SOLUTION: A teacher is buying supplies for the school. Using the tax-exempt status, she buys highlighters for 50 cents each, three-ring binders for $3 each, and hole punchers for $10 each.

Question 282766: A teacher is buying supplies for the school. Using the tax-exempt status, she buys highlighters for 50 cents each, three-ring binders for $3 each, and hole punchers for $10 each. She buys a total of 100 items and spent exactly $100. If she bought at least one of each item, the number of hole punchers she bought was:
(A)1 (B) 2 (C) 3 (D) 4 (E) 5

Found 2 solutions by Edwin McCravy, Theo:
Answer by Edwin McCravy(20055) (Show Source): You can put this solution on YOUR website!

Let x = the number of highlighters Let y = the number of three-ring binders Let z = the number of hole punchers We have this system Multiply the second equation by 10 to clear the decimal, and drop the dollar signs: Let's eliminate x: Multiply the first equation by -5 Add the first equation to the second: Divide the second equation through by 5 Solve for y: z must be a multiple of 5, since the fraction must equal to an integer: Also since Since z is a multiple of 5 and also the only possibility is that , choice (E). Substituting back we find that the complete solution is 94 highlighters, 1 three-ring binder, and 5 hole punchers Edwin

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
Your answer is that the number of hole punchers is 5.

You can't find it directly because you have 2 equations in 3 unknowns which means that one of the unknowns can only be found in relation to one of the other unknowns.

To find your solution, I did the following:

The two equations you have to solve simultaneously are:

x + y + z = 100
.5x + 3y + 10z = 100

Multiply the second equation by 2 to get:

x + y + z = 100
x + 6y + 20z = 200

Subtract first equation from second equation to get:

5y + 19z = 100

Solve y in terms of z to get:

y = (100 - 19z)/5

This is the same as:

y = 100/5 - 19z/5

y will not be an integer unless z = 5 or a multiple of 5 which means that the smallest z can be is 5.

Sounds like she could buy either 5 or 10 hole punchers and y would be an integer.

If z were 10, however, then she could not have bought any x or y because 10 * 10 = 100 which means zero money for x and y.

z would have to be equal to 5.

Here's how the equations work out.

x + y + 5 = 100
.5x + 3y + 50 = 100

Subtract 5 from both sides of the first equation and subtract 50 from both sides of the second equation to get:

x + y = 95
.5x + 3y = 50

Multiply the second equation by 2 to get

x + y = 95
x + 6y = 100

Subtract the first equation from the second equation to get:

5y = 5

Divide both sides of the equation by 5 to get:

y = 1

Use the first equation to solve for x to get:

x = 94

Your values for x,y,z are 94,1,5.

Plug them into your original equations to get:

x + y + z = 100 becomes 94 + 1 + 5 = 100 which is true.
.5x + 3y + 10z = 100 becomes .5*94 + 3*1 + 10*5 = 47 + 3 + 50 = 100 which is true.

Both original equations are true when you use the values for x,y,z of 94,1,5, so those values are good.

Your answer is:

She bought 5 hole punchers.

that would be selection E.

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