Let's denote the cookies, donuts, and chocolates as follows:
- ( c ) = cookie (🍪)
- ( d ) = donut (🍩)
- ( h ) = chocolate (🍫)
From the equations, we can set up the following system:
- ( c + d + c = 22 ) → ( 2c + d = 22 ) (Equation 1)
- ( h + h + c = 28 ) → ( 2h + c = 28 ) (Equation 2)
- ( d + d + d + h + h = 38 ) → ( 3d + 2h = 38 ) (Equation 3)
Now, let's solve this system step by step.
From Equation 1:
[
d = 22 - 2c \quad \text{(Substituting into other equations)}
]
Substituting ( d ) into Equation 3:
[
3(22 - 2c) + 2h = 38
]
[
66 - 6c + 2h = 38
]
[
2h = 38 - 66 + 6c
]
[
2h = 6c - 28 \quad \Rightarrow \quad h = 3c - 14 \quad \text{(Equation 4)}
]
Substituting Equation 4 into Equation 2:
[
2(3c - 14) + c = 28
]
[
6c - 28 + c = 28
]
[
7c - 28 = 28
]
[
7c = 56 \quad \Rightarrow \quad c = 8
]
Now substituting ( c = 8 ) back into Equation 1:
[
2(8) + d = 22
]
[
16 + d = 22 \quad \Rightarrow \quad d = 6
]
Now substitute ( c = 8 ) into Equation 4 to find ( h ):
[
h = 3(8) - 14
]
[
h = 24 - 14 \quad \Rightarrow \quad h = 10
]
Now we have:
- ( c = 8 )
- ( d = 6 )
- ( h = 10 )
Next, we need to calculate ( d d \times c - h h ):
[
d d \times c = 6 \times 6 \times 8 = 36 \times 8 = 288
]
[
h h = 10 \times 10 = 100
]
Now we can find:
[
d d \times c - h h = 288 - 100 = 188
]
Thus, the answer is:
[
\boxed{188}
]
