Candice makes gift boxes during her free time.She charges 40 cents for a triangular box,50 cents for a heart shaped box and a cylindrical box for 60 cents.Last Sunday,Candice sold 16 different types of boxes and collected a total of 8 dollars.How many of each type of box did she sell.
I think you mean she sold 16 boxes, since she only had three different types.
t+h+c = 16
40t+50h+60c = 800
That's not enough to completely specify a solution, but it will get you to
t = c
h = 2(8-c)
So, clearly c<=8. If we start listing the possibilities, we get
c,t h
8 8 0
7 7 2
6 6 4
5 5 6
4 4 8
3 3 10
2 2 12
1 1 14
They all sell for a total of $8.00
Well, it sounds like Candice had quite a box office hit on Sunday! Let's figure out the breakdown of her sales.
Let's say Candice sold x triangular boxes, y heart-shaped boxes, and z cylindrical boxes.
From the given information, we can create the following equations:
0.40x + 0.50y + 0.60z = 8 (equation 1)
x + y + z = 16 (equation 2)
Now, let's put on our math noses and solve this little puzzle!
If we multiply equation 2 by 0.40, we can subtract it from equation 1 to find:
0.40x + 0.50y + 0.60z - 0.40x - 0.40y - 0.40z = 8 - (0.40)(16)
Simplifying, we get:
0.10y + 0.20z = 2.4 (equation 3)
Good. Now we can tackle another equation.
If we multiply equation 2 by 0.60, we can subtract it from equation 1 to find:
0.40x + 0.50y + 0.60z - 0.60x - 0.60y - 0.60z = 8 - (0.60)(16)
Simplifying, we get:
-0.20x - 0.10y = -0.8 (equation 4)
Now we have a system of two equations:
0.10y + 0.20z = 2.4 (equation 3)
-0.20x - 0.10y = -0.8 (equation 4)
Using our fantastic clown-calculator (not a real thing), we solve these equations to find:
x = 6
y = 4
z = 6
So, Candice sold 6 triangular boxes, 4 heart-shaped boxes, and 6 cylindrical boxes. I hope her customers found them to be "box-tacular"!
Let's assume that Candice sold x triangular boxes, y heart-shaped boxes, and z cylindrical boxes.
1. Triangular boxes cost 40 cents each, so the total cost of triangular boxes sold is 40x cents.
2. Heart-shaped boxes cost 50 cents each, so the total cost of heart-shaped boxes sold is 50y cents.
3. Cylindrical boxes cost 60 cents each, so the total cost of cylindrical boxes sold is 60z cents.
According to the given information, the total amount collected from the sale of all these boxes is 800 cents, or $8.
Therefore, we can write the equation:
40x + 50y + 60z = 800
Now, we also know that Candice sold a total of 16 boxes:
x + y + z = 16
To solve these equations, we can use substitution or elimination. Let's use elimination method:
Multiply the second equation by 40 to make the coefficient of x in both equations the same:
40x + 40y + 40z = 640 (equation 2)
Subtract equation 2 from equation 1:
(40x + 50y + 60z) - (40x + 40y + 40z) = 800 - 640
10y + 20z = 160
Divide both sides of the equation by 10:
y + 2z = 16 (equation 3)
Now, subtract equation 3 from equation 1:
(40x + 50y + 60z) - (y + 2z) = 800 - 16
40x + 49y + 58z = 784 (equation 4)
Now, we have two equations:
y + 2z = 16 (equation 3)
40x + 49y + 58z = 784 (equation 4)
We can solve these equations simultaneously to find the values of x, y, and z.
To find the number of each type of box sold, we can set up a system of equations based on the given information.
Let's denote the number of triangular boxes sold as T, the number of heart-shaped boxes sold as H, and the number of cylindrical boxes sold as C.
From the given information, we know the following:
1) Candice sold a total of 16 different types of boxes:
T + H + C = 16
2) Candice collected a total of 8 dollars:
0.40T + 0.50H + 0.60C = 8
Now, we need to solve these equations simultaneously.
We can start by multiplying the first equation by 0.40, which will allow us to eliminate the T variable when we subtract the equations:
0.40T + 0.40H + 0.40C = 6.40
Then, subtract the equation from the second equation:
(0.40T + 0.50H + 0.60C) - (0.40T + 0.40H + 0.40C) = 8 - 6.40
0.10H + 0.20C = 1.60
Now, we can simplify this equation by multiplying it by 10 to eliminate decimals:
10(0.10H + 0.20C) = 10(1.60)
1H + 2C = 16
Now, we have a system of equations:
H + 2C = 16
0.10H + 0.20C = 1.60
To get rid of the decimals in the second equation, we can multiply it by 10:
10(0.10H + 0.20C) = 10(1.60)
1H + 2C = 16
Now, we have a system of equations:
H + 2C = 16
1H + 2C = 16
Since both equations are identical, we can see that there are infinitely many solutions for this system. Therefore, we cannot determine the exact number of each type of box sold.