Try this beautiful Problem on Algebra based on Set of Fractions from AMC 10 A, 2015. You may use sequential hints to solve the problem.

## Set of Fractions – AMC-10A, 2015- Problem 15

Consider the set of all fractions $\frac{x}{y}$, where $x$ and $y$ are relatively prime positive integers. How many of these fractions have the property that if both numerator and denominator are increased by $1,$ the value of the fraction is increased by $10 \% ?$

,

- $0$
- $1$
- $2$
- $3$
- $infinitely many$

**Key Concepts**

Algebra

fraction

factorization

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2015 Problem-15

#### Check the answer here, but try the problem first

$1$

## Try with Hints

#### First Hint

According to the questation we can write $\frac{x+1}{y+1}=\frac{11 x}{10 y}$

\(\Rightarrow xy +11x-10y=0\)

\(\Rightarrow (x-10)(y-11)=-110\)

Now can you finish the problem?

#### Second Hint

Here \(x\) and \(y\) must positive, so $x>0$ and $y>0$, so $x-10>-10$ and $y+11>11$

Now we have to find out the factors of \(110\) and find out the possible pairs to fulfill the condition….

Now Can you finish the Problem?

#### Third Hint

uses the factors of $110$ , we can get the factor pairs: $(-1,110),(-2,55),$ and $(-5,22)$

But we can’t stop here because $x$ and $y$ must be relatively prime.

$(-1,110 )$ gives $x=9$ and $y=99.9$ and 99 are not relatively prime, so this doesn’t work.

$(-2,55 )$ gives $x=8$ and $y=44$. This doesn’t work.

$(-5,22)$ gives $x=5$ and $y=11$. This does work.

Therefore the one solution exist

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=0qSCPw0YhUY&t=6s