All Exams Test series for 1 year @ ₹349 only
Question

In a 500 m race, P and Q have speeds in the ratio of 3 ∶ 4. Q starts the race when P has already covered 140 m.

What is the distance between P and Q (in m) when P wins the race?

The correct answer is

20

Race Problem Analysis: P and Q's Performance

This problem involves understanding relative speeds and distances in a race scenario. We are given the total race distance, the speed ratio of two runners, P and Q, and a head start condition. The goal is to find the distance between P and Q when P wins the race.

Understanding the Race Setup

  • The total length of the race is 500 m.
  • The speeds of P and Q are in the ratio of 3 ∶ 4. This means if P's speed is \(3v\), then Q's speed is \(4v\) for some unit speed \(v\).
  • Q starts the race when P has already covered 140 m. This is a head start for P.
  • When P wins the race, P will have covered the full 500 m distance.

Calculating P's Winning Time

When Q starts, P has already covered 140 m. This means P is 140 m ahead of the starting line, and Q is at the starting line (0 m). For P to win the race, P must cover the remaining distance to reach the 500 m mark.

  • P's initial distance covered = 140 m.
  • Total race distance = 500 m.
  • Distance P needs to cover from the point Q starts = \(500 \text{ m} - 140 \text{ m} = 360 \text{ m}\).

Let's assume P's speed is \(3x\) units/time and Q's speed is \(4x\) units/time. To find the time taken for P to win the race from the 140 m mark, we use the formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).

Time taken by P to cover 360 m:

$$ \text{Time}_{\text{P wins}} = \frac{360 \text{ m}}{3x \text{ m/unit time}} = \frac{120}{x} \text{ unit time} $$

Determining Q's Progress

Q starts running at the same moment P is at the 140 m mark. Both P and Q run for the same duration until P crosses the finish line. We need to calculate how much distance Q covers in the time P takes to win the race.

  • Q's speed = \(4x\) units/time.
  • Time Q runs = \( \frac{120}{x} \) unit time (same as P's winning time).

Distance covered by Q in this time:

$$ \text{Distance}_{\text{Q covers}} = \text{Speed}_{\text{Q}} \times \text{Time}_{\text{P wins}} $$

$$ \text{Distance}_{\text{Q covers}} = 4x \text{ m/unit time} \times \frac{120}{x} \text{ unit time} $$

$$ \text{Distance}_{\text{Q covers}} = 4 \times 120 = 480 \text{ m} $$

When P wins the race (P is at 500 m), Q has covered 480 m from the starting line.

Final Distance Calculation

Now we need to find the distance between P and Q at the exact moment P wins the race.

  • P's position when P wins = 500 m (the finish line).
  • Q's position when P wins = 480 m (distance covered by Q from the start).

The distance between P and Q is the difference between their positions:

$$ \text{Distance between P and Q} = \text{P's position} - \text{Q's position} $$

$$ \text{Distance between P and Q} = 500 \text{ m} - 480 \text{ m} = 20 \text{ m} $$

Aspect P Q
Starting Position (when Q starts) 140 m 0 m
Speed Ratio 3 (e.g., \(3x\)) 4 (e.g., \(4x\))
Distance to Finish (from Q's start) 360 m 500 m
Time P takes to win (from 140m mark) \( \frac{360}{3x} = \frac{120}{x} \) unit time
Distance covered in this time 360 m (reaches 500m) \(4x \times \frac{120}{x} = 480\) m
Final Position when P wins 500 m 480 m
Distance between P and Q \(500 \text{ m} - 480 \text{ m} = 20 \text{ m}\)

Therefore, the distance between P and Q when P wins the race is 20 m.

Was this answer helpful?

Important Questions from Ratio and Proportion

  1. The cost of a diamond is directly proportional to the square of its weight. The cost of a 14 gm diamond is Rs. 2560. This diamond got broken down into two pieces in the ratio of 5 ∶ 9. How much loss percent is incurred due to this breakage ? (Correct to two decimal places)

  2. Atul purchased Bread costing Rs.20 and gave a 100 rupee note to the shopkeeper. The shopkeeper gave the balance money in coins of denomination Rs.2, Rs.5 and Rs.10. If these coins are in the ratio 5 ∶ 4 ∶ 1, then how many Rs.5 coins did the shopkeeper give?

  3. A person divides a certain amount among his three sons in the ratio of 3 ∶ 4 ∶ 5. If he had divided this amount in the ratio of 1/3,1/4,1/5, his son, who had got the lowest share earlier, would get Rs.1,188 more. Find the amount (in Rs).

  4. In a school 3/8 of the number of students are girls and the rest are boys. One-third of the number of boys are below 10 years and 2/3 the number if girls are also below 10 years. If the number of students of age 10 or more years is 260. then the number of boys in the school is:

  5. If a : b : c = \(\frac{1}{4} : \frac{1}{3} : \frac{1}{2}, \)  then  \( \ \frac{a}{b} : \frac{b}{c} : \frac{c}{a} = ?\)

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App