Terry bought some gum and some candy. The number of packs of chewing gum was one more than the number of mints. The number of mints was three times the number of chocolate bars. If gum cost 6 cents a pack, mints cost 3 cents each, and chocolate bars cost 10 cents each, how many of each confection did he get for 80 cents?
Number of mints=x
Number of packs of chewing gum=x+1
Number of chocolate bars=y
The number of mints was three times the number of chocolate bars; then
x=3y
We can suggest this system of equations:
x=3y
6(x+1)+3x+10y=80
we can solve this system of equations by substitution method.
6(3y+1)+3(3y)+10y=80
18y+6+9y+10y=80
18y+9y+10y=80-6
37y=74
y=74/37
y=2
x=3y
x=3*(2)=6
Number of mints=x=6
number of packs of chewing gum=x+1=6+1=7
number of chocolate bars=y=2
Hope this Helped
Very weird!! I hope you're correct!
Let's assume the number of chocolate bars is x.
According to the given information, the number of mints is 3 times the number of chocolate bars, so the number of mints is 3x.
The number of packs of chewing gum is 1 more than the number of mints, so the number of packs of chewing gum is 3x + 1.
Now, let's calculate the total cost of the confections Terry bought:
The cost of x chocolate bars is 10x cents.
The cost of 3x mints is 3 * 3x = 9x cents.
The cost of (3x + 1) packs of chewing gum is (3x + 1) * 6 = 18x + 6 cents.
The total cost of all the confections is the sum of the costs of the individual confections:
10x + 9x + 18x + 6 = 80
Combining like terms:
37x + 6 = 80
Subtracting 6 from both sides:
37x = 74
Dividing both sides by 37:
x = 2
So, Terry bought 2 chocolate bars.
The number of mints is 3x = 3 * 2 = 6.
The number of packs of chewing gum is 3x + 1 = 3 * 2 + 1 = 7.
Therefore, Terry bought 2 chocolate bars, 6 mints, and 7 packs of chewing gum.
To solve this problem, let's use algebra to represent the number of each confection that Terry bought.
Let's assume:
G = number of packs of chewing gum
M = number of mints
C = number of chocolate bars
Based on the information given in the problem, we can create the following equations:
1) "The number of packs of chewing gum was one more than the number of mints."
G = M + 1
2) "The number of mints was three times the number of chocolate bars."
M = 3C
Now, let's calculate the cost of each confection:
Cost of gum = 6 cents per pack
Cost of mints = 3 cents each
Cost of chocolate bars = 10 cents each
The total cost of gum would be: 6G cents
The total cost of mints would be: 3M cents
The total cost of chocolate bars would be: 10C cents
Lastly, we know that the total cost of all the items is 80 cents:
Total cost = 6G + 3M + 10C = 80
Now we have a system of equations:
G = M + 1 -- Equation 1
M = 3C -- Equation 2
6G + 3M + 10C = 80 -- Equation 3
To find the values of G, M, and C, we need to solve this system of equations using substitution or elimination.
First, let's substitute Equation 2 into Equation 1:
G = (3C) + 1
Now substitute the expression for G into Equation 3:
6((3C) + 1) + 3M + 10C = 80
Expand and simplify the equation:
18C + 6 + 3M + 10C = 80
28C + 3M = 74 -- Equation 4
Now let's substitute the expression for M in terms of C from Equation 2 into Equation 4:
28C + 3(3C) = 74
Expand and simplify the equation:
28C + 9C = 74
37C = 74
C = 2
Now that we have the value of C, we can substitute it back into Equation 2 to find M:
M = 3(2)
M = 6
Finally, substitute C = 2 into Equation 1 to find G:
G = (6) + 1
G = 7
Therefore, Terry bought 7 packs of gum, 6 mints, and 2 chocolate bars for a total of 80 cents.