Question

A pet store sells cats for $50 and dogs for $100. lf one day they sell a total of 4 pets and make $300, how many cats and dogs did they sell?

Answer 1

2 cats and 2 dogs

Explanation:

There are at least 2 methods of solving this question type.

`color(blue)("Method 1 - Simultaneous equations")`

`color(magenta)("You will be expected to use this method")`

Let count of dogs be `d`
Let count of cats be `c`

Total pets sold is 4 so `" "d+c=4` ............................Equation(1)

Sale value is: `" "$100d+$50c=$300`.......................Equation(2)

From Eqn(1) `d=4-c`.................................Equation(3)

`color(white)(.)`

Using Eqn(3) substitute for `d` in Eqn(2) giving:

`100(4-c)+50c=300`

`400-100c+50c=300`

`400-50c=300`

`400-300=50c`

`c=100/50 = 2`

So 2 cats and 2 dogs sold
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Method 2 - by graph")`

There is always 4 animals. So if you are counting just the dogs then the count of cats is inferred as 'cat count' = 4 - 'dog count'. So if we plot a graph of just dogs sold we can relate this to total sale price.

Remember that the 'y-axis' is counting in units of $100. So a count of 4 represents $400.

You can either do the calculation to demonstrate that you are able to or you can simply read off the number of dogs from the graph. Thus determine the number of cats by 'cat count' = 4 - 'dog count'

`color(brown)("Calculation")`

The slope of all the graph is the same as the slope of part of the graph

Slope all the graph:

`("Change in total cost")/("change in dog count") = (400-200)/(4-0) = color(red)(200/4)`

So for calculation purposes the gradient is `2/4xx100 = 1/2xx100`

The target total cost is `$300` giving a change of `$300-$200 = color(green)($100)`

`color(red)(200/4)-=(color(green)(100))/d`

Turn the whole thing upside down

`color(red)(4/200)-=d/(color(green)(100))`

Multiply both sides by 100

`d=4xx100/200 -> 4xx 1/2=2`