Solving Speed, Distance, and Time Problems for A and B

This problem involves analyzing the relationship between speed, distance, and time for two individuals, A and B, under different scenarios. We are given conditions comparing the time taken by A and B to cover certain distances and asked to find the time B takes to cover a specific distance.

Defining Variables for the Speed Problem

Let's define the variables we will use:

• Let the usual speed of A be \(S_A\) km/hr.
• Let the usual speed of B be \(S_B\) km/hr.
• Recall the fundamental relationship: Time = Distance / Speed.

Setting up Equations from the Conditions

Condition 1: Covering 40 km

A takes 2 hours more than B to cover a distance of 40 km.

• Time taken by A to cover 40 km = \(\frac{40}{S_A}\) hours.
• Time taken by B to cover 40 km = \(\frac{40}{S_B}\) hours.

According to the condition:

\(\frac{40}{S_A} = \frac{40}{S_B} + 2\) (Equation 1)

Condition 2: Covering 80 km with A's Doubled Speed

If A doubles his speed (speed becomes \(2S_A\)), he takes \(1\frac{1}{2}\) hours (which is 1.5 hours) more than B to cover 80 km.

• A's new speed = \(2S_A\) km/hr.
• Time taken by A with doubled speed to cover 80 km = \(\frac{80}{2S_A} = \frac{40}{S_A}\) hours.
• Time taken by B to cover 80 km = \(\frac{80}{S_B}\) hours.

According to this condition:

\(\frac{40}{S_A} = \frac{80}{S_B} + 1.5\) (Equation 2)

Solving the System of Equations

We now have a system of two equations with two variables, \(S_A\) and \(S_B\):

1. \(\frac{40}{S_A} = \frac{40}{S_B} + 2\)
2. \(\frac{40}{S_A} = \frac{80}{S_B} + 1.5\)

Since the left-hand sides of both equations are equal (\(\frac{40}{S_A}\)), we can set the right-hand sides equal to each other:

\(\frac{40}{S_B} + 2 = \frac{80}{S_B} + 1.5\)

Now, we can solve for \(S_B\). Let's rearrange the equation to gather terms involving \(S_B\) on one side and constants on the other:

\(2 - 1.5 = \frac{80}{S_B} - \frac{40}{S_B}\)

\(0.5 = \frac{80 - 40}{S_B}\)

\(0.5 = \frac{40}{S_B}\)

To find \(S_B\), we can write 0.5 as \(\frac{1}{2}\):

\(\frac{1}{2} = \frac{40}{S_B}\)

Cross-multiplying gives:

\(S_B \times 1 = 40 \times 2\)

\(S_B = 80\) km/hr.

So, the usual speed of B is 80 km/hr.

Calculating Time for B to Cover 120 km

The question asks for the time B will take to cover a distance of 120 km travelling at his same speed (\(S_B\)).

• Distance = 120 km.
• B's speed = \(S_B = 80\) km/hr.

Time taken by B = \(\frac{\text{Distance}}{\text{Speed}}\)

Time taken by B = \(\frac{120}{80}\) hours

Simplifying the fraction:

Time taken by B = \(\frac{12}{8} = \frac{3}{2}\) hours.

The fraction \(\frac{3}{2}\) hours can be written as a mixed number:

\(\frac{3}{2} = 1 \frac{1}{2}\) hours.

Final Answer

B will take \(1\frac{1}{2}\) hours to cover a distance of 120 km travelling at his usual speed.

Revision Table: Key Concepts in Speed, Distance, Time

Additional Information: Solving Word Problems

Solving word problems like this involves translating the given information into mathematical equations. Here's a general approach:

• Read Carefully: Understand the problem and what is being asked. Identify the knowns and unknowns.
• Define Variables: Assign letters to the unknown quantities.
• Formulate Equations: Write down mathematical equations that represent the relationships described in the problem.
• Solve Equations: Use algebraic methods to solve the system of equations.
• Check the Answer: Make sure your solution makes sense in the context of the original problem.
• Units: Pay attention to units (e.g., km, hours) and ensure consistency throughout the calculations.

In this specific problem, setting up the correct equations based on the time differences was crucial. Substituting or equating expressions allowed us to find the unknown speed of B and then calculate the required time.