Question on ratios and proportions

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Question on ratios and proportions

by gmatrant » Fri Oct 12, 2007 9:54 pm
A question on ratios and proportions


If (x +y) varies directly as (x-y), then (x^2+y^2) will directly vary as

1) x^2-y^2 2) xy 3)Either (1) or (2) 4) Cannot be determined

Answer 3

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by Suyog » Sat Oct 13, 2007 12:43 am
is the option 2 is 2xy instead of xy?

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by camitava » Sat Oct 13, 2007 1:20 am
If (x +y) varies directly as (x-y), then (x^2+y^2) will directly vary as

1) x^2-y^2 2) xy 3)Either (1) or (2) 4) Cannot be determined
Hi gmatrant,
According to the question, (x + y) = k(x - y)
Now (x ^2 + y^2) = {(x + y)^2 + (x - y)^2} / 2
= {2(x + y)^2 - 4xy} / 2 - Right? Getting me? ... i
= {2k(x - y)^2 - 4xy} / 2 ................................ii

Multiply i and ii,
(x^2 + y^2) ^ 2 = {2(x + y)^2 - 4xy} * {2k(x - y)^2 - 4xy}/4
= [4k(x^2 - y^2)^2 - 8xy{k(x - y)^2 + (x + y) ^ 2} + 16 x^2y^2]/4

As we need not to take care of the portion - 8xy{k(x - y)^2 + (x + y) ^ 2}, so we can say that -
(x^2 + y^2) ^ 2 depends on (x^2 - y^2)^2 and on x^2y^2.

So (x^2 + y^2) depends on (x^2 - y^2) and on xy. So 3 is the best option to chose. Guys waiting for more comment on this qs! :wink:
Correct me If I am wrong


Regards,

Amitava

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by gmatrant » Sat Oct 13, 2007 3:18 am
camitava wrote:
If (x +y) varies directly as (x-y), then (x^2+y^2) will directly vary as

1) x^2-y^2 2) xy 3)Either (1) or (2) 4) Cannot be determined
Hi gmatrant,
According to the question, (x + y) = k(x - y)
Now (x ^2 + y^2) = {(x + y)^2 + (x - y)^2} / 2
= {2(x + y)^2 - 4xy} / 2 - Right? Getting me? ... i
= {2k(x - y)^2 - 4xy} / 2 ................................ii

Multiply i and ii,
(x^2 + y^2) ^ 2 = {2(x + y)^2 - 4xy} * {2k(x - y)^2 - 4xy}/4
= [4k(x^2 - y^2)^2 - 8xy{k(x - y)^2 + (x + y) ^ 2} + 16 x^2y^2]/4

As we need not to take care of the portion - 8xy{k(x - y)^2 + (x + y) ^ 2}, so we can say that -
(x^2 + y^2) ^ 2 depends on (x^2 - y^2)^2 and on x^2y^2.

So (x^2 + y^2) depends on (x^2 - y^2) and on xy. So 3 is the best option to chose. Guys waiting for more comment on this qs! :wink:
The answer is right, and your approach also seems fine. But can there be a easier way to solve this?

Anyway great work..thanks for the solution..