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.
Reciprocal Fractions
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My rephrase of the question would be
u/v/w = u/vw
x/y/z = x/yz
1/x/yz --> yz/x
what is the probability that u/vw = yz/x? ---> what is the probability that ux = vwyz?
statement 1)
says nothing about ux not sufficient
statement 2)
says nothing about vwyz not sufficient
statements 1) and 2) together
ux = median of 100 consecutive integers. I don't know if it refers to the first 100 consecutive numbers or not. I assume it does otherwise I think the answer should be E.
Assuming that ux = median of 100 first consecutive integers this means that ux = 50
so our rephrase would be what is the probability that four numbers v, w, y, and z are each randomly chosen from the first 100 positive integers have a product of 50.
I think the probability can be calculate choose C
u/v/w = u/vw
x/y/z = x/yz
1/x/yz --> yz/x
what is the probability that u/vw = yz/x? ---> what is the probability that ux = vwyz?
statement 1)
says nothing about ux not sufficient
statement 2)
says nothing about vwyz not sufficient
statements 1) and 2) together
ux = median of 100 consecutive integers. I don't know if it refers to the first 100 consecutive numbers or not. I assume it does otherwise I think the answer should be E.
Assuming that ux = median of 100 first consecutive integers this means that ux = 50
so our rephrase would be what is the probability that four numbers v, w, y, and z are each randomly chosen from the first 100 positive integers have a product of 50.
I think the probability can be calculate choose C
Is the median of a set having an even number of consecutive integers an integer?mikeCoolBoy wrote:My rephrase of the question would be
u/v/w = u/vw
x/y/z = x/yz
1/x/yz --> yz/x
what is the probability that u/vw = yz/x? ---> what is the probability that ux = vwyz?
statement 1)
says nothing about ux not sufficient
statement 2)
says nothing about vwyz not sufficient
statements 1) and 2) together
ux = median of 100 consecutive integers. I don't know if it refers to the first 100 consecutive numbers or not. I assume it does otherwise I think the answer should be E.
Assuming that ux = median of 100 first consecutive integers this means that ux = 50
so our rephrase would be what is the probability that four numbers v, w, y, and z are each randomly chosen from the first 100 positive integers have a product of 50.
I think the probability can be calculate choose C
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no it is not, I completely missed that in fact it does not make any sense what I wrote that the median is 50 I just read it.
Thank you for pointing it out.
Thank you for pointing it out.
You are still right though a part of your argument may not be. Here is the OAmikeCoolBoy wrote:no it is not, I completely missed that in fact it does not make any sense what I wrote that the median is 50 I just read it.
Thank you for pointing it out.
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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This question needs some brackets in order to remove ambiguity.dtweah wrote: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.
Is u/v/w meant to represent (u/v)/w or u/(v/w)? There's a big difference.
(u/v)/w = u/vw, where as u/(v/w) = uw/v
Cheers,
Brent
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We'd also like to know if the numbers are randomly chosen WITH or WITHOUT replacement from the first 100 natural numbers.