A bus travels 200 km at a constant speed. A car travelling 10 km/h faster than the bus, completes the same distance in 1 hour less than the bus

Question

A bus travels 200 km at a constant speed. A car travelling 10 km/h faster than the bus, completes the same distance in 1 hour less than the bus.a)How many hours does the bus take to travel the 200km.b)what is the speed of the car?

Answer 1

There might be more than one way to approach this one. Here is my way:

In a word problem, always write down the information you have, clearly identifying what any variables represent. If you don't, it could lead to confusion later.

Let's let b represent the speed of the bus. Don't use a different variable for the speed of the car. Since we know it is 10 km/hr faster, let   b+10   represent the speed of the car.

 

Let t represent the time it takes the bus to travel 200 km.  Since the car takes 1 hour less, use  t-1  to represent the time it takes the car.

 

Our basic formula for this problem is   distance = rate x time

 

For the bus,  b·t = 200 km

 

For the car,  (b+10)·(t-1) = 200 km    Be sure you're clear on these equations. Again, each equation is that vehicles speed times the amount of time it takes. Each vehicle goes 200 km.

 

If you're sure you understand the two equations, we an go on.

 

Since both expressions are equal to 200, they are equal to each other, so we can write

 

     bt = (b+10)(t-1)         Now multiply on the right side, to remove the parentheses. You probably learned the FOIL method in class.

 

     bt = bt + 10t - 1b - 10        Again, be sure you understand this step before going on. You should do the multiplication yourself to make sure you can do it.

 

There is a   bt   on both sides of the equation, so subtract  bt  from both sides, to simplify things.

 

     0 = 10t - b - 10   (I've written b instead of 1b, since 1b is the same as b)

 

Now we have an equation with two variables in it, so we can't solve it yet. If we go back, we see that bt = 200. If we divide both sides of this by  b , we get   t = 200/b. We can put this expression in place of  t  in our equation.

 

 

     0 = 10(200/b) - b - 10

 

     0 = 2000/b - b - 10

 

The fraction makes the equation very difficult to work with, so let's multiply both sides of the equation by b, remembering that we have to multiply every term in the equation by b.

 

     b·0 = b(2000/b) - b·b - b·10

 

     0 = 2000 - b² -10b

 

Let's multiply both sides of the equation by -1 and rearrange the terms. This gives us

 

     0 = b² + 10b - 2000   which we can solve by factoring.  You should be able to take it from here. Be sure to answer both questions.