Question

A train travels 360 km at a uniform speed. If the speed is increased by 5 km/h, the journey takes 1 hour less. Find the original speed. 
 

• A. 40 km/h
• B. 45 km/h
• C. 50 km/h
• D. 55 km/h

Hint:

In speed-time problems with time differences, always form the equation $\frac{d}{v_1} - \frac{d}{v_2} = \text{time difference}$ and solve for $v_1$.

Answer 1

The Correct Option is A

- Step 1: Representing unknown speed - Let the original speed be $s$ km/h. 

Then the time taken is $\frac{360}{s}$ hours (since time = distance / speed). 

- Step 2: New speed - If speed increases by 5 km/h, new speed = $s + 5$ km/h. 

Time with new speed is $\frac{360}{s + 5}$ hours. 

- Step 3: Time difference condition - The problem says the new journey takes 1 hour less, 

so: \[ \frac{360}{s} - \frac{360}{s + 5} = 1 \] 

- Step 4: Simplifying the equation - Take LCM: \[ 360\left( \frac{1}{s} - \frac{1}{s+5} \right) = 1 \] \[ 360\left( \frac{s+5 - s}{s(s+5)} \right) = 1 \] \[ \frac{1800}{s(s+5)} = 1 \] 

- Step 5: Solving the quadratic - Cross-multiplying: \[ s(s+5) = 1800 \implies s^2 + 5s - 1800 = 0 \] 

Using factorization: \[ (s - 40)(s + 45) = 0 \] 

Thus $s = 40$ (speed cannot be negative, so we ignore $-45$). 

- Step 6: Verification - Original time: $360 / 40 = 9$ hours. 

New time: $360 / 45 = 8$ hours. 

Difference = $1$ hour, which matches the problem statement.