Santos is 333 miles away from Sam. They are traveling towards each other. If Sam travels 7 mph faster than Santos, and they meet after 9 hours how fast was each traveling?

Question

Santos is traveling 333 miles away from Sam. They are traveling towards each other. If Sam travels 7mph faster than Santos and they meet after 9 hours, how fast was each traveling?

Answer 1

Let Santos’ speed be S then Sam’s is S+7.

If they meet x miles from where Sam started, they will meet after time t=x/(S+7).

Santos will meet Sam after the same time t=(333-x)/S.

We know t=9 hours.

x/(S+7)=(333-x)/S=9.

So xS=(333-x)(S+7), xS=333S+2331-xS-7x, x(2S+7)=333S+2331, x=(333S+2331)/(2S+7).

So we can substitute for x: (333S+2331)/((2S+7)(S+7))=9.

333S+2331=9(2S+7)(S+7), divided by 9: 37S+259=2S²+16S+49, 2S²-16S-210=0, S²-8S-105=(S-15)(S+7).

Therefore S=15mph, Santos’ speed, Sam’s speed=22mph.

CHECK

x=9×22=198 miles, the distance from Sam’s starting position, because he travels at 22mph for 9 hours. So this is 333-198=135 miles from Santos’ starting position, and it would take 135/15=9 hours for them to meet.