Understanding Speed, Time, and Distance Relationships

This problem involves the relationship between speed, time, and distance. When the distance between two points is constant, speed and time are inversely proportional. This means that if speed decreases, the time taken to cover the same distance increases, and vice versa.

Setting Up the Problem

Let's define the variables for this problem:

• Let the usual speed of the person be \(V\) units per minute.
• Let the usual time taken to reach the office be \(T\) minutes.
• Let the distance to the office be \(D\) units.

The relationship between speed, time, and distance is given by the formula:

\( \text{Distance} = \text{Speed} \times \text{Time} \)

So, for the usual scenario:

\( D = V \times T \quad (1) \)

Analyzing the New Scenario

According to the problem, the person walks at \( \frac{7}{9} \) of his usual speed. New Speed \( = \frac{7}{9} V \)

He reaches his office 10 minutes later than the usual time. New Time \( = T + 10 \) minutes

The distance to the office remains the same in this scenario. So, we can set up the equation for the new scenario:

\( D = \left( \frac{7}{9} V \right) \times (T + 10) \quad (2) \)

Solving for the Usual Time

Since the distance \(D\) is the same in both scenarios, we can equate equation (1) and equation (2):

\( V \times T = \left( \frac{7}{9} V \right) \times (T + 10) \)

We can cancel \(V\) from both sides of the equation, as speed \(V\) cannot be zero:

\( T = \frac{7}{9} \times (T + 10) \)

Now, we solve for \(T\):

Multiply both sides by 9 to eliminate the denominator:

\( 9T = 7 \times (T + 10) \)

Distribute 7 on the right side:

\( 9T = 7T + 70 \)

Subtract \(7T\) from both sides:

\( 9T - 7T = 70 \)

\( 2T = 70 \)

Divide by 2:

\( T = \frac{70}{2} \)

\( T = 35 \)

So, the usual time taken by the person to reach his office is 35 minutes.

Verification

Let's check if the answer makes sense. Usual Speed = \(V\), Usual Time = 35 minutes, Distance \(D = 35V\). New Speed = \( \frac{7}{9}V \), New Time = 35 + 10 = 45 minutes. New Distance \( = \left( \frac{7}{9}V \right) \times 45 = \frac{7}{9} \times 45 \times V = 7 \times 5 \times V = 35V \). The new distance equals the usual distance, so the calculation is correct.

Alternative Approach using Time Difference

Since distance is constant, Time is inversely proportional to Speed. Let \(S_1\) be the usual speed and \(T_1\) be the usual time. Let \(S_2\) be the new speed and \(T_2\) be the new time. We have \( S_2 = \frac{7}{9} S_1 \). Since \( \frac{T_1}{T_2} = \frac{S_2}{S_1} \), we get \( \frac{T_1}{T_2} = \frac{7/9 S_1}{S_1} = \frac{7}{9} \). So, \( \frac{T_1}{T_2} = \frac{7}{9} \), which means \( 9T_1 = 7T_2 \). We also know that \( T_2 = T_1 + 10 \) minutes. Substitute \(T_2\) in the equation: \( 9T_1 = 7(T_1 + 10) \). \( 9T_1 = 7T_1 + 70 \) \( 9T_1 - 7T_1 = 70 \) \( 2T_1 = 70 \) \( T_1 = 35 \) minutes. This confirms the usual time is 35 minutes.

Revision Table: Speed, Time, and Distance Concepts

Additional Information on Speed, Time, and Distance Problems

Speed, time, and distance problems are common in quantitative aptitude. They often involve scenarios where one of the quantities (speed, time, or distance) is constant, or there's a change in one quantity affecting the other two.

• Direct Proportion: If speed is constant, distance covered is directly proportional to time (\( D \propto T \)). If time is constant, distance covered is directly proportional to speed (\( D \propto S \)).
• Inverse Proportion: If distance is constant, speed is inversely proportional to time (\( S \propto \frac{1}{T} \)). This means if speed increases, time decreases, and if speed decreases, time increases.
• Relative Speed: When two objects are moving, their relative speed is used to calculate the time taken for them to meet or cross each other.
  • Objects moving in the same direction: Relative Speed = Difference of speeds.
  • Objects moving in opposite directions: Relative Speed = Sum of speeds.
• Unit Conversion: Always ensure that the units of speed, time, and distance are consistent. For example, if speed is in km/hr, time should be in hours and distance in km. If time is in minutes, convert speed or time appropriately. The conversion \( \text{m/s} \to \text{km/hr} \) is done by multiplying by \( \frac{18}{5} \), and \( \text{km/hr} \to \text{m/s} \) by multiplying by \( \frac{5}{18} \).