Cyclist A started his journey on cycle at 7:30 a.m. at a speed of 8 km/h. After 30 minutes, cyclist B started from the same place but with a speed of 10 km/h. At what time did B overtake A?
(The other tutor added 2 hours to 7:30am and got 9:30am. He should
have added it to 8:00am to get 10:00am.)
First let's do it in our heads without algebra,
then we'll do it with algebra:
In our heads:
When B starts at 8am, A has been going 8 km/h for half an hour, so A
has a 4 km head start on B. B then chases A going 2 km/h faster
than A, so B is approaching A at 2 km/h. At that catch-up rate it will
take B 2 hours to make up A's 4 km head start. So B will overtake A
at 10am.
By algebra.
Let t be the time it takes B to overtake A
Make this chart
Distance Rate Time
A
B
Fill in t for B's time
Distance Rate Time
A
B t
A's time is a half hour longer than B's time
since he started half an hour sooner, so fill in
t+.5 for A's time:
Distance Rate Time
A t+.5
B t
Their rates are given so fill those in:
Distance Rate Time
A 8 t+.5
B 10 t
Now use to
fill in the distances:
Distance Rate Time
A 8(t+.5) 8 t+.5
B 10t 10 t
They traveled the same distance so the
two distances are equal:
8(t+.5) = 10t
8t+4 = 10t
4 = 2t
2 = t
B started half an hour later than 7:30am, or
8am. 2 hours later it will be 10am.
Edwin