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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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Bunuel
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If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   16 Jan 2019, 23:04
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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?


(A) \(\frac{1}{64}\)

(B) \(\frac{1}{48}\)

(C) \(\frac{1}{24}\)

(D) \(\frac{1}{12}\)

(E) \(\frac{1}{2}\)
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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   16 Jan 2019, 23:21
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C = \(\frac{1}{2}*\frac{2}{3}*\frac{3}{4}\) = \(\frac{1}{4}\)

Bunuel, can you check the options?
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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   16 Jan 2019, 23:38
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Afc0892 wrote:
C = \(\frac{1}{2}*\frac{2}{3}*\frac{3}{4}\) = \(\frac{1}{4}\)

Bunuel, can you check the options?


Options are correct. You got tricked by the question. Try again.
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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   16 Jan 2019, 23:41
C= 1/2*1/3*1/4=1/24

Answer is

Option C

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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   16 Jan 2019, 23:50
1/2*1/3*1/4*2/3*2/4*3/2*3/4 = 1/64

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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   17 Jan 2019, 02:57
pradeep15793 wrote:
1/2*1/3*1/4*2/3*2/4*3/2*3/4 = 1/64

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The first and the fifth terms are the same
1/2=2/4
and I think the trickiness of the question lies there
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If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   17 Jan 2019, 04:56
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LevanKhukhunashvili wrote:
pradeep15793 wrote:
1/2*1/3*1/4*2/3*2/4*3/2*3/4 = 1/64

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The first and the fifth terms are the same
1/2=2/4
and I think the trickiness of the question lies there


LevanKhukhunashvili

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..
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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   17 Jan 2020, 20:10
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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.
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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   26 Jan 2020, 23:12
Hi, can some explain how the answer is 1/64 ?

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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   28 Jan 2020, 12:08
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DwipRatan wrote:
Hi, can some explain how the answer is 1/64 ?

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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)
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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   07 Sep 2020, 06:01
1*2*3 = 6 thrice i.e. 6^3
2*3*4 = 24 thrice i.e. 24^3 = (6*4)^3 = 6^3*4^3
Dividing both, you get [1][/6^3] = 1/64
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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   07 Sep 2020, 10:09
Each digit of numerator i.e. 1, 2 , 3 will have 2, 3 , 4 as denominator. So, the product will look like:
=> (1/2 x 1/3 x 1/4) x ( 2/2 x 2/3 x 2/4) x (3/2 x 3/3 x 3/4) = 1x1x1 x 2x2x2 x 3x3x3 / (2x3x4)^3
= 1/4^3= 1/64
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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink] New post   19 Jan 2024, 00:51
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Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th [#permalink]
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