SOLUTION: Ms. snow decide to get a test worth 90 points and contains 25 questions. multiple-choice questions are worth 3 points and word problems are worth 4 points. How many of each type of

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Ms. snow decide to get a test worth 90 points and contains 25 questions. multiple-choice questions are worth 3 points and word problems are worth 4 points. How many of each type of      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1007322: Ms. snow decide to get a test worth 90 points and contains 25 questions. multiple-choice questions are worth 3 points and word problems are worth 4 points. How many of each type of question are there?
Found 3 solutions by macston, tiffany222, Edwin McCravy:
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
.
M=number of multiple choice; W=word problems
.
M+W=25
M=25-W
.
3M+4W=90
3(25-W)+4W=90
75-3W+4W=90
W=15
ANSWER 1: There were 15 word problems.
.
M=25-W
M=25-15
M=10
ANSWER 2: There were 10 multiple choice problems.
.
CHECK:
3M+4W=90
3(10)+4(15)=90
30+60=90
90=90

Answer by tiffany222(56) About Me  (Show Source):
You can put this solution on YOUR website!
Let multiple choice questions be x
Let word problems be y
3x + 4y = 90
x + y =25
Multiply the second equation by 4 to eliminate y:
3x + 4y = 90
4x + 4y = 100
Subtract the two equations:
-x = -10
x = 10
Substitute x to get y:
x + y = 25
10 + y = 25
y = 15
Final Answer:
10 multiple choice questions and 15 word problems

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Let the number of multiple choices be x
Let the number of word problems be y


                                 Value        Value
Type               Number          of          of
 of                  of           EACH         ALL
question         questionss     question    questions
-------------------------------------------------------
multiple choices     x             3            3x
word problems        y             4            4y
-------------------------------------------------------
TOTALS              25          -----           90

 The first equation comes from the second column.

  %28matrix%283%2C1%2CNumber%2Cof%2Cmultiple-choices%29%29%22%22%2B%22%22%28matrix%283%2C1%2CNumber%2Cof%2Cword-problems%29%29%22%22=%22%22%28matrix%284%2C1%2Ctotal%2Cnumber%2Cof%2Cquestions%29%29
                 x + y = 25

 The second equation comes from the last column.
  %28matrix%284%2C1%2CValue%2Cof%2CALL%2Cmultiple+choices%29%29%22%22%2B%22%22%28matrix%284%2C1%2CValue%2Cof%2CALL%2Cword+problems%29%29%22%22=%22%22%28matrix%285%2C1%2CTotal%2Cvalue%2Cof%2CALL%2Cquestions%29%29

               3x + 4y = 90

 So we have the system of equations:
           system%28x+%2B+y+=+25%2C3x+%2B+4y+=+90%29.

We solve by substitution.  Solve the first equation for y:

           x + y = 25
               y = 25 - x

Substitute (25 - x) for y in 3x + 4y = 90

     3x + 4(25 - x) = 90
      3x + 100 - 4x = 90
          -1x + 100 = 90
                -1x = -10
                  x = 10 = the number of multiple choices.

Substitute in y = 25 - x
              y = 25 - (10)
              y = 15 = the number of word problems.


Checking:  10 multiple choices is 30 and 15 word problems is 60
            That's 25 questions.
            And indeed 30 + 60 = 90

Edwin