John drove from station A to station B a distance of 224 miles. On his way back, he increased his speed by 10 mph. If the journey back took him 24 minutes less, what was his original speed?

Sure, let's solve this step by step.

Step 1: Let's denote the original speed as 'x' mph. So, the time it took John to drive from station A to station B is '224/x' hours.

Step 2: On his way back, John increased his speed by 10 mph. So, his speed while returning was 'x+10' mph. Therefore, the time it took him to drive back from station B to station A is '224/(x+10)' hours.

Step 3: We know that the journey back took him 24 minutes less. We need to convert these minutes into hours since our speed is in mph. So, 24 minutes is '24/60 = 0.4' hours.

Step 4: Now, we can set up the equation. The time to go from A to B minus the time to return from B to A is equal to 0.4 hours. So, we have:

224/x - 224/(x+10) = 0.4

Step 5: To solve this equation, we can first multiply through by 'x(x+10)' to get rid of the denominators:

224(x+10) - 224x = 0.4x(x+10)

Step 6: Simplify the equation:

2240 + 224x - 224x = 0.4x^2 + 4x

Step 7: Rearrange the equation:

0.4x^2 + 4x - 2240 = 0

Step 8: To make it easier, let's multiply through by 10 to get rid of the decimal:

4x^2 + 40x - 22400 = 0

Step 9: This is a quadratic equation in the form of ax^2 + bx + c = 0. We can solve it using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a):

x = [-40 ± sqrt((40)^2 - 44(-22400))] / (2*4)

Step 10: Simplify further to find the value of 'x'. Remember that 'x' (speed) cannot be negative, so we only consider the positive root.

After solving, you will find that the original speed 'x' is approximately 56 mph.