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

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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

The OA is A.

I'm confused by this PS question. Experts, any suggestion about how to solve it? Thanks in advance.

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by GMATGuruNY » Thu Mar 22, 2018 5:43 pm
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.
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by Jeff@TargetTestPrep » Mon Mar 26, 2018 4:49 pm
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

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by BTGmoderatorRO » Thu Mar 29, 2018 11:33 am
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
$$3=\frac{15}{Carla\ speed\ \left(x\right)-30}$$ $$3x-90=15$$
$$x=\frac{105}{3}=35$$