Arithmetic

Time, Speed & Distance

Medium

Point $P$ lies between points $A$ and $B$ such that the length of $BP$ is thrice that of $AP$ . Car $1$ starts from $A$ and moves towards $B$ . Simultaneously, car $2$ starts from $B$ and moves towards $A$ . Car $2$ reaches $P$ one hour after car $1$ reaches $P$ . If the speed of car $2$ is half that of car $1$ , then the time, in minutes, taken by car $1$ in reaching $P$ from $A$ is

Entered answer:

✅ Correct Answer: 12

Let's first visualize what's happening:

We have three points on a line: A, P, and B (in that order)

Point P divides the distance AB such that BP = 3 × AP

Car 1 travels from A → P, Car 2 travels from B → P

Both cars start simultaneously

Car 2 reaches P exactly 1 hour later than Car 1

Speed of Car 2 = ½ × Speed of Car 1

This is a classic relative motion problem where we need to use the relationship between distance, speed, and time.

Let's define:

Time taken by Car 1 to reach P from A = x hours

Distance AP = d (we'll see this cancels out)

Distance BP = 3d (given that BP is thrice AP)

Using the formula: Speed = Distance ÷ Time

Speed of Car 1 = AP ÷ x = $\frac{d}{x}$

Car 2 travels distance BP = 3d

Car 2 takes time = (x + 1) hours (since it reaches 1 hour after Car 1)

Speed of Car 2 = BP ÷ (x + 1) = $\frac{3d}{x + 1}$

We're told that Speed of Car 2 = ½ × Speed of Car 1

Setting up the equation:

$\frac{3d}{x+1} = \frac{1}{2} \times \frac{d}{x}$

The left side is Car 2's speed, and the right side is half of Car 1's speed, exactly as stated in the problem.

Let's simplify by canceling d from both sides:

$\frac{3}{x+1} = \frac{1}{2x}$

We can cancel d because it appears on both sides and is non-zero.

Cross-multiplying:

$3 \times 2x = 1 \times (x+1)$

$6x = x + 1$

$6x - x = 1$

$5x = 1$

$x = \frac{1}{5}$

We found that Car 1 takes 1/5 hour to reach P from A.

Converting to minutes:

$\frac{1}{5} \text{ hour} = \frac{1}{5} \times 60 \text{ minutes} = 12 \text{ minutes}$

Therefore, Car 1 takes 12 minutes to reach P from A.

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