Question

Two points A and B are 45 km apart. Anil started from A. Sunil started from B. They met each other after 1 hour 30 mins. and after meeting, they continued towards their destinations. Time taken by Anil to reach Point B was 1 hr 15 mins more than the time taken by Sunil to reach point A. Find speed of Anil

• A. 12
• B. 16
• C. 18
• D. 24

Answer 1

The Correct Option is A

Step 1: Define Variables
Let the speed of Anil be \(v_A\) km/h.
Let the speed of Sunil be \(v_S\) km/h. 
They meet after 1 hour 30 minutes (\(1.5\) hours). At this point, the total distance they covered is:
\[45 \,km.\]
Step 2: Relate the Distances Traveled
In the time before they met:
\[\text{Distance covered by Anil} = v_A \times 1.5,\]
\[\text{Distance covered by Sunil} = v_S \times 1.5.\]
Since they meet after covering the total distance:
\[v_A \times 1.5 + v_S \times 1.5 = 45.\]
Simplify:
\[1.5(v_A + v_S) = 45,\]
\[v_A + v_S = 30. \quad \text{(Equation 1)}\]
Step 3: Relation Between Times After Meeting
After meeting, Anil takes 1 hour 15 minutes (\(1.25\) hours) more than Sunil to reach their destinations:
\[\text{Time taken by Anil after meeting} = \text{Time taken by Sunil after meeting} + 1.25.\]
The distances remaining after meeting are:
\[\text{For Anil: Distance remaining} = 45 - 1.5v_A,\]
\[\text{For Sunil: Distance remaining} = 45 - 1.5v_S.\]
Using the formula \(\text{time} = \frac{\text{distance}}{\text{speed}}\), the equation becomes:
\[\frac{45 - 1.5v_A}{v_A} = \frac{45 - 1.5v_S}{v_S} + 1.25. \quad \text{(Equation 2)}\]
Step 4: Solve the Equations
From Equation 1:
\[v_S = 30 - v_A.\]
Substitute \(v_S = 30 - v_A\) into Equation 2:
\[\frac{45 - 1.5v_A}{v_A} = \frac{45 - 1.5(30 - v_A)}{30 - v_A} + 1.25.\]
Simplify \(1.5(30 - v_A)\):
\[1.5(30 - v_A) = 45 - 1.5v_A.\]
Thus, the equation becomes:
\[\frac{45 - 1.5v_A}{v_A} = \frac{1.5v_A}{30 - v_A} + 1.25.\]
Multiply through by \(v_A(30 - v_A)\) to eliminate the denominators:
\[(45 - 1.5v_A)(30 - v_A) = 1.5v_A^2 + 1.25v_A(30 - v_A).\]
Expand both sides:
\[1350 - 45v_A - 1.5v_A \cdot 30 + 1.5v_A^2 = 1.5v_A^2 + 37.5v_A - 1.25v_A^2.\]
Simplify:
\[1350 - 90v_A + 1.5v_A^2 = 0.25v_A^2 + 37.5v_A.\]
Combine terms:
\[1350 + 1.25v_A^2 - 127.5v_A = 0.\]
Divide through by 1.25:
\[v_A^2 - 102v_A + 1080 = 0.\]
Solve the quadratic equation using the quadratic formula:
\[v_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\]
where \(a = 1\), \(b = -102\), and \(c = 1080\).
\[v_A = \frac{-(-102) \pm \sqrt{(-102)^2 - 4(1)(1080)}}{2(1)}.\]
Simplify:
\[v_A = \frac{102 \pm \sqrt{10404 - 4320}}{2},\]
\[v_A = \frac{102 \pm \sqrt{6084}}{2},\]
\[v_A = \frac{102 \pm 78}{2}.\]
Two solutions:
\[v_A = \frac{102 + 78}{2} = 90, \quad v_A = \frac{102 - 78}{2} = 12.\]
Since \(v_A = 90 \, \text{km/h}\) is unrealistic for the scenario, we take:
\[v_A = 12 \, \text{km/h}.\]
Final Answer
The speed of Anil is:
\[\boxed{12 \, \text{km/h}}.\]