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A car averages 40 miles per hour for the first 6 hours of a

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A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post Updated on: 23 Jun 2014, 13:19
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A
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E

Difficulty:

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Question Stats:

69% (02:08) correct 31% (02:19) wrong based on 342 sessions

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A car averages 40 miles per hour for the first 6 hours of a trip and averages 60 miles per hour for each additional hour of travel time. If the average speed for the entire trip is 55 miles per hour, how many hours long is the trip?

A. 8
B. 12
C. 16
D. 18
E. 24

Originally posted by Summer3 on 14 Mar 2007, 02:05.
Last edited by Bunuel on 23 Jun 2014, 13:19, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 05 Jul 2018, 09:27
1
Summer3 wrote:
A car averages 40 miles per hour for the first 6 hours of a trip and averages 60 miles per hour for each additional hour of travel time. If the average speed for the entire trip is 55 miles per hour, how many hours long is the trip?

A. 8
B. 12
C. 16
D. 18
E. 24


This is a MIXTURE problem.
Two speeds (40mph and 60mph) are combined to form a mixture with an average speed of 55mph.
To determine how much time must be spent at each speed, we can use ALLIGATION.
Let S = the slower speed and F = the faster speed.

Step 1: Plot the 3 speeds on a number line, with F and S on the ends and the speed for the mixture (55mph) in the middle.
S 40-----------55----------60 F

Step 2: Calculate the distances between the values on the number line.
S 40----15----55----5-----60 F

Step 3: Determine the ratio of the two given speeds.
The ratio of S to F is equal to the RECIPROCAL of the distances in red.
S:F = 5:15 = 1:3

The resulting ratio implies the following:
Of every 4 hours of travel, 1 hour must be traveled at the slower speed (40mph), and 3 hours must be traveled at the faster speed (60mph).
Thus, the 6 hours traveled at 40mph must be equal to \(\frac{1}{4}\) of the total travel time:
\(6 = \frac{1}{4}t\)
\(24 = t\)



The reasoning behind the ratio of the two speeds for every 4 hours of travel time:
Since the average speed (55) is closer to the faster speed (60) than to the slower speed (40), more time must be spent at the faster speed (3 hours) than at the slower speed (1 hour).
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 14 Mar 2007, 02:12
(40*6+60*x)/x+6=55
240+60x=55x+330
5x=90
x=18

6+18=24
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 14 Mar 2007, 02:22
Oh....I don't know how I could not see it!

Somehow I wrote it 240+60x=55+330 :oops: and got it all wrong.....
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 14 Mar 2007, 02:24
I realized now my mistakes are limited to some silly stuff, solving the Qs correctly but being careless to write all the equations properly.....I must be very vary careful :)
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 23 Jun 2014, 13:00
1
Total distance = (40 * 6 + 60 + t)
Total time = (6 + t)
Solving Average speed = Total distance/Total time, for t,
t = 18
Total time = 18 + 6 = 24
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 23 Jun 2014, 15:41
total distance = rate * time

if total time taken is x, this equation can be written as
40(x-6) + 60(x) = 55x

48+12x-72 = 11x

x = 24

Ans: E
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 23 Jun 2014, 19:50
1
Refer diagram below:

Setting up the equation to find "t"

240 + 60(t-6) = 55t

t = 24

Answer = E
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 23 Jun 2014, 22:47
40x6 + 60xt = 55(6 + t)
t = 18
Total time = 18 + 6 = 24
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 07 Nov 2015, 11:04
for the first part, the car travels 240 miles
then the car makes 60miles per hour, and drives for X hours.

the average speed is 55 miles/h we need to find the total number of hours.

we can establish the equations:
distance driven after the first 240 miles is Rate*Time. Rate=60mph time = xh or 60x miles
total time traveled is 6h + x h
average velocity for the entire trip - 55 mph.
rewrite everything as:
240 + 60x = 55(6+x) or
240+60x = 330 + 55x - substract 55 x and 240
5x = 90
x = 18
this is not the final answer. D is a trap!!!
total time for the trip is 18+6 = 24.
thus, 24 hours. E.
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 30 Jun 2018, 07:44
Another way to solve it can be a way how we sold water solution problems:
if time were equal for both part of the trip, then average would be = 40 + 60 / 2 = 50 mph
Average speed was higher (55 mph)
That means that car traveled more time with higher speed during second part of the trip
(60 - 55) / (55 - 40) = 5 / 15 = 1/3
so ratio of time traveled during first part to second part was 1:3
If duration of first part is 6 hours, than duration of second part should be 18 hrs
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 04 Jul 2018, 18:28
Summer3 wrote:
A car averages 40 miles per hour for the first 6 hours of a trip and averages 60 miles per hour for each additional hour of travel time. If the average speed for the entire trip is 55 miles per hour, how many hours long is the trip?

A. 8
B. 12
C. 16
D. 18
E. 24


We can let the additional number of hours = t and create the following equation using the average rate formula:

Average rate = total distance/total time

55 = (40(6) + 60t)/(6 + t)

55(6 + t) = 240 + 60t

330 + 55t = 240 + 60t

90 = 5t

18 = t

Thus the total number of hours for the trip is 6 + 18 = 24.

Answer: E
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Re: A car averages 40 miles per hour for the first 6 hours of a  [#permalink]

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New post 05 Jul 2018, 08:24
alligation method :
[40] [60]
[55]
=> 5 : 15
=> 1 : 3
=> 6 : 18 ( 40 km speed is for 6 hrs ,hence 60 km speed is for 18 hrs)

=> Total 24 hrs
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Re: A car averages 40 miles per hour for the first 6 hours of a &nbs [#permalink] 05 Jul 2018, 08:24
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