You need to set up a system of equations, which means we need two different equations. (You were on the right track by modeling the situation with equations, though.)
Let a = the number of 20-cent stamps.
Let b = the number of 33-cent stamps.
a + b = 75
.20a + .33b = 19.95
We can use one of two methods to solve this system: substitution or elimination. Let's use substitution for this.
From a + b = 75, we know that a = 75 - b. Plug that into our other equation.
.20a + .33b = 19.95
.20(75 - b) + .33b = 19.95
15 - .20b + .33b = 19.95
15 + .13b = 19.95
.13b = 4.95
b = 4.95/.13 = 38.07
That's the same issue you had... let's just round b to 38, and see what we come up with for a.
a + b = 75
a + 38 = 75
a = 37
Now let's check with the other equation.
.20a + .33b = 19.95
.20(37) + .33(38) = 19.95
7.40 + 12.54 = 19.95
19.94 = 19.95
(Actually, it doesn't... but I guess we just have to say it does.)
He sold 38 33-cent stamps and 37 20-cent stamps.
The question: A postal cler sold 75 stamps for $19.95. Some were 20-cent and some were 33-cent stamps. How many of each kind did he sell?
What I did so far:
1995=33n+20(75-n)
1995=33n+1500-20n
495=13n
but 13 doesn't go into 495 evenly. What did I do wrong?
2 answers
The question: A postal cler sold 75 stamps for $19.95. Some were 20-cent and some were 33-cent stamps. How many of each kind did he sell?
Lets explore this ignoring the fact that there are supposed to be 75 stamps.
1--.20a + .33b = 19.95 or 20a + 33b = 1995
2--Dividing thrugh by the lowest coefficient yields 1a + 1b + 13b/20 = 99 + 15/20
3--(13b - 15)/20 must be an integer.
4--We want the coefficient of b to be 1
5--Multiplying the numerator by 13 yields (221b - 255)/20
6--Dividing by 20 again yields 11b + b/20 - 12 - 15/20
7--Now, (b - 15)/20 = k making b = 20k + 15
8--Substituting back into (1) yields a = 77 - 33k
9--k.....0......1......2......3
...a....75.....42......9......-
...b....15.....35.....55.....75
Sum.....90.....77.....64.....75
10--Conclusion: No mixture of 75 20 cent and 33 cent stamps can total $19.95 evenly. 77 stamps is the closet total that will sum to $19.95 evenly.
Lets explore this ignoring the fact that there are supposed to be 75 stamps.
1--.20a + .33b = 19.95 or 20a + 33b = 1995
2--Dividing thrugh by the lowest coefficient yields 1a + 1b + 13b/20 = 99 + 15/20
3--(13b - 15)/20 must be an integer.
4--We want the coefficient of b to be 1
5--Multiplying the numerator by 13 yields (221b - 255)/20
6--Dividing by 20 again yields 11b + b/20 - 12 - 15/20
7--Now, (b - 15)/20 = k making b = 20k + 15
8--Substituting back into (1) yields a = 77 - 33k
9--k.....0......1......2......3
...a....75.....42......9......-
...b....15.....35.....55.....75
Sum.....90.....77.....64.....75
10--Conclusion: No mixture of 75 20 cent and 33 cent stamps can total $19.95 evenly. 77 stamps is the closet total that will sum to $19.95 evenly.
1 answer