SOLUTION: At football games last year, the snack shop took in $1200 in soft drink sales. This year the price per drink was raised 15 cents, 300 fewer drinks were sold, and $1365 was brought

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: At football games last year, the snack shop took in $1200 in soft drink sales. This year the price per drink was raised 15 cents, 300 fewer drinks were sold, and $1365 was brought       Log On


   



Question 1033906: At football games last year, the snack shop took in $1200 in soft drink sales. This year the price per drink was raised 15 cents, 300 fewer drinks were sold, and $1365 was brought in. Find the price of the soft drinks and the numbers sold
Found 2 solutions by algebrapro18, MathTherapy:
Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
What follows is a very lengthy explanation on how to do this problem where I walk you through everything. Here are the steps we will take to solve this equation with out all the detail:

1)Create a system of two equations and two unknowns.
2)Solve one of the equations for an unknown and substituting that into the other equation.
3) Solve the resulting quadratic equation using the quadratic formula. Realizing we can only have one solution because the second solution would be negative and you can't have a negative price or negative number of soda's sold.
4)Substitute the value found in (3) into the first the equation we found in (2) and solve for the remaining variable.


To solve this lets let p be the price per drink and let n be the number of drinks sold.

p = price per drink
n = number of drinks sold

If we take the price per drink times the number of drinks sold we will get the total sales. Converting this into an equation we get:

pn=1200

Now we now that the price per drink went up 15 cents and there were 300 fewer drinks sold. Writing these as equations we get:

p+0.15
n-300

Again if we take the product we get the following equation which we can manipulate a little to simplify it:

(p+0.15)(n-300)=1365 Foil out the left hand side
pn - 300p + 0.15n - 45 = 1365 Add 45 to both sides
pn - 300p + 0.15n = 1410

Now we have the following system of equations:

pn=1200
pn - 300p + 0.15n = 1410

Its best to solve this system using Substitution. So we need to solve for one of our variables in one of our equations. Lets solve the top equation for p.

pn=1200 Divide by n on both sides
p = 1200/n

Now we can substitute that value into the second equation and solve for n.

pn - 300p + 0.15n = 1410 Substitute p = 1200/n into the equation
(1200/n)n-300(1200/n)+0.15n = 1410 Multiply on the left
1200 - 360,000/n + 0.15n = 1410 Multiply both sides by n
1200n - 360,000 + 0.15n^2 = 1410n Subtract 1410n from both sides and write in standard form
0.15n^2 - 210y - 360,000 = 0 Divide both sides by 0.15 to clear the decimal
n^2 - 1400n - 2,400,000 = 0

Here we need to use the quadratic formula which is n+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ to solve for n.

Plugging in a = 1 b = -1400 and c = 2,400,000 you get

n+=+%281400+%2B-+sqrt%28+%28-1400%29%5E2-4%2A1%2A2%2C400%2C000+%29%29%2F%282%2A1%29+ simplify the inside of the square root.
n+=+%28%281400+%2B-+sqrt%2811560000%29%29%2F2%29+ take the square root
n+=+%281400+%2B-+3400%29%2F%282%29+

Now here we need to think about the answers we will get for n. If we take 1400-3400 we will get a negative number. Since n is the number of drinks sold we can't sell a negative number of drinks so we only have to add.

n+=+%281400+%2B+3400%29%2F%282%29+
n = 2400

So we now know that 2400 drinks were sold that first year. We can now substitute this back into our top equation and solve for the price per drink.

np = 1200 Substitute n = 2400 in for n
2400p = 1200 divide by 2400
p = 0.50

So we know that 2400 drinks were sold that first year at 50 cents per drink. The next year the price per drink went up 15 cents. So our new price per drink is 65 cents. And they sold 300 less drinks so they sold 2100 drinks.

Thank you for hanging in there with me I know that was ALOT.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
At football games last year, the snack shop took in $1200 in soft drink sales. This year the price per drink was raised 15 cents, 300 fewer drinks were sold, and $1365 was brought in. Find the price of the soft drinks and the numbers sold
Let price, per drink, last year be P
Then number sold last year = matrix%281%2C1%2C+%221%2C200%22%2FP%29
Increasing price by .15, this year, makes the new price, P + .15
Since 300 fewer were sold this year, then number sold this year = matrix%281%2C1%2C+%221%2C200%22%2FP+-+300%29
We then get:
------- FOILing left side
matrix%281%2C1%2C+%221%2C155%22%29+-+300P+%2B+180%2FP+=+matrix%281%2C1%2C+%221%2C365%22%29
1155P+-+300P%5E2+%2B+180+=+1365P ------- Multiplying by LCD, P
300P%5E2+-+1155P+%2B+1365P+-+180+=+0
300P%5E2+%2B+210P+-+180+=+0
10P%5E2+%2B+7P+-+6+=+0 ------- Factoring out GCF, 30
(2P – 1)(5P + 6) = 0 ------- Factoring the above trinomial
2P – 1 = 0 OR 5P + 6 = 0 (ignore)
2P = 1
P, or price last year = ½, or $0.50
Price this year: $0.50 + .15 = highlight_green%28%22%24%220.65%29
Number sold this year: matrix%281%2C1%2C+%221%2C200%22%2F.5%29+-+300, or 2,400 – 300, or highlight_green%28matrix%281%2C1%2C+%222%2C100%22%29%29