Question

An exam worth 378 points contains 66 questions. Some questions are worth 7 points, and the others are worth 4 points. How many 7 point and 4 point questions are on the test?

Answer 1

There are 38 7 point questions and 28 4 point questions.

Explanation:

This is a problem that requires two equations since there are two variables.

The first equation is easy it is the total number of questions.

x + y = 66 where x = 7 point questions and y = 4 point questions

The second equation is the total number of points.

7(x) + 4(y) = 378 7x is the points from the 7 point questions.
4y is the points from the 4 point questions.
378 is the total number of points.

To solve the two equations substitute the values from the first equation into the second equation.

x + y = 66 subtract x from both sides

x - x + y = 66 - x which gives

y  =  ( 66 - x )    now substitute the value into the second equation. 
                         in place of y.

7x +4 ( 66-x) = 378 Now there is an equation with only one variable.

Now multiply 4 across the parenthesis using the distributive property. This gives

7x + 264 - 4x = 378 now combine the x values using the associate property this results in

7x - 4x + 264 = 378 Subtract the 4x from the 7x to get

3x + 264 = 378 now subtract 264 from both sides

3x + 264 - 264 = 378 - 264 which results in

3x = 114 divide both sides by 3

`3x/3 = 114/3` the answer is

x = 38 Now to solve for y put x back into the first equation

38 + y = 66 Subtract 38 from both sides.

38 - 38 + y = 66 - 38 which gives

y =  28

So the 7 point questions = 38 and
the 4 point questions = 28

To solve a problem with two variables requires two equations.
Then solve one equation for one of the variables and substitute this value into the second equation.

Answer 2

`38` questions are valued at 7 points each.
`28` questions are valued at 4 points each.

Explanation:

This is a system of linear equations question

Let `x` = `7` point questions
Let `y` = `4` point questions

Total Number of question equation

`x+y=66`

Total Point value equation

`7x+4y=378`

We have several methods that could be used

1)Graphing
2)Elimination
3)Substitution
4)Matrices (I did not include this method yet)

Graphing

Use a graphing tool and input the 2 equations. Look for where the 2 equations intersect. The intersection point (the ordered pair) is the solution to the problem.

Elimination Method:

We need to eliminate one of the variables. Let's eliminate the `y` variable by multiplying the first equation by `-4`.

`(-4)x+(-4)y=(-4)66`

Simplify

`-4x-4y=-264`

Now add this equation with the other equation to eliminate the `y` variable.

`-4x-4y=-264`
`7x+4y=378`

`3x=114`

Now isolate the `x` variable by dividing

`cancel3x/cancel3=114/3`

`x=38`

Now substitute this value for `x` in the original version of the first equation.

`color(red)38+y=66`

Subtract 38 from both sides

`color(red)(-38)+38+y=66color(red)(-38)`

`y=28`

`38` questions are valued at 7 points each.
`28` questions are valued at 4 points each.

Substitution Method:

Take either equation and solve for either variable. We will take the first equation because it is the easiest to work with. We will solve for `y`.

`x+y=66`

Subtract x from both sides

`color(red)(-x+)x+y=66color(red)(-x)`

y=66-x#

Now substitute this into the second equation

`7x+4(color(red)(66-x))=378`

Distribute

`7x+(color(red)(264-4x))=378`

Combine like terms

`3x+264=378`

Subtract 264 from both sides

`3x+264color(red)(-264)=378color(red)(-264)`

Simplify

`3x=114`

Isolate `x` by dividing by `3`

`cancel3x/cancel3=114/3`

`x=38`

Now substitute this value for `x` in the original version of the first equation.

`color(red)38+y=66`

Subtract 38 from both sides

`color(red)(-38)+38+y=66color(red)(-38)`

`y=28`

`38` questions are valued at 7 points each.
`28` questions are valued at 4 points each.

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