Question

A man can cover a distance in 1 hour 24 minutes by covering 2/3 of the distance at 4 km/h and the rest at 5 km/h. The total distance is

• A. 2 km
• B. 5 km
• C. 6 km
• D. 10 km
• E. None of these

Hint:

For mixed-speed problems, break down the time for each part of the journey and then solve for the total distance using the total time.

Answer 1

The Correct Option is C

Let the total distance be $D$ km. He covers $\frac{2}{3}$ of the distance at 4 km/h, and $\frac{1}{3}$ at 5 km/h.  

So:

• Distance 1: $\frac{2}{3}D$ at 4 km/h ⇒ Time = $\frac{\frac{2}{3}D}{4} = \frac{2D}{12} = \frac{D}{6}$ hours
• Distance 2: $\frac{1}{3}D$ at 5 km/h ⇒ Time = $\frac{\frac{1}{3}D}{5} = \frac{D}{15}$ hours

Total time taken = $\frac{D}{6} + \frac{D}{15}$ 

Find LCM of 6 and 15 = 30: $$ \frac{D}{6} + \frac{D}{15} = \frac{5D + 2D}{30} = \frac{7D}{30} \text{ hours} $$ Given total time = 1 hour 24 minutes = $\frac{84}{60} = \frac{7}{5}$ hours So: $$ \frac{7D}{30} = \frac{7}{5} $$ Multiply both sides by 30: $$ 7D = 42 \Rightarrow D = \frac{42}{7} = \mathbf{6 \text{ km}} $$ 
Answer: (c) 6 km

Answer 2

Let the total distance be \( d \). The man covers \( \frac{2}{3} \) of the distance at 4 km/h and \( \frac{1}{3} \) at 5 km/h. Time taken for first part: \[ \text{Time} = \frac{\frac{2}{3}d}{4} = \frac{d}{6} \] Time taken for second part: \[ \text{Time} = \frac{\frac{1}{3}d}{5} = \frac{d}{15} \] Total time: \[ \frac{d}{6} + \frac{d}{15} = \frac{1}{6}d + \frac{1}{15}d = \frac{5d + 2d}{30} = \frac{7d}{30} \] We are told the total time is 1 hour 24 minutes, which is \( \frac{7}{5} \) hours. Thus: \[ \frac{7d}{30} = \frac{7}{5} \] Solving for \( d \): \[ d = \frac{30}{5} = 6 \, \text{km} \]