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:

do you have a copy of the table?

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.