Answer:
450 km
Explanation:
Equations
We can define 3 variables: a, b, d. Let "a" and "b" represent the speeds of the cars leaving cities A and B, respectively. Let "d" represent the distance between the two cities. We can write three equations in these three variables:
1. The relation between "a" and "b":
a = b -10 . . . . . . . the speed of car A is 10 kph less than that of car B
2. The relation between speed and distance when the cars leave at the same time:
d = (a +b)·5 . . . . . . distance = speed × time
3. Note that the time it takes car B to travel 150 km to the meeting point is (150/b). (time = distance/speed) The total distance covered is ...
distance covered by car A in 4 1/2 hours + distance covered by both cars (after car B leaves) = total distance
4.5a + (150/b)(a +b) = d
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Solution
Substituting for d, we have ...
4.5a + 150/b(a +b) = 5(a +b)
4.5ab +150a +150b = 5ab +5b^2 . . . . . . multiply by b, eliminate parentheses
5b^2 +0.5ab -150(a +b) = 0 . . . . . . . . . . subtract the left side
Now, we can substitute for "a" and solve for b.
5b^2 + 0.5b(b-10) -150(b -10 +b) = 0
5.5b^2 -5b -300b +1500 = 0 . . . . . . . . eliminate parentheses
11b^2 -610b +3000 = 0 . . . . . . . . . . . . . multiply by 2
(11b -60)(b -50) = 0 . . . . . . . . . . . . . . . . factor
The solutions to this equation are ...
b = 60/11 = 5 5/11 . . . and . . . b = 50
Since b must be greater than 10, the first solution is extraneous, and the values of the variables are ...
- b = 50
- a = b-10 = 40
- d = 5(a+b) = 5(90) = 450
The distance between A and B is 450 km.
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Check
When the cars leave at the same time, their speed of closure is the sum of their speeds. They will cover 450 km in ...
(450 km)/(40 km/h +50 km/h) = 450/90 h = 5 h
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When car A leaves 4 1/2 hours early, it covers a distance of ...
(4.5 h)(40 km/h) = 180 km
before car B leaves. The distance remaining to be covered is ...
450 km - 180 km = 270 km
When car B leaves, the two cars are closing at (40 +50) km/h = 90 km/h, so will cover that 270 km in ...
(270 km)/(90 km/h) = 3 h
In that time, car B has traveled (3 h)(50 km/h) = 150 km away from city B, as required.