SOLUTION: In one month a store sold pads of paper, some for $14, some for $5 and some for 50¢. A total of 54 pads were sold for a total of $54. Evaluate the total number of the 50¢ pads so

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: In one month a store sold pads of paper, some for $14, some for $5 and some for 50¢. A total of 54 pads were sold for a total of $54. Evaluate the total number of the 50¢ pads so      Log On

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Question 1209265: In one month a store sold pads of paper, some for $14, some for $5 and some for 50¢. A total of 54 pads were sold for a total of $54. Evaluate the total number of the 50¢ pads sold minus the number of $5 pads sold.

Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

There's probably a much more efficient way to do this, but here's my approach.

x = number of $14 pads
y = number of $5 pads
z = number of 50 cent pads
The goal is to evaluate z-y

54 pads were sold
x+y+z = 54
which can be rearranged to
z = 54-x-y

14x = money from just the $14 pads
5y = money from just the $5 pads
0.5z = money from just the 50 cent pads
These must total to the $54 mentioned.

14x+5y+0.5z = 54
14x+5y+0.5*(54-x-y) = 54
14x+5y+27-0.5x-0.5y = 54
13.5x+4.5y+27 = 54
13.5x+4.5y = 54-27
13.5x+4.5y = 27
135x+45y = 270
45y = -135x+270
y = (-135x+270)/45
y = (-135x)/45+270/45
y = -3x+6

If x and y are positive integers, then the only possibility is when x = 1
If x = 2 or larger, then y = 0 or smaller.
I'll assume that at least one of each item was sold.

If x = 1 then y = -3x+6 = -3*1+6 = 3
and z = 54-x-y = 54-1-3 = 50

x = 1 copy of the $14 pad of paper was sold
y = 3 copies of the $5 pad of paper were sold
z = 50 copies of the 50 cent pad of paper were sold

Check:
x+y+z = 1+3+50 = 54
and
14x+5y+0.50z = 14*1+5*3+0.50*50 = 54
Both conditions are confirmed.

The final answer is therefore z-y = 50-3 = 47

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


x = # of $14 pads
y = # of $5 pads
z = # of 50-cent pads

The total number of pads was 54:
x%2By%2Bz=54

The total cost was $54:
14x%2B5y%2B.5z=54

With the equations in this form, I find it easiest to solve using elimination.

Multiply the second equation by 2 and compare to the first using subtraction:

28x%2B10y%2Bz=108
x%2By%2Bz=54
27x%2B9y=54
3x%2By=6

There are formal methods for finding the complete set of solutions to this equation. However, since the numbers are small and all unknowns have positive integer values, the solution is simple -- the only solution in positive integers is x=1 and y=3, which leads to z=50.

# of $14 pads: 1
# of $5 pads: 3
# of 50-cent pads: 50

ANSWER: 50-3 = 47