Richard began driving from home on a trip averaging 30 miles per hour.

Catch up -> same distance, different rate and different time

Richard: r = 30, t = Carla's t + 0.5
Carla: r = x, t = 3
30 * (3 + 0.5) = x * 3 => x = 35

A

distance travelled by richard when carla started => 30 *(30/60) = 15 miles
therefore didtance between richard and carla is 15 miles.

15/(S - 30) = 3 => S = 35 miles/hrs

Ans - A

A. 35
So in half an hour Richard will travel 15miles(30m/1hour so 1/2hour=15m), then after that he'll travel for another 3 hours which is 3x30 miles(carla catch up to him in exactly 3 hours). Total distance traveled by richard = 15+90=105.
Now for carla to cover the same distance(=105miles) in 3 hours, speed = distance/time = 105/3 = 35miles/hr

Answer = A = 35

Refer diagram below:



Richard already travels 15 miles @ 30 mph when Carla started.

Lets say total distance = 15+x

Total time required by Richard \(= \frac{180}{60} + \frac{30}{60} = \frac{210}{60}\)

Speed = 30mph

Setting up the equation

\(15+x = 30 * \frac{210}{60}\)

x = 90

Speed of Carla \(= \frac{105}{3} = 35\)

Here we go,

catch here is to remember that we need to consider the relative speed.

In 30 minutes, Richard would have traveled (Distance = Speed * Time) 15 miles.

Let the speed of Carla = C

As per the question, Carla has to travel the distance i.e. 15 in 3 hours (keep in mind the concept of relative speed)

so Time = Distance/Speed

3 = 15 / (c - 30)

Solve for c

c = 35

Option A is correct

The correct answer is A.

After 30 mins, Richard travels 15 miles ahead Carla.
x = mph Carla need to catch up Richard in 3 hours. --> relative speed of them = x-30
Carla needs to catch up 15 miles in 3 hours --> \((x-30)3=15\) --> \(x=35\)
Answer: A

If Richard leaves 30 minutes prior to Carla and is traveling 30 mph, he will have travelled 15 miles by the time Carla starts. Carla then has 3 hours to make up that 15 miles, so she must travel at 5 mph faster than Richard. Richard's 30 mph plus the 5 mph to make up gives us 35 mph. Answer A.

the question is testing the basic concept of relative speed as we can see that in 30 mins richard would have covered 15 miles so to they have met on the same distance which is constant now their relative speed =x-30 and that 15 mins should be covered in 3 hrs. now Distance/speed=time such as 15/x-30=3 solve for x is 35 speed of carala .
hence A option must be the right answer
thanks
hope it helps

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

Answer: A

Let the distance between her home and her place of work be 120miles

Time spent travelling from home to work = 120/60 = 2hrs
Time spent travelling from work to home = 120/40 = 3 hours
Total time spent travelling = 5hours
Total distance covered = 240 miles
Average speed = 240/5 = 48mph.
Correct Answer is D.

C has to cover 15 miles over duration of 3 hours( because R has already got a lead of 15miles in his 30 minutes)
So that means 30 + 5 will help C achieve that.

Hence A

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