Walking at 3/4 of his original speed, a man is able to reach his destination by 10 minutes late. Find his original time

Questions : Walking at 3/4 of his original speed, a man is able to reach his destination by 10 minutes late. Find his original time.

Answer:  

 Let the man's original speed be 

$S$, and let his original time be $T$. Then, the distance to the destination is:

$$\text{Distance} = S \times T$$

Now, when he walks at $\frac{3}{4}$ of his original speed, his new speed is:

$$\frac{3}{4} S$$

Let the new time taken be $T'$. Since distance remains the same:

$$S \times T = \left(\frac{3}{4} S\right) \times T'$$

Canceling $S$ from both sides:

$$T = \frac{3}{4} T'$$

Rearrange for $T'$:

$$T' = \frac{4}{3} T$$

The problem states that he is 10 minutes late, meaning the new time is 10 minutes more than the original:

$$T' = T + 10$$

Substituting $T' = \frac{4}{3} T$:

$$\frac{4}{3} T = T + 10$$

Multiply everything by 3 to clear the fraction:

$$4T = 3T + 30$$

Solving for $T$:

$$4T - 3T = 30$$ $$T = 30$$

Final Answer:

The man's original time was 30 minutes.

Alternative 

Here's how to solve this problem:

Understanding the Relationship

• Speed and Time: When speed decreases, time increases to cover the same distance.
• Proportion: The ratio of the speeds is inversely proportional to the ratio of the times.

Setting up the Equation

Let the original time be 't' minutes.

• Original speed: Let's call it 's'.
• Original time: t minutes
• New speed: (3/4)s
• New time: t + 10 minutes

Since the distance is the same in both cases:

Original speed * Original time = New speed * New time

s * t = (3/4)s * (t + 10)

Solving for t

1. Cancel 's' from both sides:

   t = (3/4)(t + 10)

2. Distribute the (3/4):

   t = (3/4)t + (30/4)

3. Subtract (3/4)t from both sides:

   (1/4)t = (30/4)

4. Multiply both sides by 4:

   t = 30

Answer:

The man's original time to reach his destination was 30 minutes.

Short Cut 1

You can use the shortcut formula:

$$\text{Original Time} = \frac{\text{Late Time}}{\left(\frac{\text{Reduced Speed}}{\text{Original Speed}} - 1\right)}$$

Given:

• Reduced speed = $\frac{3}{4}$ of original speed
• Late time = 10 minutes

Applying the formula:

$$T = \frac{10}{\left(\frac{3}{4} - 1\right)}$$ $$T = \frac{10}{\left(\frac{3}{4} - \frac{4}{4}\right)}$$ $$T = \frac{10}{\left(-\frac{1}{4}\right)}$$ $$T = 10 \times 4 = 30 \text{ minutes}$$

Shortcut Rule:

$$T = \frac{\text{Late Time} \times \text{Original Speed}}{\text{Original Speed} - \text{Reduced Speed}}$$

This works for similar speed-time problems!

Short Cut 2

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