Back to: Further Mathematics SS1

Examples:

- Out of the 400 final year students in a secondary school, 300 are offering Biology and 190 are offering Chemistry. If only 70 students are offering neither Biology nor Chemistry. How many students are offering (i) both Biology and Chemistry? (ii) At least one of Biology or Chemistry?

Let the number of students who offered both Biology and Chemistry be X i.e (B∩C)= X. from the information given in the question

n(E)= 400

n(B)= 300

n(C)= 190

n(BƲC)^{|}= 70

since the sum of the number of elements in all region is equal to the total number of elements in the universal sets, then:

300 – x + x +190 – x + 70 =400

560 – x= 400

-x= 400 – 560

X= 160

Number of students offer both Biology and Chemistry= 160

(ii)no of students offering at least one of biology and chemistry from the Venn diagram this includes those who offered biology only, chemistry only and those whose offered both i.e

300 – x + 190 – x + x= 490

490 – 160 (from (i) above)= 330

- In a youth club with 90 members, 60 likes modern music and 50 likes traditional music. The member of them who like both traditional and modern music are three times those who do not like any type of music. How many members like only one type of music

Solution

Let the members who do not like any type of music = X

Then,

n(TnM)= 3X

Also,

n(E)= 94

n(M)=60

n(T)= 50

n(MƲT)^{|}= X

110 – 2X= 94

16= 2X

Divide both sides by 2

16= 2X

2 2

X= 8

Therefore number of member who likes only one type of music are those who like modern music only + those who like traditional music only

60 -3x + 3X + 50 – 3X= 110

110 – 6 x 8 (from above)

= 110 – 48

110 – 2X= 94

16= 2X

Divide both sides by 2

16= 2X

2 2

X= 8

Therefore number of member who likes only one type of music are those who like modern music only + those who like traditional music only

60 -3x + 3X + 50 – 3X= 110

110 – 6 x 8 (from above)

= 110 – 48

= 62= 62

Where 1 = AnB^{I} , 2 = AnB, 3 = A^{I}nB, 4 = ( AUB)^{I}

Therefore, µ = 1 + 2 + 3 + 4

µ = n( AnB^{I}) + n( AnB) + n( A^{I}nB) + n( AUB)^{I}

** Example 1:** In a class of 40 students, every student had to study French or Russian or both subjects. 25 students studied French and 20 studied Russian. Find the number of students who studied both languages.

**Solution:**

Let µ = {All the students}

F = {French students}, R = {Students studying Russian}

µ = 40, n(F)= 25, n(R) = 20

n( Fn R)= x

n(FnR^{I}) = 25-x

n (Rn F^{I})= 20- x

µ = 25 –x +x + 20-x

40= 45 –x

x = 45- 40

x=5, n(FnR) = 5 students.

**Evaluation**

Two questions A and B were given to 50 students as class work 23 of them could answer question A but not B. 15 of them could answer B but not A. If 2x of them could answer none of the two questions and 2 could answer both questions.

- a)Represent the information in a Venn diagram. (b) Find the value of x

**General evaluation**

- In a senior secondary school, 90 students play hockey or football. The numbers that play football is 5 more than twice the number that play hockey. If 5 students play both games and every students in the school plays at least one of the game. Find:
- a)The number of students that play football
- b)The number of student that play football but not hockey
- c)The number of students that play hockey but not football
- A, B and C are subset of the universal set U such that

Where 1 = AnB^{I} , 2 = AnB, 3 = A^{I}nB, 4 = ( AUB)^{I}

Therefore, µ = 1 + 2 + 3 + 4

µ = n( AnB^{I}) + n( AnB) + n( A^{I}nB) + n( AUB)^{I}

** Example 1:** In a class of 40 students, every student had to study French or Russian or both subjects. 25 students studied French and 20 studied Russian. Find the number of students who studied both languages.

** **

**Solution:**

Let µ = {All the students}

F = {French students}, R = {Students studying Russian}

µ = 40, n(F)= 25, n(R) = 20

n( Fn R)= x

n(FnR^{I}) = 25-x

n (Rn F^{I})= 20- x

µ = 25 –x +x + 20-x

40= 45 –x

x = 45- 40

Examples:

- Out of the 400 final year students in a secondary school, 300 are offering Biology and 190 are offering Chemistry. If only 70 students are offering neither Biology nor Chemistry. How many students are offering (i) both Biology and Chemistry? (ii) At least one of Biology or Chemistry?

Let the number of students who offered both Biology and Chemistry be X i.e (B∩C)= X. from the information given in the question

n(E)= 400

n(B)= 300

n(C)= 190

n(BƲC)^{|}= 70

since the sum of the number of elements in all region is equal to the total number of elements in the universal sets, then:

300 – x + x +190 – x + 70 =400

560 – x= 400

-x= 400 – 560

X= 160

Number of students offer both Biology and Chemistry= 160

(ii)no of students offering at least one of biology and chemistry from the Venn diagram this includes those who offered biology only, chemistry only and those whose offered both i.e

300 – x + 190 – x + x= 490

490 – 160 (from (i) above)= 330

- In a youth club with 90 members, 60 likes modern music and 50 likes traditional music. The member of them who like both traditional and modern music are three times those who do not like any type of music. How many members like only one type of music

Solution

Let the members who do not like any type of music = X

Then,

n(TnM)= 3X

Also,

n(E)= 94

n(M)=60

n(T)= 50

n(MƲT)^{|}= X

110 – 2X= 94

16= 2X

Divide both sides by 2

16= 2X

2 2

X= 8

Therefore number of member who likes only one type of music are those who like modern music only + those who like traditional music only

60 -3x + 3X + 50 – 3X= 110

110 – 6 x 8 (from above)

= 110 – 48

110 – 2X= 94

16= 2X

Divide both sides by 2

16= 2X

2 2

X= 8

60 -3x + 3X + 50 – 3X= 110

110 – 6 x 8 (from above)

= 110 – 48

= 62= 62

Where 1 = AnB^{I} , 2 = AnB, 3 = A^{I}nB, 4 = ( AUB)^{I}

Therefore, µ = 1 + 2 + 3 + 4

µ = n( AnB^{I}) + n( AnB) + n( A^{I}nB) + n( AUB)^{I}

**Example 1:** In a class of 40 students, every student had to study French or Russian or both subjects. 25 students studied French and 20 studied Russian. Find the number of students who studied both languages.

**Solution:**

Let µ = {All the students}

F = {French students}, R = {Students studying Russian}

µ = 40, n(F)= 25, n(R) = 20

n( Fn R)= x

n(FnR^{I}) = 25-x

n (Rn F^{I})= 20- x

µ = 25 –x +x + 20-x

40= 45 –x

x = 45- 40

x=5, n(FnR) = 5 students.

**Evaluation**

Two questions A and B were given to 50 students as class work 23 of them could answer question A but not B. 15 of them could answer B but not A. If 2x of them could answer none of the two questions and 2 could answer both questions.

- a) Represent the information in a Venn diagram. (b) Find the value of x

**General evaluation**

- In a senior secondary school, 90 students play hockey or football. The numbers that play football is 5 more than twice the number that play hockey. If 5 students play both games and every students in the school plays at least one of the game. Find:
- a) The number of students that play football
- b) The number of student that play football but not hockey
- c) The number of students that play hockey but not football
- A, B and C are subset of the universal set U such that

Where 1 = AnB^{I} , 2 = AnB, 3 = A^{I}nB, 4 = ( AUB)^{I}

Therefore, µ = 1 + 2 + 3 + 4

µ = n( AnB^{I}) + n( AnB) + n( A^{I}nB) + n( AUB)^{I}

**Example 1:** In a class of 40 students, every student had to study French or Russian or both subjects. 25 students studied French and 20 studied Russian. Find the number of students who studied both languages.

** **

**Solution:**

Let µ = {All the students}

F = {French students}, R = {Students studying Russian}

µ = 40, n(F)= 25, n(R) = 20

n( Fn R)= x

n(FnR^{I}) = 25-x

n (Rn F^{I})= 20- x

µ = 25 –x +x + 20-x

40= 45 –x

x = 45- 40

x=5, n(FnR) = 5 students.x=5, n(FnR) = 5 students.

**Evaluation**

Two questions A and B were given to 50 students as class work 23 of them could answer question A but not B. 15 of them could answer B but not A. If 2x of them could answer none of the two questions and 2 could answer both questions.

- a)Represent the information in a Venn diagram. (b) Find the value of x

** **

**General evaluation**

- In a senior secondary school, 90 students play hockey or football. The numbers that play football is 5 more than twice the number that play hockey. If 5 students play both games and every students in the school plays at least one of the game. Find:
- a)The number of students that play football
- b)The number of student that play football but not hockey
- c)The number of students that play hockey but not football
- A, B and C are subset of the universal set U such that

U={0,1,2,3,4………….12}U={0,1,2,3,4………….12}

Examples:

- Out of the 400 final year students in a secondary school, 300 are offering Biology and 190 are offering Chemistry. If only 70 students are offering neither Biology nor Chemistry. How many students are offering (i) both Biology and Chemistry? (ii) At least one of Biology or Chemistry?

Let the number of students who offered both Biology and Chemistry be X i.e (B∩C)= X. from the information given in the question

n(E)= 400

n(B)= 300

n(C)= 190

n(BƲC)^{|}= 70

since the sum of the number of elements in all region is equal to the total number of elements in the universal sets, then:

300 – x + x +190 – x + 70 =400

560 – x= 400

-x= 400 – 560

X= 160

Number of students offer both Biology and Chemistry= 160

(ii)no of students offering at least one of biology and chemistry from the Venn diagram this includes those who offered biology only, chemistry only and those whose offered both i.e

300 – x + 190 – x + x= 490

490 – 160 (from (i) above)= 330

- In a youth club with 90 members, 60 likes modern music and 50 likes traditional music. The member of them who like both traditional and modern music are three times those who do not like any type of music. How many members like only one type of music

Solution

Let the members who do not like any type of music = X

Then,

n(TnM)= 3X

Also,

n(E)= 94

n(M)=60

n(T)= 50

n(MƲT)^{|}= X

110 – 2X= 94

16= 2X

Divide both sides by 2

16= 2X

2 2

X= 8

60 -3x + 3X + 50 – 3X= 110

110 – 6 x 8 (from above)

= 110 – 48

110 – 2X= 94

16= 2X

Divide both sides by 2

16= 2X

**Evaluation**

- a)Represent the information in a Venn diagram. (b) Find the value of x

** **

**General evaluation**

- a)The number of students that play football
- b)The number of student that play football but not hockey
- c)The number of students that play hockey but not football
- A, B and C are subset of the universal set U such that

U={0,1,2,3,4………….12}U={0,1,2,3,4………….12}

Examples:

n(E)= 400

n(B)= 300

n(C)= 190

n(BƲC)^{|}= 70

300 – x + x +190 – x + 70 =400

560 – x= 400

-x= 400 – 560

X= 160

Number of students offer both Biology and Chemistry= 160

300 – x + 190 – x + x= 490

490 – 160 (from (i) above)= 330

Solution

Let the members who do not like any type of music = X

Then,

n(TnM)= 3X

Also,

n(E)= 94

n(M)=60

n(T)= 50

n(MƲT)^{|}= X

110 – 2X= 94

16= 2X

Divide both sides by 2

16= 2X

2 2

X= 8

60 -3x + 3X + 50 – 3X= 110

110 – 6 x 8 (from above)

= 110 – 48

110 – 2X= 94

16= 2X

Divide both sides by 2

16= 2X

2 2

X= 8

60 -3x + 3X + 50 – 3X= 110

110 – 6 x 8 (from above)

= 110 – 48

= 62= 62

Where 1 = AnB^{I} , 2 = AnB, 3 = A^{I}nB, 4 = ( AUB)^{I}

Therefore, µ = 1 + 2 + 3 + 4

µ = n( AnB^{I}) + n( AnB) + n( A^{I}nB) + n( AUB)^{I}

**Example 1:** In a class of 40 students, every student had to study French or Russian or both subjects. 25 students studied French and 20 studied Russian. Find the number of students who studied both languages.

**Solution:**

Let µ = {All the students}

F = {French students}, R = {Students studying Russian}

µ = 40, n(F)= 25, n(R) = 20

n( Fn R)= x

n(FnR^{I}) = 25-x

n (Rn F^{I})= 20- x

µ = 25 –x +x + 20-x

40= 45 –x

x = 45- 40

x=5, n(FnR) = 5 students.

**Evaluation**

- a) Represent the information in a Venn diagram. (b) Find the value of x

**General evaluation**

- a) The number of students that play football
- b) The number of student that play football but not hockey
- c) The number of students that play hockey but not football
- A, B and C are subset of the universal set U such that

Where 1 = AnB^{I} , 2 = AnB, 3 = A^{I}nB, 4 = ( AUB)^{I}

Therefore, µ = 1 + 2 + 3 + 4

µ = n( AnB^{I}) + n( AnB) + n( A^{I}nB) + n( AUB)^{I}

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