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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 Distance Calculation: P vs Q Speed Ratio

This solution details the calculation for the distance between runners P and Q when P finishes a 500 m race, considering their speed ratio and a staggered start for Q.

Step 1: Define Speeds Based on Ratio

Let the speeds of runner P and runner Q be $3x$ m/s and $4x$ m/s, respectively, based on the given ratio of 3:4.

Step 2: Calculate Time for P to Cover Initial Distance

P has already covered 140 m when Q starts. The time taken by P to cover this initial distance is:

$t_{P_{140}} = \frac{\text{Distance}}{\text{Speed}_P} = \frac{140 \text{ m}}{3x \text{ m/s}} = \frac{140}{3x} \text{ seconds}$

Step 3: Determine Time P Takes to Finish the Race

The total race distance is 500 m. The total time P takes to complete the race from the start is:

$T_P = \frac{\text{Total Distance}}{\text{Speed}_P} = \frac{500 \text{ m}}{3x \text{ m/s}} = \frac{500}{3x} \text{ seconds}$

Step 4: Calculate Duration Q Runs Until P Finishes

Q starts at time $t_{P_{140}}$. P finishes at time $T_P$. The duration Q runs is the difference between these times:

$ \text{Duration}_Q = T_P - t_{P_{140}} = \frac{500}{3x} - \frac{140}{3x} = \frac{360}{3x} = \frac{120}{x} \text{ seconds} $

Step 5: Calculate Distance Covered by Q

In the duration calculated above, Q covers a distance using their speed:

$ \text{Distance}_Q = \text{Speed}_Q \times \text{Duration}_Q = (4x \text{ m/s}) \times \left(\frac{120}{x} \text{ seconds}\right) = 480 \text{ m} $

Step 6: Find the Distance Between P and Q at P's Finish

When P finishes the race, P is at the 500 m mark. Q has covered 480 m. The distance between them is:

$ \text{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.

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Important Questions from Speed Distance and Time

  1. A car is moving on a horizontal surface in a straight line with a constant velocity of 3 m/s. A ball is thrown vertically upwards at time $t = 0$ from the top of the moving car with a velocity of 20 m/s. The acceleration due to gravity is 10 m/s$^2$.
    At what value(s) of time $t$ in second(s), the ball is at a height of 15 m from the top of the moving car?
  2. Velocity of an object fired directly in upward direction is given by $V = 80 – 32 \ t$, where $t$ (time) is in seconds. When will the velocity be between 32 m/sec and 64 m/sec?
  3. An automobile travels from city A to city B and returns to city A by the same route. The speed of the vehicle during the onward and return journeys were constant at 60 km/h and 90 km/h, respectively. What is the average speed in km/h for the entire journey?
  4. Two cars start at the same time from the same location and go in the same direction. The speed of the first car is 50 km/h and the speed of the second car is 60 km/h. The number of hours it takes for the distance between the two cars to be 20 km is ___________.

  5. Trucks (10 m long) and cars (5 m long) go on a single lane bridge. There must be a gap of at least 20 m after each truck and a gap of at least 15 m after each car. Trucks and cars travel at a speed of 36 km/h. If cars and trucks go alternately, what is the maximum number of vehicles that can use the bridge in one hour?
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