Understanding the Worker's Travel Problem

This question is a classic example of a distance, speed, and time problem where a journey is divided into parts with different speeds. We are given the total distance, total time, and the speeds for each part of the journey (on foot and on bicycle). Our goal is to find the specific distance covered on the bicycle.

Setting up the Equations

Let's break down the given information:

• Total distance covered = 81 km
• Total time taken = 11 hours
• Speed on foot = 4.5 km/h
• Speed on bicycle = 15 km/h

We need to find the distance covered on the bicycle. Let's use variables:

• Let \(D_f\) be the distance covered on foot (in km).
• Let \(D_c\) be the distance covered on the bicycle (in km).
• Let \(T_f\) be the time spent traveling on foot (in hours).
• Let \(T_c\) be the time spent traveling on the bicycle (in hours).

From the problem statement, we know the total distance and total time:

• Equation 1 (Total Distance): \(D_f + D_c = 81\)
• Equation 2 (Total Time): \(T_f + T_c = 11\)

We also know the relationship between distance, speed, and time: Distance = Speed \(\times\) Time, which can be rearranged to Time = Distance / Speed.

• For the journey on foot: \(T_f = \frac{D_f}{\text{Speed on foot}} = \frac{D_f}{4.5}\)
• For the journey on bicycle: \(T_c = \frac{D_c}{\text{Speed on bicycle}} = \frac{D_c}{15}\)

Solving for the Distance on Bicycle

Now we can substitute the expressions for \(T_f\) and \(T_c\) into the total time equation (Equation 2):

\[ \frac{D_f}{4.5} + \frac{D_c}{15} = 11 \]

We have two variables, \(D_f\) and \(D_c\), in this equation. We can use Equation 1 (\(D_f + D_c = 81\)) to express \(D_f\) in terms of \(D_c\):

\[ D_f = 81 - D_c \]

Now substitute this expression for \(D_f\) into the equation involving time:

\[ \frac{81 - D_c}{4.5} + \frac{D_c}{15} = 11 \]

To clear the denominators, find the least common multiple (LCM) of 4.5 and 15. Note that 4.5 is \( \frac{9}{2} \). The LCM of \( \frac{9}{2} \) and 15 is 45. Multiply the entire equation by 45:

\[ 45 \left( \frac{81 - D_c}{4.5} \right) + 45 \left( \frac{D_c}{15} \right) = 45 \times 11 \]

\[ 10 (81 - D_c) + 3 D_c = 495 \]

Now, distribute and simplify the equation:

\[ 810 - 10 D_c + 3 D_c = 495 \]

\[ 810 - 7 D_c = 495 \]

Subtract 495 from both sides and add \(7 D_c\) to both sides:

\[ 810 - 495 = 7 D_c \]

\[ 315 = 7 D_c \]

Divide by 7 to find the value of \(D_c\):

\[ D_c = \frac{315}{7} \]

\[ D_c = 45 \]

So, the distance covered on the bicycle is 45 km.

Verification

Let's check if our answer is correct. If \(D_c = 45\) km, then \(D_f = 81 - D_c = 81 - 45 = 36\) km.

Time taken on foot \(T_f = \frac{D_f}{4.5} = \frac{36}{4.5} = \frac{360}{45} = 8\) hours.

Time taken on bicycle \(T_c = \frac{D_c}{15} = \frac{45}{15} = 3\) hours.

Total time taken = \(T_f + T_c = 8 + 3 = 11\) hours. This matches the total time given in the problem.

Final Answer

The distance covered on the cycle is 45 km.

Revision Table: Distance, Speed, Time

Additional Information: Solving Multi-Part Journey Problems

Problems where a journey is covered in parts with different speeds and/or times are common in quantitative aptitude. Here's a general approach:

• Identify Variables: Assign variables to unknown distances and times for each part of the journey.
• Formulate Equations:
  • Write an equation for the total distance (sum of distances of each part).
  • Write an equation for the total time (sum of times for each part).
  • Use the fundamental relationship (Distance = Speed \(\times\) Time) for each part to relate distance, speed, and time for that specific part.
• Solve the System of Equations: Use substitution or elimination methods to solve the system of equations you've created. Often, expressing time in terms of distance and speed for each part and substituting into the total time equation is a useful strategy, as shown in this problem.
• Verify the Solution: Plug the calculated values back into the original equations or the problem statement to ensure they satisfy all conditions.

This systematic approach helps break down complex problems into manageable steps.