Question

Roger can run one mile in 8 minutes. jeff can run one mile in 6 minutes. if jeff gives roger a 1 minute head​ start, how long will it take before jeff catches up

Answer 1

Recall your d = rt, distance = rate * time.

now, if Roger can do 1 mile in 8 minutes, so in 1 minute, he has done then 1/8 of a mile, so his rate is 1/8 miles per minute.

if Jeff can do 1 mile in 6 minutes, he's faster, in 1 minute he has done 1/6 of a mile, so his rate is 1/6 miles per minute.

now, when Jeff catches up with Roger, the distance covered by both will be the same, say "d" miles, because, at that millisecond, Jeff will be neck and neck with Roger, and their covered distance will be the same.

now, Jeff is generous and let Roger roll on for 1 minute before him, so, by the time time Roger has covered "d" miles, he has been running for say "t" minutes.

however, since Jeff started later by 1 minute, he hasn't been running for "t" minutes, but for "t - 1" minutes.

$\bf \begin{array}{lccclll} &\stackrel{miles}{distance}&\stackrel{mpm}{rate}&\stackrel{minutes}{time}\\ &------&------&------\\ Roger&d&\frac{1}{8}&t\\\\ Jeff&d&\frac{1}{6}&t-1 \end{array} \\\\\\ \begin{cases} \boxed{d}=\frac{1}{8}t\\\\ d=\frac{1}{6}(t-1)\\ ------\\ \boxed{\frac{1}{8}t}=\cfrac{t-1}{6} \end{cases} \\\\\\ \cfrac{t}{8}=\cfrac{t-1}{6}\implies 6t=8t-8\implies 8=2t\implies \cfrac{8}{2}=t\implies \boxed{\stackrel{mins}{4}=t}$