Richard began driving from home on a trip averaging...

LUANDATO wrote:Richard began driving from home on a trip averaging 30 miles per hour. How many miles per hour must Carla drive on average to catch up to him in exactly 3 hours if she leaves 30 minutes after Richard?

A. 35
B. 55
C. 39
D. 40
E. 60

Since Richard travels alone at 30 mph for 1/2 hour, the distance by which Richard gets ahead = rt = (30)(1/2) = 15 miles.
Since Carla is 15 miles behind Richard -- and she must catch up to Richard in 3 hours -- she must catch up by 5 miles each hour.
To catch up by 5 miles each hour, Carla must travel 5 mph FASTER than Richard.
Since Richard's rate = 30 mph, the required rate for Carla = 30+5 = 35 mph.

The correct answer is A.

LUANDATO wrote:Richard began driving from home on a trip averaging 30 miles per hour. How many miles per hour must Carla drive on average to catch up to him in exactly 3 hours if she leaves 30 minutes after Richard?

A. 35
B. 55
C. 39
D. 40
E. 60

We can classify this problem as a “catch-up” rate problem, for which we use the formula:

distance of Richard = distance of Carla

We are given that Richard began driving from home on a trip averaging 30 miles per hour and that Carla leaves 30 minutes after Richard. We need to determine at what rate Carla will have to drive to catch Richard in 3 hours. We can let Carla’s rate = r.

Since Richard started 30 minutes before Carla, Richard’s time = 1/2 + 3 = 3.5 hours and Carla's time = 3 hours.

Since distance = rate x time, we can calculate each person’s distance.

Richard’s distance = 30 x 3.5 = 105 miles

Carla’s distance = r x 3 = 3r miles

We can equate the two distances and determine r.

105 = 3r

r = 35 mph

Alternate Solution:

At the time when Carla begins driving, Richard has been driving for 30 minutes = 0.5 hours; thus he has covered a distance of 30 x 0.5 = 15 miles. So, when Carla begins driving, the distance between them is 15 miles. If Carla is to catch Richard 3 hours after she begins driving, the distance between the two must decrease by 15 / 3 = 5 miles every hour. This means that Carla should drive 5 mph faster than Richard, i.e. 30 + 5 = 35 mph.

Answer: A

Richard speed=30miles/hour
R(S)= distance / time = 30miles / 60minutes
The total time of the trip= 3hours
since Carla started 30 minutes late before Richard, then Richard must have covered 15 miles in 30 minutes.
Therefore, Time = Distance/Speed
$$3=\frac{15}{Carla\ speed\ \left(x\right)-30}$$ $$3x-90=15$$
$$x=\frac{105}{3}=35$$