Joline collects hockey cards. She needs 5 cards to complete a set. This is 20 percent of the set. How many cards are in the set?
and how do you figure this out ? please help:(
thank you !!
20% = 0.2
x = total number of cards
0.2x = 5
x = 5/0.2
x = 25
You're welcome.
20% = 0.2
0.2x = 5
5/0.2
= 25
Yes, that is correct!
Well, trying to figure out how many cards are in the set is an interesting problem. But before I give you the answer, let me ask you something: Have you ever heard the saying, "Why did the hockey card go to school? Because it wanted to get a good 'a-card-emic' education!" Ah, sorry, I couldn't resist throwing in some card-themed humor. Now, back to the question at hand.
To find out how many cards are in the set, we can use a simple equation. We know that Joline needs 5 cards to complete the set, and this is 20% of the total set. So, let's represent the total number of cards in the set with the variable "x". We can set up the equation:
20% of x = 5
To solve for "x", we need to isolate it on one side of the equation. We can start by converting 20% to a decimal, which is 0.2. Now, our equation becomes:
0.2x = 5
To solve for "x", we can divide both sides of the equation by 0.2:
x = 5 / 0.2
Calculating that, we find:
x = 25
So, there are 25 cards in the set. But remember, this answer is no joke!
To figure out the total number of cards in the set, we can set up a proportion using the information given.
Let's assume "x" represents the total number of cards in the set.
We are told that Joline needs 5 cards to complete the set, which is 20% of the set. This means that 20% of x is equal to 5.
We can write this information as a proportion:
20/100 = 5/x
To solve this proportion, we can cross-multiply:
20x = 5 * 100
20x = 500
Finally, divide both sides of the equation by 20:
x = 500 / 20
x = 25
So, there are 25 cards in the set.
To figure this out, we used the proportion method where we set up a ratio between the known quantity (5) and the unknown quantity (x). Then, we cross-multiplied and solved for x by dividing both sides of the equation. This method can be applied to solve various types of proportion problems.