5? median of consecutive integers is an integer. correct? Still that does not tell us any thing about product vwyz

Re: reciprocal probability [#permalink]
30 Sep 2008, 19:20

icandy wrote:

aim2010 wrote:

yes, it should be C.

I - v w y z could be any of the first 100 positive integers. no information about u and x thus insufficient

II - ux could be any number of the form x.5 where x is any integer. for example, 0.5, 1.5, 2.5, 3.5 etc.. like if we take "first" 100 consecutive integers, ux is 50.5. doesn't tell anything about v w y z

I and II - we have to find out probability of us being equal to vwyz; product of v w y z (all integers) can never be a fraction of the form x.5. probability is 0.

thus C.

The product of U X is the median of 100 consecutive integers.

How can the median of 100 consecutive integers take the form of x.5? median of consecutive integers is an integer. correct? Still that does not tell us any thing about product vwyz

median of an odd number of consecutive integers is an integer. for even number of consecutive integers median is the mean of the middle 2 which would give you a fraction. like for { 1, 2, 3, 4 } median is 2.5

Re: reciprocal probability [#permalink]
30 Sep 2008, 19:24

aim2010 wrote:

icandy wrote:

aim2010 wrote:

yes, it should be C.

I - v w y z could be any of the first 100 positive integers. no information about u and x thus insufficient

II - ux could be any number of the form x.5 where x is any integer. for example, 0.5, 1.5, 2.5, 3.5 etc.. like if we take "first" 100 consecutive integers, ux is 50.5. doesn't tell anything about v w y z

I and II - we have to find out probability of us being equal to vwyz; product of v w y z (all integers) can never be a fraction of the form x.5. probability is 0.

thus C.

The product of U X is the median of 100 consecutive integers.

How can the median of 100 consecutive integers take the form of x.5? median of consecutive integers is an integer. correct? Still that does not tell us any thing about product vwyz

median of an odd number of consecutive integers is an integer. for even number of consecutive integers median is the mean of the middle 2 which would give you a fraction. like for { 1, 2, 3, 4 } median is 2.5

Shit! I need some caffeine. Thanks for making it obvious

gmatclubot

Re: reciprocal probability
[#permalink]
30 Sep 2008, 19:24