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Solve..

by neya » Sun Jul 17, 2011 8:33 pm
Two numbers x and y (x>y) are such that their sum is equal to three times their difference.Then the value of 3xy/2(x^2-y^2) ?
1.2/3
2.1
3.4/3
43/2
5.none of these
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by beatthegmat.garry » Sun Jul 17, 2011 8:44 pm
x+y=3(x-y) ....given

(x+y)^2=3(x^2-y^2) .....multiplyin both sides by (x+y)

x^2 + y^2 +2xy=3x^2-3y2;

2xy=2x^2-2y^2;
xy/x^2-y^2=1

3xy/2(x^2-y^2)=3/2

Hence answer is 3/2 which is number 4.
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by neya » Sun Jul 17, 2011 9:40 pm
But the correct answer given is option 2 (i.e) 1
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by Anurag@Gurome » Sun Jul 17, 2011 9:51 pm
neya wrote:Two numbers x and y (x>y) are such that their sum is equal to three times their difference.Then the value of 3xy/2(x^2-y^2) ?
(x + y) = 3(x - y) ---> x = 2y

Hence, 3xy/2(x^2 - y^2) = 3*(2y)*y/[2*((2y)^2 - y^2)] = 6y^2/6y^2 = 1

The correct answer is B.

beatthegmat.garry' wrote:...
x^2 + y^2 +2xy=3x^2-3y2;
2xy=2x^2-2y^2;
...
Check it. It should be 2xy = 2x^2 - 4y^2
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by dodgeforgmat » Sun Jul 17, 2011 9:58 pm
X+Y = 3(X-Y)
=> X = 2Y

Substitute
[3(2Y)(Y)]/[2(4Y^2-Y^2)]
=> 6Y^2/6Y^2
= 1
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by Vivek Murari » Mon Jul 18, 2011 8:05 pm
The answer is option B i.e. Number 1.
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