Question

Points A, P, Q and B lie on the same line such that P, Q and B are, respectively, 100 km, 200 km and 300 km away from A. Cars 1 and 2 leave A at the same time and move towards B. Simultaneously, car 3 leaves B and moves towards A. Car 3 meets car 1 at Q, and car 2 at P. If each car is moving in uniform speed then the ratio of the speed of car 2 to that of car 1 is

• A. 1:2
• B. 2:9
• C. 1:4
• D. 2:7

Answer 1

The Correct Option is A

We are given that points A, P, Q, and B lie on a line with specific distances from A to P (100 km), Q (200 km), and B (300 km). Cars 1 and 2 travel from A to B, while car 3 travels from B to A. Car 3 meets car 1 at Q and car 2 at P. We need to find the ratio of the speed of car 2 to car 1.

Since car 3 meets car 1 at Q and car 2 at P, we can deduce the following:

• Distance A to Q for car 1 = 200 km
• Distance A to P for car 2 = 100 km

Let's denote the speeds of cars as follows:

• Speed of car 1 = v₁
• Speed of car 2 = v₂
• Speed of car 3 = v₃

Since car 3 meets car 1 at Q:

Time for car 1 to reach Q: \( t_1 = \frac{200}{v_1} \)

Similarly, time for car 3 from B to Q is the same as it meets at the same point.

Distance B to Q = 100 km (300 km - 200 km):

Time for car 3 to reach Q: \( t_3 = \frac{100}{v_3} \)

Setting the times equal:

\( \frac{200}{v_1} = \frac{100}{v_3} \)

Solving for v₃ in terms of v₁:

\( v_3 = 2v_1 \)

Next, consider car 3 meeting car 2 at P:

Time for car 2 to reach P: \( t_2 = \frac{100}{v_2} \)

Time for car 3 to reach P from B (300 km - 100 km = 200 km):

Time: \( t_3' = \frac{200}{v_3} \)

Setting times equal again gives:

\( \frac{100}{v_2} = \frac{200}{v_3} \)

Substituting \( v_3 = 2v_1 \):

\( \frac{100}{v_2} = \frac{200}{2v_1} \)

\( \frac{100}{v_2} = \frac{100}{v_1} \)

Simplifying:

\( v_2 = \frac{v_1}{2} \)

Thus, the ratio of the speed of car 2 to car 1 is:

\( v_2 : v_1 = \frac{v_1}{2} : v_1 = 1 : 2 \)

The correct answer is: 1:2