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# If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...

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Director
Joined: 11 Feb 2015
Posts: 719
If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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Updated on: 11 Jun 2018, 11:31
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00:00

Difficulty:

65% (hard)

Question Stats:

58% (02:48) correct 42% (02:52) wrong based on 59 sessions

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If x > 0, y > 0, $$\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}$$ = 4 and what is the value of $$\frac{{x+y}}{y}$$?

A) $$\frac{3}{4}$$

B) $$\frac{4}{3}$$

C) $$\frac{10}{7}$$

D) $$\frac{7}{4}$$

E) $$\frac{7}{3}$$

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Originally posted by CAMANISHPARMAR on 11 Jun 2018, 10:26.
Last edited by CAMANISHPARMAR on 11 Jun 2018, 11:31, edited 1 time in total.
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3340
Location: India
GPA: 3.12
If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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11 Jun 2018, 11:41
3
2
CAMANISHPARMAR wrote:
If x > 0, y > 0, $$\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}$$ = 4 and what is the value of $$\frac{{x+y}}{y}$$?

A) $$\frac{3}{4}$$

B) $$\frac{4}{3}$$

C) $$\frac{10}{7}$$

D) $$\frac{7}{4}$$

E) $$\frac{7}{3}$$

If $$\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}$$ = 4,

We can simplify this equation as follows

$$7x^2 + 72xy + 4y^2 = 4(4x^2 + 12xy + 5y^2) = 16x^2 + 48xy + 20y^2$$

-> $$16x^2 - 7x^2 + 48xy - 72xy + 20y^2 - 4y^2 = 9x^2 - 24xy + 16y^2 = (3x)^2 - 2(3x)(4y) + (4y)^2 = 0$$

-> $$(3x - 4y)^2 = 0$$ -> $$3x = 4y$$ -> $$\frac{x}{y} = \frac{4}{3}$$

Adding one on both sides,

$$\frac{x}{y} + 1 = \frac{4}{3} + 1$$ -> $$\frac{{x+y}}{y} = \frac{7}{3}$$ (Option E)

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##### General Discussion
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3340
Location: India
GPA: 3.12
Re: If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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11 Jun 2018, 11:29
2
CAMANISHPARMAR wrote:
If x > 0, y > 0, $$\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}$$ and what is the value of $$\frac{{x+y}}{y}$$?

A) $$\frac{3}{4}$$

B) $$\frac{4}{3}$$

C) $$\frac{10}{7}$$

D) $$\frac{7}{4}$$

E) $$\frac{7}{3}$$

CAMANISHPARMAR, Is the question complete??
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Director
Joined: 11 Feb 2015
Posts: 719
Re: If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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11 Jun 2018, 11:34
1
pushpitkc wrote:
CAMANISHPARMAR wrote:
If x > 0, y > 0, $$\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}$$ and what is the value of $$\frac{{x+y}}{y}$$?

A) $$\frac{3}{4}$$

B) $$\frac{4}{3}$$

C) $$\frac{10}{7}$$

D) $$\frac{7}{4}$$

E) $$\frac{7}{3}$$

CAMANISHPARMAR, Is the question complete??

Thanks pushpitkc - I don't know by mistake I missed typing "= 4", I have done the needful. Thanks for informing
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VP
Joined: 09 Mar 2018
Posts: 1001
Location: India
If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...  [#permalink]

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06 Feb 2019, 21:37
CAMANISHPARMAR wrote:
If x > 0, y > 0, $$\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}$$ = 4 and what is the value of $$\frac{{x+y}}{y}$$?

A) $$\frac{3}{4}$$

B) $$\frac{4}{3}$$

C) $$\frac{10}{7}$$

D) $$\frac{7}{4}$$

E) $$\frac{7}{3}$$

Better to invest time and get this question correct

$$\frac{{7x^2 + 72xy + 4y^2}}{{4x^2 + 12xy + 5y^2}}$$ = 4

$$7x^2 + 72xy + 4y^2$$ = 4 $$4x^2 + 12xy + 5y^2$$

just solve this expression to get the inline
9x^2 - 24xy +16y^2 = 0
3x-4y * 3x-4y = 0

x = 4/3 y

$$\frac{{x+y}}{y}$$ = 7/3

E
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If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...   [#permalink] 06 Feb 2019, 21:37
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# If x > 0, y > 0, {7x^2+72xy+4y^2}/{4x^2+12xy+5y^2}...

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