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If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th
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16 Jan 2019, 22:04
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Difficulty:
35% (medium)
Question Stats:
73% (02:01) correct 27% (02:13) wrong based on 88 sessions
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If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all the fractions whose numerators are in A and whose denominators are in B, what is the product of all of the numbers in C?
question says product of all the numbers in c : so IMO 1/64 would be correct.. I agree that 1/2 is repetitive but no where does the question say that the set has distinct numbers..
Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th
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17 Jan 2020, 19:10
The publisher of this question got it wrong. By definition a set in mathematics cannot have duplicate elements. 1/2 and 2/4 are the same fraction, and therefore Set C should not include it twice. The answer is really 1/32. If the question had said that C is a list than yes, the answer would be 1/64 because a list allows for duplicate elements.
Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th
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28 Jan 2020, 11:08
DwipRatan wrote:
Hi, can some explain how the answer is 1/64 ?
Posted from my mobile device
You use each number from A with each number from B to form fractions. Thus you use 1 three times, 2 three times and 3 three times (Set A) and 2 three times, 3 three times and 4 three times (Set B).
You multiply the fractions and get:
\(1^3\)*\(2^3\)*\(3^3\) / \(2^3\)*\(3^3\)*\(4^3\)
We can cancel out \(2^3\) and \(3^3\) and are left with
\(\frac{1}{{4^3\)}
which is
\(\frac{1}{64}\)
(Sorry, I don't know how to write it better)
gmatclubot
Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th
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28 Jan 2020, 11:08
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73% (02:01) correct 
