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

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 $10 \% ?$

,

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

Algebra

fraction

factorization

Pre College Mathematics

AMC-10A, 2015 Problem-15

$1$

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?

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?

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

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

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.

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 $10 \% ?$

,

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

Algebra

fraction

factorization

Pre College Mathematics

AMC-10A, 2015 Problem-15

$1$

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?

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?

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

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

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