Dan drove home from college at an average rate of 60 miles per hour.

Solution

• Dan’s speed while going to home from college = 60 miles/hour
• While coming back to college from home, his speed was 10 miles per hour slower.

o So, Dan’s speed while coming back to college from home \(= 60-10 = 50\) miles/hour
• Let’s say Dan took \(t\) hours to complete his trip from college to home.

o Then, time taken by Dan to complete his trip back to college \(= t+1\) hours
• Let us assume that Dan’s home is at a \(d\) miles distance from his college.

o Then, \(60*t = 50*(t+1) = d\)
o Or, \(60*t = 50*t + 50\)
o Or,\( t = 5\) hours
• Hence, \(d = 60*5 = 300 \) miles

Alternately,

• Let us assume that the Dan’s home is d miles away from his college.

• Dan’s speed on his trip to home \(= 60 \)miles/hour
• Since on his trip back to college his speed was 10 miles per hour slower,

o So, his speed while going back to college from home \(= 60-10 = 50\) miles/hour
• It is given that on his trip back to college, he took one hour more than his trip to home.

o So, \(\frac{d}{50} – \frac{d}{60 }=1\)
o Or, \(\frac{6d – 5d}{300} =1\)
o Or, \(6d – 5d =300\)
o Or, \(d = 300\) miles.