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28 Jun 2020, 12:53
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A
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Difficulty:
45%
(medium)
Question Stats:
75% (02:39) correct
25%
(02:33)
wrong
based on 527
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A car traveled the first 1/3 of the distance from Town X to Town Y at an average speed of 20 miles per hour, it traveled the next 1/3 of the distance from Town X to Town Y at an average speed of 30 miles per hour, and it traveled the final 1/3 of the distance from Town X to Town Y at an average speed of 36 mile per hour. What was the car's average speed, in miles per hour, when it traveled from Town X to Town Y?
A) 27
B) 28
C) 28 2/3
D) 29
E) 30 1/2
A) 27
B) 28
C) 28 2/3
D) 29
E) 30 1/2
28 Jun 2020, 13:04
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Total Distance is D &
Average speed = Total Distance/ Total Time
T1 for 1st 1/3rd = Speed x Distance
= \(D/(3*20)\)
Similarly T2 =\(D/3*30\)
& T3 = \(D/3*36\)
Therefore Average Speed will be = \(D/[D/(3*20)+D/(3*30)+D/(3*36)]\)
= \(3/(1/20+1/30+1/36)\)
=\((3*180)/20\)
=27
A is the Answer
Average speed = Total Distance/ Total Time
T1 for 1st 1/3rd = Speed x Distance
= \(D/(3*20)\)
Similarly T2 =\(D/3*30\)
& T3 = \(D/3*36\)
Therefore Average Speed will be = \(D/[D/(3*20)+D/(3*30)+D/(3*36)]\)
= \(3/(1/20+1/30+1/36)\)
=\((3*180)/20\)
=27
A is the Answer
28 Jun 2020, 13:26
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coreyander
Average speed = Total distance/total time
Another important aspect for Speed Distance problems: each set of trip needs to have a separate equation.
In this case, we have three set of trips covered with three different speeds. So we will make equation for each trip and calculate time taken for each part of trip as distance and speed is given for each part of the trip.
We can assume the total distance to be D and then calculate the time t1, t2, t3 for each part of trip by using speeds 20 mph, 30 mph, 36 mph.
t1=(D/3)/20
t2=(D/3)/30
t3=(D/3)/36
T=t1+t2+t3
Average speed = D/T
27 mph (by putting the values and calculating)
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