Question:
<p>A and B are friends. A is elder to B by 5 years. B’s sister C is half the age of B while A’s father D is 8 years older than twice the age of B. If the present age of D is 48 years, then find the present ages of A, B and C respectively.</p>
<p style="display:inline">40 years, 20 years, 15 years</p>
<p style="display:inline">20 years, 15 years, 10 years</p>
<p style="display:inline">25 years, 20 years, 10 years</p>
<p style="display:inline">50 years, 25 years, 20 years</p>
Step-by-Step Solution
Key Concept: Express all ages as linear functions of a single variable to solve the system through substitution once one numerical value is identified.
<p>Let the present ages of A, B, C and D are x, y, z and t respectively.<br />
Since, present age of D = t = 48 years.<br />
According to question,<br />
x = y + 5<br />
<span class="math-tex">$z=\frac{1}{2} y$</span><br />
f = 2y + 8<br />
From (iii), 48 = 2y + 8<br />
<span class="math-tex">$\Rightarrow$</span> From (iii), 48 = 2y + 8<br />
From (ii), z = <span class="math-tex">$\frac{1}{2}$</span> x 20 = 10 years<br />
From (i), x = 20 + 5 = 25 years<br />
So, present ages of A, B and C are 25 years, 20 years and 10 years respectively.</p>
Correct Answer: C