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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] New post 26 Feb 2012, 05:30
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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
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Re: If x and y are positive integers and 21x=12y [#permalink] New post 26 Feb 2012, 05:34
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Expert's post
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] New post 26 Feb 2012, 05:37
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] New post 26 Feb 2012, 09:07
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] New post 26 Feb 2012, 11:19
Expert's post
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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COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If x and y are positive integers and 21x=12y [#permalink] New post 20 Aug 2013, 04:06
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] 20 Aug 2013, 04:06
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