If Aman travels with 12.5 percent less speed of his usual speed, then he will reach 9 minutes late. What is the usual time taken by him?

Question

Prepp Team · Accepted Answer

Solving Speed, Time, and Distance Problems

This problem involves the relationship between speed, time, and distance. A key concept here is that when the distance traveled is constant, speed and time are inversely proportional. This means if speed decreases, time increases proportionally, and vice versa.

Understanding the Problem Statement

Aman travels a certain distance. When he reduces his speed by 12.5 percent, he takes 9 minutes longer to cover the same distance. We need to find the usual time he takes to travel this distance at his usual speed.

Step-by-Step Solution

1. Calculate the New Speed in Terms of Usual Speed

Aman's speed is reduced by 12.5 percent. Let the usual speed be $S_u$. The percentage reduction is 12.5%. 12.5% can be written as a fraction: $$12.5\% = \frac{12.5}{100} = \frac{125}{1000} = \frac{1}{8}$$ So, the speed is reduced by $ \frac{1}{8} $ of the usual speed. The new speed, $S_n$, will be the usual speed minus the reduction: $$S_n = S_u - \frac{1}{8} S_u = \left(1 - \frac{1}{8}\right) S_u = \frac{7}{8} S_u$$ The new speed is $ \frac{7}{8} $ of the usual speed.

2. Relate Speed and Time When Distance is Constant

The distance Aman travels is the same in both cases (usual speed and reduced speed). Let this distance be $D$. The relationship between distance, speed, and time is: $D = \text{Speed} \times \text{Time}$.

Let the usual time taken be $T_u$. Using usual speed and usual time: $D = S_u \times T_u$.

Let the new time taken be $T_n$. Using new speed and new time: $D = S_n \times T_n$.

Since the distance is constant, we can equate the two expressions for $D$: $$S_u \times T_u = S_n \times T_n$$ Substitute the expression for $S_n$ from Step 1: $$S_u \times T_u = \left(\frac{7}{8} S_u\right) \times T_n$$ We can cancel $S_u$ from both sides (assuming $S_u \neq 0$): $$T_u = \frac{7}{8} T_n$$ This equation shows the relationship between the usual time ($T_u$) and the new time ($T_n$).

3. Use the Given Information About Time Difference

Aman reaches 9 minutes late when traveling at the new speed. This means the new time ($T_n$) is 9 minutes more than the usual time ($T_u$). $$T_n = T_u + 9$$

4. Solve for the Usual Time ($T_u$)

Now we have a system of two equations:

$T_u = \frac{7}{8} T_n$
$T_n = T_u + 9$

Substitute the second equation into the first equation:

$$T_u = \frac{7}{8} (T_u + 9)$$

Now, solve for $T_u$:

$$T_u = \frac{7}{8} T_u + \frac{7}{8} \times 9$$ $$T_u = \frac{7}{8} T_u + \frac{63}{8}$$

Subtract $ \frac{7}{8} T_u $ from both sides:

$$T_u - \frac{7}{8} T_u = \frac{63}{8}$$ $$\left(1 - \frac{7}{8}\right) T_u = \frac{63}{8}$$ $$\left(\frac{8 - 7}{8}\right) T_u = \frac{63}{8}$$ $$\frac{1}{8} T_u = \frac{63}{8}$$

Multiply both sides by 8:

$$T_u = \frac{63}{8} \times 8$$ $$T_u = 63$$

The usual time taken by Aman is 63 minutes.

Using Ratio Method (Alternative Approach)

Since speed and time are inversely proportional for a constant distance, the ratio of times is the inverse of the ratio of speeds.

Usual Speed : New Speed = $S_u : S_n = S_u : \frac{7}{8} S_u$

Dividing by $S_u$, the ratio is $1 : \frac{7}{8}$. Multiplying by 8, the ratio is $8 : 7$. So, Usual Speed : New Speed = $8 : 7$.

Inverse ratio for time: Usual Time : New Time = $T_u : T_n = 7 : 8$.

Let the usual time be $7x$ and the new time be $8x$, where $x$ is a constant.

The problem states the new time is 9 minutes more than the usual time: $$T_n - T_u = 9$$ $$8x - 7x = 9$$ $$x = 9$$

The usual time is $T_u = 7x$. $$T_u = 7 \times 9 = 63$$ The usual time taken is 63 minutes.

Both methods yield the same result.

Variable	Usual Case	New Case
Speed	$S_u$	$S_n = \frac{7}{8} S_u$
Time	$T_u$	$T_n = T_u + 9$
Distance	$D$	$D$

Conclusion

When Aman travels at 12.5 percent less than his usual speed, he reaches 9 minutes late. His usual time taken is 63 minutes.

Revision Table: Speed, Time, and Distance Concepts

Concept	Formula/Relationship	Notes
Distance	$D = S \times T$	Speed times Time
Speed	$S = \frac{D}{T}$	Distance divided by Time
Time	$T = \frac{D}{S}$	Distance divided by Speed
Constant Distance	$S_1 \times T_1 = S_2 \times T_2$	Speed is inversely proportional to Time ($S \propto 1/T$)
Constant Speed	$D_1 / T_1 = D_2 / T_2$	Distance is directly proportional to Time ($D \propto T$)
Constant Time	$D_1 / S_1 = D_2 / S_2$	Distance is directly proportional to Speed ($D \propto S$)

Additional Information: Percentage Change and Fractions

Understanding percentages and their fractional equivalents is very useful in quantitative problems, especially those involving speed, time, distance, profit/loss, etc.

A 12.5 percent reduction means the new value is (100 - 12.5)% = 87.5% of the original value.

Converting percentage to fraction:

100% = 1
50% = 1/2
25% = 1/4
12.5% = 1/8
87.5% = 7/8

In this problem, the new speed is 87.5% of the usual speed, which is $ \frac{7}{8} $ of the usual speed. This aligns with our calculation $S_n = \frac{7}{8} S_u$.

Similarly, if a quantity increases by a percentage, you add that fraction/percentage to the original value. For example, a 20% increase means the new value is $1 + 20\% = 1 + \frac{1}{5} = \frac{6}{5}$ times the original value.

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