A car travels from P to Q at a constant speed. If its speed were increased by 10 km/h, it would have taken one hour less to cover the distance. It would have taken further 45 min lesser if the speed were further increased by 10 km/h. What is the distance between two cities?

Answer

To solve the problem step by step, let's denote the following:

- Let the original speed of the car be \( S \) km/h.
- Let the distance between cities P and Q be \( D \) km.

### Step 1: Set up the equations based on the information given.

1. When the speed is increased by 10 km/h, the time taken is 1 hour less.
- The time taken at speed \( S \) is \( \frac{D}{S} \).
- The time taken at speed \( S + 10 \) is \( \frac{D}{S + 10} \).
- According to the problem, we have:
\[ \frac{D}{S} - \frac{D}{S + 10} = 1 \quad \text{(1)} \]

2. When the speed is increased by another 10 km/h (total increase of 20 km/h), the time taken is 45 minutes (or 0.75 hours) less.
- The time taken at speed \( S + 20 \) is \( \frac{D}{S + 20} \).
- According to the problem, we have:
\[ \frac{D}{S + 10} - \frac{D}{S + 20} = 0.75 \quad \text{(2)} \]

### Step 2: Solve Equation (1)

From Equation (1):
\[ \frac{D}{S} - \frac{D}{S + 10} = 1 \]
Multiplying through by \( S(S + 10) \) to eliminate the denominators:
\[ D(S + 10) - DS = S(S + 10) \]
This simplifies to:
\[ 10D = S^2 + 10S \quad \text{(3)} \]

### Step 3: Solve Equation (2)

From Equation (2):
\[ \frac{D}{S + 10} - \frac{D}{S + 20} = 0.75 \]
Multiplying through by \( (S + 10)(S + 20) \):
\[ D(S + 20) - D(S + 10) = 0.75(S + 10)(S + 20) \]
This simplifies to:
\[ 10D = 0.75(S^2 + 30S + 200) \quad \text{(4)} \]

### Step 4: Equate Equations (3) and (4)

From (3), we have \( D = \frac{S^2 + 10S}{10} \).

Substituting \( D \) into (4):
\[ 10\left(\frac{S^2 + 10S}{10}\right) = 0.75(S^2 + 30S + 200) \]
This simplifies to:
\[ S^2 + 10S = 0.75S^2 + 22.5S + 150 \]
Rearranging gives:
\[ 0.25S^2 - 12.5S - 150 = 0 \]
Multiplying through by 4 to eliminate the decimal:
\[ S^2 - 50S - 600 = 0 \]

### Step 5: Solve the quadratic equation

Using the quadratic formula \( S = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -50, c = -600 \):
\[ S = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 1 \cdot (-600)}}{2 \cdot 1} \]
\[ S = \frac{50 \pm \sqrt{2500 + 2400}}{2} \]
\[ S = \frac{50 \pm \sqrt{4900}}{2} \]
\[ S = \frac{50 \pm 70}{2} \]
Calculating the two possible values:
1. \( S = \frac{120}{2} = 60 \) km/h
2. \( S = \frac{-20}{2} = -10 \) km/h (not valid)

Thus, the speed \( S = 60 \) km/h.

### Step 6: Calculate the distance \( D \)

Substituting \( S \) back into Equation (3):
\[ D = \frac{60^2 + 10 \cdot 60}{10} \]
\[ D = \frac{3600 + 600}{10} = \frac{4200}{10} = 420 \text{ km} \]

### Final Answer
The distance between the two cities P and Q is **420 km**.

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