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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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If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th  [#permalink]

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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]

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New post 16 Jan 2019, 23:21
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]

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New post 16 Jan 2019, 23:41
C= 1/2*1/3*1/4=1/24

Answer is

Option C

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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]

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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]

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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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If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th   [#permalink] 17 Jan 2019, 04:56
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