I'll be using \(d = rt\) to represent his original time to get to the office.


Now let's think about how much time he actually spent driving on that day he was late.

"Mr. Richards started 30 minutes late from home and reached his office 50 minutes late..."

Say he usually leaves for work at 8:00am and get's there at 9:00am. So it takes him 1 hour to get to work.
On the day he was late, he left for work 30 minutes late, which means he left his house at 8:30am. He arrived at work 50 minutes late, which means he arrived at 9:50am. So in total, on the day he was late, he travelled 1 hour and 20 minutes. So it took him 20 minutes longer than usual, or \(t+\frac{1}{3}\) hours to get to the office.

"...while driving 25% slower than his usual speed"

Since we let his original speed be \(r\), the speed he was going on the day he was late was 25% slower, so it would be 75% of \(r\).

Or in other other words: \(\frac{3}{4}\)\(r\).


Putting this all together, we get two equations:

\(d = rt\)

\(d = \frac{3}{4}r(t+\frac{1}{3})\)

The distance he travels to work is the same in both scenarios, so \(d\) will be the same in both. So we can do the following:

\(rt = \frac{3}{4}r(t+\frac{1}{3})\)

\(rt = \frac{3}{4}rt+\frac{1}{4}r\)

\(\frac{1}{4}rt = \frac{1}{4}r\)

The \(\frac{1}{4}\) and \(r\) on both sides cancel, so we're left with:

\(t = 1\)

\(t\) is in hours, and we need to convert this to minutes. 1 hour = 60 minutes.

So the answer is C