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Mechanics Question

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by Lifetron » Sun Sep 02, 2012 11:24 pm
We get that Richard drives for 3.5 hrs and Carla for 3 hrs

So, Richard goes 105 miles. Carla has to cover that in x mph.

x = 105/3 = 35 mph !

Answer is A !
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by chris558 » Mon Oct 01, 2012 7:33 am
Rate * Time = Distance

Carla drives for 3 hours but leaves 1/2 hr after Rich leaves, which means Rich was driving for 3.5 hours. The question is asking what rate Carla will have to drive at to "catch up to" Rich, meaning when will their distance traveled both be the same?

Set up the equations...
Let r = Carla's rate

Rich: (30 m/hr)*(7/2 hrs)
Carla: (r) * (3)

Make them equal to each other...
(30)(7/2)=r(3)
Solve for r...
(15)(7)=r3
5(7)=r=35

Answer is A.
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by LalaB » Mon Oct 01, 2012 8:33 am
30(3+1/2)=3x
x=35
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by rajeshsinghgmat » Thu Apr 11, 2013 1:31 am
(A) 35
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by Java_85 » Sun Nov 03, 2013 8:26 am
A it is
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by pareekbharat86 » Mon Nov 04, 2013 1:19 am
Always try and make a matrix in the following format whenever we have a rate problem with 2 scenarios.


Image

In Case of Richard-
d=30*3.5 (Distance= Speed*Time)

In Case of Carla-
d=s*3

Since d is same in each case, we can equate the 2 equations,

30*3.5=s*3

s=35. Answer is A.
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Bharat.
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by jaspreetsra » Sun Dec 28, 2014 6:42 pm
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
IMO: A
Speed Time Distance
R 30/h 3.5 105 miles
C ? 3.0 105 miles

Speed = 105/3 = 35miles/hour
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by GMATGuruNY » Sun Dec 28, 2014 10:11 pm
N:Dure 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?

- 35
- 55
- 39
- 40
- 60
At a rate of 30mph, the distance traveled by Richard in 1/2 hour = (30)(1/2) = 15 miles.
For Carla to catch-up 15 miles in 3 hours, the required CATCH-UP rate = 15/3 = 5 miles per hour.
Implication:
To catch up 15 miles in 3 hours, every hour Carla must travel 5 more miles than Richard.
Thus:
Carla's rate = (Richard rate) + 5 = 30+5 = 35 miles per hour.

The correct answer is A.
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by nikhilgmat31 » Fri Aug 28, 2015 2:12 am
Answer is A

30 * 3.5 m /hours = 105 miles.

105 miles / 3 hours = 35 mph
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by nikhilgmat31 » Tue Sep 08, 2015 2:40 am
Simple way of doing it is 30 mph goes for 3.5 hours = 105 miles

So to catch him in 3 hours , second driver must drive at 105/3 = 35 mph

Answer A
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by Scott@TargetTestPrep » Tue Dec 12, 2017 9:53 am
N:Dure 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?

- 35
- 55
- 39
- 40
- 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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