Two automobiles left simultaneously from cities a and b heading towards each other and met in 5 hours. the speed of the automobile that left city a was 10 km/hour less than the speed of the other automobile. if the first automobile had left city a 4 1/2 hours earlier than the other automobile left city b, then the two would have met 150 km away from b. find the distance between a and b.

  450 km

Step-by-step 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 ...

The distance between A and B is 450 km.

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.

step-by-step explanation:

hey there!

if she's saving for six months, let's divide the total number we need for the trip by six, dividing it into the amount that needs to be saved for each month.

500 ÷ 6 =

she needs to save at least this amount each month, so we should use this symbol: ≥

let x be the amount she saves each month.

your final answer is x ≥

hope this !