
Solve the following problem and verify the data by drawing Venn diagrams. In a class of a certain school, 50 students offered mathematics, 42 offered biology, and 24 offered both the subjects. Find the number of students, who offer biology only.
Answer
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Hint: We have a total number of students who offered Biology and the number of students is 42. We have 24 such students who offered both subjects that are Mathematics and Biology. We have to get the number of such students who offered only biology. So, we have to deduct the number of students who offered both Mathematics and Biology from the number of students who offered Biology. Now, solve it further.
Complete step-by-step answer:
According to the question, it is given that in a class of a certain school, 50 students offered mathematics, 42 offered biology, and 24 offered both the subjects. We have to find the number of students, who offer biology only.
The number of students who offered mathematics = 50 ……………..(1)
The number of students who offered biology = 42 ………………(2)
The number of students who offered both subjects that is mathematics and biology = 24 ……………..(3)
The total number of students is the summation of the number of students who were offered biology and the number of students who were offered mathematics.
The total number of students = 50 students + 42 students. = 92 students.
Let \[n(M)\] , \[n(B)\] be the number of students who were offered mathematics and biology respectively.
Now, equation (1) can be written as,
\[n(M)=50\] ………………(4)
\[n(B)=42\] ………………..(5)
We have 24 students who offered both subjects that are mathematics and biology. It means that we have 24 common students in mathematics and biology. Also, we know that the intersection is the common value.
\[n\left( M\cap B \right)=24\] …………………..(6)
The number of students who offered only Mathematics = \[n(M)-n\left( M\cap B \right)=50-24=26\] .
The number of students who offered only Biology = \[n(B)-n\left( M\cap B \right)=42-24=18\] .
Using the Venn diagram, we can also verify this question. 24 students are common for both of the subjects. So, it is shown in the common part shared by both of the circles. The total number of biology students is 42. The part which is not common will give us the number of students who offered the only biology. So, it should be 42-24 students = 18 students.
Note: In this question, one may deduct 24 from 50 which is wrong. 24 is the number of students who offered both Mathematics and Biology. 50 is the number of students who offered Mathematics. If we deduct 24 from 50, we will get the number of students who offered only mathematics. But, in the question, it is asked to find the number of students who offered only Biology.
Complete step-by-step answer:
According to the question, it is given that in a class of a certain school, 50 students offered mathematics, 42 offered biology, and 24 offered both the subjects. We have to find the number of students, who offer biology only.
The number of students who offered mathematics = 50 ……………..(1)
The number of students who offered biology = 42 ………………(2)
The number of students who offered both subjects that is mathematics and biology = 24 ……………..(3)
The total number of students is the summation of the number of students who were offered biology and the number of students who were offered mathematics.
The total number of students = 50 students + 42 students. = 92 students.
Let \[n(M)\] , \[n(B)\] be the number of students who were offered mathematics and biology respectively.
Now, equation (1) can be written as,
\[n(M)=50\] ………………(4)
\[n(B)=42\] ………………..(5)
We have 24 students who offered both subjects that are mathematics and biology. It means that we have 24 common students in mathematics and biology. Also, we know that the intersection is the common value.
\[n\left( M\cap B \right)=24\] …………………..(6)
The number of students who offered only Mathematics = \[n(M)-n\left( M\cap B \right)=50-24=26\] .
The number of students who offered only Biology = \[n(B)-n\left( M\cap B \right)=42-24=18\] .
Using the Venn diagram, we can also verify this question. 24 students are common for both of the subjects. So, it is shown in the common part shared by both of the circles. The total number of biology students is 42. The part which is not common will give us the number of students who offered the only biology. So, it should be 42-24 students = 18 students.
Note: In this question, one may deduct 24 from 50 which is wrong. 24 is the number of students who offered both Mathematics and Biology. 50 is the number of students who offered Mathematics. If we deduct 24 from 50, we will get the number of students who offered only mathematics. But, in the question, it is asked to find the number of students who offered only Biology.
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