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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Mechanics Question
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chris558
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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.
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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LalaB
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30(3+1/2)=3x
x=35
x=35
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rajeshsinghgmat
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pareekbharat86
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Always try and make a matrix in the following format whenever we have a rate problem with 2 scenarios.
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.
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.
Thanks,
Bharat.
Bharat.
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jaspreetsra
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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
IMO: A
Speed Time Distance
R 30/h 3.5 105 miles
C ? 3.0 105 miles
Speed = 105/3 = 35miles/hour
(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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At a rate of 30mph, the distance traveled by Richard in 1/2 hour = (30)(1/2) = 15 miles.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
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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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
So to catch him in 3 hours , second driver must drive at 105/3 = 35 mph
Answer A
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We can classify this problem as a "catch-up" rate problem, for which we use the formula: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
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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