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

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

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

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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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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

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

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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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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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New post 19 Jan 2015, 01:21
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Answer = A = 35

Refer diagram below:

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

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New post 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.


The correct answer is A.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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

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New post 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 --> \((x-30)3=15\) --> \(x=35\)
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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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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New post 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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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

Answer: A
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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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