A bike travels a distance of 200 km at a constant speed, If the speed of the bike is increased by 5 km/hr, the journey would have taken 2 hours less. What is the speed of the bike?

Options

A. 30 km/hr
B. 25 km/hr
C. 20 km/hr
D. 15 km/hr

Answer

To solve the problem, we need to find the speed of the bike (let's denote it as \( u \)). We know the following:

1. The distance traveled by the bike is 200 km.
2. If the speed is increased by 5 km/hr, the journey takes 2 hours less.

Let's break this down step by step:

### Step 1: Set up the equations

We know that:
- Distance = Speed × Time

Let \( t \) be the time taken to cover 200 km at speed \( u \). Therefore, we can write:
\[ 200 = u \times t \quad \text{(1)} \]

When the speed is increased by 5 km/hr, the new speed is \( u + 5 \) and the time taken becomes \( t - 2 \). The equation for this scenario is:
\[ 200 = (u + 5) \times (t - 2) \quad \text{(2)} \]

### Step 2: Express \( t \) in terms of \( u \)

From equation (1), we can express \( t \):
\[ t = \frac{200}{u} \quad \text{(3)} \]

### Step 3: Substitute \( t \) in equation (2)

Now, substitute equation (3) into equation (2):
\[ 200 = (u + 5) \left( \frac{200}{u} - 2 \right) \]

### Step 4: Simplify the equation

Expanding the right-hand side:
\[ 200 = (u + 5) \left( \frac{200 - 2u}{u} \right) \]
\[ 200 = \frac{(u + 5)(200 - 2u)}{u} \]

Multiply both sides by \( u \) to eliminate the fraction:
\[ 200u = (u + 5)(200 - 2u) \]

### Step 5: Expand and rearrange

Expanding the right side:
\[ 200u = 200u + 1000 - 2u^2 - 10u \]
Now, rearranging gives:
\[ 0 = 1000 - 2u^2 - 10u \]
Rearranging further:
\[ 2u^2 + 10u - 1000 = 0 \]

### Step 6: Simplify the quadratic equation

Dividing the entire equation by 2:
\[ u^2 + 5u - 500 = 0 \]

### Step 7: Solve the quadratic equation

We can use the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = 5, c = -500 \):
\[ u = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-500)}}{2 \cdot 1} \]
\[ u = \frac{-5 \pm \sqrt{25 + 2000}}{2} \]
\[ u = \frac{-5 \pm \sqrt{2025}}{2} \]
\[ u = \frac{-5 \pm 45}{2} \]

Calculating the two possible values:
1. \( u = \frac{40}{2} = 20 \) (valid speed)
2. \( u = \frac{-50}{2} = -25 \) (not valid since speed cannot be negative)

### Conclusion

Thus, the speed of the bike is:
\[ \boxed{20 \text{ km/hr}} \]