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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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Math Expert
Joined: 02 Sep 2009
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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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16 Jan 2019, 22:04
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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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16 Jan 2019, 22:21
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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16 Jan 2019, 22:38
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]

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16 Jan 2019, 22:41
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]

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16 Jan 2019, 22:50
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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17 Jan 2019, 01:57
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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17 Jan 2019, 03:56
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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]

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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.
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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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26 Jan 2020, 22:12
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]

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28 Jan 2020, 11:08
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)
Re: If A = {1, 2, 3}, B = {2, 3, 4}, and C is the set consisting of all th   [#permalink] 28 Jan 2020, 11:08
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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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