Question

A man takes \(4\) hours \(20\) minutes in walking to a certain place and riding back. If he walks on both sides, he loses \(1\) hour. The time he would take by riding both ways is

• A. \(2\) hours \(20\) min
• B. \(2\) hours
• C. \(3\) hours \(20\) min
• D. \(4\) hours \(40\) min

Hint:

Shortcut: \((\text{Ride both ways}) = 2 \times (\text{Walk + Ride}) - (\text{Walk both ways})\).
In minutes: $2 \times 260 - 320 = 520 - 320 = 200 \text{ min} = 3 \text{ h } 20 \text{ min}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Let \(W\) represent the time taken to walk one way and \(R\) represent the time taken to ride one way. We can form linear equations to solve for the required sum.
Step 2: Detailed Explanation:
1. From the first condition: Walking one way + Riding one way = \(4\) h \(20\) min.
\[ W + R = 4 \text{ h } 20 \text{ min} \quad \text{---(Eq. 1)} \] 2. From the second condition: Walking both ways = \(W + W = 2W\).
"Losing \(1\) hour" means it takes \(1\) hour more than the original time.
\[ 2W = 4 \text{ h } 20 \text{ min} + 1 \text{ h} = 5 \text{ h } 20 \text{ min} \] Divide by 2: \(W = 2 \text{ h } 40 \text{ min}\).
3. Substitute \(W\) in Eq. 1:
\[ (2 \text{ h } 40 \text{ min}) + R = 4 \text{ h } 20 \text{ min} \] \[ R = 4 \text{ h } 20 \text{ min} - 2 \text{ h } 40 \text{ min} = 1 \text{ h } 40 \text{ min} \] 4. Time for riding both ways = \(2R\):
\[ 2R = 2 \times (1 \text{ h } 40 \text{ min}) = 2 \text{ h } 80 \text{ min} = 3 \text{ h } 20 \text{ min} \] Step 3: Final Answer:
The time taken for riding both ways is \(3\) hours \(20\) minutes.