Difference between revisions of "2015 AMC 10A Problems/Problem 15"

Latest revision as of 13:41, 14 September 2024

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 $10\%$ ?

$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad\textbf{(E) }\text{infinitely many}$

Solution 1

You can create the equation $\frac{x+1}{y+1}=\frac{11x}{10y}$

Cross multiplying and combining like terms gives $xy + 11x - 10y = 0$ .

This can be factored into $(x - 10)(y + 11) = -110$ .

$x$ and $y$ must be positive, so $x > 0$ and $y > 0$ , so $x - 10> -10$ and $y + 11 > 11$ .

Using 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.

We found one valid solution so the answer is $\boxed{\textbf{(B) }1}$ .

Solution 2

The condition required is $\frac{x+1}{y+1}=\frac{11}{10}\cdot\frac{x}{y}$ .

Observe that $x+1 > \frac{11}{10}\cdot x$ so $x$ is at most $9.$

By multiplying by $\frac{y+1}{x}$ and simplifying we can rewrite the condition as $y=\frac{11x}{10-x}$ . Since $x$ and $y$ are integer, this only has solutions for $x\in\{5,8,9\}$ . However, only the first yields a $y$ that is relative prime to $x$ .

There is only one valid solution so the answer is $\boxed{\textbf{(B) }1}$

Video Solution

https://youtu.be/p7g0hTxE9I8

~savannahsolver

@@ Line 11: / Line 11: @@
 Cross multiplying and combining like terms gives <math>xy + 11x - 10y = 0</math>.
-This can be factored into <math>(x - 10)(y + 11) = 110</math>.
+This can be factored into <math>(x - 10)(y + 11) = -110</math>.
 <math>x</math> and <math>y</math> must be positive, so <math>x > 0</math> and <math>y > 0</math>, so <math>x - 10> -10</math> and <math>y + 11 > 11</math>.

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

Page

Toolbox

Search