P and Q run a race of 2 km. First, P gives Q a head starts of 300 m and beats him by 20 seconds. While coming back, P gives Q a lead of 1 minute and gets beaten by 500 m. Find the time taken by Q to cover a distance of 1450 m.

Let's break down the problem step by step:

### 1. Key Information:
- Total race distance: 2 km or 2000 m.
- In the first race:
  - \( P \) gives \( Q \) a head start of 300 m.
  - Despite this, \( P \) still wins by 20 seconds.
  
- In the second race:
  - \( P \) gives \( Q \) a time advantage of 1 minute (60 seconds).
  - \( Q \) beats \( P \) by 500 m.

We need to determine the time \( Q \) takes to cover 1450 m.

### 2. Define Variables:
Let:
- \( v_P \) = speed of \( P \) (in m/s),
- \( v_Q \) = speed of \( Q \) (in m/s).

#### First Race (P wins by 20 seconds):
- \( Q \)'s head start is 300 m, so \( Q \) runs 1700 m while \( P \) runs 2000 m.
- Let the time taken by \( P \) to run 2000 m be \( t_P \), and by \( Q \) to run 1700 m be \( t_Q \).

From the problem, we know:
\[
t_Q = t_P + 20 \text{ seconds}
\]
Using the basic formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \), we have:
\[
t_P = \frac{2000}{v_P}, \quad t_Q = \frac{1700}{v_Q}
\]
Thus,
\[
\frac{1700}{v_Q} = \frac{2000}{v_P} + 20
\]

#### Second Race (Q wins by 500 meters):
- \( P \) gives \( Q \) a lead of 1 minute (60 seconds).
- By the time \( P \) covers 1500 m, \( Q \) has covered the full 2000 m.

Using similar reasoning as above, we have:
\[
\frac{1500}{v_P} = \frac{2000}{v_Q} - 60
\]

### 3. Solve the System of Equations:
We now have two equations:
1. \( \frac{1700}{v_Q} = \frac{2000}{v_P} + 20 \)
2. \( \frac{1500}{v_P} = \frac{2000}{v_Q} - 60 \)

Solving this system will give us \( v_P \) and \( v_Q \).

Let's solve these equations to find the speeds first.The solution to the system of equations gives the following speeds:
- \( v_P = \frac{725}{31} \approx 23.39 \, \text{m/s} \)
- \( v_Q = \frac{145}{9} \approx 16.11 \, \text{m/s} \)

### 4. Calculate the Time for Q to Cover 1450 m:
Now, we can calculate the time taken by \( Q \) to cover 1450 m using the formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \):
\[
t = \frac{1450}{v_Q} = \frac{1450}{\frac{145}{9}} = \frac{1450 \times 9}{145} = 90 \, \text{seconds}
\]

Thus, the time taken by \( Q \) to cover 1450 m is 90 seconds.

Would you like further details or have any other questions?

Here are 5 related questions:
1. What if the distance was reduced to 1 km? How would it affect the race outcomes?
2. How much faster would \( Q \) need to run to tie the first race?
3. How long would it take for \( P \) to cover 1450 m at his current speed?
4. What would the outcome be if \( P \) gave a 2-minute head start instead of 1 minute?
5. How can we calculate the distance \( P \) covers in 20 seconds?

**Tip:** Speed is distance divided by time; maintaining consistent units ensures accurate calculations.

P and Q run a race of 2 km. First, P gives Q a head start of 300 m and beats him by 20 seconds. While coming back, P gives Q a lead of 1 minute and gets beaten by 500 m. Find the time taken by Q to cover a distance of 1450 m.

Solve a race problem where P and Q compete in a 2 km race with head starts and time advantages. Learn how to calculate the time taken by Q to cover 1450 meters using algebraic equations and speed-distance relationships. This problem is suitable for students in Grades 10-12.

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