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

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Math Expert
Joined: 02 Sep 2009
Posts: 43893
Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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15 Jan 2015, 06:42
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82% (01:19) correct 18% (01:31) wrong based on 273 sessions

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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

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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15 Jan 2015, 07:09
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A. 35,
Richard arn 15 miles in 1/2 hour. Now to meet him in 3 hours, Carla must run 5 miles per hour extra. Hence, she should run 35 miles per hour ( 15 mile extra in 3 hours).
Intern
Joined: 27 Mar 2014
Posts: 26
Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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16 Jan 2015, 02:39
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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
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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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16 Jan 2015, 03:02
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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

Carla starts 30 minutes later and it takes 3 hr for carla to meet richard

So richad total time travelled = 3hr + 30 minutes

Richard distance = 30 * (3 1/2) = 105

so carla need to travle 105 to meet richard in 3 hrs

speed of carla = 105/3 = 35 miles per hour

Consider kudos , if helped
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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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16 Jan 2015, 11:23
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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
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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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16 Jan 2015, 12:06
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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
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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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19 Jan 2015, 01:21
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Refer diagram below:

Attachment:

diag.png [ 2.89 KiB | Viewed 3062 times ]

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 $$= \frac{105}{3} = 35$$
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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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19 Jan 2015, 02:55
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
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Posts: 43893
Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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19 Jan 2015, 03:58
Bunuel 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

Kudos for a correct solution.

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Joined: 22 Dec 2014
Posts: 43
Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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16 Jul 2015, 09:26
Bunuel 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

Kudos for a correct solution.

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 --> $$(x-30)3=15$$ --> $$x=35$$
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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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16 Jul 2015, 15:28
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Expert's post
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.
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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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13 Oct 2016, 01:11
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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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13 Feb 2017, 03:58
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
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Re: Richard began driving from home on a trip averaging 30 miles per hour. [#permalink]

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14 Feb 2017, 15:56
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Expert's post
Bunuel 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

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Re: Richard began driving from home on a trip averaging 30 miles per hour.   [#permalink] 14 Feb 2017, 15:56
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