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What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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30 Oct 2014, 09:46
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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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30 Oct 2014, 18:20
Bunuel wrote: Tough and Tricky questions: Probability. What is the probability that u/v/w and x/y/z are reciprocal fractions? (1) v, w, y, and z are each randomly chosen from the first 100 positive integers. (2) The product (u)(x) is the median of 100 consecutive integers. I got this one wrong. Please correct my reasoning: 1  we only know about v,w,y,z. No information about u and x  Insufficient 2. u*x is the median of 100 consecutive integers. Firstly, the only information provided is about u and x and secondly, the the consecutive integers can start from anywhere. They may be negative positive, some negative some positive, so Insufficient Combining the two, We have some information about v,w,y,z but info about u and x is unclear So E
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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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31 Oct 2014, 12:00
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@ThoughttoSphere: the question stem can be rephrased as : Is (v)(w)(y)(z)=(u)(x) ? Statement 1: really doesn't tells us anything except that v,w , y and z are integers chosen from the first 100 positive integers. Clearly not sufficient. Statement 2: tells us that product of x and u is the median of 100 consecutive integers. Now whatever be the set of 100 consecutive integers, the median is always going to be the mean of the 50th and 51st term and since the terms are consecutive, the median will be x.5 which is not an integer. Clearly not sufficient by itself. Combining 1 and 2  we can surely tell that since v,w,y and z are integers their product can never be equal to a fraction. Hence C.



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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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01 Nov 2014, 06:10
romitsn wrote: @ThoughttoSphere: the question stem can be rephrased as : Is (v)(w)(y)(z)=(u)(x) ? Statement 1: really doesn't tells us anything except that v,w , y and z are integers chosen from the first 100 positive integers. Clearly not sufficient. Statement 2: tells us that product of x and u is the median of 100 consecutive integers. Now whatever be the set of 100 consecutive integers, the median is always going to be the mean of the 50th and 51st term and since the terms are consecutive, the median will be x.5 which is not an integer. Clearly not sufficient by itself. Combining 1 and 2  we can surely tell that since v,w,y and z are integers their product can never be equal to a fraction. Hence C. Please explain how did you get the question stem can be rephrased as : Is (v)(w)(y)(z)=(u)(x) ?



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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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05 Nov 2014, 13:46
u/v/w & x/y/z are the two fractions given.
I can interpret it in two ways
1. [u/v] / w & [x/y] / z ....in this case ...we get the equation ux = yzvw ...and hence ans C.
2. But if i interpret the fraction as
U / [v/w] & X / [y/z] ....in this case....we get the equation uwxz = vy ...i.e. ux = vy / wz i.e. 50.5 = vy / wz [50.5 if we consider 1 to 100]
So this equation can give somenumber point five and no way we can determine if reciprocals or not . Hence here ans E.
@@bunuel ....should the question have more clarity . Please suggest.



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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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12 Sep 2015, 05:34
Not sure about the answer as the expression in the question is not clear. I could not get an expression which I could take as a premise. Different interpretations, different expressions. Bunuel.. a little help please!



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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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19 Oct 2016, 18:10
Question: (u*w)/v = y/(x*z) ? [1] v, w, y, z E {1,2, ... 99, 100} Not sufficient... [2] ux = (50+51)/2 = 101/2 = 50.5 Not sufficient... [1]+[2]: ux = 50.5 = (v/w)*(y/z) Now no pair of random values from 1 to 100 can form (101/2) as 101 is a prime number. So ux = (v/w)*(y/z) => (u*w)/v = y/(x*z) Sufficient. Answer C



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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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26 Nov 2016, 04:24
This can be simplified as follows: What is the probability that ux/vy/wz = 1 ? What is the probability that ux/vy = wz ? Finally: What is the probability that ux = vywz ? Statement (1) tells us that vywz is an integer, since it is the product of integers. However, this gives no information about u and x and is herefore not sufficient to answer the question. Statement (2) tells us that ux is NOT an integer. This is because the median of an even number of consecutive integers is NOT an integer. (For example, the median of 4 consecut ive integers  9, 10, 11, 12  equals 10.5.) However, this gives us no information about vywz and is therefore not sufficient to answer the question. Taking both statements together, we know that vywz IS an integer and that ux is NOT an integer. Therefore vywz CANNOT be equal to ux. The probability that the fractions are reciprocals is zero. The correct answer is C: Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, but NEITHER statement ALONE is sufficient
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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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28 Nov 2016, 11:03
if we know we can calculate the probability that all x,z,u,w,v, and y are integers, we get the answer.



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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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06 Jan 2017, 07:06
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Please find the solution of this problem attached.
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Re: What is the probability that u/v/w and x/y/z are reciprocal fractions? [#permalink]
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22 Nov 2017, 12:43
anairamitch1804 wrote: Please find the solution of this problem attached. Hi anairamitch1804 Bunuel chetan2u can we call 1/2.5 as a fraction or fraction needs to be strictly reduced form of integers? Thanks




Re: What is the probability that u/v/w and x/y/z are reciprocal fractions?
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