In a class of 50 student, 25 offer mathematics, 22 offer physcis, 30 offer chemistry and all the student take at least one of these subjects , 10 offer physics and mathematics, 8 offer chemistry and mathematics , 16 offer physics and chemistry. (a) draw a venn diagram to illustrate the information. (b) find the number of students who offer all three subjects
I'm sure you can do the diagram. Then it should be clear that if x take all three subjects,
(25+22+30)-(10+8+16)+x = 50
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To draw the Venn diagram, first, draw three overlapping circles to represent the subjects: mathematics, physics, and chemistry. Label each circle accordingly.
Next, refer to the given information and fill in the appropriate numbers within the diagram.
Since all the students take at least one of the subjects, the total number of students (50) will be placed outside of all the circles.
Now, let's find the number of students who offer all three subjects.
We can start by finding the number of students who offer physics and mathematics (10) and subtract the number of students who offer physics and mathematics but not chemistry. From the information given, we know that 16 students offer physics and chemistry, so we can subtract this number.
10 - 16 = -6
However, we cannot have a negative value here, so this means that there are no students who offer physics and mathematics but not chemistry.
Therefore, the number of students who offer all three subjects is 16.
To draw a Venn diagram to illustrate the given information, we start by drawing three overlapping circles, one for each subject: mathematics, physics, and chemistry.
(a) Let's label the circles as follows:
- M for mathematics
- P for physics
- C for chemistry
Now let's place the numbers we know in their corresponding regions in the Venn diagram:
- 10 students offer both physics and mathematics (inside the overlapping region of M and P)
- 8 students offer both chemistry and mathematics (inside the overlapping region of M and C)
- 16 students offer both physics and chemistry (inside the overlapping region of P and C)
To find the number of students who offer all three subjects, we need to look for the center region where all three circles overlap. Let's label this region as "MPC" for mathematics, physics, and chemistry.
(b) Since the Venn diagram does not provide specific numbers, we need to calculate the number of students who offer all three subjects by using the information from the diagram.
Given that all the students take at least one of the subjects, we can use this formula to find the number of students who offer all three subjects:
Number of students who offer all three subjects = Total number of students - (Number of students who only take mathematics + Number of students who only take physics + Number of students who only take chemistry + Number of students who take exactly two subjects)
To calculate this:
Total number of students = 50 (as given)
Number of students who only take mathematics = Number of students taking mathematics (25) - Number of students taking both mathematics and physics (10) - Number of students taking both mathematics and chemistry (8) + Number of students taking all three subjects (MPC) (unknown)
= 25 - 10 - 8 + MPC
Number of students who only take physics = Number of students taking physics (22) - Number of students taking both mathematics and physics (10) - Number of students taking both physics and chemistry (16) + Number of students taking all three subjects (MPC) (unknown)
= 22 - 10 - 16 + MPC
Number of students who only take chemistry = Number of students taking chemistry (30) - Number of students taking both mathematics and chemistry (8) - Number of students taking both physics and chemistry (16) + Number of students taking all three subjects (MPC) (unknown)
= 30 - 8 - 16 + MPC
From the given information, we know that all students take at least one subject. Therefore, the sum of the students who only take mathematics, who only take physics, and who only take chemistry should be equal to the total number of students:
Number of students who only take mathematics + Number of students who only take physics + Number of students who only take chemistry = Total number of students
By substituting in the values we have:
(25 - 10 - 8 + MPC) + (22 - 10 - 16 + MPC) + (30 - 8 - 16 + MPC) = 50
Simplifying the equation:
(7 + MPC) + (-4 + MPC) + (6 + MPC) = 50
19 + 3MPC = 50
3MPC = 50 - 19
3MPC = 31
MPC = 31 / 3
MPC ≈ 10.33
Since we cannot have a fraction of a student, we round down to the nearest whole number. Therefore, the number of students who offer all three subjects is 10.
In conclusion, the number of students who offer all three subjects is 10.