Match List-I with List-II.
| List-I | List-II | ||
| (A) | The largest of three consecutive integers whose sum is 72 | (I) | 12 |
| (B) | The number whose three-fifths exceeds its one-third by 16 | (II) | 16 |
| (C) | The length (in cm) of a rectangle of perimeter 56 cm whose length is 4 cm more than its breadth | (III) | 25 |
| (D) | The numerator of a fraction whose denominator exceeds the numerator by 5 and which becomes \(\displaystyle\frac{3}{4}\) when 3 is added to both numerator and denominator | (IV) | 60 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Each problem becomes a linear equation in one unknown.
(A) The largest of three consecutive integers whose sum is 72
This matches (III) 25. Let the integers be \(n - 1\), \(n\), \(n + 1\).
\(3n = 72\), so \(n = 24\) and the largest is 25
(B) The number whose three-fifths exceeds its one-third by 16
This matches (IV) 60. Let the number be \(x\).
\(\displaystyle \frac{3x}{5} - \frac{x}{3} = 16\)
\(\displaystyle \frac{4x}{15} = 16\), so \(x = 60\)
(C) The length (in cm) of a rectangle of perimeter 56 cm whose length is 4 cm more than its breadth
This matches (II) 16. Let the breadth be \(b\) cm.
\(2(b + 4 + b) = 56\)
\(2b + 4 = 28\), so \(b = 12\) and the length is 16 cm
(D) The numerator of a fraction whose denominator exceeds the numerator by 5 and which becomes \(\displaystyle\frac{3}{4}\) when 3 is added to both numerator and denominator
This matches (I) 12. Let the numerator be \(n\), so the denominator is \(n + 5\).
\(\displaystyle \frac{n + 3}{n + 8} = \frac{3}{4}\)
\(4n + 12 = 3n + 24\), so \(n = 12\)
The fraction is \(\displaystyle\frac{12}{17}\), and \(\displaystyle\frac{15}{20} = \frac{3}{4}\).