in a class of 30 students, 17 have computers at home and 20 have cassettes. If 4 students in the class have neither a computer or a cassette, how many of the 30 students have both?
If you count all the students with computers, and then count all those with cassettes, you count twice the ones that have both. So, adjust the total accordingly.
30 students. 4 with nothing, leaves 26 with either computer or cassette or both
17+20 = 37, but there are only 26 students available, so there must be 37-26=11 with both. Just adding up the numbers counted 11 of them twice.
To determine how many students have both a computer and a cassette, we need to find the intersection between the two sets.
In this case, we have:
- The set of students with computers at home, denoted as C, has 17 members.
- The set of students with cassettes, denoted as S, has 20 members.
- The set of students with neither a computer nor a cassette, denoted as N, has 4 members.
We can use the principle of inclusion-exclusion to find the number of students who have both a computer and a cassette.
Step 1: Find the total number of students who have either a computer or a cassette:
- The total number of students who have either a computer or a cassette is given by the sum of the sizes of the two sets, minus the size of their intersection:
|C ∪ S| = |C| + |S| - |C ∩ S|
|C ∪ S| = 17 + 20 - |C ∩ S|
Step 2: Find the total number of students who have either a computer or a cassette, including those with neither:
- The total number of students who have either a computer or a cassette, including those with neither, is given by the sum of the sizes of the sets of students with either a computer or a cassette, and those with neither:
|C ∪ S ∪ N| = |C ∪ S| + |N|
|C ∪ S ∪ N| = 17 + 20 - |C ∩ S| + 4
Step 3: Find the number of students who have both a computer and a cassette:
- The number of students who have both a computer and a cassette is given by the difference between the total number of students who have either a computer or a cassette, and those with neither:
|C ∩ S| = |C ∪ S ∪ N| - |N|
|C ∩ S| = (17 + 20 - |C ∩ S| + 4) - 4
|C ∩ S| + |C ∩ S| = 17 + 20
2|C ∩ S| = 37
Step 4: Calculate the number of students who have both a computer and a cassette:
|C ∩ S| = 37 / 2
|C ∩ S| = 18.5
Since we cannot have a fraction of a student, we can conclude that 18 students in the class have both a computer and a cassette.
To find out how many students have both a computer and a cassette, we need to subtract the number of students who have neither from the total number of students in the class.
We are given that there are 30 students in the class, 17 have computers at home, and 20 have cassettes. Let's calculate how many students have neither a computer nor a cassette.
Number of students with neither = Total number of students - Number of students with a computer - Number of students with a cassette
Number of students with neither = 30 - 17 - 20
Number of students with neither = 30 - 37
Number of students with neither = -7
Uh-oh! We have a problem. The number of students with neither a computer nor a cassette is negative, which doesn't make sense. This means that there is an error in the given data. Please check the numbers again and make sure they are accurate.