In a class of 120 students, 100 students participate in either Golf or Skating or both. Among them, total 60 students participate in Golf. A total of 56 students participate in Skating. How many students participate only in Skating?
40
This problem involves analyzing data about students participating in two activities: Golf and Skating. We are given the total number of students in the class, the number of students involved in at least one activity (Golf or Skating or both), the total number of students participating in Golf, and the total number of students participating in Skating. Our goal is to find out how many students participate only in Skating.
We can use set theory concepts, often visualized with Venn diagrams, to solve this type of problem. Let G represent the set of students participating in Golf and S represent the set of students participating in Skating.
The principle of inclusion-exclusion for two sets states:
$$|G \cup S| = |G| + |S| - |G \cap S|$$
We can plug in the values we know:
$$100 = 60 + 56 - |G \cap S|$$
Now, we solve for $|G \cap S|$:
$$100 = 116 - |G \cap S|$$
$$|G \cap S| = 116 - 100$$
$$|G \cap S| = 16$$
So, 16 students participate in both Golf and Skating.
Students who participate only in Skating are those who are in the Skating set but not in the Golf set. In set notation, this is $|S \setminus G|$, which can be calculated as the total number of students in Skating minus the number of students in both Golf and Skating:
$$\text{Only Skating} = |S| - |G \cap S|$$
Using the values we have:
$$\text{Only Skating} = 56 - 16$$
$$\text{Only Skating} = 40$$
Therefore, 40 students participate only in Skating.
Let's check our numbers:
The calculation for students participating only in Skating is consistent with all the given data.
| Category | Number of Students |
|---|---|
| Total Students | 120 |
| Participate in Golf or Skating or Both ($|G \cup S|$) | 100 |
| Participate in Golf ($|G|$) | 60 |
| Participate in Skating ($|S|$) | 56 |
| Participate in Both Golf and Skating ($|G \cap S|$) | 16 |
| Participate Only in Golf ($|G| - |G \cap S|$) | 44 |
| Participate Only in Skating ($|S| - |G \cap S|$) | 40 |
| Participate in Neither | 20 |
Based on the calculations using set theory, the number of students who participate only in Skating is 40.
Reviewing the key terms and formulas used in this problem:
| Concept | Explanation | Formula Used |
|---|---|---|
| Set Union ($A \cup B$) | Elements in set A OR set B OR both. | $|A \cup B|$ |
| Set Intersection ($A \cap B$) | Elements in set A AND set B (in both). | $|A \cap B|$ |
| Principle of Inclusion-Exclusion (for 2 sets) | Relates the sizes of the union, intersection, and individual sets. | $|A \cup B| = |A| + |B| - |A \cap B|$ |
| Elements Only in Set B | Elements in set B but not in set A. | $|B| - |A \cap B|$ |
Problems involving students participating in activities can often be solved effectively using either set theory formulas or by drawing a Venn diagram. Both methods rely on the same underlying logic.
Both approaches lead to the same correct answer for problems like this one, helping you find the number of students only in Skating or other specific groups.
Dates of birth of some persons are given below. Find out the date of birth of the oldest person:
A. 12.08.1989
B. 13.09.1991
C. 19.06.1991
D. 20.02.1989
E. 22.03.1991
F. 20.01.1991
G. 20.12.1989Who takes a banana?
A. Hu
B. Ku
C. Moo
D. Cannot be determined
Which fruit does Ku take?
A. Orange
B. Plum
C. Banana
D. Mango
Which is the correct combination?
A. Moo - Banana
B. Hu - Plum
C. Hu - Orange
D. Moo - Plum
There are eight girls A to H was need to go for dance sessions in two batches of four girls each. Following are the criteria:
(1) B and H have to go together
(2) D and F do not go together
(3) A and C never go together
If B and C go in the first batch, then who of the following can be in the second batch?