The yearbook club is having a bake sale to raise money for the senior class. Large cupcakes are sold for $1.25 each and small cupcakes are sold for $0.75 each. If 105 cupcakes were sold for a total amount of $109.75, how many large cupcakes did the yearbook club sell? A) 43 B) 55 C) 62 D) 16
So I know the answer is 62, but isn't this a question that can only be answered because we have the 4 multiple choice answers to test? Can this be figured out if we didn't have the answers?
This can be done with simple algebra, but I don't know if you have studied that already. If so ...
let the number of large cupcakes be x
let the number of small ones be y
x+y = 105 -----> y = 105-x
cost equation:
1.25x + .75y = 109.75
or, multiply each term by 100
125x + 75y = 10975
divide each term by 25
5x + 3y = 439
sub in the above value for y
5x + 3(105-x) = 439
5x + 315 - 3x = 439
2x = 124
x = 62 , which you had indicated was the correct answer.
Yes, you can solve this problem even if you don't have the multiple-choice answers given to you. To do so, let's work through the problem step by step.
Let's assume the number of large cupcakes sold is "x" and the number of small cupcakes sold is "y". We know that the total number of cupcakes sold is 105, so we can write the equation: x + y = 105.
We also know the total amount of money raised from the sale of cupcakes is $109.75. Since each large cupcake costs $1.25 and each small cupcake costs $0.75, we can write another equation to represent the total amount of money raised: 1.25x + 0.75y = 109.75.
Now, we have a system of two equations:
(1) x + y = 105
(2) 1.25x + 0.75y = 109.75
To solve this system, we can use substitution or elimination. Let's use substitution here. From equation (1), we have y = 105 - x. We can substitute this value of y into equation (2):
1.25x + 0.75(105 - x) = 109.75
Simplifying and solving for x:
1.25x + 78.75 - 0.75x = 109.75
0.50x = 31
x = 62
Therefore, the yearbook club sold 62 large cupcakes.
In this case, the answer choices confirm the solution we found through solving the equations. However, even without the answer choices, we could solve the problem by using algebraic techniques.