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28 votes
28 votes
2. A car leaves Cleveland for Albuquerque and travels at 55 mph. Simultaneously, a sec-ond car leaves Albuquerque for Cleveland and travels at 50 mph. If the total distanceis 1,575 miles, how much time will go by before the two cars meet?Answer3. If the Cleveland to Albuquerque car (from problem 2) left at 9 A.M., how far fromAlbuquerque would it be at 6 P.M.?Answer

User James Westgate
by
2.6k points

1 Answer

15 votes
15 votes
Answer:

2) 15 hours

3) 1080 miles

Explanations:

Let the distance traveled by the first car be x

Let the distance traveled by the second car be y

The total distance = 1575 miles

x + y = 1575..........................(1)

Speed of the first car = 55 mph

Speed of the second car = 50 mph

Time = Distance/speed

Time spent by the first car = x/55

Time spent by the second car = y/50

Both cars meet after spending the same time on the road

x/55 = y/50

50x = 55y....................(2)

From equation (1)

x = 1575 - y..........(3)

Substitute equation (3) into equation (2)

50(1575 - y) = 55y

78750 - 50y = 55y

50y + 55y = 78750

105y = 78750

y = 78750/105

y = 750

Time spent = y/50

Time spent = 750/50

Time spent = 15 hours

The two cars meet after 15 hours

3)

There are 9 hours from 9AM to 6 PM

Time spent by the first car = x/55

9 = x/55

x = 9(55)

x = 495 miles

Since the total distance is 1575

The distance left for the car to reach Albuquerque = 1575 - 495

The distance left for the car to reach Albuquerque = 1080 miles

If the Cleveland to Albuquerque car (from problem 2) left at 9 A.M, it will be 1080 miles far from Albuquerque at 6 PM

User Suffii
by
3.6k points
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