Video Transcript
Mark wants to buy 21 candies. Chocolate-flavored candies cost 70 cents each, strawberry-flavored candies cost seven cents each, and Mark will buy some of each. If he does not wish to spend more than seven dollars in total, what is the maximum number of chocolate-flavored candies he can buy?
What do we know here? Mark wants to buy chocolate-flavored candies and strawberry-flavored candies. We can let 𝑐 represent the number of chocolate-flavored candies Mark buys. And 𝑠 represents the number of strawberry-flavored candies. We know that 𝑐 plus 𝑠 must equal 21, since Mark wants to buy 21 candies. We know that the chocolate candies cost 70 cents, and the strawberry candies cost seven cents. We need to be careful here because the cost of each of these candies is given in cents. But the maximum Mark can spend is given in dollars, seven dollars. But we can turn that seven dollars into 700 cents. Or, if you prefer to leave it in dollars, you would need to convert the cost in cents to dollars. And it would look like this, 0.70 times 𝑐 plus 0.07 times 𝑠 is less than or equal to seven.
In this case, I’m gonna use the cents because that eliminates the need of using a decimal for solving. We’ll be using the equation 70𝑐 plus seven 𝑠 is less than or equal to 700. We’re using the inequality less than or equal to because Mark can spend less than seven dollars, but he can’t spend more. The money that Mark spends on chocolate candies plus the money he spends on strawberry candies should be less than or equal to 700 cents. We also have this sentence that says Mark will buy some of each. This means for both the chocolate candy and the strawberry candy, the minimum amount he would buy is one. So we say that 𝑐 must be greater than or equal to one, and 𝑠 must be greater than or equal to one.
As we get ready to solve for our variables, let’s label our equations equation one and equation two. In equation two, I see that 70, seven, and 700 all have a factor of seven. We can simplify this equation by dividing through by seven. When we do that, we get a simplified form of equation two that says 10𝑐 plus 𝑠 must be less than or equal to 100. At this point, we’ll want to rearrange equation one. If we’re solving for 𝑐, the amount of chocolate-flavored candies Mark wants to buy, then we want to rewrite equation one to write about 𝑠 in terms of 𝑐. To do that, we will subtract 𝑐 from both sides of equation one.
This tells us that 𝑠 equals 21 minus 𝑐. The number of strawberry candies he buys will be equal to 21 minus the number of chocolate candies he buys. We then substitute 21 minus 𝑐 in for 𝑠 in our second equation. We know that 𝑠 is equal to 21 minus 𝑐. So 10𝑐 plus 𝑠 is the same thing as 10𝑐 plus 21 minus 𝑐. From here, we see that we have like terms, 10𝑐 minus 𝑐 equals nine 𝑐. Nine 𝑐 plus 21 must be less than or equal to 100. From there, we subtract 21 from both sides. We have nine 𝑐 on the left to calculate 100 minus 21. I think 100 minus 20 equals 80, and 80 minus one equals 79.
So we know that nine 𝑐 must be less than or equal to 79. If nine 𝑐 must be less than or equal to 79, we find 𝑐 by dividing both sides by nine. 𝑐 must be less than or equal to 79 over nine. And how do we simplify that? I know that nine times eight equals 72, and I know that 72 plus seven equals 79. And that means we can break up this fraction into two pieces, 72 over nine plus seven over nine. But 72 over nine equals eight. And so, we say that 𝑐 must be less than or equal to eight and seven-ninths. But we can’t buy seven-ninths of a chocolate candy. And that means the maximum number of chocolate-flavored candies that he can buy will be the largest whole number that is less than eight and seven-ninths, which is eight.
If you wanted to check whether or not this is true, we would first need to find the number of strawberry candies Mark would buy. If he bought eight chocolate candies, the number of strawberry candies will be 21 minus eight. If Mark buys eight chocolate candies, he buys 13 strawberry candies. And then, we plug that in to the equation to see how much money he would spend. 70 times eight plus seven times 13 must be less than or equal to 700. 70 times eight equals 560. Seven times 13 equals 81. Is 560 plus 81 less than or equal to 700? It is, 641 is less than or equal to 700. Which means eight is a valid solution.
