Bill spent less than $23 on lunch for his co-workers. If burgers cost $5.25 each and sodas cost $2.25 each, which of the following is not a possible combination of burgers and sodas that Bill could have purchased?
60 burgers and a dozen sodas.
To find the combination that is not possible, we should consider the total cost of burgers and sodas.
Let's start by assuming Bill bought x burgers and y sodas.
The cost of x burgers would be 5.25 * x dollars.
The cost of y sodas would be 2.25 * y dollars.
According to the given information, Bill spent less than $23. Therefore, we can set up the following inequality:
5.25 * x + 2.25 * y < 23
Now, let's check the options to find the one that doesn't satisfy this inequality:
A. 3 burgers and 3 sodas:
5.25 * 3 + 2.25 * 3 = 15.75 + 6.75 = 22.5 < 23
B. 5 burgers and 2 sodas:
5.25 * 5 + 2.25 * 2 = 26.25 + 4.5 = 30.75 > 23
C. 2 burgers and 5 sodas:
5.25 * 2 + 2.25 * 5 = 10.5 + 11.25 = 21.75 < 23
D. 4 burgers and 1 soda:
5.25 * 4 + 2.25 * 1 = 21 + 2.25 = 23.25 > 23
Based on the calculations, the combination that is not possible for Bill to have purchased is option B: 5 burgers and 2 sodas.
To find the combinations of burgers and sodas that Bill could have purchased, we need to consider that he spent less than $23. Let's start by finding the maximum number of burgers and sodas he can buy with $23.
First, let's assume he spends all his money on burgers. We divide $23 by the cost of a burger ($5.25) to determine the maximum number of burgers he can buy.
$23 / $5.25 ≈ 4.38
Since we can't buy a fraction of a burger, the maximum number of burgers he can buy is 4. Therefore, the maximum cost of the burgers would be 4 x $5.25 = $21.
Next, let's find the maximum number of sodas he can buy with the remaining amount. Subtracting the cost of the burgers from $23 gives us the remaining amount.
$23 - $21 = $2
Now, we divide $2 by the cost of a soda ($2.25) to determine the maximum number of sodas he can buy.
$2 / $2.25 ≈ 0.89
Again, we can't buy a fraction of a soda, so the maximum number of sodas he can buy is 0. Therefore, the maximum cost of the sodas would be 0 x $2.25 = $0.
From the calculations, we know that he can buy up to 4 burgers and 0 sodas within his budget. Let's consider the given options and check which one of them is impossible.
A. 3 burgers and 5 sodas: This option costs 3 x $5.25 + 5 x $2.25 = $15.75 + $11.25 = $27. This combination exceeds his budget, so it is not possible.
B. 2 burgers and 7 sodas: This option costs 2 x $5.25 + 7 x $2.25 = $10.50 + $15.75 = $26.25. Once again, this combination exceeds his budget, so it is not possible.
C. 4 burgers and 2 sodas: This option costs 4 x $5.25 + 2 x $2.25 = $21 + $4.50 = $25.50. Once again, this combination exceeds his budget, so it is not possible.
D. 1 burger and 9 sodas: This option costs 1 x $5.25 + 9 x $2.25 = $5.25 + $20.25 = $25.50. Similar to the previous options, this combination exceeds his budget, so it is not possible.
Therefore, the combination that is not possible is option B: 2 burgers and 7 sodas.