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Re: On the first day of her vacation, Louisa traveled 216 miles. [#permalink]
let v be the speed of travel; let time be x and x-3; since the speed is the same you can do 216/x-3=778/x

you got x =7 and you can find the avg speed for the entire trip simply by doing 594(total distance)/(4+7)= 54 ans e
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Re: On the first day of her vacation, Louisa traveled 216 miles. [#permalink]
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Let, Speed be x
(378/x) –(216/x) = 3 , x= 54
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Re: On the first day of her vacation, Louisa traveled 216 miles. [#permalink]
Just to elaborate,

Lets take average speed of Louisa is v miles per hour

Total time taken on first day to travel 216 miles = 216/v

Total time taken on second day to travel 378 miles = 378/v

Travelling time on first day is 3 hour less than travelling time on second day

travel time on first day = travel time on second day -3

216/v = 378/v - 3

162/v = 3
or v = 162/3 = 54

Hence Answer is D
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Re: On the first day of her vacation, Louisa traveled 216 miles. [#permalink]
tonebeeze wrote:
On the first day of her vacation, Louisa traveled 216 miles. On the second day, traveling at the same average speed, she traveled 378 miles. If the 216-mile trip took 3 hours less than the 378-mile trip, what was the average speed, in miles per hour?

A. 31
B. 38
C. 50
D. 54
E. 56

Hey All,

This question is not difficult. I got the question correct rather quickly by using the Plug in method. I want to make sure I understand how to quickly solve this problem using algebra. Can someone please walk me through this. Thanks!


hi

378/x - 216/x = 3
or, x = 54

thanks
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Re: On the first day of her vacation, Louisa traveled 216 miles. [#permalink]
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tonebeeze wrote:
On the first day of her vacation, Louisa traveled 216 miles. On the second day, traveling at the same average speed, she traveled 378 miles. If the 216-mile trip took 3 hours less than the 378-mile trip, what was the average speed, in miles per hour?

A. 31
B. 38
C. 50
D. 54
E. 56


We can let the time to travel the 216 miles = t and the time to travel the 378 miles = t + 3. Since the rates are the same and rate = distance/time, we have:

216/t = 378/(t + 3)

216(t + 3) = 378t

216t + 648 = 378t

648 = 162t

4 = t

Thus, the average speed was 216/4 = 54 mph.

Answer: D
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Re: On the first day of her vacation, Louisa traveled 216 miles. [#permalink]
VeritasKarishma wrote:
tonebeeze wrote:
Hey All,

This question is not difficult. I got the question correct rather quickly by using the Plug in method. I want to make sure I understand how to quickly solve this problem using algebra. Can someone please walk me through this. Thanks!

"On the first day of her vacation, Louisa traveled 216 miles. On the second day, traveling at the same average speed, she traveled 378 miles. If the 216-mile trip took 3 hours less than the 378-mile trip, what was the average speed, in miles per hour?"

1. 31
b. 38
c. 50
d. 54
3. 56


Rather than plugging numbers, the easier way would be to use logic here...
216 mile trip took 3 hrs less because Louisa traveled 162 miles (= 378 - 216) less. That means her average speed must be 162/3 = 54 miles/hr.



Karishma,

Thankuverymuch, reallyfastwayofsolving. can u give please any theory on this kind of application of logic?

THANK UUUUUUU
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Re: On the first day of her vacation, Louisa traveled 216 miles. [#permalink]
Rate * Time = Distance

R * T-3 = 216
RT -3R = 216

and we know in the second leg R*T = 378

sub in

378 - 3R = 216
3R = 162
R=54

We are told that she went at the same rate each leg, so the average speed will be the same.
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Re: On the first day of her vacation, Louisa traveled 216 miles. [#permalink]
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Re: On the first day of her vacation, Louisa traveled 216 miles. [#permalink]
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