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You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 5 Information Processing Ex 5.1

Question 1.
Study and complete the following pattern.

Solution:

Question 2.
Find next three numbers in the following number patterns,
i) 50, 51, 53, 56, 60,…
ii) 77, 69, 61, 53,…
iii) 10, 20, 40, 80, …..
iv) \(\frac{21}{33}, \frac{321}{444}, \frac{4321}{5555}\)
Solution:
i) The pattern generating these numbers is
50, 50 + 1, 51 + 2, 53 + 3, 56 + 4, 60 + 5, 65 + 6, 71 + 7,
∴ 50, 51, 53, 56, 60, 65, 71, 78, ……
∴ The next three numbers will be 65, 71, 78

ii) The pattern generating these numbers is
77, 77 – 8, 69 – 8, 61 – 8, 53 – 8, 45 – 8, 37 – 8, 29
77, 69, 61, 53, 45, 37, 29, 21,
∴ The next three numbers will be 45, 37, 29.

iii) The pattern generating these numbers is
10, 10 + 10, 20 + 20, 40 + 40, 80 + 80, 160 + 160, 320 + 320,….
10, 20, 40, 80, 160, 320, 640,….
∴ The next three numbers will be 160, 320, 640.

iv) The pattern generating these numbers is

Question 3.
Consider the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,… Observe and complete the following table by understanding the number pattern followed. After filling the table discuss the pattern followed in addition and subtraction of the numbers of the sequence.

Solution:

Question 4.
Complete the following patterns.

Solution:

Rotate the figure 90°C clockwise to get the next figure.

Solution:

* Move the circled arrow to each comer clockwise and place the square to the opposite comer of the arrow, in first row.
* Copy the first row in the second row.

Solution:

Rotate the arrow inside the square 90° anticlockwise direction so that the arrow pointing each comer anticlockwise.

Question 5.
Find HCF of the following pair of numbers by Euclid’s game.
(i) 25 and 35
(ii) 36 and 12
(iii) 15 and 29
Solution:
(i) HCF of (25, 35 – 25)
25 = 5 × 5
10 = 2 × 5
HCF of (25, 10) = 5

(ii) HCF of (36, 36 – 12)
36 = 2 × 2 × 3 × 3
24 = 2 × 2 × 2 × 3
HCF of (36, 24) = 2 × 2 × 3 = 12

(iii) HCF of (15, 29 -15)
15 = 3 × 5 × 1
14 = 2 × 7 × 1
HCF of (15, 14) = 1

Question 6.
Find HCF of 48 and 28. Also find the HCF of 48 and the number Obtained by finding their difference.
Solution:

Question 7.
Give instructions to fill in a bank withdrawal form issued in a bank.
Solution:

• The name should be written in capital letters from left to right.
• Write the date of withdrawal on the right top comer of the form.
• Write the amount (in words) to be withdrawn in the space provided.
• Write the amount (in figures) to be withdrawn in the box provided.
• Put your signature at the right bottom above the ‘signature of the depositor’.

Question 8.
Arrange the name of your classmates alphabetically.
Solution:
Name of my classmates is given below.

1. Akila
2. Akshaya
3. Bharathi
4. Divya
5. Ezhil
6. Fathima
7. Gayathri
8. Hlelen
9. Irusammal
10. Joy
11. Kaviya
12. Lakshmi
13. Monika
14. Nisha
15. Olin
16. Patsy
17. Queenlin
18. Ratha
19. Sindhu
20. Vidhya

Question 9.
Follow and execute the instructions given below?
i) Write the number 10 in the place common to the three figures
ii) Write the number 5 in the place common for square and circle only.
iii) Write the number 7 in the place common for triangle and circle only.
iv) Write the number 2 in the place common for triangle and square only.
v) Write the numbers 12,14 and 8 only in square, circle and triangle respectively.

Solution:

Question 10.
Fill in the following information.

Solution:

Objective Type Questions

Question 11.
The next term in the sequence 15, 17, 20, 22, 25, is
(a) 28
(b) 29
(c) 27
(d) 26
Solution:
(c) 27

Question 12.
What will be the 25th letter in the pattern? ABCAABBCCAAABBBCCC
(a) B
(b) C
(c) D
(d) A
Hint:
Write A, B, C with increasing number of A, B, and C
Solution:
(a) B

Question 13.
The difference between 6th term and 5th term in the Fibonacci sequence is
(a) 6
(b) 8
(c) 5
(d) 3
Solution:
(d) 3

Question 14.
The 11^th term in the Lucas sequence 1, 3, 4, 7,… is
(a) 199
(b) 76
(c) 123
(d) 47
Solution:
(a) 199

Question 15.
If the Highest Common Factor of 26 and 54 is 2, then HCF of 54 and 28 is
(a) 26
(b) 2
(c) 54
(d) 1
Solution:
(b) 2

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 5 Statistics Ex 5.3

Question 1.
Read the given Bar Graph which shows the percentage of marks obtained by Brinda in different subjects in an assessment test.

Observe the bar graph and answer the following questions.
(i) 1 unit = ___ % of marks on vertical line.
(ii) Brinda has scored maximum marks in _____ subject.
(iii) Brinda has scored minimum marks in ______ subject.
(iv) The percentage of marks scored by Brinda in Science is _____
(v) Brinda scored 60% marks in the subject ______
(vi) Brinda scored 20% more in _______ subject than _____ subject.
Solution:
(i) 10
(ii) Mathematics
(iii) Language
(iv) 65%
(v) English
(vi) Mathematics, English

Question 2.
Chitra has to buy Laddus in order to distribute to her friends as follows:

Draw a Bar Graph for this data.
Solution:

Question 3.
The fruits liked by the students of a class are as follows:

Draw a Bar Graph for this data.
Solution:

Question 4.
The pictograph below gives the number of absentees on different days of the week in class six. Draw the Bar graph for the same.

Solution:

Objective Type Questions

Question 5.
A bar graph can be drawn using
(a) Horizontal bars only
(b) Vertical bars only
(c) Both horizontal bars and Vertical bars
(d) Either horizontal bars or vertical bars
Solution:
(d) Either horizontal bars or vertical bars

Question 6.
The spaces between any two bars in a bar graph _____
(a) can be different
(b) are the same
(c) are not the same
(d) all of these
Solution:
(b) are the same

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.3

Miscellaneous Practice Problems

Question 1.
Every even number greater than 2 can be expressed as the sum of two prime numbers. Verify this statement for every even number up to 16.
Solution:
Even numbers greater then 2 upto 16 are 4, 6, 8, 10, 12, 14 and 16
4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7 (or) 5 + 5
12 = 5 + 7
14 = 7 + 7 (or) 3 + 11
16 = 5 + 11 (or) 3 + 13

Question 2.
Is 173 a prime? Why?
Solution:
yes, because it has two factors.

Question 3.
For which of the numbers, from n = 2 to 8, is 2n – 1 a prime?
Solution:

For n = 2, 3, 4, 6 and 7 it is prime.

Question 4.
Explain your answer with the reason for the following statements.
(i) A number is divisible by 9, if it is divisible by 3.
(ii) A number is divisible by 6, if it is divisible by 12.
Solution:
(i) False 42 is divisible by 3 but it is not divisible by 9
(ii) True 36 is divisible by 12. Also divisible by 6.

Question 5.
Find A as required
(i) The greatest 2 digit number 9A is divisible by 2.
(ii) The least number 567A is divisible by 3.
(iii) The greatest 3 digit number 9A6 is divisible by 6.
(iv) The number A08 is divisible by 4 and 9.
(v) The number 225A85 is divisible by 11.
Solution:
(i) A number is divisible by 2 if it is an even number.
Greatest 2 digit even number is 98.
∴ A = 8

(ii) A number is divisible by 3 if the sum of its digits is divisible by 3
Sum of digits of 567A = 5 + 6 + 7+ A = 18 + A
∴ 18 is divisible by 3
∴ A maybe 0
The number will be 5670

(iii) A number is divisible by 6 if it is divisible by both 2 and 3
9A6 is even and so divisible by 2
If A = 9 then the sum of digits will be = 24 which is divisible by 3.
The number will be 996 and A = 9

(iv) 08 is divisible by 4, so A08 is divisible by 4.
If A = 1 then the sum of digits will be 9 which is divisible by 9.
The number will be 108 and A = 1

(v) 5 + A + 2 – (8 + 5 + 2) = 7 + A – 15 = -8 + A
∴ A = 8

Question 6.
Numbers divisible by 4 and 6 are divisible by 24. Verify this statement and support your answer with an example.
Solution:
False 12 is divisible by both 4 and 6. But not divisible by 24

Question 7.
The sum of any two successive odd numbers ir always divisible by 4. Justify this statement with an example.
Solution:
True.
The sum of any two consecutive odd numbers is divisible by 4
For example 11 + 13 = 24, divisible by 4
Also, all the consecutive odd numbers are of the form 4n + 1 or 4n + 3
Their sum = 4x + 4 which is divisible by 4.

Question 8.
Find the length of the longest rope that can be used to measure exactly the ropes of length lm 20 cm, 3m 60 cm and 4 m.
Solution:
Length of ropes are 4 m, 3 m 60 cm and lm 20 cm = 400 cm, 360 cm, 120 cm
Finding HCF (400, 360, 120)
10, 9 and 3 has no common divisor
HCF (400, 360, 120) = 2 × 2 × 2 × 5 = 40
The length of the rope will be 40 cm

Challenge Problems

Question 9.
The sum of three prime numbers is 80. The difference of two of them is 4. Find the numbers.
Solution:
Three prime numbers 2, 37, 41
Sum 2 + 37+ 41 = 80
The difference between two of them 41 – 37 = 4

Question 10.
Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single-digit numbers.
Solution:
Prime numbers between 10 and 20 are 11, 13, 17 and 19
Sum = 11 + 13 + 17 + 19 = 60
60 is divisible by 1, 2, 3, 4, 5 and 6.

Question 11.
Find the smallest number which is exactly divisible by all the numbers from 1 to 9.
Solution:
To find the smallest number we have to Find the LCM (1, 2, 3, 4, 5, 6, 7, 8, 9)
LCM is 2 × 3 × 2 × 5 × 7 × 2 × 3 = 2520
The required number is 2520

Question 12.
The product of any three consecutive number is always divisible by 6. Justify this statement with an example.
Solution:
2 × 3 × 4 = 24 is divisible by 6.

Question 13.
Malarvizhi, Karthiga and Anjali are friends and natives of the same village. They work in different places. Malarvizhi comes to her home once in 5 days. Similarly, Karthiga and Anjali come to their homes once in 6 days and 10 days respectively. Assuming that they met each other on the 1st of October, when will all the three again?
Solution:
Find the LCM (5, 6, 10)
LCM (15, 25, 30) = 5 × 6 = 30
They meet again after 30 days
∴ They met on 1st October
They will meet again on 31st October

Question 14.
In an apartment consisting of 108 floors, two lifts A & B starting from the ground floor, stop at every 3rd and 5 th floors respectively. On which floors, will both of them stop together?
Solution:
LCM of 3 and 5 = 3 × 5 = 15
The lifts stop together at floors 15, 30, 45, 60, 75, 90 and 105.

Question 15.
The product of 2 two-digit numbers is 300 and their HCF is 5. What are the numbers?
Solution:
Given that HCF of 2 numbers is 5
The numbers may like 5x and 5y
Also given their product = 300
5x × 5y = 300
⇒ 25xy = 300
⇒ \(x y=\frac{300}{25}\)
⇒ xy = 12
The possible values of x andy be (1, 12) (2, 6) (3, 4)
The numbers will be (5x, 5y)
⇒ (5 × 1, 5 × 12) = (5, 60)
⇒ (5 × 2, 5 × 6) = (10, 30)
⇒ (5 × 3, 5 × 4) = (15, 20)
(5, 60) is impossible because the given the numbers are two digit numbers.
The remaining numbers are (10, 30) and (15, 20)
But given that HCF is 5
(10, 30) is impossible, because its HCF = 10
The numbers are 15, 20

Question 16.
Find whether the number 564872 is divisible by 88?
Solution:
564872 Divisibility by 8
564872 It is divisible by 8
Divisibility by 11
5 + 4 + 7 = 16
6 + 8 + 2 = 16
16 – 16 = 0
It is divisible by both 8 and 11 and hence divisible by 88.

Question 17.
Wilson, Mathan and Guna can complete one round of a circular track in 10, 15 and 20 minutes respectively. If they start together at 7 a.m from the starting point, at what time will they meet together again at the starting point?
Solution:
LCM (10, 15, 20) = 60
∴ They will meet at 8 a.m

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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
Completing the unknown values in the table.

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

Solution:
Perimeter of a square = (4 × side) units
Area of a square = (side × side) unit²
(i) s = 6 cm
Perimeter = 4s units = 4 × 6 cm = 24 cm
P = 24 cm
Area = s × s unit² = 6 × 6 cm² = 36 cm²
A = 36 cm²

(ii) Perimeter = 4 × s unit
100 = (4 × s) m
\(\frac{100}{4}\) = s
s = 25 m
Area = s × s unit²= 25 × 25 m² = 625m²
A = 625m²

(iii) Area = s × s unit²
49 = s × s square feet
s² = 7²
s = 7 feet
Perimeter = 4 × s unit = 4 × 7 feet = 28 feet
Perimeter = 28 feet
Completing the unknown values in the table

Question 3.
The table given below contains some measures of the 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
Tabulating the unknown values

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

Solution:
Perimeter of a triangle = sum of three sides.
(i) Perimeter = 6 + 5 + 2 cm = 13 cm
p = 13 cm

(ii) Perimeter = (side 1 + side 2 + side 3) m
17 = (side 1 + 8 + 3) m
17 m = (side 1 + 11) m
side 1 = 17 – 11 = 6m

(iii) Perimeter = side 1 + side 2 + side 3
28 feet = 11 feet + side 2 + 9 feet
28 ft = 20 feet + side 2
28 – 20 = side 2
side = 8 feet
Tabulating the unknowns.

Question 5.
Fill in the blanks.
i) 5 cm² = mm²
Hint: 1 cm² = 100 mm²
ii) 26 m² = cm²
Hint: 1 m² = 10000
iii) 8 km² = m²
Hint 4 1 km²– 1000000 m²
Solution:
(i) 500
(ii) 2,60,000
(iii) 80,00,000

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) cm = 48 cm
Perimeter = 48 cm
Area of 5 squares of side 4 cm
Area of a square = (side × side) unit²
∴ A = 5 × (4 × 4) cm² = 5 × 16 cm2 = 80 cm²
80 cm²

(ii) Perimeter = (4 + 5 + 4 + 5 + 4 + 5 + 4 + 5)cm = 36cm
Perimeter = 36 cm
Area of a square of side 3cm + Area of 4 right triangles
= (3 × 3) + [4 × \(\frac{1}{2}\) × 4 × 3] cm² = (9 + 24) cm² = 33 cm²
Area = 33 cm²

(iii) Perimeter = (50 + 12 + 13 + 40 + 10 + 10 + 10 + 5) cm = 150 cm
Perimeter = 150 cm
Area = Area of a rectangle + Area of a square + Area of a right triangle.
= (l × b) + (s × s) + ( \(\frac{1}{2}\) × b × h) cm²
= (50 × 5) + (10 × 10) + \(\frac{1}{2}\) × 12 × 5) cm²
= (250 + 100 + 30) cm² = 380 cm²
Area = 380 cm²

Question 7.
Find the perimeter and area of the rectangle whose length is 6 m and breadth 4 m.
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 the area of the square whose side is 8 cm.
Solution:
Perimeter of a square = (4 × side) units
Side = 8 cm
∴ Perimeter = 4 × 8 cm = 32 cm
Perimeter = 32 cm
Area of a square = (side × side) unit² = (8 × 8) cm² = 64 cm²
Area = 64 cm²

Question 9.
Find the perimeter and area of the 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 side 6 cm.
Solution:
i) Perimeter of a scalene triangle = (7 + 8 + 10) m = 25 m
ii) The three sides of the isosceles triangle are 10 cm, 10 cm and 7 cm
∴ Perimeter = (10 + 10 + 7) cm = 27 cm
iii) An equilateral triangle with side 6 cm.
The sides of equilateral triangle are 6 cm, 6 cm and 6 cm
∴ Perimeter = (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:
Given perimeter = 40 m
Perimeter of a square = 4 × Length of a side
40 = 4 × Length of a side
∴ Length of its side = \(\frac{40}{4}\) m = 0 m
∴ Side of the park = 10m
Area of a square = (Side × side) unit² = (10 × 10) m² = 100 m²
∴ Area of the Park = 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 right angled triangle whose base is 25 m and height 20 m. Find the cost of levelling the field at the rate of ₹ 45/- per sq. m.
Solution:
Area of a right angled triangle = \(\frac{1}{2}\) × (base × height) unit²
base = 25 m
height = 20 m
∴ Area = \(\frac{1}{2}\) × (25 × 20)
Area = 250 m²
Cost of levelling per m² = ₹ 45.
∴ Cost of levelling 250 m² = 250 × 45 = ₹ 11,250
Cost of levelling = ₹ 11,250

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:

Hint:
(a) 12 units
(b) 10 units
(c) 12 units
(d) 12 units

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
Hint:

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

Question 19.
The side of a 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
SolutionL
(d) 3 times
30 × 4 = 120 = 3 × 40

Question 20.
The length and breadth of a rectangular sheet of a 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:
(c) There will be a change in both area and perimeter
Hint:

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 1 Numbers Ex 1.1

Question 1.
Fill in the blanks.
(i) The number of prime numbers between 11 and 60 is _______
(ii) The numbers 29 and _______ are twin primes.
(iii) 3753 is divisible by 9 and hence divisible by _______
(iv) The number of distinct prime factors of the smallest 4 digit number is______
(v) The sum of distinct prime factors of 30 is ________
Solution:
(i) 12
(ii) 31
(iii) 3
(iv) 2
(v) 10

Question 2.
Say True or False
(i) The sum of any number of odd numbers is always even.
(ii) Every natural number is either prime or composite.
(iii) If a number is divisible by 6, then it must be divisible by 3.
(iv) 16254 is divisible by each of 2, 3, 6 and 9.
(v) The number of distinct prime factors of 105 is 3.
Solution:
(i) False
(ii) False
(iii) True
(iv) True
(v) True

Question 3.
Write the smallest and the biggest two digit prime number.
Solution:
Smallest is 11
Biggest is 97

Question 4.
Write the smallest and the biggest three-digit composite number.
Solution:
Smallest three-digit composite number – 100
Biggest three-digit composite number – 999

Question 5.
The sum of any three odd natural numbers is odd. Justify this statement with an example.
Solution:
True, as we know, that,“the sum of any three odd numbers is always an odd number”.
Example: 3 + 7 + 9 = 19 is odd.

Question 6.
The digits of the prime number 13 can be reversed to get another prime number 31. Find if any such pairs exist up to 100.
Solution:
(17, 71), (37, 73) and (79, 97)

Question 7.
Your friend says that every odd number is prime. Give an example to prove him/her wrong.
Solution:
15 is an odd number not prime.

Question 8.
Each of the composite numbers has at least three factors. Justify this statement with an example.
Solution:
True. The composite number 4 has 3 factors namely 1, 2 and 4.

Question 9.
Find the dates of any month of a calendar which are divisible by both 2 and 3.
Solution:
Every month the dates 6, 12, 18, 24 and 30 (excluding February) are divisible by both 2 and 3.

Question 10.
I am a two-digit prime number and the sum of my digits is 10.1 am also one of the factors of 57. Who am I?
Solution:
19

Question 11.
Find the prime factorisation of each number by factor tree method and division method.
(a) 60
(b) 128
(c) 144
(d) 198
(e) 420
(f) 999
Solution:
(a) 60

∴ 60 = 2 × 2 × 3 × 5
Also

∴ 60 = 2 × 2 × 3 × 5
(b) 128

∴ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Also

∴ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
(c) 144

∴ 144 = 2 × 2 × 2 × 2 × 3 × 3
Also

∴ 144 = 2 × 2 × 2 × 2 × 3 × 3
(d) 198

∴ 198 = 2 × 3 × 3 × 11
Also

∴ 198 = 2 × 3 × 3 × 11
(e) 420

∴ 420 = 2 × 2 × 3 × 5 × 7
Also

∴ 420 = 2 × 2 × 3 × 5 × 7
(f) 999

∴ 999 = 3 × 3 × 3 × 37
Also

∴ 999 = 3 × 3 × 3 × 37

Question 12.
If there are 143 math books to be arranged in equal numbers in all the stacks, then find the number of books in each stack and also the number of stacks.
Solution:
Total number of books = 143
Factorizing 143 = 11 × 13

A number of stacks and number of books in each stack may be (11, 13) or (13, 11).

Objective Type Questions

Question 13.
The difference between two successive odd number is
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(b) 2

Question 14.
The only even prime number is
(a) 4
(b) 6
(c) 2
(d)
Solution:
(c) 2

Question 15.
Which of the following numbers is not prime?
(a) 53
(b) 92
(c) 97
(d) 71
Solution:
(b) 92

Question 16.
The sum of the factors of 27 is
(a) 28
(b) 37
(c) 40
(d) 31
Solution:
(c) 40

Question 17.
The factors of a number are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80. What is the number?
(a) 80
(b) 100
(c) 128
(d) 160
Solution:
(a) 80

Question 18.
The prime factorisation of 60 is 2 × 2 × 3 × 5. Any other number which has the same prime factorisation as 60 is
(a) 30
(b) 120
(c) 90
(d) impossible
Solution:
(d) impossible

Question 19.
If the number 6354*97 is divisible by 9, then the value * is
(a) 2
(b) 4
(c) 6
(d) 7

Solution:
(a) 2

Question 20.
The number 87846 is divisible by
(a) 2 only
(b) 3 only
(c) 11 only
(d) all of these
Solution:
(d) all of these

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 4 Geometry 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) A Scalene triangle has no line of symmetry

(ii) An isosceles triangle has only one

(iii) An equilateral triangle has three lines of symmetry.

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:
Number of lines of symmetry is 0; order of rotation is 2

ii)
Number of lines of symmetry is 1.
Order of rotation is 0, because it is an isosceles triangle

iii) Number of lines of symmetry are 2
Order of rotation is 2

iv) Number of lines of symmetry 8 and order of rotation 8

v)

Number of line of symmetry is 1 and order of rotation 0

Question 4.
The three digit number 101 has rotational and reflection symmetry. Give five more examples of three digit number which have both rotational and reflection symmetry.
Solution:
181, 111, 808, 818, 888

Question 5.
Translate the given pattern and com ilete the design in rectangular strip?

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 have reflection symmetry (any three ways).
Solution:
The required combination of squares are given below.

Question 8.
Draw the following:
Solution:
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.

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 rational symmery is _____.

Question 10.
Color the boxes in such a way that it possess translation symmetry.

Solution:
The following figures coloured to posses translation symmetry.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 3 Ratio and Proportion Ex 3.5

Miscellaneous Practice Problems

Question 1.
The maximum speed of some of the animals are given below:
the Elephant = 20 km/h, the Lion = 80 km/hr; the Cheetah = 100 km/h.
Find the following ratios of their speeds in simplified form and find which ratio is the least?
(i) the Elephant and the Lion
(ii) the Lion and the Cheetah
(iii) the Elephant and the Cheetah.
Solution:

∴ The ratio of Elephant to Cheetah is the least.

Question 2.
A particular high school has 1500 students, 50 teachers and 5 administrators. If the school grow s to 1800 students and the ratios are mentioned, then find the number of teachers and administrators.
Solution:
The ratio of Students : teachers : administrators = 1500 : 50 : 5.
The new ratio of Students : teachers : administrators = 1800 : new teachers : new administrators.
Given the ratios are maintained. i.e., they are proportional.

∴ For 1800 students 60 teachers and 6 administrators are needed.

Question 3.
I have a box which has 3 green, 9 blue, 4 yellow, 8 orange coloured cubes in it.
(a) What is the ratio of orange to yellow cubes?
(b) What is the ratio of green to blue cubes?
(c) How many different ratios can be formed, when you compare each colour to any one of the other colours?

Solution:
Number of green cubes = 3
Number of blue cubes = 9
Number of yellow cubes = 4
Number of orange cubes = 8
(a) Ratio of orange to yellow cubes \(\frac{\text { Number of orange cubes }}{\text { Number of yellow cubes }}=\frac{8}{4}=\frac{2}{1}=2: 1\)
Ratio of orange to yellow cubes = 2 : 1
(b) \(\frac{\text { Number of green cubes }}{\text { Number of blue cubes }}=\frac{3}{9}=\frac{1}{3}\)
Ratio of green to blue cubes = 1 : 3
(c) The ratios can be Orange : Yellow, Orange: blue, Orange : green, Yellow : Orange, yellow : blue, yellow : green, blue : green, blue : orange, blue : yellow, green : orange, green : yellow, green : blue. Thus 12 ratios can be formed.

Question 4.
A gets double of what B gets and B gets double of what C gets. Find A : B and B : C and verify whether the result is in proportion or not.
Solution:
A : B = 2 : 1
B : C = 2 : 1
They are in proportion

Question 5.
The ingredients required for the preparation of Ragi Kali, a healthy dish of Tamilnadu is given below.

(a) If one cup of ragi flour is used then, what would be the amount of raw rice required?
(b) If 16 cups of water are used, then how much of ragi flour should be used?
(c) Which of these ingredients cannot be expressed as a ratio? Why?
Solution:
(a) Given Ragi flour: raw rice = 4 : 1
If one cup of ragi flour used then the ratio of ragi flour : raw rice = \(\frac{4}{4}: \frac{1}{4}=1: \frac{1}{4}\)
Raw rice required = \(\frac{1}{4}\) cup
(b) Ratio of water : ragi flour = 8 : 4
If 16 cups of water used then ratio of water : ragi flour = 8 × 2 : 4 × 2 = 16 : 8
8 cups of ragi flour are used.
(c) We know that the quantities of the same units can be compared. Here Ragi flour, Raw rice and water are in one unit, Sesame oil and salt are in different units, these different units cannot be compared and cannot be expressed as a ratio.

Challenging Problem (Textbook Page No. 74)

Question 6.
Antony brushes his teeth in the morning and night on all days in a week. Shabeen brushes her teeth only in the morning. What is the ratio of the number of times they brush their teeth in a week?
Solution:
Number of times = 14 : 7 = 2 : 1

Question 7.
Thirumagal’s mother wears a bracelet made of 35 red beads and 30 blue beads. Thirumagal wants to make smaller bracelets using the same two coloured beads in the same ratio. In how many different ways can she make the bracelets?

Solution:
Ratio of red beads : blue beads = 35 : 30 = 7 : 6
Also given that the bracelet made is smaller in size
∴ The possible ways are red beads : blue beads = 7 : 6 (or) 14 : 12 (or) 21 : 18 (or) 28 : 24
In 4 different way can she made.

Question 8.
Team A wins 26 matches out of 52 matches. Team B wins three-fourth of 52 matches played. Which team has a better winning record?
Solution:
Team A = \(\frac{26}{52}\) = \(\frac{1}{2}\)
Team B = \(\frac{3}{4}\) × 52 = 39
Team B has a better winning record.

Question 9.
In a school excursion, 6 teachers and 12 students from 6th standard and 9 teachers and 27 students from 7th standard, 4 teachers and 16 students from 8th standard took part. Which class has the least teacher to student ratio?
Solution:
Teacher to Student Ratio of 6th Std = 6 : 12 = 1 : 2
Teacher to Student Ratio of 7th Std = 9 : 27 = 1 : 3
Teacher to Student Ratio of 8th Std = 4 : 16 = 1 : 4
In standard 8th the ratio of teacher to student is 1 : 4 and which is the least.

Question 10.
Fill the blanks using any set of suitable numbers 6 : …… :: ……. : 15
Solution:
6 : ……. = …….. : 15
Product of the extremes = 6 × 15 = 90
Set of suitable numbers
1 and 90, 2 and 45, 3 and 30, 5 and 18, 6 and 15

Question 11.
From your school diary, write the ratio of the number of holidays to the number of working days in the current academic year.
Solution:
Number of working days in an academic year = 220
Number of holidays = 365 – 220 = 145
\(\frac{\text { Number of holidays }}{\text { Number of working days }}=\frac{145}{220}=\frac{29}{44}\)
Number of holidays : No of working days = 29 : 44

Question 12.
If the ratio of Green, Yellow and Black balls in a bag is 4 : 3 : 5, then
(a) Which is the most likely ball that you can choose from the bag?
(b) How many balls in total are there in the bag if you have 40 black balls in it?
(c) Find the number of green and yellow balls in the bag.
Solution:
Green : Yellow : Black = 4 : 3 : 5
(i) Blackballs;
(ii) 96 balls (32 + 24 + 40);
(iii) green balls = 32
yellow balls = 24

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 2 Introduction to Algebra Ex 2.1

Question 1.
Fill in the blanks:
(i) The letters a, b, c, .., x, y, z are used to represent _____
(ii) A quantity that takes _____ values is called a variable.
(iii) If there are 5 students on a bench, then the number of students in ‘n’ benches is ‘5 × n’. Here _____ is a variable.
Solution:
(i) Variables
(ii) Different
(iii) n

Question 2.
Say True or False:
(i) The length of part B in the pencil shown is ‘a – 6’.

(ii) If the cost of an apple is ‘x’ and cost of banana is ₹ 5, then the total cost of fruits is ₹ ‘x + 5′
(iii) If there are 11 players in a team, then there will be ’11 + q’ players in ‘q’ teams.
Solution:
(i) False
Length of B is 6 – a
(ii) True
(iii) False
There will be 11q players.

Question 3.
Draw the next two patterns and complete the table.

Solution:

Question 4.
Use a variable to write the rule, which gives the number of ice candy sticks required to make the following patterns.
(a) a pattern of letter C as

(b) the pattern of letter M as

Solution:
(a) A number of sticks used for 1  is 3.

If the number of  formed is ‘n’ then
the number of ice candy sticks required = 3 × n = 3n.
(b) A number of sticks used for one’ is 4.

The number of ice candy sticks required = 4n.

Question 5.
The teacher forms a group of five students in a class. How many students will be there in ‘p’ groups?
Solution:

5p students will be there in groups.

Question 6.
Arivazhagan is 30 years younger to his father. Write Arivazhagan’s age in terms of his father’s age.
Solution:
Let Arivazhagan’s father’s age be x years
According to the problem,
Arivazhagan’s age = (x – 30) years

Question 7.
If ‘u’ is an even number, how would you represent?
(i) the next even number?
(ii) the previous even number?
Solution:
(i) Difference between two even numbers = 2
Given that ‘u’ is an even number.
Next, even number is u + 2.
(ii) Previous even number is u – 2.

Objective Type Questions

Question 8.
Variable means that it
(a) can take only a few values
(b) has a fixed value
(c) can take different values
(d) can take only 8 values
Solution:
(c) can take different values

Question 9.
‘6y’ means.
(a) 6 + y
(b) 6 – y
(c) 6 × y
(d) \(\frac{6}{y}\)
Solution:
(c) 6 × y

Question 10.
Radha is ‘x’ years of age now 4 years ago, her age was
(a) x – 4
(b) 4 – x
(c) 4 + x
(d) 4x
Solution:
(a) x – 4

Question 11.
The number of days in ‘w’ weeks is
(a) 30 + w
(b) 30w
(c) 7 + w
(d) 7w
Solution:
(d) 7w

Question 12.
The value of ‘x’ in the circle is

(a) 6
(b) 8
(c) 21
(d) 22
Solution:
(d) 22
2 + 2 = 4, 4 + 3 = 7, 7 + 4 = 11, 11 + 5 = 16, 16 + 6 = 22

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.3

Miscellaneous Practice Problems

Question 1.
Two pipes whose lengths are 7 m 25 cm and 8 m 13 cm joined by welding and then a small piece 60 cm is cut from the whole. What is the remaining length of the pipe?
Solution:

Question 2.
The saplings are planted at a distance of 2 m 50 cm in the road of length 5 km by saravanan. If he has 2560 saplings, how many saplings will be planted by him? how many saplings are left?
Solution:
Distance between two saplings = 2 m 50 cm = 250 cm
Total length of the road = 5000 m = 500000 cm

Question 3.
Put ✓ a mark in the circles which adds up to the given measure.

Solution:

Question 4.
Make a calendar for the month of February 2020 (Hint: January 1st 2020 is Wednesday)
Solution:
Given Jan 1st 2020 is Wednesday.
1st + 7 = 8th Wednesday
8th + 7 = 15th Wednesday
15th + 7 = 22nd Wednesday
22nd + 7 = 29th Wednesday
Jan = 30th Thursday
Jan 31st Friday
Feb 1st Saturday
February has 29 days as 2020 is a leap year.

Question 5.
Observe and Collect the data for a minute:

Solution:

Challenge Problems

Question 6.
A squirrel wants to eat the grains quickly. Help the squirrel to find the shortest way to reach the grains. (Use your scale to measure the length of the line segments)

Solution:
The squirrel is in A and the grains are at E
The ways are (i) A → B → C → D → E
(ii) A → G → F → K → E
(iii) A → H → I → J → E
Distance
(i) AB + BC + CD + DE = 2 cm + 2.5 cm + 2.5 cm + 2 cm = 9 cm
(ii) AG + GF + FK + KE = (2.6 + 1.7 + 1.8 + 3) cm = 9.1 cm
(iii) AH + HI + IJ + JE = 3 cm + 2.3 cm + 1 cm + 3.2 cm = 9.5 cm.
1st way ABCDE is the shortest one.

Question 7.
A room has a door whose measures are 1 m wide and 2 m 50 cm high.
(i) Can we make a bed of 2 m and 20 cm length and 90 cm wide into the room?
Solution:
Measures of the door length 2 m 50 cm width and lm (100 cm)
(i) Measures of the bed 2 m 20 cm width and 90 cm
Measures of the bed < measures of the door
∴ Yes, we can take the bed into the room.

Question 8.
A post office functions from 10 a.m to 5.45 pm with a lunch break of 1 hour. If the post office works for 6 days a week. Find the total duration of working hours in a week.
Solution:
Working hours in a day = 6 hrs 45 min
= (6 × 60 min) + 45 min
= (360 + 45) min
= 405 min
Total duration of working hours in a week
= 6 × 405 min
= 2430 min
= \(\frac{2430}{60}\)
= \(\frac{810}{20}\)
= 40 \(\frac{1}{2}\)
= 40 hours 30 minutes

Question 9.
Seetha wakes up at 5.20 a.m. She spends 35 minutes to get ready and travels 15 minutes to reach the railway station. If the train departs exactly at 6 : 00 a.m, will Seetha catch the train?
Solution:
Time of wake up = 5.20 a.m = 5 hour 20 minutes
Time spends = 35 minutes
Then the time = 5 hour 55 minutes
Travelling time to reach railway station = 15 minutes
Now the time will be = 5 hours 70 minutes
= 5 hours (60 + 10) minutes
= 5 hours + (1 hour 10 minutes)
= 6 hours 10 minutes
= 6.10 a.m.
The departure time of the train to get ready = 6.0 a.m.
∴ She will not be able to catch the train.

Question 10.
A doctor advised Vairavan to take one tablet every 6 hours once in the 1st day and once every 8 hours on the 2nd and 3rd day. If he starts to take 9.30 a.m first dose. Prepare a time chart to take the tablet in railway time.
Solution:
The first dose is taken at 9.30 a.m. = 09:30 hours
Duration of every dose in 1st day = 6 hours
Duration of every dose in 2nd and 3rd day = 8 hours
Time Chart.

You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.4

Miscellaneous Practice Problems

Question 1.
Find the type of lines marked in thick lines (Parallel, intersecting or perpendicular)

Solution:
(i) Parallel lines
(ii) Parallel lines
(iii) Parallel lines and Perpendicular lines
(iv) Intersecting lines

Question 2.
Find the parallel and intersecting line segments in the picture given below.

Solution:
(a) Parallel line segments

• \(\overline{\mathrm{YZ}} \text { and } \overline{\mathrm{DE}}\)
• \(\overline{\mathrm{EA}} \text { and } \overline{\mathrm{ZV}}\)
• \(\overline{\mathrm{VW}} \text { and } \overline{\mathrm{AB}}\)
• \(\overline{\mathrm{WX}} \text { and } \overline{\mathrm{BC}}\)
• \(\overline{\mathrm{YX}} \text { and } \overline{\mathrm{DC}}\)
• \(\overline{\mathrm{YD}} \text { and } \overline{\mathrm{XC}}\)
• \(\overline{\mathrm{XC}} \text { and } \overline{\mathrm{WB}}\)
• \(\overline{\mathrm{WB}} \text { and } \overline{\mathrm{VA}}\)
• \(\overline{\mathrm{VA}} \text { and } \overline{\mathrm{ZE}}\)
• \(\overline{\mathrm{ZE}} \text { and } \overline{\mathrm{YD}}\)

(b) Intersecting line segments

• DE and ZV
• WX and DC

Question 3.
Name the following angles as shown in the figure.

Solution:
(i) ∠1 = ∠DBC or ∠CBD
(ii) ∠2 = ∠DBE or ∠EBD
(iii) ∠3 = ∠ABE or ∠EBA
(iv) ∠1 + ∠2 = ∠EBC or ∠CBE
(v) ∠2 + ∠3 = ∠ABD or ∠DHA
(vi) ∠1 + ∠2 + ∠3 = ∠ABC or ∠B or ∠CBA

Question 4.
Measure the angles of the given figures using a protractor and identify the type of angle as acute, obtuse, right or straight.

Solution:
(i) 90° – Right Angle
(ii) 45° – Acute Angle
(iii) 180° – Straight Angle
(iv) 105° – Obtuse Angle

Question 5.
Draw the following angles using the protractor.
(i) 45°
(ii) 120°
(iii) 65°
(iv) 135°
(v) 0°
(vi) 180°
(vii) 38°
(viii) 90°
Solution:
(i) 45°

Construction:
1. Drawn the base ray PQ.
2. Placed the centre of the protractor at the vertex P. Lined up the ray \(\overrightarrow{\mathrm{PQ}}\) with the 0° line. Then drawn and labelled a pointed (R) at the 45° mark on the inner scale (a) anticlockwise and (b) outer scale (clockwise)
3. Removed the protractor and drawn at \(\overrightarrow{\mathrm{PR}}\) to complete the angle
Now ∠P = ∠QPR – ∠RPQ = 45°.

(ii) 120°

Construction:
1. Placed the centre of the protractor at the vertex X. Lined up the ray \(\overline{\mathrm{XY}}\) with the 0° Line. Then draw and label a point Z at 120° mark on the (a) inner scale (anti-clockwise) and (b) outer scale (clockwise).
2. Removed the protractor and draw \(\overline{\mathrm{XZ}}\) to complete the angle.
Now, ∠X = ∠ZXY = ∠YXZ = 120°.

(iii) 65°

Construction:
1. Placed the centre of the protractor at the vertex A. Line up the ray \(\overrightarrow{\mathrm{AB}}\) with the 0° line. Then draw and label a point C at the 65° mark on the (a) inner scale (anti-clockwise) (b) outer scale (clockwise).
2. Removed the protractor and draw \(\overrightarrow{\mathrm{AB}}\) to complete the angle.
Now ∠A = ∠BAC = ∠CAB = 65°.

(iv) 135°

Construction:
1. Placed the centre of the protractor at the Vertex A. Lined up the ray \(\overrightarrow{\mathrm{EG}}\) with the 0° line. Then draw and label a point F at the 135° mark on the (a) inner scale (anti-clockwise) and (b) outer scale (clockwise)
2. Removed the protractor and draw \(\overrightarrow{\mathrm{EF}}\) to complete the angle.
Now ∠E = ∠FEG = ∠GEF = 135°.

(v) 0°

Construction:
1. Placed the centre of the protractor at the vertex G. Lined up the ray \(\overrightarrow{\mathrm{GH}}\) with the 0° line. Then draw and label a point I at the 0° mark on the
(a) inner scale (anti-clockwise)
(b) outer scale (clockwise)
2. Removed the protractor and seen \(\overrightarrow{\mathrm{GI}}\) lies exactly on \(\overrightarrow{\mathrm{GH}}\)
Now ∠G = ∠HGI = ∠IGH = 0°, which is a zero angle.

(vi) 180°

Construction:
1. Placed the centre of the protractor at the vertex I. Lined up the ray \(\overrightarrow{\mathrm{IJ}}\) with the 0° line. Then draw and labelled a point K at the 180° mark on the (a) inner scale (anticlockwise) (b) outer scale (clockwise)
2. Removed the protractor and draw \(\overrightarrow{\mathrm{IK}}\) to complete the angle.
Now ∠I = ∠JHK = ∠KIJ = 180°, which is a straight Angle.

(vii) 38°

Construction:
1. Placed the centre of the protractor at the vertex L. Lined up the ray \(\overrightarrow{\mathrm{LM}}\) with the 0° line. Then draw and label a point N at 38° mark on the (a) inner scale (anticlockwise) and (b)*huter scale (clockwise).
2. Removed the protractor and draw \(\overrightarrow{\mathrm{LN}}\) to complete the angle.
Now ∠L = ∠MLN = ∠NLM = 38°.

(viii) 90°

Construction:
1. Placed the centre of the protractor at the vertex ‘O’. Lined up the ray \(\overrightarrow{\mathrm{OP}}\) with the 0° line. Then draw and label a point Q at 90° mark on the (a) inner scale (anticlockwise) and (b) outer scale (clockwise)
2. Removed the protractor and draw \(\overrightarrow{\mathrm{OQ}}\) to complete the angle.
Now ∠O = ∠POQ = ∠QOP = 90°.

Question 6.
From the figures given below, classify the following pairs of angles into complementary and non-complementary.

Solution:
We know that the two angles are complementary if they add up to 90°.
Therefore (a) (i) is complementary.
In (v) ∠ABC and ∠CBD are complementary
(b) (ii), (iii), (iv) and (v) are non-complementary

Question 7.
From the figures given below, classify the following pairs of angles into supplementary and non-supplementary.

Solution:
If two angles add up to 180°, then they are supplementary angles.
(a) In (ii) ∠AOB and ∠BOD are supplementary. In (iv) the pair is supplementary
(b) (i) and (iii) are not supplementary.

Question 8.
From the figure.
(i) name a pair of complementary angles
(ii) name a pair of supplementary angles

Solution:
(i) ∠FAE and ∠DAE are complementary
(ii) ∠FAD and ∠DAC are supplementary

Question 9.
Find the complementary angle of
(i) 30°
(ii) 26°
(iii) 85°
(iv) 0°
Solution:
When we have an angle, how far we need to go to reach the right angle is called the complementary angle.
(i) Complementary angle of 30° is 90° – 30° = 60°
(ii) Complementary angle of 26° is 90° – 26° = 64°
(iii) Complementary angle of 85° is 90° – 85° = 5°
(iv) Complementary angle of 0° is 90° – 0° = 90°
(v) Complementary angle of 90° is 90° – 90° = 0°

Question 10.
Find the supplementary angle of
(i) 70°
(ii) 35°
(iii) 165°
(iv) 90°
(v) 0°
(vi) 180°
(vii) 95°
Solution:
How far we should go in the same direction to reach the straight angle (180°) is called supplementary angle.
(i) Supplementary angle of 70° = 180° – 70° = 110°
(ii) Supplementary angle of 35° is 180° – 35° = 145°
(iii) Supplementary angle of 165° is 180° – 165° = 15°
(iv) Supplementary angle of 90° is 180° – 90° = 90°
(v) Supplementary angle of 0° is 180° – 0° = 180°
(vi) Supplementary angle of 180° is 180° – 180° = 0°
(vii) Supplementary angle of 95° is 180° – 95° = 85°

Challenging Problems

Question 11.
Think and write and object having.
(i) Parallel Lines
1. _____________
2. _____________
3. _____________
(ii) Perpendicular lines
1. _____________
2. _____________
3. _____________
(iii) Intersecting lines
1. _____________
2. _____________
3. _____________
Solution:
(i) 1. Opposite edges of a Table.
2. Path traced by the wheels of a car on a straight road
3. Opposite edges of a black board
(ii) 1. Adjacent edges of a Table.
2. Hands of the block when it shows 3.30
3. Strokes of the letter ‘L’
(iii) 1. Sides of a triangle
2. Strokes of letter ‘V’
3. Hands of a scissors

Question 12.
Which angle is equal to twice of its complement?
Solution:
We know that the sum of complementary angles 90°
Given Angle = 2 × Complementary angle

By trial and error, we find that Angle = 2 × Complement for 60°
The required angle = 60°
Another method:
Let the angle be x given
x = 2 (90 – x)
⇒ x = 180 – 2x
⇒ x + 2x = 180
⇒ 3x = 180
⇒ x = 60°

Question 13.
Which angle is equal to two-thirds of its supplement.
Solution:
Supplementary angles sum upto 180°
Given Angle = \(\frac{2}{3}\) × Supplement.
Forming the Table.

By trial and error, we find that angle = \(\frac{2}{3}\) × supplement for 72°.
The required angle 72°.

Question 14.
Given two angles are supplementary and one angle is 20° more than other. Find the two angles.
Solution:
Given two angles are supplementary i.e. their sum = 180°.
Let the angle be x
Then another angle = x + 20 (given)

The two angles are 80° and 100°.

Question 15.
Two complementary angles are in the ratio 7 : 2. Find the angles.
Solution:
Let the angles be 7x, 2x
According to the problem,
7x + 2x = 90
9x = 90
x = \(\frac{90}{9}\)
x = 10
7x = 7 × 10
= 70
2x = 2 × 10
= 20
∴ Two angles are 70° and 20°

Question 16.
Two supplementary angles are in ratio 5 : 4. Find the angles.
Solution:
Total of two supplementary angles = 180°
Given they are in the ratio 5 : 4
Dividing total angles to 5 + 4 = 9 equal parts.
One angle \(=\frac{5}{9} \times 180=100^{\circ}\)
Another angle \(=\frac{4}{9} \times 180=80^{\circ}\)
Two angles are 100° and 80°.