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Leviafan [203]
3 years ago
8

In a class of 30 students (x+10) study algebra, (10x+3) study statistics, 4 study both algebra and statistics. 2x study only alg

ebra and 3 study neither algebra nor statistics. Each student studies at least one of the two subjects. 1. Illustrate the information on a venn diagram. 2. How many students study a. Algebra b. Statistics

Mathematics
1 answer:
Vladimir [108]3 years ago
6 0

Answer:

1. The Venn diagrams are attached

2. When the statistics students number = 10·x + 3, we have;

The number of students that study

a. Algebra = 128/11

b. Statistic = 213/11

When the statistics students number = 2·x + 3, we have;

The number of students that study

a. Algebra = 16

b. Statistic = 15

Step-by-step explanation:

The parameters given are;

Total number of students = 30

Number of students that study algebra n(A) = x + 10

Number of students that study statistics n(B) = 10·x + 3

Number of student that study both algebra and statistics n(A∩B) = 4

Number of student that study only algebra n(A\B) = 2·x

Number of students that study neither algebra or statistics n(A∪B)' = 3

Therefore;

The number of students that study either algebra or statistics = n(A∪B)

From set theory we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 10·x + 3 - 4 = 27

11·x+13 = 27 + 4 = 31

11·x = 18

x = 18/11

The number of students that study

a. Algebra

n(A) = 18/11 + 10 = 128/11

b. Statistic

n(B) = 213/11

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 128/11 - 4 = 84/11

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 213/11 - 4 = 169/11

However, assuming n(B) = (2·x + 3), we have;

n(A∪B) = n(A) + n(B) - n(A∩B)

n(A∪B) = 30 - 3 = 27

Therefore, we have;

n(A∪B) = x + 10 + 2·x + 3 - 4 = 27

2·x+3 + x + 10= 27 + 4 = 31

3·x = 18

x = 6

Therefore, the number of students that study

a. Algebra

n(A) = 16

b. Statistics

n(B) = 15

Hence, we have;

n(A - B) = n(A) - n(A∩B) = 16 - 4 = 12

Similarly, we have;

n(B - A) = n(B) - n(A∩B) = 15 - 4 = 11

The Venn diagrams can be presented as follows;

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