To determine the speed of A, we can start by defining the variables and using the given information.

Let the speed of A be $v_A$ km/hr and the speed of B be $v_B$ km/hr.

According to the problem:

1. $v_A + v_B = 7$ km/hr (the sum of their speeds is 7 km/hr)
2. The time taken by A to walk 24 km is $\frac{24}{v_A}$ hours.
3. The time taken by B to walk 24 km is $\frac{24}{v_B}$ hours.
4. The sum of their times is 14 hours, so $\frac{24}{v_A} + \frac{24}{v_B} = 14$.

We now have the following system of equations: $v_A + v_B = 7$ $\frac{24}{v_A} + \frac{24}{v_B} = 14$

First, we can simplify the second equation by factoring out the 24: $\frac{24}{v_A} + \frac{24}{v_B} = 14$ $24 \left( \frac{1}{v_A} + \frac{1}{v_B} \right) = 14$ $\frac{1}{v_A} + \frac{1}{v_B} = \frac{14}{24}$ $\frac{1}{v_A} + \frac{1}{v_B} = \frac{7}{12}$

Now, we can use the relationship $v_B = 7 - v_A$ from the first equation and substitute it into the second equation: $\frac{1}{v_A} + \frac{1}{7 - v_A} = \frac{7}{12}$

To solve this equation, find a common denominator: $\frac{7 - v_A + v_A}{v_A (7 - v_A)} = \frac{7}{12}$ $\frac{7}{v_A (7 - v_A)} = \frac{7}{12}$

We can now cancel out the 7 from both sides: $\frac{1}{v_A (7 - v_A)} = \frac{1}{12}$

Cross-multiply to solve for $v_A$: $12 = v_A (7 - v_A)$ $12 = 7v_A - v_A^2$ $v_A^2 - 7v_A + 12 = 0$

This is a quadratic equation. Solving this using the quadratic formula $v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = -7$, and $c = 12$: $v_A = \frac{7 \pm \sqrt{49 - 48}}{2}$ $v_A = \frac{7 \pm 1}{2}$ $v_A = \frac{8}{2} \quad \text{or} \quad v_A = \frac{6}{2}$ $v_A = 4 \quad \text{or} \quad v_A = 3$

Since A is faster than B, $v_A = 4$ km/hr and $v_B = 3$ km/hr.

Thus, the speed of A is $4$ km/hr.

Would you like further details or have any questions?

Here are 8 related questions to expand your understanding:

1. What is the speed of B if A is faster and their combined speed is 10 km/hr?
2. How would the solution change if the distance was 30 km instead of 24 km?
3. Can you derive the formula used to solve the quadratic equation?
4. What would be the time taken by each if the distance was doubled?
5. How do you interpret the solutions of the quadratic equation in this context?
6. If A and B were to race over 10 km, who would win and by how much time?
7. What would the speeds be if the sum of their speeds was 10 km/hr but the sum of their times remained the same?
8. How can we graphically represent the speeds of A and B?

Tip: Always check the conditions given in the problem to determine which solution of the quadratic equation is applicable.