Solving the Speed, Distance, and Time Problem

This question involves two travelers, X and Y, covering the same distance but at different speeds. We are given the distance, the sum of their speeds, and the total time they took combined. We need to find the ratio of their speeds.

Let's denote the speed of X as \(S_X\) km/h and the speed of Y as \(S_Y\) km/h.

We are given the following information:

• Distance traveled by X = 90 km
• Distance traveled by Y = 90 km
• Sum of their speeds: \(S_X + S_Y = 100\) km/h
• Total time taken by both = 3 hours 45 minutes
• Speed of Y is greater than the speed of X (\(S_Y > S_X\))

First, let's convert the total time into hours. 45 minutes is equal to \(45/60\) hours, which simplifies to \(3/4\) hours. So, the total time is \(3 + 3/4 = 15/4\) hours.

The formula connecting speed, distance, and time is:

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

Using this, the time taken by X to travel 90 km is \(T_X = \frac{90}{S_X}\) hours.

The time taken by Y to travel 90 km is \(T_Y = \frac{90}{S_Y}\) hours.

The total time taken by both is \(T_X + T_Y\), which is given as \(15/4\) hours.

So, we have the equation:

\(\frac{90}{S_X} + \frac{90}{S_Y} = \frac{15}{4}\)

From the given sum of speeds, we know \(S_X + S_Y = 100\). We can express \(S_Y\) in terms of \(S_X\): \(S_Y = 100 - S_X\).

Now, substitute this into the time equation:

\(\frac{90}{S_X} + \frac{90}{100 - S_X} = \frac{15}{4}\)

We can factor out 90 from the left side:

\(90 \left( \frac{1}{S_X} + \frac{1}{100 - S_X} \right) = \frac{15}{4}\)

Now, combine the fractions inside the parenthesis:

\(90 \left( \frac{100 - S_X + S_X}{S_X (100 - S_X)} \right) = \frac{15}{4}\)

\(90 \left( \frac{100}{100S_X - S_X^2} \right) = \frac{15}{4}\)

\(\frac{9000}{100S_X - S_X^2} = \frac{15}{4}\)

Cross-multiply:

\(9000 \times 4 = 15 \times (100S_X - S_X^2)\)

\(36000 = 1500S_X - 15S_X^2\)

Divide the entire equation by 15 to simplify:

\(\frac{36000}{15} = \frac{1500S_X}{15} - \frac{15S_X^2}{15}\)

\(2400 = 100S_X - S_X^2\)

Rearrange this into a quadratic equation:

\(S_X^2 - 100S_X + 2400 = 0\)

We can solve this quadratic equation by factoring. We need two numbers that multiply to 2400 and add up to -100. These numbers are -40 and -60.

So, the equation becomes:

\((S_X - 40)(S_X - 60) = 0\)

This gives two possible values for \(S_X\): \(S_X = 40\) or \(S_X = 60\).

Now, let's find the corresponding values for \(S_Y\) using \(S_Y = 100 - S_X\):

• If \(S_X = 40\) km/h, then \(S_Y = 100 - 40 = 60\) km/h.
• If \(S_X = 60\) km/h, then \(S_Y = 100 - 60 = 40\) km/h.

The problem states that the speed of Y is greater than that of X (\(S_Y > S_X\)).

• In the first case (\(S_X = 40\), \(S_Y = 60\)), \(60 > 40\). This condition is satisfied.
• In the second case (\(S_X = 60\), \(S_Y = 40\)), \(40 > 60\). This condition is not satisfied.

Therefore, the correct speeds are \(S_X = 40\) km/h and \(S_Y = 60\) km/h.

Finally, we need to find the ratio of the speed of X to that of Y, which is \(S_X : S_Y\).

\(S_X : S_Y = 40 : 60\)

To simplify the ratio, divide both parts by their greatest common divisor, which is 20.

\(40 \div 20 : 60 \div 20 = 2 : 3\)

The ratio of the speed of X to that of Y is 2:3.

Revision Table: Key Problem Details

Additional Information: Speed, Distance, and Time Concepts

Understanding the relationship between speed, distance, and time is fundamental to solving these types of problems. The core formula is \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \). This can be rearranged to find distance (\(\text{Distance} = \text{Speed} \times \text{Time}\)) or time (\(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)).

When dealing with problems involving multiple travelers or varying conditions, it's often helpful to:

• Define variables for unknown quantities (like speeds or times).
• Write down all given information as equations.
• Ensure all units are consistent (e.g., km and hours).
• Use substitution or elimination methods to solve the system of equations.
• Always check if the solution satisfies any additional conditions given in the problem (like one speed being greater than another).

Quadratic equations often appear when setting up equations involving time or distance in more complex scenarios like this one. Remember how to solve them either by factoring, completing the square, or using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) for an equation in the form \( ax^2 + bx + c = 0 \).