Sample Question 15:

Consider the set of all fractions \(\tfrac{x}{y},\) 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 \(10\%\)?

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

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Answer Keys

Question 15: B

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Solutions

Question 15
Step 1: Create the equation \(\frac{x+1}{y+1}=\frac{11x}{10y}\).

Step 2: Cross multiply and combine like terms to achieve the new equation \(xy + 11x - 10y = 0\).

Step 3: Factor this equation into \((x - 10)(y + 11) = -110\).

Step 4: Since \(x\) and \(y\) must be positive, it means \(x - 10 > -10\) and \(y + 11 > 11\).

Step 5: Write down the factor pairs of 110: \((-1, 110)\), \((-2, 55)\), and \((-5, 22)\).

Step 6: We now need to test each possible pair since \(x\) and \(y\) must be relatively prime.

Step 7: \((-1, 110)\) gives \(x = 9\) and \(y = 99\). \(9\) and \(99\) are not relatively prime, so this pair doesn't work.

Step 8: \((-2, 55)\) gives \(x = 8\) and \(y = 44\). This pair doesn't work either.

Step 9: \((-5, 22)\) gives \(x = 5\) and \(y = 11\). This is the only pair that works, so we have found one valid solution.

Step 10: Therefore, the answer is \(\textbf{(B) }1\).