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# A natural number is divided into two positive unequal parts

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A natural number is divided into two positive unequal parts  [#permalink]

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Updated on: 08 Jun 2013, 04:01
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Difficulty:

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Question Stats:

66% (02:58) correct 34% (02:48) wrong based on 213 sessions

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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)

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Originally posted by atalpanditgmat on 08 Jun 2013, 02:40.
Last edited by Bunuel on 08 Jun 2013, 04:01, edited 2 times in total.
Edited the question.
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Posts: 54369
Re: A natural number is divided into two positive unequal parts  [#permalink]

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08 Jun 2013, 04:00
8
2
atalpanditgmat 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. (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)

Say a natural number (integer) is n, then given that n=a+b, where a>b.

The ratio of the original number to the larger (divided) part is equal to the ratio of the larger part to the smaller part --> $$\frac{n}{a}=\frac{a}{b}$$.

Question: $$\frac{a}{b}=?$$

Now, since $$n=a+b$$, then $$\frac{a+b}{a}=\frac{a}{b}$$ --> $$1+\frac{b}{a}=\frac{a}{b}$$ --> $$1+\frac{1}{x}=x$$, where $$\frac{a}{b}=x$$.

Solving: $$x=\frac{a}{b}=\frac{1\pm\sqrt{5}}{2}$$ --> $$\frac{a}{b}=\frac{1+\sqrt{5}}{2}$$ (since a/b must be positive).

Hope it's clear.

P.S. Please format the questions properly.
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Re: A natural number is divided into two positive unequal parts  [#permalink]

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08 Jun 2013, 08:51
1
Bunuel wrote:
atalpanditgmat 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. (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)

Say a natural number (integer) is n, then given that n=a+b, where a>b.

The ratio of the original number to the larger (divided) part is equal to the ratio of the larger part to the smaller part --> $$\frac{n}{a}=\frac{a}{b}$$.

Question: $$\frac{a}{b}=?$$

Now, since $$n=a+b$$, then $$\frac{a+b}{a}=\frac{a}{b}$$ --> $$1+\frac{b}{a}=\frac{a}{b}$$ --> $$1+\frac{1}{x}=x$$, where $$\frac{a}{b}=x$$.

Solving: $$x=\frac{a}{b}=\frac{1\pm\sqrt{5}}{2}$$ --> $$\frac{a}{b}=\frac{1+\sqrt{5}}{2}$$ (since a/b must be positive).

Hope it's clear.

P.S. Please format the questions properly.

Hi,
Can you please explain how the numerical values were assigned?
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Posts: 54369
Re: A natural number is divided into two positive unequal parts  [#permalink]

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09 Jun 2013, 03:06
1
navigator123 wrote:
Bunuel wrote:
atalpanditgmat 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. (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)

Say a natural number (integer) is n, then given that n=a+b, where a>b.

The ratio of the original number to the larger (divided) part is equal to the ratio of the larger part to the smaller part --> $$\frac{n}{a}=\frac{a}{b}$$.

Question: $$\frac{a}{b}=?$$

Now, since $$n=a+b$$, then $$\frac{a+b}{a}=\frac{a}{b}$$ --> $$1+\frac{b}{a}=\frac{a}{b}$$ --> $$1+\frac{1}{x}=x$$, where $$\frac{a}{b}=x$$.

Solving: $$x=\frac{a}{b}=\frac{1\pm\sqrt{5}}{2}$$ --> $$\frac{a}{b}=\frac{1+\sqrt{5}}{2}$$ (since a/b must be positive).

Hope it's clear.

P.S. Please format the questions properly.

Hi,
Can you please explain how the numerical values were assigned?

$$1+\frac{1}{x}=x$$ --> $$\frac{x+1}{x}=x$$ --> $$x^2-x-1=0$$ --> $$x=\frac{1\pm\sqrt{5}}{2}$$.

Hope it helps.
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Re: A natural number is divided into two positive unequal parts  [#permalink]

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11 Jul 2014, 10:52
cant we do this by assuming numbers???????

say n=20 , a=18 , b =2

kindly explain.

thanks
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Re: A natural number is divided into two positive unequal parts  [#permalink]

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11 Jul 2014, 11:11
riskygurpreet wrote:
cant we do this by assuming numbers???????

say n=20 , a=18 , b =2

kindly explain.

thanks

No. The question is about finding numerical value of the golden ratio: two quantities are in the golden ratio if their ratio is the same as the ratio of their sum to the larger of the two quantities. As you can see from the solution above this ratio is irrational number and thus cannot be written as the ratio of two integers.
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Re: A natural number is divided into two positive unequal parts  [#permalink]

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26 Apr 2015, 21:16
yes, poor wording, it seems n=ab rather then n = a+b
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Re: A natural number is divided into two positive unequal parts  [#permalink]

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24 Jul 2017, 11:49
atalpanditgmat 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. (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)

Let N be the natural number...
Let N be divided into 2 parts p (larger part) and m (smaller part)

So, N = m+p........(i)

Also Given N/p = p/m
-> Nm = p^2 ......(ii)

Multiplying m on both sides of (i), we get
Nm = m^2 + pm
Putting value from (ii), we get

P^2 = m^2 + pm
Dividing both sides by m^2 , we get
(p/m)^2 = 1+ p/m

Let p/m = x
SO, x^2 = 1+x
X^2 - x - 1 = 0
$$x = (1+\sqrt{5})/2, (1-\sqrt{5})/2$$
Ratio must be +ve
So, x =$$(1+\sqrt{5})/2$$

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Re: A natural number is divided into two positive unequal parts   [#permalink] 24 Jul 2017, 11:49
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