If a freight train travels at an average speed of 30 mph, it will...

Question

If a freight train travels at an average speed of 30 mph, it will arrive at its destination 1.5 hours late. If the train instead travels at an average speed of 45 mph, it will arrive 1 hour early. At what average speed, in miles per hour, should the train travel in order to arrive at its destination exactly on time?
A) 33

B) 35

C) 36

D) 37.5

E) 39

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ScottTargetTestPrep · Accepted Answer

We are given that if a freight train travels at an average speed of 30 mph, it will arrive at its destination 1.5 hours late. We can let T = the time it would take the train to arrive on time, and since distance = rate x time, the distance traveled by the train can be expressed as:

Distance = 30(T + 1.5) = 30T + 45

We are also given that if the train instead travels at an average speed of 45 mph, it will arrive 1 hour early. Thus, the distance traveled by the train can be also expressed as:

Distance = 45(T - 1) = 45T - 45

Since the distance traveled by the train, whether it travels at 30 mph or at 40 mph, is the same, we can now set our two distance values to each other and determine T:

30T + 45 = 45T - 45

15T = 90

T = 6

We can now determine the distance:

Distance = 30(6) + 45 = 180 + 45 = 225

Thus, the rate needed to arrive on time is 225/6 = 37.5 mph.

Answer: D

alados14 · Answer

The distance is the same whether you arrive early or late.

30(t+1.5)=45(t-1)
 2t+3=3t-3
 t=6
So distance is 45(5)=225
And t=6
Avg speed =225/6=37.5

D.

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kapilsingal27 · Answer

30(X+1.5)=D
45(X-1)=D
30X+45=45X-45
15X=90
X=6
D=30(6+1.5)=225
6*Y=225
Y=37.5

Fdambro294 · Answer

Distance is Constant over the 2 scenarios. Speed is Inversely Proportional to Time.

Case 1 Speed = 30 mph

Case 2 Speed = 45 mph

Speed Increases by ——-> (45 - 30) / (30) = 1/2 Increase by

Concept: when 2 Quantities are Inversely Proportional, if 1 Quantity Increases by (n / d)
———> then the OTHER Quantity Decreases by (n)/(d + n)

Speed Increased by 1/2

Time will DECREASE by 1/3

Time taken in Case 1: Normal Time (T) + 1.5

Time taken in Case 2: T - 1

Actual Time has Decreased from Case 1 to Case 2 by -2.5 hours

This corresponds to a -(1/3) Decrease in Time

Let Time taken in Case 1 = X hours

X - 2.5 = X - (1/3)X

X = 7.5 hours = Time taken in Case 1

Constant Distance = (30 mph) * (7.5 hr) = 225 miles

In case 1, he is + 1.5 hours late.

Thus the Usual Time to get to the destination on Time =

T + 1.5 = 7.5

T = 6 hours

To travel 225 miles and get there on time in 6 hours, the Speed Required is:

Speed = (225 miles) / (6 hr) = 37.5 mph

-D-
37.5

Posted from my mobile device

Foreheadson · Answer

30*(T+1.5)=45*(T-1) => t=6

So Distance is 225 (plug in any of the above)

so normally train needs to move at 225/6=37.5 mph. answer D.

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