Five years ago, Adil was thrice as old as Bharat. Ten years later Adil shall be twice as old as Bharat. To know the present ages of Adil and Bharat : (i) form the linear equations representing the above information. (ii) show that the system of equations is consistent with unique solution. (iii) find the present ages of Adil and Bharat.
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding rag
Model Answer
(i) Forming the equations:
Let present age of Adil = x years, present age of Bharat = y years.
Five years ago: Adil's age = (x – 5), Bharat's age = (y – 5)
$$x - 5 = 3(y - 5) \implies x - 3y = -10 \quad \text{...(1)}$$
Ten years later: Adil's age = (x + 10), Bharat's age = (y + 10)
$$x + 10 = 2(y + 10) \implies x - 2y = 10 \quad \text{...(2)}$$
(ii) Checking consistency:
From equations (1) and (2): $a_1 = 1,\ b_1 = -3,\ c_1 = -10$ and $a_2 = 1,\ b_2 = -2,\ c_2 = 10$
$$\frac{a_1}{a_2} = \frac{1}{1} = 1, \quad \frac{b_1}{b_2} = \frac{-3}{-2} = \frac{3}{2}$$
Since $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the system is consistent with a unique solution.
(iii) Finding the ages:
Subtracting (1) from (2):
$$(x - 2y) - (x - 3y) = 10 - (-10)$$
$$y = 20 \text{ years}$$
Substituting in (2): $x - 2(20) = 10 \implies x = 50$ years.
∴ Present age of Adil = 50 years; Present age of Bharat = 20 years.
Source: Chapter 3, Exercise 3.3 (Q2-ii) and Section 3.2
---
Explanation
- Part (i): Carefully translate "five years ago" and "ten years later" into algebraic expressions before equating. A common error is using wrong signs.
- Part (ii): The consistency condition from the textbook is: if $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the pair is consistent with a unique solution (intersecting lines). Always state this rule explicitly for marks.
- Part (iii): Elimination by subtraction is the cleanest method here. Always verify: 5 years ago Adil = 45, Bharat = 15 (45 = 3 × 15 ✓); 10 years later Adil = 60, Bharat = 30 (60 = 2 × 30 ✓).