In a class of 45 students, 24 do arts, 20 to chemistry and 22 to biology. All the students do at least one of 3 subjects.3 do all the 3 subjects while 7 do art and biology, 6 do art and chemistry but not biology. How many do biology only, chemistry only or art only
That took a while :)
Draw Venn diagram
arts alone = a
chem alone = c
bio alone = b
art and chem alone = 6
art and bio alone = 7-3 = 4
bio and chem alone = bc
all 3 = 3
so
a + 6+3+4 = a+13 = 24 so a = 11
b + 7 + bc = 22 or b +bc = 15
11 + b + c +bc +13 = 45
those two are
b + bc =15
11 + 13 + 15 + c = 45
c = 6
so far a = 11 and c = 6
now
6 + bc = 11
so bc = 5
so
b = 15 - 5 = 10
so
a = 11
b = 10
c = 6
Please show the Venn diagram for the question mentioned above
Venn diagram show
In a class of 50 student .it's know that 30 do art, 24 do chemistry and 22 do biology.All the student do at least one of the three subject.4 do all the student while 8 do art and biology.7 do art and chemistry but not biology and 10 do chemistry and biology.how many students do
biology only, chemistry only or art only?
To create the Venn diagram we start by making three circles that intersect, representing art, chemistry, and biology.
We know that 4 students do all three, so we can put that in the intersection of all three circles:
```
A
/ | \
/ | \
/___|___\
B C Bio
```
We also know that 8 students do art and biology, but not chemistry. We can add that to the intersection of the art and biology circles, but not in the intersection with the chemistry circle:
```
A
/ | \
/ | \
/___|_8_\
B C Bio
```
Similarly, 7 students do art and chemistry, but not biology:
```
A
/ | \ 7
/ | \
/___|_8_\
B C Bio
```
And 10 students do chemistry and biology, but not art:
```
A
7 / | \ 8
/ | \
/___|___\
B C Bio 10
```
From here, we can start filling in the remaining values. We know that 30 students do art, and we've already accounted for 8 + 7 + 4 students who do art and another subject. So the number of students who do art only is:
```
A
/ | \ 7
/ | \
/___|___\
B C Bio 10
|
|
13
```
30 - 8 - 7 - 4 = 11 students who do art only.
Similarly, we can fill in the values for the number of students who do chemistry only and biology only:
```
A
/ | \ 7
/ | \
/___|___\
B C Bio 10
| |
| 3|
|_______|
|
|
6
```
Adding up the numbers in each sector, we get:
- Art only: 11
- Chemistry only: 3
- Biology only: 6
- Art and chemistry only: 7
- Art and biology only: 8
- Chemistry and biology only: 10
- All three: 4
And we can check that this adds up to the total number of students:
11 + 3 + 6 + 7 + 8 + 10 + 4 = 49
One student is missing, but we know that all students do at least one of the three subjects. This means that there must be one student who does all three subjects, in addition to the four we've already accounted for:
```
A
/ | \ 7
/ | \
/___|___\
B C Bio 10
| |
| 3|
|_______|
| 1
|
1|
|
```
So the final answer is:
- Art only: 11
- Chemistry only: 3
- Biology only: 6
- Art and chemistry only: 7
- Art and biology only: 8
- Chemistry and biology only: 10
- All three: 5