A natural number is divided into two positive unequal parts

Question

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)

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)

Bunuel · Accepted Answer

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).

Answer: B.

Hope it's clear.

P.S. Please format the questions properly.

navigator123 · Answer

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

Bunuel · Answer

1+\frac{1}{x}=x --> \frac{x+1}{x}=x --> x^2-x-1=0 --> x=\frac{1\pm\sqrt{5}}{2} . Solving and Factoring Quadratics: https://www.purplemath.com/modules/solvquad.htm https://www.purplemath.com/modules/factquad.htm Hope it helps.

riskygurpreet · Answer

cant we do this by assuming numbers???????

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

kindly explain.

thanks

Bunuel · Answer

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.

vipulgoel · Answer

yes, poor wording, it seems n=ab rather then n = a+b

shashankism · Answer

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

Answer B

Regor60 · Answer

Set the parts equal to:

KX: original
X: larger
X/K: smaller

Where K is the ratio being sought.

This will satisfy the ratio requirements because:

KX/X = X/(X/K) = K

Since the original is equal to the larger plus the smaller:

KX = X + X/K

Multiplying by K:

K^2X -KX - X = 0

Solving for K using quadratic formula:

K =  /2X =

X/2X + X√5/2X =

1/2 + √5/2

Posted from my mobile device

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=85250 --- Harsha GMAT-Club-Forum-mx3c346t.png

A natural number is divided into two positive unequal parts

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A natural number is divided into two positive unequal parts

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