SOLUTION: A TEST IS WORTH 150 PTS. THERE ARE 46 3 AND 5 PT QUESTIONS. HOW MANY OF EACH IS ON THE TEST?

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: A TEST IS WORTH 150 PTS. THERE ARE 46 3 AND 5 PT QUESTIONS. HOW MANY OF EACH IS ON THE TEST?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 124214: A TEST IS WORTH 150 PTS. THERE ARE 46 3 AND 5 PT QUESTIONS. HOW MANY OF EACH IS ON THE TEST?
Answer by edjones(8007) About Me  (Show Source):
You can put this solution on YOUR website!
Let t be the number of 3 pt queations.
Let f be the number of 5 pt queations.
.
t+f=46
3t+5f=150
.
40 three point questions and 6 five pointers.
Ed
.
Solved by pluggable solver: Linear Systems by Addition
We'll solve the system:
1%2At+%2B+1%2Af+=+46
3%2At+%2B+5%2Af+=+150
by elimination by addition.To eliminate by addition, we need to set both coefficients of x to numbers with changed signs, i.e a and -a. Since in the second equation we have 3 as our coefficient for x, to get -1 we have to multiply all terms of the second equation by -1%2F3 which is equal to -0.333333333333333.

Multiplying, we get on our second equation:%283%2A-0.333333333333333%29t+%2B+%285%2A-0.333333333333333%29f+=+150%2A-0.333333333333333
-1%2At+%2B+-1.66666666666667%2Af+=+-50

Adding both equations we get:

%281%2B-1%29t+%2B+%281%2B-1.66666666666667%29f+=+%2846%2B-50%29

Since 1 and -1 cancel out, we have a linear equation:Therefore, we know that f = 6.

Plugging that in into the first equation gives us:

1%2At+%2B+1%2Af+=+46
1%2At+%2B+1%2A6+=+46
1%2At+%2B+6+=+46
1%2At+=+46+-+6
t+=+%2846+-+6%29%2F1
t+=+40%2F1
t+=+40

Therefore, our answer is:

system%28+t=40%2C+f=6+%29