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15 June, 06:14

Two planes flew between two cities at speeds of 200 mph and 260 mph respectively. If the slower plane required 3/2 hours longer, how far apart are the cities

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  1. 15 June, 07:08
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    The cities are 207 miles far apart

    Step-by-step explanation:

    -To solve this problem, we can use the rule of three. This mathematical rule is used to solve problems with proportions. To use it we must have three numbers. In this case we have: the time (1,5 hours) and two-speed data (200 mph and 260 mph). So, we solve the problem in this way:

    With a speed of 200mph--> the plane took 1,5 to reach the first city (city A).

    With a speed of 260mph<- - How long the other plane took to reach the other city (city B) ?

    We first multiply:

    260mph*1,5 hours = 390mph*hours

    Then, we divide:

    390 mph*hours/200mph = 1,95 hours

    The faster plane took 1,95 to reach the city B.

    -Now, the slower plane traveled 200 miles/hour*1,5 hours = 300 miles

    The faster plane travelled 260 miles/hour*1,95 hours = 507 miles

    The difference between the distance that both planes travelled is the distance between the two cities (if they flew in a straight line).

    The difference is: 507 miles-300 miles = 207 miles

    The cities are 207 miles far apart
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