Question

r is a rational number such that if 2 is added to its denominator its value is \( \frac{1}{2} \) and if 5 is subtracted from its numerator, its value is \( \frac{1}{3} \). Then \( \frac{1+r}{1-r} = \)

• A. \( \frac{35}{12} \)
• B. \( \frac{25}{9} \)
• C. \( \frac{37}{11} \)
• D. \( \frac{32}{9} \)

Hint:

Whenever a rational number problem involves changes in numerator or denominator, assume the number as \( \frac{x}{y} \), convert every condition into an equation, solve for \(x\) and \(y\), and then substitute into the required expression.

Answer 1

The Correct Option is C

Concept: When a rational number is represented as a fraction \( \frac{x}{y} \), the conditions given in the question can be converted into algebraic equations. Solving the simultaneous equations gives the numerator and denominator, from which the required expression can be evaluated.

Step 1: Represent the rational number in fractional form.
Let \[ r=\frac{x}{y} \] where \(x\) is the numerator and \(y\) is the denominator. According to the first condition, if 2 is added to the denominator, the value becomes \( \frac12 \). Therefore, \[ \frac{x}{y+2}=\frac12 \] Cross multiplying, \[ 2x=y+2 \] \[ y=2x-2 \] This is our first equation.

Step 2: Use the second condition to form another equation.
If 5 is subtracted from the numerator, the value becomes \( \frac13 \). Hence, \[ \frac{x-5}{y}=\frac13 \] Cross multiplying, \[ 3(x-5)=y \] \[ y=3x-15 \] This is our second equation.

Step 3: Solve the simultaneous equations.
From the two equations, \[ 2x-2=3x-15 \] Subtracting \(2x\) from both sides, \[ -2=x-15 \] \[ x=13 \] Substituting \(x=13\) into \[ y=2x-2 \] we get \[ y=2(13)-2 \] \[ y=24 \] Therefore, \[ r=\frac{13}{24} \]

Step 4: Evaluate the required expression.
We need to find \[ \frac{1+r}{1-r} \] Substituting \(r=\frac{13}{24}\), \[ \frac{1+\frac{13}{24}}{1-\frac{13}{24}} \] Taking LCM in numerator and denominator, \[ =\frac{\frac{24+13}{24}}{\frac{24-13}{24}} \] \[ =\frac{\frac{37}{24}}{\frac{11}{24}} \] \[ =\frac{37}{24}\times\frac{24}{11} \] \[ =\frac{37}{11} \] Hence, \[ \boxed{\frac{37}{11}} \]