A car travels from P to Q at a constant speed. If its speed were increased by 7 km/h, it would have taken one hour less to cover the distance. It would have taken 1 hour more if the speed were decreased by 5 km/h. What is the distance between two cities and find the speed of car?

Answer

To solve the problem step by step, we will denote the usual speed of the car as \( S \) (in km/h) and the usual time taken as \( T \) (in hours). The distance between the two cities, P and Q, can be expressed as:

\[ \text{Distance} = S \times T \]

### Step 1: Set Up the Equations

1. **When the speed is increased by 7 km/h:**
- New speed = \( S + 7 \)
- New time = \( T - 1 \)
- Distance remains the same:
\[ S \times T = (S + 7) \times (T - 1) \]

2. **When the speed is decreased by 5 km/h:**
- New speed = \( S - 5 \)
- New time = \( T + 1 \)
- Distance remains the same:
\[ S \times T = (S - 5) \times (T + 1) \]

### Step 2: Expand the Equations

From the first equation:
\[ S \times T = (S + 7)(T - 1) \implies S \times T = S \times T - S + 7T - 7 \]
This simplifies to:
\[ 0 = -S + 7T - 7 \implies S = 7T - 7 \tag{1} \]

From the second equation:
\[ S \times T = (S - 5)(T + 1) \implies S \times T = S \times T + S - 5T - 5 \]
This simplifies to:
\[ 0 = S - 5T - 5 \implies S = 5T + 5 \tag{2} \]

### Step 3: Equate the Two Expressions for S

From equations (1) and (2):
\[ 7T - 7 = 5T + 5 \]
Rearranging gives:
\[ 7T - 5T = 5 + 7 \implies 2T = 12 \implies T = 6 \text{ hours} \]

### Step 4: Substitute T Back to Find S

Substituting \( T = 6 \) back into either equation (1) or (2):
Using equation (1):
\[ S = 7(6) - 7 = 42 - 7 = 35 \text{ km/h} \]

### Step 5: Calculate the Distance

Now, we can find the distance:
\[ \text{Distance} = S \times T = 35 \times 6 = 210 \text{ km} \]

### Final Answers

- Speed of the car: \( 35 \text{ km/h} \)
- Distance between cities P and Q: \( 210 \text{ km} \)

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