Understanding the Speed, Distance, and Time Problem

This problem involves calculating the time taken for a journey based on different speeds. We are given information about how a change in speed affects the time taken to cover a certain distance. We need to use this information to find the original speed and then calculate the time for a different distance at a different reduced speed.

Setting Up the Equations for Akhil's Journey

Let's define the variables for Akhil's usual journey:

• Let the usual speed of Akhil be \(v\) km/h.
• Let the usual time taken to cover 150 km be \(t\) hours.

The relationship between distance, speed, and time is: Distance = Speed × Time.

For the usual journey of 150 km:

\[150 = v \times t \quad \text{(Equation 1)}\]

In the first scenario described, Akhil drives 10 km/h slower than his usual speed. This means his new speed is \((v - 10)\) km/h.

He takes 30 minutes extra. 30 minutes is equal to 0.5 hours. So, the new time taken is \((t + 0.5)\) hours.

The distance covered in this scenario is still 150 km.

Using the distance formula for this scenario:

\[150 = (v - 10) \times (t + 0.5) \quad \text{(Equation 2)}\]

Solving for Akhil's Usual Speed

We now have two equations with two unknowns, \(v\) and \(t\). We can solve these equations simultaneously.

From Equation 1, we can express \(t\) in terms of \(v\):

\[t = \frac{150}{v}\]

Substitute this expression for \(t\) into Equation 2:

\[150 = (v - 10) \left(\frac{150}{v} + 0.5\right)\]

Now, let's expand the right side of the equation:

\[150 = v \times \frac{150}{v} + v \times 0.5 - 10 \times \frac{150}{v} - 10 \times 0.5\]

\[150 = 150 + 0.5v - \frac{1500}{v} - 5\]

Subtract 150 from both sides:

\[0 = 0.5v - \frac{1500}{v} - 5\]

Multiply the entire equation by \(v\) to eliminate the denominator:

\[0 = 0.5v^2 - 1500 - 5v\]

Rearrange the terms to form a quadratic equation:

\[0.5v^2 - 5v - 1500 = 0\]

Multiply by 2 to get rid of the decimal:

\[v^2 - 10v - 3000 = 0\]

We can solve this quadratic equation for \(v\) using factoring. We need two numbers that multiply to -3000 and add up to -10. These numbers are -60 and 50.

\[(v - 60)(v + 50) = 0\]

This gives two possible values for \(v\):

• \(v - 60 = 0 \implies v = 60\)
• \(v + 50 = 0 \implies v = -50\)

Since speed cannot be negative, Akhil's usual speed \(v\) is 60 km/h.

Calculating Time for the Second Scenario

Now we need to find the time Akhil will take to drive 90 km if he drives 15 km per hour slower than his usual speed.

His usual speed is 60 km/h.

The new speed is 15 km/h slower, so the new speed is \(60 - 15 = 45\) km/h.

The distance to cover is 90 km.

Using the distance formula, Time = Distance / Speed:

\[\text{Time} = \frac{90 \text{ km}}{45 \text{ km/h}}\]

\[\text{Time} = 2 \text{ hours}\]

The time taken to drive 90 km at 45 km/h is 2 hours.

Summary of Calculations

The calculation shows that Akhil will take 2 hours to drive 90 km at a speed of 45 km/h.

Revision Table: Key Concepts Review

Additional Information on Speed, Distance, and Time Problems

Speed, distance, and time problems are common in quantitative aptitude. They often involve scenarios where one variable changes, affecting the others. It's crucial to set up the relationships correctly using the fundamental formula: Distance = Speed × Time.

Key strategies for solving these problems include:

• Defining variables clearly for unknown quantities (like usual speed or usual time).
• Translating the given information into mathematical equations. Pay attention to units (km, hours, minutes) and convert them if necessary (e.g., minutes to hours).
• Solving the system of equations to find the unknown variables. This might involve substitution, elimination, or solving quadratic equations.
• Once the base values (like usual speed) are found, use them to calculate the required quantity in the second part of the problem.
• Always check if the calculated values make sense in the context of the problem (e.g., speed cannot be negative).

These problems test your ability to apply algebraic skills to practical situations.