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If x and y are positive integers and 21x=12y

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If x and y are positive integers and 21x=12y [#permalink]

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If x and y are positive integers and 21x=12y what is the least possible value of xy?

(A) 14
(B) 28
(C) 63
(D) 84
(E) 252
[Reveal] Spoiler: OA

Kudos [?]: 56 [1], given: 14

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Re: If x and y are positive integers and 21x=12y [#permalink]

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New post 26 Feb 2012, 06:34
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tom09b wrote:
If x and y are positive integers and 21x=12y what is the least possible value of xy?

(A) 14
(B) 28
(C) 63
(D) 84
(E) 252


\(21x=12y\) --> \(\frac{x}{y}=\frac{12}{21}=\frac{4}{7}\). Now, since x and y are positive integers then the least value of x is 4 and the least value of y is 7 --> \(xy=4*7=28\).

Answer: B.
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Re: If x and y are positive integers and 21x=12y [#permalink]

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New post 26 Feb 2012, 06:37
tom09b wrote:
If x and y are positive integers and 21x=12y what is the least possible value of xy?

(A) 14
(B) 28
(C) 63
(D) 84
(E) 252


B....

21x=12y
=> x/y = 4/7
=> 7x=4y
7(3)=4(3) => x*y=9 But it is not given
7(4)=4(7) => x*y=28
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Re: If x and y are positive integers and 21x=12y [#permalink]

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New post 26 Feb 2012, 10:07
GMATD11 wrote:
tom09b wrote:
If x and y are positive integers and 21x=12y what is the least possible value of xy?

(A) 14
(B) 28
(C) 63
(D) 84
(E) 252


B....

21x=12y
=> x/y = 4/7
=> 7x=4y
7(3)=4(3) => x*y=9 But it is not given
7(4)=4(7) => x*y=28


Sorry, I don't understand your explanation...
Could you please explain again??

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Re: If x and y are positive integers and 21x=12y [#permalink]

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New post 26 Feb 2012, 12:19
tom09b wrote:
GMATD11 wrote:
tom09b wrote:
If x and y are positive integers and 21x=12y what is the least possible value of xy?

(A) 14
(B) 28
(C) 63
(D) 84
(E) 252


B....

21x=12y
=> x/y = 4/7
=> 7x=4y
7(3)=4(3) => x*y=9 But it is not given
7(4)=4(7) => x*y=28


Sorry, I don't understand your explanation...
Could you please explain again??


This red part is not correct x and y can not equal to 3, since their ratio in this case would be 3/3=1 and not 4/7 as it should be (or simply 7*3=4*3 is not correct).
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Re: If x and y are positive integers and 21x=12y [#permalink]

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New post 20 Aug 2013, 05:06
21x=14y ==> 7x=4y ===> x=4y/7

Now we have to calculate xy =
I can substitute value of x from above calculated relationship.

I got ==> (4y^2)/7

Now test for given values
A. 14 ===> Y^2 = 14*7/4 = 24.5
B. 28 ===> Y^2 = 28*7/4 = 49 Thus Y=7 answer is B.
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Re: If x and y are positive integers and 21x=12y [#permalink]

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Re: If x and y are positive integers and 21x=12y [#permalink]

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New post 11 Mar 2017, 07:28
tom09b wrote:
If x and y are positive integers and 21x=12y what is the least possible value of xy?

(A) 14
(B) 28
(C) 63
(D) 84
(E) 252


\(21x = 12y = 84\)

So, \(x = 4\) & \(y = 7\)

Thus, \(xy = 28\)

Hence, correct answer must be (B) 28
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Re: If x and y are positive integers and 21x=12y [#permalink]

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New post 11 Mar 2017, 09:45
tom09b wrote:
If x and y are positive integers and 21x=12y what is the least possible value of xy?

(A) 14
(B) 28
(C) 63
(D) 84
(E) 252


21x=12y which is 7x=4y after dividing by 3 on both sides.
7 and 4 are co-primes that is neither of them is factor of each other. So the first common multiple will be 7*4 = 28
hence x=4 and y=7 and hence x*y=28

Answer B

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Re: If x and y are positive integers and 21x=12y   [#permalink] 11 Mar 2017, 09:45
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