Bhagya

Students can download Maths Chapter 4 Symmetry Ex 4.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 4 Symmetry Ex 4.2

Miscellaneous Practice Problems

Question 1.
Draw and answer the following.
(i) A triangle which has no line of symmetry.
(ii) A triangle which has only one line of symmetry
(iii) A triangle which has three lines of symmetry.
Solution:
(i) Scalene triangle
(ii) Isosceles triangle
(iii) Equilateral triangle

Question 2.
Find the alphabets in the box which have

(i) No line of symmetry
(ii) Rotational symmetry
(iii) Reflection symmetry
(iv) Reflection and rotational symmetry.
Solution:
i) The alphabets which have no line of symmetry are P, N, S, Z
ii) The alphabets which have Rotational symmetry are I, O, N, X, S, H, Z
iii) The alphabets which have reflection symmetry are A, M, E, D, I, K, O, X, H, U, V, W.
iv) The alphabets which has reflection and rotational symmetry are I, O, X, H.

Question 3.
For the following pictures, find the number of lines of symmetry and also find the order of rotation.

Solution:
(i) 0, 2
(ii) 1, 0
(iii) 2, 2
(iv) 8, 8
(v) 1, 0

Question 4.
The three-digit number 101 has rotational and reflection symmetry. Give five more examples of three-digit numbers that have both rotational and reflection symmetry
Solution:
The digits 0, 1, 8 have rotational and reflection symmetry.
∴ The three digits numbers 181, 111, 808, 818, 888 have both rotational and reflection symmetry.

Question 5.
Translate the given pattern and complete the design in a rectangular strips?

Solution:

Challenge Problems

Question 6.
Shade one square so that it possesses

(i) One line of symmetry
(ii) Rotational symmetry of order 2
Solution:

Question 7.
Join six identical squares so that atleast one side of a square fits exactly with any other side of the square and has reflection symmetry (any three ways).
Solution:

Question 8.
Draw the following
(i) A figure which has reflection symmetry but no rotational symmetry.
(ii) A figure which has rotational symmetry but no reflection symmetry.
(iii) A figure which has both reflection and rotational symmetry.
Solution:

Question 9.
Find the line of symmetry and the order of rotational symmetry’ of the given regular polygons and complete the following table and answer the questions given below.

i) A regular polygon of 10 sides will have ______ lines of symmetry.
ii) If a regular polygon has 10 lines of symmetry then its order of rotational symmetry is ______
iii) A regular polygon of ‘n’ sides has lines of symmetry and the order of rotational symmetry is ______
Solution:
(i) 10
(ii) 10
(iii) n, n

Question 10.
Colour the boxes in such a way that it possesses translation symmetry.

Solution:

Students can download Maths Chapter 4 Symmetry Ex 4.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 4 Symmetry Ex 4.1

Question 1.
Fill in the blanks
(i) The reflected image of the letter ‘q’ is …….
(ii) A rhombus has ………… lines of symmetry.
(iii) The order of rotational symmetry of the letter ‘Z’ is ……….
(iv) A figure is said to have rotational symmetry, if the order of rotation is atleast ……….
(v) ……… symmetry occurs when an object slides to new position.
Solution:
(i) P
(ii) two
(iii) 2
(iv) two
(v) Translation

Question 2.
Say True or False
(i) A rectangle has four lines of symmetry.
(ii) A shape has reflection symmetry if it has a line of symmetry.
(iii) The reflection of the name RANI is INAЯ.
(iv) Order of rotation of a circle is infinite.
(v) The number 191 has rotational symmetry.
Solution:
(i) False
(ii) True
(iii) False
(iv) True
(v) False

Question 3.
Match the following shapes with their number of lines of symmetry.

Solution:
(i) d
(ii) a
(iii) b
(iv) c

Question 4.
Draw the lines of symmetry of the following.

Solution:

Question 5.
Using the given horizontal line/ vertical line as a line of symmetry, complete each alphabet to discover the hidden word.

Solution:
(i) DECODE
(ii) KICK
(iii) BED
(iv) WAY
(v) MATY
(vi) TOMATO

Question 6.
Draw a line of symmetry of the given figures such that one hole coincides with the other hole(s) to make pairs.

Solution:

Question 7.
Complete the other half of the following figures such that the dotted line is the line of symmetry.

Solution:

Question 8.
Find the order of rotation for each of the following.

Solution:
(i) 2
(ii) 2
(iii) 4
(iv) 8
(v) 2

Question 9.
A standard die has six faces which are shown below. Find the order of rotational symmetry of each face of a die?

Solution:
(i) 4
(ii) 2
(iii) 2
(iv) 4
(v) 4
(vi) 2

Question 10.
What pattern is translated in the given border kolams?

Solution:

Objective Type Questions

Question 11.
Which of the following letter does not have a line of symmetry?
(a) A
(b) P
(c) T
(d) U
Hint: A, T, U have one line of symmetry
Solution:
(b) P

Question 12.
Which of the following is a symmetrical figure?

Solution:
(c)

Question 13.
Which word has a vertical line of symmetry?
(a) DAD
(b) NUN
(c) MAM
(d) EVE
Hint: D, N, E have no vertical line of symmetry
Solution:
D, N, E have no vertical line of symmetry

Question 14.
The order of rotational symmetry of 818 is ………
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2

Question 15.
The order of rotational symmetry ★ is ___
(a) 5
(b) 6
(c) 7
(d) 8
Solution:
(a) 5

Students can download Maths Chapter 3 Perimeter and Area Ex 3.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.2

Miscellaneous Practice Problems

Question 1.
A piece of wire is 36 cm long. What will be the length of each side if we form
(i) a square
(ii) an equilateral triangle
Solution:

Given the length of the wire = 36 cm
i) When a square is formed out of it
The perimeter of the square = 36 cm
4 × side = 36
side = \(\frac{36}{4}\) = 9 cm
Side of the square

ii) When an equilateral triangle is formed out of it, its perimeter = 36 cm
i.e., side + side + side = 36 cm .
3 × side = 36 cm
side = \(\frac{36}{3}\) = 12 cm
One side of an equilateral triangle = 12 cm

Question 2.
From one vertex of an equilateral triangle with a side of 40 cm, an equilateral triangle with 6 cm side is removed. What is the perimeter of the remaining portion? The perimeter of the remaining portion
Solution:

= (40 + 34 + 6 + 34) cm
= 114 cm

Question 3.
Rahim and Peter go for a morning walk, Rahim walks around a. square path of side 50 m and Peter walks around a rectangular path with a length of 40 m and a breadth of 30 m. If both of them walk 2 rounds each, who covers more distance and by how much?
Solution:
Distance covered by Rahim
= 50 × 4 m
= 200 m
If he walks 2 rounds, distance covered = 2 × 200 m
= 400 m
Distance covered by peter
= 2 (40 + 30) m
= 2(70)m
= 140 m
If he walks 2 rounds, distance covered = 2 × 140 m
= 280 m
∴ Rahim covers more distance by (400 – 280) = 120 m

Question 4.
The length of a rectangular park is 14 m more than its breadth. If the perimeter of the park is 200 m, what is its length? Find the area of the park.
Solution:
Let the length be b + 14 m
breadth = b
perimeter = 200
2 (l + b) = 200
2 (b + 14 + b) = 200
2 (2b + 14) = 200
28 + 4b = 200
4b = 200 – 28
4b = 172 m
b = \(\frac{172}{4}\)
b = 43 m
Length = b + 14
= 43 + 14
Length l = 57 m
Area = l × b units
= 57 × 43 m²
= 2451 m²

Question 5.
Your garden is in the shape of a square of side 5 m. Each side is to be fenced with 2 rows of wire. Find how much amount is needed to fence the garden at ₹ 10 per metre.
Solution:
a = 5 m
The perimeter of the garden
= 4 a units
= 4 × 5 m
= 20 m
For 1 row
Amount needed to fence l m= Rs 10
Amount needed to fence 20 m
= Rs 10 × 20
= Rs 200
For 2 rows
Total amount needed = 2 × Rs 200 = Rs 400

Challenge Problems

Question 6.
A closed shape has 20 equal sides and one of its sides is 3 cm. Find its perimeter.
Solution:
Number of equal sides in the shape = 20
One of its side = 3 cm
Perimeter = length of one side × Number of equal sides
∴ Perimeter = (3 × 20) cm = 60 cm
∴ Perimeter = 60 cm

Question 7.
A rectangle has length 40 cm and breadth 20 cm. How many squares with side 10 cm can be formed from it.
Solution:
Area of rectangle = (length × breadth) units²
Length = 40 cm
Breadth = 20 cm
∴ Area = (40 × 20) cm² = 800 cm²
Area of rectangle = 800 cm²
Area of square = (side × side) units²
side = 10 cm
∴ Area of square = (10 × 10) cm² = 100 cm²
Required number of squares = \(\frac{\text { Area of Rectangle }}{\text { Area of } 1 \text { square }}=\frac{800 \mathrm{cm}^{2}}{100 \mathrm{cm}^{2}}\) = 8
8 squares can be formed.

Question 8.
The length of a rectangle is three times its breadth. If its perimeter is 64 cm, find the sides of the rectangle.
Solution:
Given perimeter of a rectangle = 64 cm
Also given length is three times its breadth.
Let the breadth of the rectangle = b cm
∴ Length = 3 × b cm
Perimeter = 64 m
i.e., 2 × (l + b) = 64 m
2 × (3b + b) = 64 m
2 × 4b = 64m
4b = \(\frac{64}{2}\) = 32 m
b = \(\frac{32}{4}\) = 8 m
l = 3 × b = 3 × 8 = 24 m
∴ Breadth of the rectangle = 8 m
Length of the rectangle = 24 m

Question 9.
How many different rectangles can be made with a 48 cm long string? Find the possible pairs of length and breadth of the rectangles.
Solution:
Length of the string to be made into rectangle = 48 cm
∴ Perimeter of the rectangle = 48 cm
2 × (l + b) = 48 cm
l + b = \(\frac{48}{2}\)
l + b = 24 cm
Possible pairs of length and breadth are (1, 23), (2, 22) (3, 21), (4, 20), (5, 19),
(6, 18), (7, 17), (8, 16), (9, 15), (10, 14), (11, 13), (12, 12)
Number of different rectangles = 12.

Question 10.
Draw a square B whose side is twice of the square A. Calculate the perimeters of the squares A and B.
Solution:

Perimeter of A = s + s + s + s units = 4 s units
Perimeter of B = (2s + 2s + 2s + 2s) units
= 8s units = 2 (4s) units.
∴ The perimeter of B is twice the perimeter of A

Question 11.
What will be the area of a new square formed if the side of a square is made one – fourth?
Solution:
Let the side of square is s units then area = (s × s) units²
If the side of the new square is made one fourth then side = \(\left(\frac{1 \times s}{4}\right)\) units
Then area = \(\left(\frac{1 \times s}{4} \times \frac{1 \times s}{4}\right)\) units² = \(\frac{s \times s}{16}=\frac{1}{16}\) (s × s) units²
Area of the new square is reduced to \(\frac{1}{16}\) times to that of original area.

Question 12.
Two plots have the same perimeter. One . is a square of side 10 m and another is a rectangle of breadth 8 m. Which plot has the greater area and by how much?
Solution:
a = 10 m, b = 8 m
Perimeter of the square plot
= 4 a units
= 4 × 10 m
= 40 m
Perimeter of the rectangular plot
40 = 2 (l + b) units
40 = 2 (l + 8) m
40 = 2 l + 16
2 l = 40 – 16
2 l = 24
l = \(\frac{24}{2}\)
l = 12 m
Area of the square plot
= a × a sq units
= 10 × 10 m²
= 100 m²
Area of the rectangular plot
= l × b sq units
= 8 × 12 m²
= 96 m²
Square plot has the greater area by 100 m² – 96 m² – 4 m²

Question 13.
Look at the picture of the house given and find the total area of the shaded portion.

Solution:
Total area of the shaded region = Area of a right triangle + Area of a rectangle
= (\(\frac{1}{2}\) × b × h) + (l × b) cm²
= [(\(\frac{1}{2}\) × 3 × 4) + (9 × 6)] cm²
= (6 + 54) cm² = 60 cm²

Question 14.
Find the approximate area of the flower in the given square grid.

Solution:
No of full squares = 11
No of half squares = 9
Area of 11 full squares
= 11 x 1 cm²
= 11 cm²
Area of 9 half squares
= 9 × \(\frac{1}{2}\) cm²
= 4.5 cm²
Area of the flower = (11 + 4.5) cm²
= 15.5 cm²

Students can download Maths Chapter 3 Perimeter and Area Ex 3.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.1

Question 1.
The table given below contains some measures of the rectangle. Find the unknown values.

Solution:

(i) Area of the rectangle = (length × breadth) sq unit.
Perimeter of a rectangle = 2(1 + b) units.
l = 5 cm
b = 8 cm
∴ p = 2 (l + b) cm = 2 (5 + 8) cm = 2 × 13 cm
p = 26 cm
Area = (l × b) cm² = (5 × 8) cm²
A = 40 cm²

(ii) l = 13 cm
p = 54 cm
Perimeter = 2 (l + b) units
54 = 2 (13 + b) cm
\(\frac{54}{2}\) = 13 + b
27 = 13 + b
b = 27 – 13
b = 14 cm
Area = l × b sq. unit = 13 × 14 cm²
A = 182 cm²

(iii) b = 15 cm
p = 60 cm
p = 2 (l + b) units
60 = 2 (l + 15) cm
\(\frac{60}{2}\) = l + 15
30 = l + 15
l = 30 – 15 .
l = 15 cm
Area = l × b unit² = 15 × 15 cm² = 225 cm²
A = 225 cm²

(iv) l = 10 m
Area = 120 sq metre
Area = l × b sq.m
120 = 10 × 6
b = \(\frac{120}{10}\)
b = 12 m
Perimeter =2 (l + b) units = 2(10 + 12) units = 2 × 22 m
A = 44 m

(v) b = 4 feet.
Area = 20 sq. feet
Area = l × b sq .feet
20 = l × 4
l = \(\frac{20}{4}\) feet
l = 5 feet
Perimeter = 2 (l + b) units.
p = 2 (5 + 4) feet = 2 × 9
p = 18 feet

Question 2.
The table given below contains some measures of the square. Find the unknown values.

Solution:
(i) 24 cm, 36 cm²
(ii) 25 m, 625 m²
(iii) 7 feet, 28 feet

Question 3.
The table given below contains some measures of the right angled triangle. Find the unknown values.

Solution:

Area of the right triangle = \(\frac{1}{2}\) × (base × height) unit²
(i) b = 20 cm
h = 40 cm
Area = \(\frac{1}{2}\) (b × h) cm² = \(\frac{1}{2}\) × 20 × 40 = 400 cm²
A = 400 cm²

(ii) b = 5 feet
Area = \(\frac{1}{2}\) × b × h unit²
= 20 = \(\frac{1}{2}\) × 5 × h sq. feet
\(\frac{20 \times 2}{5}\) = h
h = 8 feet

(iii) Area = \(\frac{1}{2}\) × (base × height) unit²
24 = \(\frac{1}{2}\) × b × 12 m²
base = \(\frac{24 \times 2}{12}\) m = 4 m
Base = 4m

Question 4.
The table given below contains some measures of the triangle. Find the unknown values.

Solution:
(i) 13 cm
(ii) 6 m
(iii) 8 feet

Question 5.
Fill in the blanks.
(i) 5 cm² = ______ mm²
(ii) 26 m² = ______ cm²
(iii) 8 km² = ______ m²
Solution:
(i) 500
(ii) 260000
(iii) 8000000

Question 6.
Find the perimeter and area of the following shapes.

Solution:
(i) Perimeter = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4)
= 48 cm
a = 4 cm
Area of 1 square = 4 × 4 cm²
= 16 cm²
Area of 5 squares = 5 × 16 cm²
= 80 cm²

(ii) Perimeter = (4 + 5 + 4 + 5 + 4 + 5 + 4 + 5)
= 36 cm
Area of 1 triangle = \(\frac{1}{2}\) × b × h sq units
= \(\frac{1}{2}\) × 4 × 5 cm²
= 10 cm²
Area of 4 triangles= 4 × 10 cm²
= 40 cm²
Area of the square = 3 × 3 cm²
= 9 cm²
Total area = (40 + 9) cm²
= 49 cm²

(iii) Perimeter = (15 + 50 + 12 + 13 + 10 + 10 + 40)
= 150 cm
Area of the square = 10 × 10 cm²
= 100 cm²
= 250 cm²
Area of the triangle = \(\frac{1}{2}\) × 12 × 5 cm²
= \(\frac{1}{2}\) × 12⁶ x 5 cm²
= 30 cm²
Total area = (100 + 250 + 30) cm²
= 380 cm²

Question 7.
Find the perimeter and the area of the rectangle whose length is 6 m and breadth is 4m?
Solution:
l = 6 m, b = 4 m Perimeter of the rectangle
= 2 (l + b) units
= 2 (6 + 4) m
= 2 (10) m
= 20 m
Area of the rectangle = l × b sq units
= 4 × 6 m²
= 24 m²

Question 8.
Find the perimeter and area of a square whose side is 8 cm.
Solution:
a = 8 cm
The perimeter of a square
= 4a units
= 4 × 8 cm
= 32 cm
Area of the square = a × a sq units
= 8 × 8 cm²
= 64 cm²

Question 9.
Find the perimeter and the area of right angled triangle whose sides are 6 feet, 8 feet and 10 feet.
Solution:
Perimeter of the triangle
= (a + b + c) units
= (6 + 8 + 10) feet
= 24 feet
Area of the triangle = \(\frac{1}{2}\) × b × h sq units
\(\frac{1}{2}\) × 6³× 8 feet square = 24 sq. feet

Question 10.
Find the perimeter of
(i) A scalene triangle with sides 7 m, 8 m, 10 m.
(ii) An isosceles triangle with equal sides 10 cm each and third side is 7 cm.
(iii) An equilateral triangle with a side of 6 cm.
Solution:
(i) Perimeter of the triangle
= (a + b + c) units
= (7 + 8 + 10) m
= 25

(ii) Perimeter of the triangle
= (10 + 10 + 7) cm
= 27 cm

(iii) Perimeter of the triangle
= (6 + 6 + 6) cm
= 18 cm

Question 11.
The area of a rectangular shaped photo is 820 sq. cm. and its width is 20 cm. What is its length? Also find its perimeter.
Solution:
Given Area = 820 cm²
Width = 20 cm
Area of the rectangle
= l × b sq. units
820 = l × 20
\(\frac{820}{20}\) = l
41 = l
length l = 41 cm
Perimeter = 2(l + b) units
= 2(41 + 20) cm
= 2(61) cm
= 122 cm

Question 12.
A square park has 40 m as its perimeter. What is the length of its side? Also find its area.
Solution:
perimeter = 40 m
4a = 40 m
a = \(\frac{40}{4}\)
Side a = 10 m
Area = a × a sq units
= 10 × 10 m²
= 100 m²

Question 13.
The scalene triangle has 40 cm as its perimeter and whose two sides are 13 cm and 15 cm, find the third side.
Solution:
Let the third side be C
perimeter = (a + b + c) units
40 = 13 + 15 + C
40 = 28 + C
C = 40 – 28
C = 12 units
C = 12 cm

Question 14.
A field is in the shape of a right angled triangle whose base is 25 m and height 20 m. Find the cost of levelling the field at the rate of Rs 45/- per sq.m²
Solution:
b = 25 m, h = 20 m
Area of the triangle = \(\frac{1}{2}\) × bh sq.units
= \(\frac{1}{2}\) × 25 × 20 m²
= 250 m²
Cost of levelling 1 m² = Rs 45
Cost of levelling 250 m² = Rs 45 × 250
= Rs. 11250

Question 15.
A square of side 2 cm is joined with a rectangle of length 15 cm and breadth 10 cm. Find the perimeter of the combined shape.
Solution:

Perimeter of the combined shape = Lengths of the outer boundaries
= (15 + 10 + 2 + 2 + 2 + 13 + 10) cm = 54 cm
Perimeter = 54 cm

Objective Type Questions

Question 16.
The following figures are of equal area. Which figure has the least perimeter?

Solution:
(b)

Question 17.
If two identical rectangles of perimeter 30 cm are joined together, then the perimeter of the new shape will be
(a) equal to 60 cm
(b) less than 60 cm
(c) greater than 60 cm
(d) equal to 45 cm
Solution:
(b) less than 60 cm

Question 18.
If every side of a rectangle is doubled, then its area becomes _____ times
(a) 2
(b) 3
(c) 4
(d) 6
Solution:
(c) 4
2l × 2b = 4l × b

Question 19.
The side of the square is 10 cm. If its side is tripled, then by how many times will its perimeter increase?
(a) 2 times
(b) 4 times
(c) 6 times
(d) 3 times
Solution:
(d) 3 times

Question 20.
The length and breadth of a rectangular sheet of paper are 15 cm and 12 cm respectively. A rectangular piece is cut from one of its corners. Which of the following statement is correct for the remaining sheet?
(a) Perimeter remains the same but the area changes
(b) Area remains the same but the perimeter changes
(c) There will be a change in both area and perimeter
(d) Both the area and perimeter remains the same.
Solution:
(a) Perimeter remains the same but the area changes

Students can download Maths Chapter 5 Statistics Ex 5.4 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.4

Miscellaneous Practice Problems

Question 1.
The heights (in centimeters) of 40 children are.

Prepare a tally marks table.
Solution:

Question 2.
There are 1000 students in a school. Data regarding the mode of transport of the students is given below. Draw a pictograph to represent the data.

Solution:
Mode of transport of the students
Scale: 1 Unit=100 Students

Question 3.
The following pictograph shows the total savings of a group of friends in a year. Each .picture represents a saving of Rs.100. Answer the following questions.

(i) What is the Ratio of Ruby’s saving to that of Thasnim’s?
(ii) What is the ratio of Kuzhali’s savings to that of others?
(iii) How much is Iniya’s savings?
(iv) Find the total amount of savings of all the friends?
(v) Ruby and Kuzhali save the same amount. Say True or False.
Solution:
(i) Ratio of Ruby’s saving to that of Thasnim’s
\(=\frac{\text { Ruby’s saving }}{\text { Thasnim’s saving }}=\frac{5 \times 100}{4 \times 100}=\frac{5}{4}=5: 4\)
Ratio of Ruby’s saving to that of Thasnims = 5 : 4
(ii) Ratio of Kuzhali’s savings to that of others
\(=\frac{\text { kuzhali’s saving }}{\text { others saving }}=\frac{5 \times 100}{19 \times 100}=\frac{5}{19}=5: 19\)
Ratio of Kuzhali’s saving to that of others = 5 : 19
(iii) Iniya’s saving = 3 × 100 = ₹ 300
(iv) Saving of all the friends = (5 + 7 + 4 + 5 + 3) × 100 = 24 × 100 = ₹ 2400.
Total savings = ₹ 2400
(v) True.

Challenging Problems

Question 4.
The table shows the numbers of moons that orbit each of the planets in oar solar system.

Make a Bar graph for the above data.
Solution:
Number of moons that orbit each of the planets in our solar system
Scale: 1 Unit = 2 moons

Question 5.
The prediction of the weather in the month of September is given below.

(i) Make a frequency table of the types of weather by reading the calendar.
(ii) How many days are either cloudy or partly cloudy?
(iii) How many days do not have rain? Give two ways to find the answer?
(iv) Find the ratio of the number of Sunny days to Rainy days.
Solution:
(i)
(ii) 14 days
(iii) 24 days (30 – 6 = 24 days)
(iv) 10 : 6

Question 6.
26 students were interviewed to find out what they want they to become in future. Their responses are given in the following table.

Represent this data using pictograph.
Solution:
Students responses in an interview about their future profession
Scale: 1 Unit = 1 student

Question 7.
Yasmin of class VI was given a task to count the number of books which are biographies, in her school library. The information collected by her is represented as follows.

Observe the pictograph and answer the following questions.
(i) Which title has the maximum number of biographies?
(ii) Which title has the minimum number of biographies?
(iii) Which title has exactly half the number of biographies as Novelists?
(iv) How many biographies are there on the title of sportspersons?
(v) What is the total number of biographies in the library?
Solution:
(i) ‘The title Novelists’ have the maximum number of biographies
(ii) ‘The title Scientists’ have a minimum number of biographies.
(iii) ‘Sportspersons’ title has exactly half the number of biographies as Novelist.
(iv) \((1 \times 20)+\frac{20}{4}=20+5=25\) biographies are there in the title sportsperson.
(v) 8 × 20 = 160 biographies are there in the library.

Question 8.
The bar graph illustrates the results of a survey conducted on vehicles crossing over a Toll Plaza in one hour.

Observe the bar graph carefully and fill up the following table.

Solution:
Vans = 50; Buses = 40; Cars = 65; Others = 15
Total Vehicles = 245

Question 9.
The lengths (in the nearest centimeter) of 30 drumsticks are given as follows.

Draw the bar graph showing the same information.
Solution:
The lengths (in nearest cm) of drumsticks
Scale : 1 Unit = 1 drumstick

Students can download Maths Chapter 5 Coordinate Geometry Ex 5.1 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 5 Coordinate Geometry Ex 5.1

Question 1.
Find the area of the triangle formed by the points
(i) (1,-1), (-4, 6) and (-3, -5)
(ii) (-10, -4), (-8, -1) and (-3, -5)
Solution:
(i) Let the vertices A (1, -1), B (-4, 6) and C (-3, -5)


= \(\frac { 1 }{ 2 } \) [(6 + 20 + 3) – (4 – 18 – 5)] = \(\frac { 1 }{ 2 } \) [29 – (-19)] = \(\frac { 1 }{ 2 } \) [29 + 19]
= \(\frac { 1 }{ 2 } \) × 48 = 24 sq. units.
Area of ∆ABC = 24 sq. units

(ii) Let the vertices be A(-10, -4), B(-8 -1) and C(-3, -5)

Area of ∆ABC = \(\frac { 1 }{ 2 } \)[(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(50 + 3 + 32) – (12 + 40 + 10)]

= \(\frac { 1 }{ 2 } \) [85 – (62)] = \(\frac { 1 }{ 2 } \) [23] = 11.5
Area of ∆ACB = 11.5 sq.units

Question 2.
Determine whether the sets of points are collinear?
(i) (-\(\frac { 1 }{ 2 } \),3)
(ii) (a,b + c), (b,c + a) and (c,a + b)
Solution:
(i) Let the points be A (-\(\frac { 1 }{ 2 } \),3), B (-5, 6) and C(-8, 8)
Area of ∆ABC = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₃ + x₁y₃)]
= \(\frac { 1 }{ 2 } \) [(- 3 – 40 – 24) – (-15 – 48 – 4)]

= \(\frac { 1 }{ 2 } \) [-67 + 67] = \(\frac { 1 }{ 2 } \) × 0 = 0
Area of a ∆ is 0.
∴ The three points are collinear.

(ii) Let the points be A (a, b + c), B (b, c + a) and C (c, a + b)
Area of the triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]

Since the area of a triangle is 0.
∴ The given points are collinear.

Question 3.
Vertices of given triangles are taken in order and their areas are provided aside. In each case, find the value of ‘p’

Solution:
(i) Let the vertices be A (0,0) B (p, 8), c (6, 2)
Area of a triangle = 20 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 20

\(\frac { 1 }{ 2 } \) [(0 + 2p + 0) – (0 + 48 + 0)] = 20
\(\frac { 1 }{ 2 } \) [2p – 48] = 20
2p – 48 = 40 ⇒ 2p = 40 + 48
p = \(\frac { 88 }{ 2 } \) = 44
The value of p = 44

(ii) Let the vertices be A (p, p), B (5, 6) and C (5, -2)
Area of a triangle = 32 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 32

\(\frac { 1 }{ 2 } \) [6p – 10 + 5p) – (5p + 30 – 2p)] = 32
\(\frac { 1 }{ 2 } \) [11 p – 10 – 3p – 30] = 32
11p – 10 – 3p – 30 = 64
8p – 40 = 64
8p = 64 + 40 ⇒ 8p = 104
p = \(\frac { 104 }{ 8 } \) = 13
The value of p = 13

Question 4.
In each of the following, find the value of ‘a’ for which the given points are collinear.
(i) (2,3), (4, a) and (6, -3)
(ii) (a, 2 – 2a), (-a + 1, 2a) and (-4 -a, 6 – 2a).
Solution:
(i) Let the points be A (2, 3), B(4, a) and C(6, -3).
Since the given points are collinear.
Area of a triangle = 0
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)] = 0)

\(\frac { 1 }{ 2 } \) [(2a – 12 + 18) – (12 + 6a – 6)] = 0
2a + 6 – (6 + 6a) = 0
2a + 6 – 6 – 6a = 0
-4a = 0 ⇒ a = \(\frac { 0 }{ 4 } \) = 0
The value of a = 0

(ii) Let the points be A (a, 2 – 2a), B (-a + 1, 2a) C (-4 -a, 6 – 2a).
Since the given points are collinear.
Area of a ∆ = 0

6a² – 2a – 2 – (-2a² – 6a + 2) = 0
6a² – 2a – 2 + 2a² + 6a – 2 = 0
8a² + 4a – 4 = 0 (Divided by 4)
2a² + a – 1 = 0
2a² + 2a – a – 1 = 0
2a (a + 1) – 1 (a + 1) = 0

(a + 1) (2a – 1) = 0
a + 1 = 0 (or) 2a – 1 = 0
a = -1 (or) 2a = 1 ⇒ a = \(\frac { 1 }{ 2 } \)
The value of a = -1 (or) \(\frac { 1 }{ 2 } \)

Question 5.
Find the area of the quadrilateral whose vertices are at
(i) (-9, -2), (-8, -4), (2, 2) and (1, -3)
(ii) (-9, 0), (-8,6), (-1, -2) and (-6, -3)
Solution:
(i) Let the vertices A (-9, -2), B(-8, -4), C(2, 2) and D(1, -3).
Plot the vertices in a graph.

[Note: Consider the points in counter clock wise order]

Area of the Quadrilateral ABDC = \(\frac { 1 }{ 2 } \) [36 + 24 + 2 – 4 – (16 – 4 – 6 – 18)]
= \(\frac { 1 }{ 2 } \) [58 – (-12)] – \(\frac { 1 }{ 2 } \)[58 + 12]
= \(\frac { 1 }{ 2 } \) × 70 = 35 sq. units 2
Area of the Quadrilateral = 35 sq. units

(ii) Let the vertices A(-9, 0), B(-8, 6), C(-1, -2) and D(-6, -3)
Plot the vertices in a graph and take them in counter – clock wise order.
Area of the Quadrilateral DCB

= \(\frac { 1 }{ 2 } \) [33 + 35] = \(\frac { 1 }{ 2 } \) × 68 = 34 sq. units
Area of the Quadrilateral = 34 sq. units

Question 6.
Find the value of k, if the area of a quadrilateral is 28 sq. units, whose vertices are (-4, -2), (-3, k), (3, -2) and (2, 3)
Solution:
Let the vertices A (-A, -2), B (-3, k), C (3, -2) and D (2, 3)
Area of the Quadrilateral = 28 sq. units
\(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)] = 28

-7k + 21 = 56
-7k = 56 – 21
-7k = 35 ⇒ 7k = – 35
k = – \(\frac { 35 }{ 7 } \) = -5
The value of k = -5

Question 7.
If the points A(-3, 9), B(a, b) and C(4, -5) are collinear and if a + b = 1, then find a and b.
Solution:
Since the three points are collinear
Area of a ∆ = 0

-3b – 5a + 36 – 9a – 4b – 15 = 0
-7b – 14a + 21 = 0
(÷ by 7) – b – 2a + 3 = 0
2a + b – 3 = 0

Substitute the value of a = 2 in (2) ⇒ 2 + b = 1
b = 1 – 2 = -1
The value of a = 2 and b = -1

Question 8.
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC . Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
Solution:
Let the vertices of the ∆ABC be A(x₁,y₁), B(x₂,y₂), C(x₃,y₃)

Question 9.
In the figure, the quadrilateral swimming pool shown is surrounded by concrete patio. Find the area of the patio.

Solution:

= \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₄ + x₄y₁) – (x₂y₁ + x₃y₂ + x₄y₃ + x₁y₄)]
= \(\frac { 1 }{ 2 } \) [(16 + 80 + 36 + 80) – (-64 – 24 – 100 – 24)]
= \(\frac { 1 }{ 2 } \) [212 – (-212)]

= \(\frac { 1 }{ 2 } \) [212 + 212] = \(\frac { 1 }{ 2 } \) [424] = 212 sq. units

= \(\frac { 1 }{ 2 } \) [90 – (-90)]

= \(\frac { 1 }{ 2 } \) [90 + 90]
= \(\frac { 1 }{ 2 } \) × 180 = 90 sq. units
Area of the patio = Area of the Quadrilateral ABCD – Area of the Quadrilateral EFGH
= (212 – 90) sq. units
Area of the patio = 122 sq. units

Question 10.
A triangular shaped glass with vertices at A(-5, -4), B(l, 6) and C(7, -4) has to be painted. If one bucket of paint covers 6 square feet, how many buckets of paint will be required to paint the whole glass, if only one coat of paint is applied.
Solution:
Given the vertices of the triangular glass is A (-5, -4), B (1, 6), and C (7, -4)


= \(\frac { 1 }{ 2 } \) [(20 + 42 – 4) – (-28 – 4 – 30)]
= \(\frac { 1 }{ 2 } \) [58 – (-62)]
= \(\frac { 1 }{ 2 } \) [58 + 62]
= \(\frac { 1 }{ 2 } \) × 120 = 60 sq. feet
Number of cans to paint 6 square feet = 1
∴ Number of cans = \(\frac { 60 }{ 6 } \) = 10 ⇒ Number of cans = 10

Question 11.
In the figure, find the area of
(i) triangle AGF
(ii) triangle FED
(iii) quadrilateral BCEG.

Solution:
Area of a triangle = \(\frac { 1 }{ 2 } \) [(x₁y₂ + x₂y₃ + x₃y₁) – (x₂y₁ + x₃y₂ + x₁y₃)]
(i) Area of ∆AGF = \(\frac { 1 }{ 2 } \) [(-2.5 – 13.5 – 6) – (-13.5 – 1 – 15)]
= \(\frac { 1 }{ 2 } \) [-22 – (-29.5)]

= \(\frac { 1 }{ 2 } \) [-22 + 29.5]
= \(\frac { 1 }{ 2 } \) × 7.5 = 3.75 sq.units

(ii) Area of ∆FED = \(\frac { 1 }{ 2 } \) [(-2 + 4.5 + 3) – (4.5 + 1 – 6)]
= \(\frac { 1 }{ 2 } \) [5.5 – (-0.5)]

= \(\frac { 1 }{ 2 } \) [5.5 + 0.5] = \(\frac { 1 }{ 2 } \) × 6 = 3 sq.units

(iii)

= \(\frac { 1 }{ 2 } \) [(4 + 2 + 0.75 + 9) – (-4 -1.5 – 4.5 -2)]
= \(\frac { 1 }{ 2 } \) [15.75 + 12]
= \(\frac { 1 }{ 2 } \) [27.75] = 13.875
= 13.88 sq. units

Students can download Maths Chapter 6 Information Processing Ex 6.3 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.3

Question 1.
How many Triangles are there in each of the following figures?

Solution:
(i) 12 triangles
(ii) 16 triangles
(iii) 32 triangles
(iv) 35 triangles

Question 2.
Find the number of dots in the tenth figure of the following patterns.

Solution:
(i) 55
(ii) 100

Question 3.

(i) Draw the next pattern.
(ii) Prepare a table for the number of dots used for each pattern.
(iii) Explain the pattern.
(iv) Find the number of dots in the 25th pattern.
Solution:

(iv) 350

Question 4.

Solution:
(i) 20 squares
(ii) 10 squares

Question 5.
How many circles are there in the following figure?

Solution:
7 circles

Question 6.
Find the minimum number of straight lines used in forming the following figures.

Solution:
(i) 10
(ii) 12

Students can download Maths Chapter 6 Information Processing Ex 6.2 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.2

Question 1.
In the following magic triangle, arrange the numbers from 1 to 6, so that you get the same sum on all its sides.

Solution:
One of the answers is.

Question 2.
Using the numbers from 1 to 9
(i) Can you form a magic triangle?
(ii) How many magic triangles can be formed?
(iii) What are the sums of the sides of the magic triangle?

Solution:
(i) Yes
(ii) 5
(iii) 17, 19, 20, 21, 23

Question 3.
Arrange the odd numbers from 1 to 17 without repetition to get a sum of 30 on each side of the magic triangle.

Solution:

Question 4.
Put the numbers 1, 2, 3, 4, 5, 6 & 7 in the circles so that each straight line of three numbers add up to the same total.

Solution:

Question 5.
Place the number 1 to 12 in the 12 circles so that the sum of the numbers in each of the six lines of the star is 26. Use each number from 1 to 12 exactly once. Find more possible ways?

Solution:

Students can download Maths Chapter 6 Information Processing Ex 6.1 Questions and Answers, Notes, Samacheer Kalvi 6th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.1

Question 1.
Suppose, you have two pairs of shorts, one is black and the other one is blue; three shirts which are white, blue and red. You again wish to make different combinations, but you always want to make sure that the shorts and shirt that you wear are of different colours. List and check how many combinations are possible now.
Solution:
6 combinations are possible
Black-White
Black-Blue
Black-Red
Blue-White
Blue-Blue
Blue-Red

Question 2.
You have two red and two blue blocks. How many different towers can you build that are four blocks high using these blocks? List all the possibilities.
Solution:
6 Possibilities,
R – B – R – B
R – R – B – B
B – R – R – B
B – R – B – R
B – B – R – R
R – B – B – R

Students can download Maths Chapter 1 Relations and Functions Ex 1.4 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.4

Question 1.
Determine whether the graph given below represent functions. Give reason for your answers concerning each graph.




Answer:

The vertical line cuts the graph at A and B. The given graph does not represent a function.

The vertical line cuts the graph at most one point P. The given graph represent a function.

The vertical line cuts the graph at three points S,T and U. The given graph does not represent a function.

The vertical line cuts the graph at most one point D. The given graph represents a function.

Question 2.
Let f: A → B be a function defined by
f(x) = \(\frac { x }{ 2 } \) – 1, where A = {2, 4,6,10,12},
B = {0,1,2,4,5,9}. Represent f by
(i) set of ordered pairs
(ii) a table
(iii) an arrow diagram
(iv) a graph
Answer:
A = {2,4,6, 10, 12}
B = {0,1, 2, 4, 5, 9}
f(x) = \(\frac { x }{ 2 } \) – 1
f(2) = \(\frac { 2 }{ 2 } \) – 1 = 1 – 1 = 0
f(4) = \(\frac { 4 }{ 2 } \) – 1 = 2 – 1 = 1
f(6) = \(\frac { 6 }{ 2 } \) – 1 = 3 – 1 = 2
f(10) = \(\frac { 10 }{ 2 } \) – 1 = 5 – 1 = 4
f(12) = \(\frac { 12 }{ 2 } \) – 1 = 6 – 1 = 5

(i) Set of ordered pairs
f = {(2, 0) (4, 1) (6, 2) (10, 4) (12, 5}

(ii) Table

X	2	4	6	10	12
f(x)	0	1	2	4	5

(iii) Arrow diagram

(iv) Graph

Question 3.
Represent the function f = {(1,2), (2,2), (3,2), (4,3),(5,4)} through (i) an arrow diagram (it) a table form (iii) a graph.
Answer:
f = {(1, 2) (2, 2) (3, 2) (4, 3) (5,4)}
Let A = {1,2, 3, 4, 5}
B = {2, 3, 4}

(i) Arrow diagram

(ii) Table form

X	1	2	3	4	5
f(x)	2	2	2	3	4

(iii) Graph

Question 4.
Show that the function f : N → N defined by f(x) = 2x – 1 is one-one but not onto.
Answer:
f: N → N
N = {1,2,3,4,5,… }
f(x) = 2x – 1
f(1) = 2(1) – 1 = 2 – 1 = 1
f(2) = 2(2) – 1 = 4 – 1 = 3
f(3) = 2(3) – 1 = 6 – 1 = 5
f(4) = 2(4) – 1 = 8 – 1 = 7
f(5) = 2(5) – 1 = 10 – 1 = 9
f = {(1,1) (2, 3) (3, 5) (4, 7) (5,9) …..}

(i) Different elements has different images. This function is one to one function.
(ii) Here Range is not equal to co-domain. This function not an onto function.
∴ The given function is one-one but not an onto.

Question 5.
Show that the function f: N ⇒ N defined by f(m) = m² + m + 3 is one-one function.
Answer:
N = {1,2,3, 4,5, ….. }
f(m) = m² + m + 3
f(1) = 1² + 1 + 3 = 5
f(2) = 2² + 2 + 3 = 9
f(3) = 3² + 3 + 3 = 15
f(4) = 4² + 4 + 3 = 23
f = {(1,5) (2, 9) (3, 15) (4, 23)}

From the diagram we can understand different elements in (N) in the domain, there are different images in (N) co-domain.
∴ The function is a one-one function.

Question 6.
Let A = {1, 2, 3, 4) and B = N. Letf: A → B be
defined by f(x) = x³ then,
(i) find the range off
(ii) identify the tpe of function
Solution:
A = {1, 2, 3, 4}
B = N
f: A → B,f(x) = x³
(i) f(1) = 1³ = 1
f(2) = 2³ = 8
f(3) = 3³ = 27
f(4) = 4³ = 64
(ii) Therange of f = {1, 8, 27, 64 )
(iii) It is one-one and into function.

Question 7.
In each of the following cases state whether the function is bijective or not. Justify your answer.
(i) f: R → R defined by f (x) = 2x + 1
(ii) f: R → R defined by f(x) = 3 – 4x²
Answer:
(i) f(x) = 2x + 1
f(0) = 2(0) + 1 = 0 + 1 = 1
f(1) = 2(1) + 1 = 2 + 1 = 3
f(2) = 2(2) + 1 = 4 + 1 = 5
f(3) = 2(3) + 1 = 6 + 1 = 7
Different elements has different images
∴ It is an one-one function.
It is also an onto function. The function is one-one and onto function.
∴ It is a bijective function.

(ii) f(x) = 3 – 4x²
f(1) = 3 – 4(1)²
= 3 – 4 = -1
f(2) = 3 – 4(2)² = 3 – 16 = – 13
f(3) = 3 – 4(3)² = 3 – 36 = – 33
f(4) = 3 – 4(4²) = 3 – 64 = – 61
It is not a bijective function. The positive numbers “R” do not have negative pre – image in X in R.

Question 8.
Let A= {-1,1}and B = {0,2}.
If the function f: A → B defined by
f(x) = ax + b is an onto function? Find a and b.
Answer:
A = {-1, 1}; B = {0,2}
f(x) = ax + b
f(-1) = a(-1) + b
0 = -a + b
a – b = 0 ….(1)
f(1) = a(1) + b
2 = a + b
a + b = 2 ….(2)
Solving (1) and (2) we get

Substitute a = 1 in (1)
The value of a = 1 and b = 1

Question 9.
If the function f is defined by

find the value of
(i) f(3)
(ii) f(0)
(iii) f(1. 5)
(iv) f(2) + f(-2)
Answer:
f(x) = x + 2 when x = {2,3,4,……}
f(x) = 2
f(x) = x – 1 when x = {-2}
(i) f(x) = x + 2
f(3) = 3 + 2 = 5

(ii) f(x) = 2
f(0) = 2

(iii) f(x) = x – 1
f(-1.5) = -1.5 – 1 = -2.5

(iv) f(x) = x + 2
f(2) = 2 + 2 = 4
f(x) = x – 1
f(-2) = – 2 – 1 = – 3
f(2) + f(-2) = 4 – 3
= 1

Question 10.
A function f: [-5, 9] → R is defined as follows:

Answer:
f(x) = 6x + 1 ; x = {-5,-4,-3,-2,-1,0,1}
f(x) = 5x² – 1 ; x = {2, 3, 4, 5}
f(x) = 3x – 4 ; x = {6, 7, 8, 9}

(i) f(-3) + f(2)
f(x) = 6x + 1
f(-3) = 6(-3) + 1 = -18 + 1 = -17
f(x) = 5x² – 1
f(2) = 5(2)² – 1 = 20 – 1 = + 19
f(-3) + f(2) = – 17 + 19
= 2

(ii) f(7) – f(1)
f(x) = 3x – 4
f(7) = 3(7) – 4 = 21 – 4 = 17
f(x) = 6x + 1
f(1) = 6(1) + 1 = 6 + 1 = 7
f(7) – f(1) = 17 – 7
= 10

(iii) 2f(4) + f(8)
f(x) = 5x² – 1
f(4) = 5(4)2 – 1 = 5(16) – 1
= 80 – 1 = 79
f(x) = 3x – 4
f(8) = 3(8) – 4 = 24 – 4 = 20
2f(4) + f(8) = 2(79) + 20
= 158 + 20
= 178

f(x) = 6x + 1
f(-2) = 6(-2) + 1 = -12 + 1 = -11
f(x) = 3x – 4
f(6) = 3(6) – 4 = 18 – 4 = 14
f(x) = 5x² – 1
f(4) = 5(4)² – 1 = 5(16) – 1
= 80 – 1 = 79
f(x) = 6x + 1
f(-2) = 6(-2) + 1 = -12 + 1 = -11

Question 11.
The distance S an object travels under the influence of gravity in time t seconds is given by S(t) = \(\frac{1}{2}\) gt² + at + b where, (g is the acceleration due to gravity), a, b are constants. Check if the function S (t)is one-one.
Solution:
S(t) = \(\frac{1}{2}\) gt² + at + b
Let t be 1, 2, 3, ………, seconds
S(1) = \(\frac{1}{2}\) g(1²) + a(1) + b = \(\frac{1}{2}\) g + a + b
S(2) = \(\frac{1}{2}\) g(2²) + a(2) + b = 2g + 2a + b
Yes, for every different values of t, there will be different values as images. And there will be different preimages for the different values of the range. Therefore it is one-one function.

Question 12.
The function ‘t’ which maps temperature in Celsius (C) into temperature in Fahrenheit (F) is defined by
t(C) = F where F = \(\frac { 9 }{ 5 } \) C + 32. Find,
(i) t(0)
(ii) t(28)
(iii) t(-10)
(iv) the value of C when t(C) = 212
(v) the temperature when the Celsius value is equal to the Fahrenheit value.
Answer:
Given t(C) = \(\frac { 9C }{ 5 } \) + 32

(i) t(0) = \(\frac { 9(0) }{ 5 } \) + 32
= 32° F

(ii) t(28) = \(\frac { 9(28) }{ 5 } \) + 32
= \(\frac { 252 }{ 5 } \) + 32
= 50.4 + 32
= 82.4° F

(iii) t(-10) = \(\frac { 9(-10) }{ 5 } \) + 32
= -18 + 32
= 14° F

(iv) t(C) = 212
\(\frac { 9C }{ 5 } \) + 32 = 212
\(\frac { 9C }{ 5 } \) = 212 – 32
= 180
9C = 180 × 5
C = \(\frac{180 \times 5}{9}\)
= 100° C

(v) consider the value of C be “x”
t(C) = \(\frac { 9C }{ 5 } \) + 32
x = \(\frac { 9x }{ 5 } \) + 32
5x = 9x + 160
-160 = 9x – 5x
-160 = 4x
x = \(\frac { -160 }{ 4 } \) = -40
The temperature when the Celsius value is equal to the fahrenheit value is -40°

Composition of two Functions

Let f: A → B and g: B → C be two functions. Then the composition of f and g denoted by gof is defined as the function gof (x) = g[f(x)] for all x ∈ A.

Composition of three Functions

Let A, B, C, D be four sets and let f: A → B; g : B → C and h : C → D be three functions, using composite functions fog and goh, we get two new functions like (fog) oh and fo (goh).
Note: Composition of three function is always associative.