Q3 Both the numerator and the denominator of 3/4 are increased by a positive integer x, and those of 15/17 are decreased by the same integer. This operation results in the same value for both fractions.
What is the value of x?
x = 3 is the correct value
Both modified fractions equal 6/7 when x = 3, satisfying the condition that 3+x/4+x = 15−x/17−x.
Problem Setup
The equation arises from equating the modified fractions: 3+x/4+x = 15−x/17−x. Cross-multiplying yields (3+x)(17−x)=(4+x)(15−x), which simplifies to 51+14x−x²=60+11x−x². The −x² terms cancel, leaving 14x+51=11x+60, so 3x=9 and x=3.
Option Analysis
Testing each option confirms only x = 3 works, as verified computationally.
| Option | Modified 3/4 | Modified 15/17 | Equal? |
|---|---|---|---|
| (A) 1 | 4/5 = 0.8 | 14/16 = 0.875 | No |
| (B) 2 | 5/6 ≈ 0.833 | 13/15 ≈ 0.867 | No |
| (C) 3 | 6/7 ≈ 0.857 | 12/14 = 6/7 | Yes |
| (D) 4 | 7/8 = 0.875 | 11/13 ≈ 0.846 | No |
Increasing the numerator and denominator of 3/4 by positive integer x while decreasing those of 15/17 by x results in equal fractions. This fraction modification problem tests algebraic manipulation for competitive exams like CSIR NET.
Key Equation Derivation
Start with 3+x
4+x = 15−x
17−x. Cross-multiply: (3+x)(17−x)=(4+x)(15−x). Expand to 51+14x−x²=60+11x−x², simplify by canceling −x², subtract 11x and 51: 3x=9, so x = 3. Verification: Both become 6/7.
Why Options Fail Except x=3
Options 1, 2, and 4 yield unequal decimals, as shown in the table above. Only x=3 ensures denominators remain positive (17-3=14>0) and matches exactly.
