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

Question

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}

Chethan92 · Answer

C =  \frac{1}{2}*\frac{2}{3}*\frac{3}{4}  =  \frac{1}{4}

Bunuel , can you check the options?

Bunuel · Answer

Options are correct. You got tricked by the question. Try again.

Ana4001 · Answer

C= 1/2*1/3*1/4=1/24

Answer is

Option C

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pradeep15793 · Answer

1/2*1/3*1/4*2/3*2/4*3/2*3/4 = 1/64

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LevanKhukhunashvili · Answer

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

Archit3110 · Answer

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..

chacinluis · Answer

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.

DwipRatan · Answer

Hi, can some explain how the answer is 1/64 ?

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InaKi20 · Answer

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)

gaiknil · Answer

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    = 1/64

Gauravji21 · Answer

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

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