SOLUTION: the ratio of pickups to cars sold at a dealership is 2 to 3. If the dealership sold 142 more cars than pickups in 1999, then how many of each did it sell

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: the ratio of pickups to cars sold at a dealership is 2 to 3. If the dealership sold 142 more cars than pickups in 1999, then how many of each did it sell      Log On


   



Question 1201392: the ratio of pickups to cars sold at a dealership is 2 to 3. If the dealership sold 142 more cars than pickups in 1999, then how many of each did it sell
Found 3 solutions by ikleyn, Theo, greenestamps:
Answer by ikleyn(52838) About Me  (Show Source):
You can put this solution on YOUR website!
.

The number of pickups 2x

The number of cars    3x


Equation

    3x - 2x = 142


You do the rest.

Solved.



Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
let x = number of pickups.
let y = number of cars.
you get:
x/y = 2/3.
the dealership sold 142 more cars than pickups.
you get y = x + 142
in the equation of x/y = 2/3, replace y with x + 142 to get:
x/(x+142) = 2/3
cross multiply to get:
3x = 2*(x+142)
simplify to get:
3x = 2x + 284
subtract 2x from both sides of the equation to get:
x = 284.
this tells you that the number of pickups is 284.
the number of cars is 142 more than that = 426.
x/y becomes equal to 284/426.
simplify that fraction to get:
x/y = 2/3, confirming the values of x and y are correct.

Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


A general comment on this kind of problem -- as demonstrated by the two responses you have received to now.

Given a ratio of 2:3, you can let the two numbers be x and y and then solve the problem using two variables and a bunch of operations involving ugly fractions, as one tutor did.

Or you can take the given ratio of 2:3 and let the two numbers be 2x and 3x -- leading to a solution using one variable and very simple operations.

The lesson:

Given a ratio A:B at the beginning of a problem, let the two numbers be Ax and Bx. Virtually always (if not always) that will make solving the problem much easier.