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# stumped... retired test question is killing me, please help

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Intern
Joined: 23 Sep 2007
Posts: 14

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23 Sep 2007, 00:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

this question is killing me...

if (2/x) + (3/y) = 4 and xy = 5, then 3x + 2y = ?

a. 1/5
b. 1/4
c. 4/5
d. 4
e. 20

-------

Correct answer is supposedly E. I can't for the life of me figure out how to solve this in less than two minutes.

Every time I work it, I end up with a nasty quadratic that takes forever to factor, or solve with the quad formula.

I'm sure it is in my approach... I'm using substitution

if xy = 5, then x = 5/y, then sub into the first equation...

(2/(5/y)) + (3/y) = 4
(2y/5) + (3/y) = 4
2y + (15/y) = 20
2y^2 + 15 = 20y
2y^2 -20y + 15 = 0

This is where it gets really ugly... Quad formula, nope you have to solve something like the sqrt(280), that takes too long to solve in under two minutes.

Factor quickly? What factors of -30 add up to 15?

I must have done something wrong up to this point. I'm just not sure what it is. Thanks!

Any help, step by step, would be VERY much appreciated! Thanks in advance.
Manager
Joined: 09 Jul 2007
Posts: 178

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23 Sep 2007, 00:48
its very simple prob..

2/x + 3/y= 4
2y+3x/xy=4 (taking lcm)

xy= 5--given

2y+3x/5=4

2y+3x=4*5

2y+3x= 20 i.e. 3x+2y=20
hence it is E
CEO
Joined: 29 Mar 2007
Posts: 2559
Re: stumped... retired test question is killing me, please h [#permalink]

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23 Sep 2007, 00:50
kz wrote:
this question is killing me...

if (2/x) + (3/y) = 4 and xy = 5, then 3x + 2y = ?

a. 1/5
b. 1/4
c. 4/5
d. 4
e. 20

-------

Correct answer is supposedly E. I can't for the life of me figure out how to solve this in less than two minutes.

Every time I work it, I end up with a nasty quadratic that takes forever to factor, or solve with the quad formula.

I'm sure it is in my approach... I'm using substitution

if xy = 5, then x = 5/y, then sub into the first equation...

(2/(5/y)) + (3/y) = 4
(2y/5) + (3/y) = 4
2y + (15/y) = 20
2y^2 + 15 = 20y
2y^2 -20y + 15 = 0

This is where it gets really ugly... Quad formula, nope you have to solve something like the sqrt(280), that takes too long to solve in under two minutes.

Factor quickly? What factors of -30 add up to 15?

I must have done something wrong up to this point. I'm just not sure what it is. Thanks!

Any help, step by step, would be VERY much appreciated! Thanks in advance.

Very common GMAT trick question here. Your used to solving the question by using the ol Two equations two unknowns. This doesn't always work to your advantage on the GMAT.

The question is asking for 3x+2y. So if you can find what 3x+2y = then you know the answer. No need to solve for both X and Y.

here: 2/x+3/y=4 and xy=5. 2/x+3/y=4 (make denom same)

(2y+3x)/xy=4 ---> 2y+3x=4xy ----> xy=5 so---> 2y+3x =4(5)

2y+3x= 20. just change 2y and 3x around and you have:

3x+2y=20.

Ans E.

Remember you don't always have to solve for both X and Y, just 3x + 2y. You must know this concept to do well on the quant section.
Intern
Joined: 23 Sep 2007
Posts: 14
Re: stumped... retired test question is killing me, please h [#permalink]

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23 Sep 2007, 00:54
Thank you both!

I won't even tell you how many hours I spent trying to solve this the "hard way". I figured I was missing something VERY simple. Sure enough.

Thanks again...
Intern
Joined: 23 Sep 2007
Posts: 14

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23 Sep 2007, 01:04
12345678 wrote:
its very simple prob..

2/x + 3/y= 4
2y+3x/xy=4 (taking lcm)

I don't know why... But I stared at that thing forever and couldn't figure out how to get the two fractions to have a common denominator.

Must be the hours quant/verbal studying or something, but I just blanked on the fact that you can multiply fractions independently if you are multiplying by "1"...

(2/x)(y/y) + (3/y)(x/x) = 4, hence
(2y + 3x) / xy = 4
then, as you state it is VERY simple!

Thanks again!
23 Sep 2007, 01:04
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# stumped... retired test question is killing me, please help

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