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Ratios

by marcusking » Thu Feb 12, 2009 11:46 am
A natural number is divided into two positive unequal parts such that the ratio of the original number to the larger (divided) part is equal to the ratio of the larger part to the smaller part. What is the value of this ratio?

A (5^1/2 − 1)

B (5^1/2 + 1) / 2

C (5^1/2 + 1) / 4

D (5^1/2 + 1) / (5^1/2 − 1)

E (5^1/2 + 3) / (5^1/2 − 1)

OA is B.
Last edited by marcusking on Fri Feb 13, 2009 5:29 am, edited 3 times in total.
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Re: Ratios

by Stuart@KaplanGMAT » Thu Feb 12, 2009 11:51 am
marcusking wrote:A natural number is divided into two positive unequal parts such that the ratio of the original number to the larger (divided) part is equal to the ratio of the larger part to the smaller part. What is the value of this ratio?

A (51/2 − 1)

B (51/2 + 1) / 2

C (51/2 + 1) / 4

D (51/2 + 1) / (51/2 − 1)

E (51/2 + 3) / (51/2 − 1)
I don't understand the answer choices.

On the GMAT, numbers get reduced. So "51/2 + 1" should just be "53/2" and "51/2 - 1" should just be "49/2". "(51/2 + 1)/(51/2 - 1)" should be "(53/2)/(49/2)" = "53/49".

Are the answer choices reproduced correctly? Where did you get this question? In future, please cite all your sources.
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by marcusking » Thu Feb 12, 2009 12:07 pm
Sorry Stuart, I forgot the ^ after the 5's. I got this question off of the www.Syvum.com/gmat quat questions
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Ans is B ...

by ksharada » Thu Feb 12, 2009 6:35 pm
Is the answer B ?

ie (root(5) +1) /2
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by Uri » Fri Feb 13, 2009 9:16 am
let the number be "N" and the larger part of the number be x
then, then question says,
N/x = (x/N-x)
i.e. N^2 -Nx -x^2 = 0
we have to find out N/x. I could not decipher how that can be done. Can someone show the steps please?
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Solution ...

by ksharada » Fri Feb 13, 2009 9:28 am
Let the number be 2 for eg. Take the larger part of the number as p, so the smaller number will be 2-p

Taking ratios : Original/Larger = Larger/Smaller

2/p = p/2-p This will be p^2 + 2 * p - 4 = 0

solving it p = -1 +(or -) root(5)

Since the numbers are positive, p = -1 + root(5)

Hence, ratio is 2/(-1+root(5)) .. soving this using conjugates will give you ( root(5) + 1 )/ 2 ie option B

Taking any other number for eg 4 will give the same ratio.
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by BuckeyeT » Fri Feb 13, 2009 12:02 pm
For those that have taken the GMAT before or teach about it,
2/p = p/2-p This will be p^2 + 2 * p - 4 = 0

solving it p = -1 +(or -) root(5)
(1) Are you expected to use the quadratic equation to find roots?
Hence, ratio is 2/(-1+root(5)) .. soving this using conjugates will give you ( root(5) + 1 )/ 2
(2) Would the GMAT expect us to rationalize an answer in this way? From other practice, I would guess it's a typical expectation.

I love the challenge of many of these problems, but I want to make sure I'm focusing on applicable practice. [/quote]
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by cjb » Fri Feb 13, 2009 12:21 pm
BuckeyeT wrote: (1) Are you expected to use the quadratic equation to find roots?
No idea, but I used it. It may not be on the tested syllabus, but they're hardly going to check your working!

Still, I'd like to see a simpler way of solving this one.
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by sureshbala » Fri Feb 13, 2009 12:55 pm
Folks, this can be solved using a quadratic equation but as pointed by some of the users if you want to eliminate the quadratic here, you can do it by go through the options...In fact this can be done fast...

Let N be the number, L the larger part and S the smaller part.

Straight away options C, D and E are eliminated.

Option C: Givne N/L = (5^1/2 + 1)/4.

This means L>N which is a contradiction

Option D: Given N/L = (5^1/2 +1)/(5^1/2-1)

Since L = 5^1/2 -1, S = 2.

Thus S>L which is again a contradiction.

So D eliminated...Similarly E is also eliminated.

Left with A and B.....

Let us check option A.

Givne N/L = (5^1/2 -1)/1

This means L/S = 1/(5^1/2 -2)

We can observe that N/L is not equal to L/S

So B is the answer...

Folks, since I want to eliminate all the options I have done this.

One can start with A and then B and finalize the answer as B.

................................................................................

I am sure if you can solve a quadratic you can answer this fast....

Instead of taking the number as 2, why don't u consider it as 1.

So 1/x = x/1-x (where x is the larger part)

Hence x^2+x-1 = 0

Solving this quadratic we get x = (5^1/2 -1)/2 (taking only the +ve root)

So 1/x = 2/(5^1/2-1) = (5^1/2 + 1)/2
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