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Indra-Maths Problem


On Tue, Feb 9, 2010 at 6:59 PM, indra priyadarsini wrote:

Dear Sir/Madam

I received your reply. I’ll send my solutions as soon as possible to the mentioned address.

As I was going through some of the sample papers I had, I came across the following problem. I could not solve it for a long time. Finally I got the answer as 42.666. Since the solutions were not given, I request you to kindly check my answer and let me know. If my answer is wrong, please let me know the right answer along with the working.

Problem :

If 5x – 11y = 2x + 5y, then find the value of 3x^2 + 2y^2 / 3x^2 – 2y^2

My Answer:

3x = 16y
x = 16 / 3y ………………[eq. 1]

(3x^2 + 2y^2) + (3x^2 – 2y^2) / (3x^2 + 2y^2) – (3x^2 – 2y^2) [By Componendo & Dividendo]
= (3x^2 + 2y^2 + 3x^2 – 2y^2) / (3x^2 + 2y^2 – 3x^2 + 2y^2)
= 6x^2 / 4y^2
= 3x^2 / 2y^2
= 3(16 / 3y)^2 / 2y^2 [On substituting x from eq.1]
= 3*256*y^2 / 2*9*y^2
= 42.66

Thanking you

Yours faithfully
Indra

Our Reply:

Hi Indra,

You have solved it right.

Perfect!!

Well Done!!

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6 Responses to “Indra-Maths Problem”

  1. APRAJITA says:

    GREAT

  2. raj says:

    why is the exponent^ used???

  3. admin says:

    please read ^ as 2 (square)

  4. anmol says:

    I THINK SO THAT X SHOULD BE 16Y/3????

  5. shivdutt says:

    how can we ask our doubts

  6. Ishan says:

    Here, x=16y/3 & not 16/3y. Shouldn’t it be ??????

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