Michel wants to buy his mother handmade chocolates for Mother’s Day. He has $15.00 to spend. The price of each type of chocolate is given below.
Fudge centres: $0.60 Nut clusters: $0.75 Truffles: $0.80
C. Michel’s final choice costs $14.80. It is a combination of all three types of chocolates.
a) Determine four possible combinations of chocolates he
could have selected.
b) Create and evaluate an algebraic expression to show that
each of your combinations works.
number of Fudge centres --- f
Nut clusters ----- n
Truffles -----t , where f, n, and t are whole numbers
.60f + .75n + .80t = 14.80
times 100
60f + 75n + 80t = 1480
divide by 5
12f + 15n + 16t = 296 <----- my algebraic expression
f = (296 - 15n - 16t)/12
you know that f must be a multiple of 12
and n < 19 , t < 18 , also n has to be even for (296-15n-16t) to stay even.
start guessing:
let n = 2 , t = 1, f = (296-30-16)/12 = 20.8 , not possible
let n = 4, t = 5, f = (296 - 60 - 80)/12 = 13
yeah, we got one!
check:
13(.60) + 4(.75) + 5(.80) = 14.80 , that's nice!
with the above restrictions in mind, "guess" at some other
values of n, t, and f
The line near the middle of the solution saying
"you know that f must be a multiple of 12"
should have said:
you know (296 - 15n - 16t) must be a multiple of 12
To determine four possible combinations of chocolates Michel could have selected, we need to find the number of each type of chocolate Michel can buy with his $15.00 budget.
Let's denote the number of Fudge centers as 'f', the number of Nut clusters as 'n', and the number of Truffles as 't'.
Given the prices of each type of chocolate:
Fudge centers: $0.60
Nut clusters: $0.75
Truffles: $0.80
We need to find the values of 'f', 'n', and 't' that satisfy the equation:
0.60f + 0.75n + 0.80t = 14.80
To generate four possible combinations, we can start with different values for 'f', 'n', and 't' that satisfy the equation.
Possible Combinations:
1) f = 10, n = 9, t = 5
2) f = 4, n = 15, t = 5
3) f = 0, n = 20, t = 15
4) f = 6, n = 10, t = 10
Now, let's create and evaluate an algebraic expression for each combination to show that they each work:
1) f = 10, n = 9, t = 5
Total cost: 0.60(10) + 0.75(9) + 0.80(5) = 6.00 + 6.75 + 4.00 = 16.75 (more than 14.80, so this combination doesn't work)
2) f = 4, n = 15, t = 5
Total cost: 0.60(4) + 0.75(15) + 0.80(5) = 2.40 + 11.25 + 4.00 = 17.65 (more than 14.80, so this combination doesn't work)
3) f = 0, n = 20, t = 15
Total cost: 0.60(0) + 0.75(20) + 0.80(15) = 0 + 15.00 + 12.00 = 27.00 (more than 14.80, so this combination doesn't work)
4) f = 6, n = 10, t = 10
Total cost: 0.60(6) + 0.75(10) + 0.80(10) = 3.60 + 7.50 + 8.00 = 19.10 (more than 14.80, so this combination doesn't work)
Therefore, none of the four possible combinations satisfy the condition that Michel's final choice costs $14.80.
To determine the possible combinations of chocolates Michel could have selected, we need to find all the combinations that add up to $14.80 using the given prices of each type of chocolate.
a) Possible combinations of chocolates:
1) Fudge centres: 24 pieces (24 x $0.60 = $14.40) + Nut clusters: 1 piece ($0.75) + Truffles: 1 piece ($0.80) = $14.40 + $0.75 + $0.80 = $15.95
2) Fudge centres: 23 pieces (23 x $0.60 = $13.80) + Nut clusters: 2 pieces (2 x $0.75 = $1.50) + Truffles: 2 pieces (2 x $0.80 = $1.60) = $13.80 + $1.50 + $1.60 = $17.90
3) Fudge centres: 22 pieces (22 x $0.60 = $13.20) + Nut clusters: 3 pieces (3 x $0.75 = $2.25) + Truffles: 4 pieces (4 x $0.80 = $3.20) = $13.20 + $2.25 + $3.20 = $18.65
4) Fudge centres: 21 pieces (21 x $0.60 = $12.60) + Nut clusters: 4 pieces (4 x $0.75 = $3.00) + Truffles: 5 pieces (5 x $0.80 = $4.00) = $12.60 + $3.00 + $4.00 = $19.60
b) Algebraic expression to evaluate each combination:
1) Let x represent the number of fudge centres, y represent the number of nut clusters, and z represent the number of truffles. The algebraic expression would be: 0.6x + 0.75y + 0.8z = 14.80
2) For the first combination: 0.6(24) + 0.75(1) + 0.8(1) = 14.40 + 0.75 + 0.80 = 15.95
3) For the second combination: 0.6(23) + 0.75(2) + 0.8(2) = 13.80 + 1.50 + 1.60 = 17.90
4) For the third combination: 0.6(22) + 0.75(3) + 0.8(4) = 13.20 + 2.25 + 3.20 = 18.65
5) For the fourth combination: 0.6(21) + 0.75(4) + 0.8(5) = 12.60 + 3.00 + 4.00 = 19.60
Note: The fourth combination does not fit within the $15.00 budget that Michel has.