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:
$$\boxed{12 \, \text{km/h}}.$$