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maks197457 [2]
2 years ago
11

A certain 90-mile trip took 2 hours. Exactly 1/3 of the distance traveled was by rail and this part of the trip took 1/5 of the

travel time. What was the average rate, in miles per hour, of the rail portion of the trip?
A. 71 B. 73 C. 74 D. 75
Mathematics
1 answer:
antiseptic1488 [7]2 years ago
7 0

The average rate, in miles per hour, of the rail portion of the trip is; D: 75

<h3>How to calculate the average speed?</h3>

Since exactly 1/3 of the distance travelled was by rail, then we can say that;

(1/3) * (90) = 30 miles

We are told that part of the trip took 1/5 of the travel time. Thus;

Time it takes to travel that distance is:

(1/5) * (2) = 2/5 of an hour

Average rate is;

30/(2/5) = 150/2 = 75 mph

Read more about Average speed at; brainly.com/question/4931057

#SPJ1

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x+y+z=20\\\\\Rightarrow x=20-y-z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\0.05x+0.10y+0.25z=3.35\\\\\Rightarrow 5x+10y+25z=335\\\\\Rightarrow x+2y+5z=67~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)\\\\0.05y+0.25x+0.10z=2.75\\\\\Rightarrow 5y+25x+10z=275\\\\\Rightarrow y+2z+5x=55~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)

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(20-y-z)+2y+5z=67\\\\\Rightarrow y+4z=47\\\\\Rightarrow y=47-4z~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iv)

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y+2z+5(20-y-z)=55\\\\\Rightarrow -4y-3z=-45\\\\\Rightarrow y=\dfrac{45-3z}{4}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(v)

Comparing the values of y from equations (iv) and (v), we get

47-4z=\dfrac{45-3z}{4}\\\\\Rightarrow 188-16z=45-3z\\\\\Rightarrow 16z-3z=188-45\\\\\Rightarrow 13z=143\\\\\Rightarrow z=\dfrac{143}{13}\\\\\Rightarrow z=11.

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