Class 7

Students can Download Maths Chapter 1 Number System Ex 1.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Ex 1.3

Question 1.
Compare the following decimal numbers and find out the smaller number.
(i) 2.08,2.086
(ii) 0.99,1.9
(iii) 3.53,3.35
(iv) 5.05,5.50
(v) 123.5,12.35
Solution:
(i) 2.08, 2.086
Let us take 2.080, 2.086.
Comparing 2.08 and 2.086 the whole number part, tenths place digit and the digit in the hundredths place are equal.
Comparing the digits at thousandths place we get 0 < 6.
Therefore 2.08 < 2.086.
Smallest number is 2.08

(ii) 0.99, 1.9
Comparing 0.99 and 1.9.
First when we compare the digit in the whole number parts we get 0 < 1.
∴ 0.99 < 1.9 Smallest number is 0.99

(iii) 3.53,3.35
Comparing 3.53 and 3.35
Here the whole number parts of the given two numbers are equal.
Comparing the digits at tenths place, we get 3 < 5
∴ 3.35 < 3.53
Smallest number is 3.35

(iv) 5.05, 5.50
Comparing 5.05 and 5.50
Here the whole number parts of the given two numbers are equal.
Comparing the digits at tenths place, we get 0 < 5.
∴ 5.5 < 5.50
Smallest number is 5.05

(v) 123.5, 12.35
Comparing 123.5 and 12.35.
Comparing the whole number parts, we get 12 < 123
∴ 12.35 < 123.5
Smallest number is 12.35

Question 2.
Arrange the following in ascending order.
(i) 2.35, 2.53, 5.32, 3.52, 3.25
(ii) 123.45, 123.54,125.43, 125.34,125.3
Solution:
(i) 2.35, 2.53, 5.32, 3.52, 3.25
Comparing the whole number parts of all the numbers 5 is the greatest and 5 > 3 > 2.
∴ Greatest number is 5.32
Next 3.52 and 3.25 are equal in their whole number.
So comparing their digits in tenths place, we get 5 > 2
So 3.52 > 3.25
Now comparing 2.35 and 2.53 their whole number parts also equal.
∴ Comparing the digit in tenths place we get
2.53 > 2.35 ……(2)
Ascending order :
2.35 < 2.53 < 3.25 < 3.52 < 5.32

(ii) 123.45, 123.54, 125.43, 125.34, 125.3
Comparing the whole number parts we have 123 is the smallest number and two numbers 123.45 and 123.54 have same whole number part.
So in 123.45 and 123.54 comparing their digits in the tenths place we get 4 < 5
∴ 123.45 < 123.54 …(1)
Now comparing the remaining numbers
125.43, 125.34, 125.3 they all have the same whole number part.
Comparing the numbers in the tenths place we have 3 < 4
∴ 125.43 is the greatest …(2)
Also tenths place value 3 = 3 in 125.34 and 125.3
Again comparing the hundredths place value in 125.34 and 125.3, we get
125.3 < 125.34 …(3)
From (1), (2) and (3) we have,
123.45 < 123.54 < 125.3 < 125.34 < 125.43

Question 3.
Compare the following decimal numbers and find the greater number.
(i) 24.5,20.32
(ii) 6.95,6.59
(iii) 17.3,17.8
(iv) 235.42,235.48
(v) 0.007,0.07
(vi) 4.571,4.578
Solution:
(i) 24.5, 20.32
Comparing the whole number part we get 24 > 20
∴ 24.5 > 20.32
greater number is 24.5

(ii) 6.95,6.59
Here the whole number part of given two numbers are equal.
Comparing the digits at tenths place we get 9 > 5.
∴ 6.95 > 6.59
Greater number is 6.95

(iii) 17.3,17.8
Here the whole number part of given two numbers are equal.
Comparing the digits at tenths place we get 8 > 3.
∴ 17.8 > 17.3
Greater number is 17.8

(iv) 235.42,235.48
Here the whole number part of given two numbers are equal.
Also the digits at tenths place also equal.
Comparing the digits at the hundredths place we get 8 > 4.
∴ 235.48 > 235.42
Greater number is 235.48

(v) 0.007,0.07
Here the whole number part of given two numbers are equal.
Also the digits at the tenths place also equal.
∴ Comparing the the digits at the hundredths place we get 7 > 0.
0. 07 > 0.007
greater number is 0.07.

(vi) 4.571,4.578
Here the whole number part of given two numbers are equal.
Also the digits at the tenths place and the hundredths place also equal.
Again comparing the digits in the thousandths place we get 8 > 1.
∴ 4.578 > 4.571
∴ Greater number is 4.578

Question 4.
Arrange the given decimal numbers in descending order.
(i) 17.35, 71.53, 51.73, 73.51,37.51
(ii) 456.73, 546.37, 563.47, 745.63 457.71
Solution:
(i) 17.35,71.53,51.73,73.51,37.51
Comparing the whole number parts of the given numbers we get
73 > 71 > 51 > 37 > 17.
Descending order:
73.51,71.53,51.73,37.51, 17.35

(ii) 456.73, 546.37, 563.47, 745.63 457.71
Comparing the whole number parts of the given numbers from left to right we get
745 > 563 > 546 > 457 > 456
Descending Order: 745.63, 563.47, 546.37, 457.71, 456.73

Objective Question

Question 5.
0.009 is equal to
(i) 0.90
(ii) 0.090
(iii) 0.00900
(iv) 0.900
Answer:
(iii) 0.00900

Question 6.
37.70 [ ] 37.7
(i) =
(ii) <
(iii) >
(iv) ≠
Answer:
(i) =

Question 7.
78.56 [ ] 78.57
(i) <
(ii) >
(iii) =
(iv) ≠
Answer:
(i) <

Students can Download Maths Chapter 1 Number System Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 1 Number System Intext Questions

Exercise 1.1
(Try These Text book Page No. 2)

Question 1.
Observe the following and write the fraction of the shaded portion and mention in decimal form also.

Solution:
(i) Total parts = 8
Shaded parts = 4

(ii)

(iii)

Question 2.
Represent the following fractions in decimal form by converting denominator into ten or powers of 10.

Solution:

Question 3.
Give any two life situations where we use decimal numbers.
Solution:
(i) Measuring weight of gold.
(ii) Weighing our height

(Try These Text book Page No. 3)

Question 1.
Represent the following decimal numbers pictorially.
(i) 5 ones and 3 tenths
(ii) 6 tenths
(iii) 7 ones and 9 tenths
(iv) 6 ones and 4 tenths
(v) Seven tenths
Solution:
(i) 5 ones and 3 tenths

(ii) 6 tenths

(iii) 7 ones and 9 tenths

(iv) 6 ones and 4 tenths

(v) Seven tenths

(Try These Text book Page No. 5 & 6)

Question 1.
Express the following decimal numbers in an expanded form and place value grid form.
(i) 56.78
(ii) 123.32
(iii) 354.56
Solution:
(i) 56.78
(a) Expanded form
56.78 = 5 × 10¹ + 6 × 10⁰ + 7 × 10^-1 + 8 × 10^-2

(b) Place value grid

(ii) 123.32
(a) Expanded form
123.32 = 1 × 10² + 2 × 10¹ + 3 × 10⁰ + 3 × 10^-1 +2 × 10^-2

(b) Place value grid

(iii) 354.56
(a) Expande form
354.56 = 3 × 10² + 5 × 10¹ + 4 × 10⁰+ 5 × 10^-1 + 6 × 10^-2

(b) Place value grid

Question 2.
Express the following measurements in terms of metre and in decimal form. One is done for you.

Solution:

Question 3.
Write the following numbers in the place value grid and find the place value of the underlined digits.
(i) 36.37
(ii) 267.06
(iii) 0.23
(iv) 27.69
(v) 53.27
Solution:

(i) Place value of 3 in 36.37 is Tenths.
(ii) Place value of 6 in 267.06 is Hundredths.
(iii) Place value of 2 in 0.23 is Tenths.
(iv) Place value of 9 in 27.69 is Hundredths.
(v) Place value of 2 in 53.27 is Tenths.

Exercise 1.2

(Try These Text book Page No. 10)

Question 1.
Convert the following fractions into the decimal numbers.
(i) \(\frac { 16 }{ 1000 } \)
(ii) \(\frac { 638 }{ 10 } \)
(iii) \(\frac { 1 }{ 20 } \)
(iv) \(\frac { 3 }{ 50 } \)
Solution:
(i) \(\frac { 16 }{ 1000 } \) = 0.016
(ii) \(\frac { 638 }{ 10 } \) = 63.8
(iii) \(\frac { 1 }{ 20 } \) = \(\frac{1 \times 5}{20 \times 5}\) = \(\frac { 5 }{ 100 } \) = 0.05
(iv) \(\frac { 3 }{ 50 } \) = \(\frac{3 \times 2}{50 \times 2}\) = \(\frac { 6 }{ 100 } \) = 0.06

Question 2.
Write the fraction for each of the following:
(i) 6 hundreds + 3 tens + 3 ones + 6 hundredths + 3 thousandths.
(ii) 3 thousands + 3 hundreds + 4 tens + 9 ones + 6 tenths.
Solution:
(i) 6 hundreds + 3 tens + 3 ones + 6 hundreds + 3 thousandths.
= 6 × 100 + 3 × 10 + 3 × 1 + 0 × \(\frac { 1 }{ 10 } \) + 6 × \(\frac { 1 }{ 100 } \) + 3 × \(\frac { 1 }{ 1000 } \)
= 600 + 30 + 3 + 0 + \(\frac { 6 }{ 100 } \) + \(\frac { 3 }{ 1000 } \)
= 633 + 0.06 + 0.003
= 633.063

(ii) 3 thousands + 3 hundreds + 4 tens + 9 ones + 6 tenths.
= 3 × 1000 + 3 × 100 + 4 × 10 + 9 × 1 + 6 × \(\frac { 1 }{ 10 } \)
= 3000 + 300 + 40 + 9 + \(\frac { 6 }{ 10 } \)
= 3349 + 0.6
= 3349.6

Question 3.
Convert the following decimals into fractions.
(i) 0.0005
(ii) 6.24
Solution:
(i) 0.0005 = \(\frac { 5 }{ 10000 } \) = \(\frac{5 \div 5}{10000 \div 5}\) = \(\frac { 1 }{ 2000 } \)
(ii) 6.24 = \(\frac { 624 }{ 100 } \) = \(\frac{624 \div 4}{100 \div 4}\) = \(\frac { 156 }{ 25 } \)

Exercise 1.4

(Try These Text book Page No. 22)

Question 1.
Mark the following decimal numbers on the number line.
(i) 0.3
(ii) 1.7
(iii) 2.3
Solution:
(i) 0.3
We know that 0.3 is more than 0, but less than 1.
There are 3 tenths in it. Divide the unit lenght between O and 1 on the number line
into 10 equal parts and take 3 parts, which represent 0.3.

(ii) 1.7
We know that 1.7 is more than 1, but less than 2.
There are one ones and 7 tenths in it. Divide the unit length between 1 and 2 on the number line into 10 equal parts and take 7 parts which represents 1.7 = 1 + 0.7

(iii) We know that 2.3 is more than 2 and less than 3.
There are 2 ones and 3 tenths in it. Divide the unit length between 2 and 3 into 10 equal parts and take 3 parts, which represents 2.3 = 2 + 0.3

Question 2.
Identify any two decimal numbers between 2 and 3.
Solution:
2.5 and 2.9

Question 3.
Write any decimal number which is greater than 1 and less than 2.
Solution:
1.7, 1.9, 1.6, ………………..

Students can Download Maths Chapter 3 Algebra Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions

Additional Questions and Answers

Exercise 3.1

Question 1.
Write any three expressions each having 4 terms:
Solution:
(i) 2x³ – 3x² + 3xy + 8
(ii) 7x³ + 9y² – 2xy² – 6
(iii) 9x² – 2x + 3xy – 1

Question 2.
Identify the co-efficients of the terms of the following expressions
(i) 2x – 2y
(ii) x + y +3
Solution:
(i) 2x – 2y
The co-efficient of x in 2x is 2
The co-efficient of y in – 2y is – 2

(ii) x + y + 3
The co-efficient of x is 1
The co-efficient ofy is 1
The constant term is 3

Question 3.
Group the like terms together from the following: 6x, 6, -5x, – 5, 1, x, 6y, y, 7y, 16x, 3
Solution:
We have 6x, -5x, x, 16x are like terms
6y, y, 7y, are like terms
6, – 5, 1, 3 are like terms

Question 4.
Give the algebraic expressions for the following cases:
(i) One half of the sum of a and b.
(ii) Numbers p and q both squared and added
Solution:
(i) \(\frac{1}{2}\) (a + b)
(ii) p² + q²

Exercise 3.2

Question 1.
If A = 2a² – 4b – 1 ; B = 5a² + 3b – 8 and C = 2a² – 9b + 3 then find the value of A – B + C.
Solution:
Given A = 2a² – 4b – 1 ; B = 5a² + 3b – 8 ; C = 2a² – 9b + 3
A – B + C = (2a² – 4b – 1) – (5a² + 3b – 8) + (2a² – 9b + 3)
= 2a² – 4b – 1 + (-5a² – 3b + 8) + 2a² – 9b + 3
= 2a² – 4b – 1 – 5a² – 3b + 8 + 2a² – 9b + 3
= 2a² – 5a² + 2a² – 4b – 3b – 9b – 1 + 8 + 3
= (2 – 5 + 2) a² + (-4 – 3 – 9) 6 + (-1 + 8 + 3)
= -a² – 16b + 10

Question 2.
How much 2x³ – 2x² + 3x + 5 is greater than 2x³ + 7x² – 2x + 7?
Solution:
The required expression can be obtained as follows.
= 2x³ – 2x² + 3x + 5 – (2x³ + 7x² – 2x + 7)
= 2x³ – 2x² + 3x + 5 + (-2x³ – 7x² + 2x – 7)
= 2x³– 2x² + 3x + 5 – 2x³ – 7x² + 2x – 7
= (2 – 2) x³ + (-2 – 7) x² + (3 + 2) x + (5 – 7)
= 0x³ + (-9x²) + 5x – 2 = -9x² + 5x – 2
∴ 2x³ – 2x² + 3x + 5 is greater than 2x³ + 7x² – 2x + 7 by -9x² + 5x – 2

Question 3.
What should be added to 2b² – a² to get b² – 2a²
Solution:
The required expression is obtained by subtracting 2b² – a² from b² – 2a²
b² – 2a² – (2b² – a²) = b² – 2a² + (-2b² + a²)
= b² – 2a² – 2b² + a²
= (1 – 2) b² + (-2 + 1) a² = -b² – a²
So -b² – a² must be added

Exercise 3.3

Question 1.
Length of one side of an equilateral triangle is 3x – 4 units. Find the perimeter.
Solution:
Equilateral triangle has three sides equal.
Perimeter = Sum of three sides
= (3x – 4) + (3x – 4) + (3x – 4) = 3x – 4 + 3x – 4 + 3x – 4
= (3 + 3 + 3)x + [(-4) + (-4) + (-4)] = 9x + (-12) = 9x – 12
∴ Perimeter = 9x – 12 units.

Question 2.
Find the perimeter of a square whose side is y – 2 units.
Solution:
Perimeter = (y – 2) + (y – 2) + (y – 2) + (y – 2)
= y – 2 + y – 2 + y – 2 + y – 2 = 4y – 8
Perimeter of the square = 4y – 8 units.

Question 3.
Simplify 3x – 5 – x + 9 if x = 3
Solution:
3x – 5 – x + 9 = 3(3) – 5 – 3 + 9
= 9 – 5 – 3 + 9 = 18 – 8 = 10

Students can Download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

Miscellaneous Practice Problems

Question 1.
Subtract – 3ab – 8 from 3ab – 8. Also subtract 3ab + 8 from -3ab – 8.
Solution:
Subtracting -3ab – 8 from 3ab + 8
= 3ab + 8 – (-3ab – 8) = 3ab + 8 + (3ab + 8)
= 3ab + 8 + 3ab + 8 = (3 + 3) ab + (8 + 8)
= 6ab + 16
Also subtracting 3 ab + 8 from – 3ab – 8
= – 3ab – 8 – (3ab + 8) = – 3ab – 8 + (-3ab – 8) = – 3ab – 8 – 3 ab – 8
= [(-3) + (- 3)] ab + [(-8) + (-8)] = – 6ab + (- 16)
= -6ab – 16

Question 2.
Find the perimeter of a triangle whose sides are x + 3y, 2x + y, x – y.
Solution:
Perimeter of a triangle = Sum of three sides
= (x + 3y) + (2x + y) + (x – y)
= x + 3y + 2x + y + x – y
= (1 + 2 + 1)x + (3 + 1 + (-1))y = 4x + 3y
∴ Perimeter of the triangle = 4x + 3y

Question 3.
Thrice a number when increased by 5 gives 44. Find the number.
Solution:
Let the required number be x.
Thrice the number = 3x.
Thrice the number increased by 4 = 3x + 5
Given 3x + 5 = 44
3x + 5 – 5 = 44 – 5
3x = 39
\(\frac{3 x}{3}=\frac{39}{3}\)
x = 13
∴ The required number = 13

Question 4.
How much smaller is 2ab + 4b – c than 5ab – 3b + 2c.
Solution:
To find the answer we have to find the difference.
Here greater number 5ab – 3ab + 2c.
∴ Difference = 5ab – 3b + 2c – (2ab + 4b – c) = 5ab – 3b + 2c + (- 2ab -4b + c)
= 5ab – 3b + 2c – 2ab – 4b + c
= (5 – 2) ab + (-3 – 4) b + (2 + 1) c = 3ab + (-7)b + 3c
= 3ab – 7b + 3c
It is 3ab – 7b + 3c smaller.

Question 5.
Six times a number subtracted from 40 gives – 8. Find the number.
Solution:
Let the required number be x. Six times the number = 6x.
Given 40 – 6x = – 8
-6x + 40 – 40 = -8 – 40
– 6x = – 48
\(\frac{-6 x}{-6}=\frac{-48}{-6}\)
x = 8
∴ The required number is 8.

Challenge Problems

Question 6.
From the sum of 5x + 7y -12 and 3x – 5y + 2, subtract the sum of 2x – 7y – 1 and – 6x + 3y + 9.
Solution:
Sum of 5x + 7y – 12 and 3x – 5y + 2 .
= 5x + 7y- 12 + 3x – 5y + 2 = (5 + 3) x + (7 – 5) y + ((- 12) + 2)
= 8x + 2y – 10.
Again Sum of 2x – 7y – 1 and – 6x + 3y + 9
= 2x – 7y – 1 + (- 6x + 3y + 9) = 2x – 7y – 1 – 6x + 3y + 9
= (2 – 6) x + (- 7 + 3) y + (- 1 + 9)
= – 4x – 4y + 8
Now 8x + 2y – 10 – (-4x – 4y + 8)
= 8x + 2y – 10 + (4x + 4y – 8)
= 8x + 2y – 10 + 4x + 4y – 8
= (8 + 4) x + (2 + 4) y + ((- 10) + (- 8))
= 12x + 6y – 18

Question 7.
Find the expression to be added with 5a – 3b – 2c to get a – 4b – 2c?
Solution:
To get the required expression we must subtract 5a – 3b + 2c from a – 4b – 2c.
∴ a – 4b – 2c – (5a – 3b + 2c) = a – 4b – 2c + (- 5a + 3b – 2c)
= a – 4b – 2c – 5a + 3b -2c
= (1 – 5) a + (- 4 + 3) b + (- 2 – 2) c
= – 4a – b – 4c.
∴ -4a – b – 4c must be added.

Question 8.
What should be subtracted from 2m + 8n + 10 to get – 3m + 7n + 16?
Solution:
To get the expression we have to subtract – 3m + 7n + 16 from 2m + 8n + 10.
(2m + 8n + 10) – (-3m + 7n + 16) = 2m + 8n + 10 + 3m – 7n – 16
= (2 + 3) m + (8 – 7) n + (10 – 16)
= 5m + n – 6

Question 9.
Give an algebraic equation for the following statement:
“The difference between the area and perimeter of a rectangle is 20”.
Solution:
Let the length of a rectangle = l and breadth = b then Area = lb; Perimeter = 2(1 + b)
Area – Perimeter = 20
∴ lb – 2(l + b)

Question 10.
Add : 2a + b + 3c and a + \(\frac{1}{3}\)b + \(\frac{2}{5}\)c
Solution:

Students can Download Maths Chapter 3 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Exercise 3.1

Try These (Text Book Page No. 51)

Question 1.
Identify the variable and constants among the following terms.
a, 11 – 3x, xy, -89, -m, -n, 5, 5ab, -5 3y, 8pqr, 18, -9t, -1, -8
Solution:
Variable : a, -3x, xy, -m, -n, 5ab, 3y, -9t, 8pqr
Constants : 11, -89, 5, -5, 18, -1, -8

Question 2.
Complete the following table.

Solution:

Try this (Text book Page No. 53)

Question 1.
Can we use the operations multiplication and division to combine terms?
Solution:
No, We can use addition and subtraction to combine terms.
If we use multiplication or division to combine then it become a single term.
Eg : xy, \(\frac{x}{y}\) are monomials.

Try This (Text book Page No. 54)

Question 1.
Complete the following table by forming expressions using the terms given. One is done for you.

Solution:

Try this (Text book Page No. 56)

Question 1.
Identify the like terms among the following and group them.
7xy, 19x, 1, 5y, x, 3yx, 15, -13y, 6x, 12xy, -5, 16y, -9x, 15xy, 23, 45y, -8y, 23x, -y, 11
Solution:
7xy, 3yx, 12xy, 15xy, are like terms
19x, x, 6x,-9x, 23x, are like terms
5y, -13y, 16y, 45y, -8y, -y, are like terms
1, 15, -5, 23, 11, are like terms.

Try This (Text book Page No. 57)

Question 1.
Try to find the value of the following expressions if p = 5 and q = 6.
(i) p + q
(ii) q – p
(iii) 2p + 2 > q
(iv) pq – p – q
(v) 5pq – 1
Solution:
(i) Given p = 5; q = 6
p + q = 5 + 6 = 11
(ii) q – p = 6 – 5 = 1
(iii) 2p + 2 > q = 2(5) + 3(6) = 10 + 18 = 28
(iv) pq – p – q = (5) (6) – 5 – 6 = 30 – 5 – 6 = 25 – 6 = 19

Exercise 3.2

Try These (Text book Page No. 59)

Question 1.
Add the terms
(i) 3p, 14p
(ii) m, 12m, 21m
(iii) 11abc, 5abc
(iv) 12y, -y
(v) 4x, 2x, -7x.
Solution:
(i) 3p + 14p = 17p
(ii) m + 12m + 21m = (1 + 12 + 21 )m
= 34 m
(iii) 11abc + 5abc = (11 + 5) abc
= 16 abc
(iv) 12y + (-y) = (12 + (-1))y
= (12 – 1 )y
= 11y
(v) 4x + 2x + (-7x) = (4 + 2+(-7))x
= (6 + (-7))x
= -1x

Ty this (Text Book Page No. 60)

Question 1.
3x; + (y – x) = 3x + y – x, but 3x – (y – x) ≠ 3x – y – x. why ?
Solution:
In the first case
LHS = 3x + (y – x) = 3x + y – x = 3x – x + y = (3 – 1)x + y
= 2x + y
RHS = 3x + y – x = 2x + y
LHS = RHS ⇒ 3x + (y – x) = 3x + y – x
But in the second case
LHS = 3x – (y – x) = 3x – y + x
= (3 + 1)x – y = 4x – y
RHS = 3x – y – x = 3x – x – y
LHS ≠ RHS
∴ 3x – (y – x) ≠ 3x – y – x

Try this (Page No. 1)

Question 1.
What will you get if twice a number is subtracted from thrice the same number?
Solution:
Let the unknown number be x.
Twice the number = 2x.
Thrice the number = 3x.
Twice the number is subtracted from thrice the number = 3x – 2x = (3 – 2)x = x

Exercise 3.3

Try These (Text book Page No. 65)

Question 1.
Try to construct algebraic equations for the following verbal statements.

Question 1.
One third of a number plus 6 to 10.
Solution:
\(\frac{1}{3}\) + 6 = 10

Question 2.
The sum of five times of x and 3 is 28
Solution:
5 (x + 3) = 28

Question 3.
Taking away 8 from y gives 11
Solution:
y – 8 = 11

Question 4.
Perimeter of a square with side a is 16 cm.
Solution:
4 × a = 16

Question 5.
Venkat’s mother’s age is 7 years more than 3 times venkat’s age. His mother’s age is 43 years.
Solution:
3x + 7 = 43, where x is venkat’s age.

Try this (Text book Page No. 65)

Question 1.
Why should we subtract 5 and not some other number ? why don’t we add 5 on both sides? Discuss.
Solution:
Given x + 5 = 12
(i) Our aim is to find the value of x. Which means we have to eliminate the other values from LHS. Since 5 is given with x it should be subtracted.
(ii) If we add 5 on both sides we cannot eliminate the numbers from LHS and we get x + 10.

Try this (Text book Page No. 66)

Question 1.
If the dogs, cats and parrots represents unknown find them. Substitute each of the values so obtained in the equations and verify the answers.
Solution:
(i) 1 dog + 1 dog + 1 dog = 24
3 dog = 24

(ii) 1 dog + 1 cat + 1 cat = 14
1 dog + 2cat = 14
8 + 2cat = 14
2cat = 14 – 8
2 cat = 6
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(iii) 1 dog + 1 cat – 1 parrot = 9
8 + 3 – 1 parrot = 9
8 + 3 – 9 = 1 parrot
11 – 9 = 1 parrot
2 = 1 parrot
1 parrot = 2

(iv) 1 dog + 1 cat + 1 parrot = ?
8 + 3 + 2 = 13
Verification:
(i) 8 + 8 + 8 = 24
(ii) 8 + 3 + 3 = 14
(iii) 8+ 3 – 2 = 9
(iv) 8 + 3 + 2 = 13

Try These (Text book Page No. 68)

Question 1.
Kandhan and kaviya are friends. Both of them are having some pen. Kandhan: If you give me one pen then, we will have equal number of pens. Will you? Kaviya: But, if you give me one of your pens, then mine will become twice as yours. Will you?
Construct algebraic equations for this situation, can you guess and find the actual number of pens, they have?
Solution:
Let the number of pens initially Kandhan and Kaviya had be x and y respectively.

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3

Question 1.
Fill in the blanks.
(i) An expressions equated to another expression is called _______.
(ii) If a = 5, the value of 2a + 5 is _______.
(iii) The sum of twice and four times of the variable x is ______.
Solution:
(i) an equation
(ii) 15
(iii) 6x

Question 2:
Say True or False
(i) Every algebraic expression is an equation.
(ii) The expression 7x + 1 cannot be reduced without knowing the value of x.
(iii) To add two like terms, its coefficients can be added.
Solution:
(i) False
(ii) True
(iii) True

Question 3.
Solve (i) x + 5 = 8
(ii) p – 3 = 1
(iii) 2x = 30
(iv) \(\frac{m}{6}\) = 5
(v) 7x + 10 = 80
Solution:
(i) Given x + 5 = 8 ; Subtracting 5 on both the sides
x + 5 – 5 = 8 – 5
x = 3

(ii) Given p – 3 = 7 ; Adding 3 on both the sides,
p – 3 + 3 = 7 + 3
p = 10

(iii) Given 2x = 30 ; Dividing both the sides by 2,
\(\frac{2 x}{2}=\frac{30}{2}\)
x = 15

(iv) Given \(\frac{m}{6}\) = 5 ; Multiplying both the sides by 6,
\(\frac{m}{6}\) × 6 = 5 × 6
m = 30

(v) Given 7x + 10 = 80 ; Subtracting 10 from both the sides,
7x + 10 – 10 = 80 – 10
7x = 70
Dividing both sides by 7,
\(\frac{7 x}{7}=\frac{70}{7}\)
x = 10

Question 4.
What should be added to 3x + 6y to get 5x + 8y?
Solution:
To get the expression we should subtract 3x + 6y from 5x + 8y
5x + 8y – (3x + 6y) = 5x + 8y + (-3x – 6y)
= 5x + 8y – 3x – 6y = (5 – 3) x + (8 – 6) y
= 2x + 2y
So 2x + 2y should be added.

Question 5.
Nine added to thrice a whole number gives 45. Find the number
Solution:
Let the whole number required be x.
Thrice the whole number = 3x
Nine added to it = 3x + 9
Given 3x + 9 = 45
3x + 9 – 9 = 45 – 9 [Subtracting 9 on both sides]
3x = 36
\(\frac{3 x}{3}=\frac{36}{3}\)
x = 12
∴ The required whole number is 12

Question 6.
Find the two consecutive odd numbers whose sum is 200
Solution:
Let the two consecutive odd numbers be x and x + 2
∴ Their sum = 200
x + (x + 2) = 200
x + x + 2 = 200
2x + 2 = 200
2x + 2 – 2 = 200 – 2 [∵ Subtracting 2 from both sides]
2x = 198
\(\frac{2 x}{2}=\frac{198}{2}\) [Dividing both sides by 2]
x = 99
The numbers will be 99 and 99 + 2.
∴ The numbers will be 99 and 101.

Question 7.
The taxi charges in a city comprise of a fixed charge of ₹ 100 for 5 kms and ₹ 16 per km for ever additional km. If the amount paid at the end of the trip was ₹ 740, find the distance traveled.
Solution:
Let the distance travelled by taxi be ‘x’ km
For the first 5 km the charge = ₹ 100
For additional kms the charge = ₹ 16(x – 5)
∴ For x kms the charge = 100 + 16(x – 5)
Amount paid = ₹ 740
∴ 100 + 16 (x – 5) = 740
100 + 16 (x – 5) – 100 = 740- 100
16 (x – 5) = 640
\(\frac{16(x-5)}{16}=\frac{640}{16}\)
x – 5 = 40
x – 5 + 5 = 45 + 5
x = 45
x = 45 km
∴ Total distance travelled = 45 km

Objective Type Questions

Question 8.
The generalization of the number pattern 3, 6, 9, 12, …………. is
(i) n
(ii) 2n
(iii) 3n
(iv) 4n
Solution:
(iii) 3n

Question 9.
The solution of 3x + 5 = x + 9 is t
(i) 2
(ii) 3
(iii) 5
(iv)4
Solution:
(i) 2
Hint: 3x + 5 = x + 9 ⇒ 3x – x = 9 – 5 ⇒ 2x = 4 ⇒ x = 2

Question 10.
The equation y + 1 = 0 is true only when y is
(i) 0
(ii) -1
(iii) 1
(iv) – 2
Solution:
(ii) -1

Students can Download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 1.
Fill in the blanks
(i) The addition of – 7b and 2b is _______
(ii) The subtraction of 5m from -3m is ______
(iii) The additive inverse of -37xyz is _____
Solution:
(i) -5b
(ii) -8m
(iii) 37xyz

Question 2.
Say True or False
(i) The expressions 8x + 3y and 7x + 2y cannot be added
(ii) If x is a natural number, then x + 1 is its predecessor.
Hint: x – 1 is its predecessor.
(iii) Sum of a – b + c and -a + b – c is zero
Solution:
(i) False
(ii) False
(iii) True

Question 3.
Add: (i) 8x, 3x
(ii) 7mn, 5mn
(iii) -9y, 11y, 2y
Solution:
(i) 8x + 3x = (8 + 3) x = 11x
(ii) 7mn + 5mn = (7 + 5)mn = 12mn
(iii) -9y + 11y + 2y =(-9 + 11 + 2 )y = (2 + 2)y = 4y

Question 4.
Subtract:
(i) 4k from 12k
(ii) 15q from 25q
(iii) 7xyz from 17xyz
Solution:
(i) 4k from 12k
12k – 4k = (12 – 4) k = 8k
(ii) 15q from 25q
25q – 15q = (25 – 15)q = 10q
(iii) 7xyz from 17xyz
17xyz – 7xyz = (17 – 7)xyz = 10xyz

Question 5.
Find the sum of the following expressions
(i) 7p + 6q, 5p – q, q + 16p
Solution:
(7p + 6q) + (5p – q) + (q + 16p) = 7p + 6q + 5p – q + q + 16p
= (7p + 5p + 16p) + (6q – q + q)
= (7 + 5 + 16) p + (6 – 1 + 1) q
= (12 + 16) p + 6q = 28p + 6q

(ii) a + 5b + 7c, 2a + 106 + 9c
Solution:
(a + 5b + 7c) + (2a + 10b + 9c) = a + 5b + 7c + 2a + 10b + 9c
= a + 2a + 5b + 10b + 7c + 9c
= (1 + 2)a + (5 + 10)b + (7 + 9)c
= 3a + 15b + 16c

(iii) mn + t, 2mn – 2t, – 3t + 3mn
Solution:
(mn + t) + (2mn – 2t) + (-3t + 3mn)
= mn + t + 2mn – 2t + (-3t) + 3mn
= (mn + 2mn + 3mn) + (t – 2t – 3t)
= (1 + 2 + 3) mn + (1 – 2 – 3) t
= 6mn + (1 – 5)t
= 6mn + (- 4) t
= 6mn – 4t

(iv) u + v, u – v, 2u + 5v, 2u – 5v
Solution:
(u + v) + (u – v) + (2u + 5v) + (2u – 5v)
= u + v + u – v + 2u + 5v + 2u – 5v
= u + u + 2u + 2u + v – v + 5v – 5v
= (1 + 1 + 2 + 2) u +(1 – 1 + 5 – 5)v = 6u + 0v
= 6u

(v) 5xyz – 3xy, 3zxy – 5yx
Solution:
5xyz – 3xy + 3zxy – 5yx = 5xyz + 3xyz – 3xy – 5xy
= (5 + 3) xyz + [(-3) + (-5)] xy = 8xyz + (-8) xy
= 8xyz – 8xy

Question 6.
Subtract
(i) 13x + 12y – 5 from 27x + 5y – 43
Solution:
27x + 5y – 43 – (13x + 12y – 5) = 27z + 5y – 43 + (-13x – 12y + 5)
= 27x + 5y – 43 – 13x – 12y + 5
= (27 – 13) x + (5 – 12)y + (- 43) + 5
= 14x + (- 7) y + (- 38) = 14x – 7y – 38

(ii) 3p + 5 from p – 2q + 7
Solution:
p – 2q + 7 – (3p + 5) = p – 2q + 7 + (- 3p – 5)
= p – 2q + 7 – 3p – 5 = p – 3p – 2q + 7 – 5
= (1 – 3)p – 2q + 2 = -2p – 2q + 2

(iii) m + n from 3m – 7n
Solution:
3m – 7n – (m + n) = 3m – 7n + (-m – n)
= 3m – 7n – m – n = (3m – m) + (-7n – n)
= (3 – 1 )m + (-7 – 1) n = 2m + (-8) n
= 2m – 8n

(iv) 2y + z from 6z – 5y
Solution:
6z – 5y – (2y + z) = 6z – 5y + (-2y – z)
= 6z – 5y – 2y – z = 6z – z – 5y – 2y
= (6 – 1) z + (-5 -2) y = 5z + (-7) y
= 5z – 7y = -7y + 5z

Question 7.
Simplify
(i) (x + y – z) + (3x – 5y + 7z) – (14x + 7y – 6z)
Solution:
(x + y – z) + (3x – 5y + 7z) – (14x – 7y – 6z)
= (x + y – z) + (3x – 5y + 7z) + (-14x – 7y + 6z)
= (x + 3x – 14x) + (y – 5y – 7y) + (-z + 7z + 6z)
= (1 + 3 – 14) x + (1 – 5 – 7)y + (-1 + 7 + 6) z
= – 10x – 11y + 12z

(ii) p + p + 2 + p + 3 + p – 4 – p – 5 + p + 10
Solution:
p + p + 2 + 3 – p – 4 – p – 5 + p + 10 = (p + p + p – p – p + p) + (2 + 3 – 4 – 5 + 10)
= (1 + 1 + 1 – 1 – 1 + 1) p + 6 = 2p + 6

(iii) n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
Solution:
n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
= n + m + 1 + n + 2 + m + 3 + n + 4 + m + 5
= n + n + n + m + m + m + 1 + 2 + 3 + 4 + 5
= (1 + 1 + 1)n + (1 + 1 + 1)m + 15
= 3n + 3m + 15 = 3m + 3n + 15

Objective Type Questions

Question 8.
The addition of 3mn, -5mn, 8mn and – 4mn is
(i) mn
(ii) – mn
(iii) 2mn
(iv) 3mn
Solution:
(iii) 2mn
Hint: = 3 mn + 8mn – 5 mn – 4 mn = 11 mn – 9 mn = 2 mn

Question 9.
When we subtract ‘a’ from ‘-a’, we get ______
(i) a
(ii) 2a
(iii) -2a
(iv) -a
Solution:
(iii) -2a
Hint: – a – a = – 2a

Question 10.
In an expression, we can add or subtract only _____
(i) like terms
(ii) unlike terms
(iii) all terms
(iv) None of the above
Solution:
(i) like terms

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 4 Direct and Inverse Proportion Additional Questions

Exercise 4.1

Question 1.
The amount of extension in an elastic spring varies directly as the weight hung on it. If a weight of 150 gm produces an extension of 2.9 cm, then what weight would produce an extension of 17.4 cm?
Solution:
To produce 2.9 cm extension weight needed = 150 gm

Question 2.
Reeta types 540 words during half on hour. How many words would she type in 12 minutes?
Solution:
In \(\frac{1}{2}\) an hour number of words typed = 540
i.e., In 30 min No. of words typed = 540

= 18
In 12 minutes number of words typed = 18 × 12
= 216
216 words can be typed in 12 min

Question 3.
A call taxi charges ₹ 130 for 100 km. How much would one travel for ₹ 390?
Solution:

Exercise 4.2

Question 1.
In the following table find out x and y vary directly or inversely?

Solution:
From the table itself we observe that as x increases y decreases.
∴ x and y are inversely proportional
∴ xy = 8 × 32 = 16 × 16 = 32 × 8 = 256 × 1 = 256

Question 2.
If x and y vary inversely as each other and x = 10 when y = 6. Find y when x = 15.
Solution:
Since x and y vary inversely as each other
xy = constant
10 × 6 = 15 xy
60 = 15y

Question 3.
If x and y vary inversely and if y = 35 find x when constant of variation is 7.
Solution:
Given x andy are inversely proportional
xy = constant
when y = 35 and constant = 7 ; x × 35 = 7

Exercise 4.3

Question 1.
Sumathi sweeps 600 m long road in 2\(\frac{1}{2}\) hrs. Ramani sweeps \(\frac{2}{3}\) rd of same road in 1\(\frac{1}{2}\) hrs. Who sweeps more speedily?
Solution:

Question 2.
Suma weaves 25 baskets in 35 days. In how many days will she weave 110 baskets?
Solution:

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

Question 1.
Fill in the blanks
(i) The variable in the expression 16x – 7 is _____
(ii) The constant term of the expression 2y – 6 is _____
(iii) In the expression 25m + 14M, the type of the terms are ______ terms
(iv) The number of terms in the expression 3ab + 4c – 9 is _____
Hint: Terms are 3ab, 4c – 9.
(v) The numerical co-efficient of the term -xy is ______
Hint: -x,y = (- 1 )xy.
Solution:
(i) x
(ii) -6
(iii) unlike
(iv) three
(v) -1

Question 2.
Say true or False
(i) x + (-x) = 0.
(ii) The co-efficient of ab in the term 15 abc is 15.
Hint: Coefficient of ab is 15c
(iii) 2pq and – 7qp are like terms.
(iv) When y = -1, the value of the expression 2y – 1 is 3.
Hint: 2(-1) – 1 = -2 – 1 = – 3
Solution:
(i) True
(ii) False
(iii) True
(iv) False

Question 3.
Fing the numerical co-efficient of each of the following terms: -3yx, 12k, y, 121bc, -x, 9pq, 2ab.
Solution:
(i) Numerical co-efficient of-3yx is – 3
(ii) Numerical co-efficient of 12k is 12
(iii) Numerical coefficient of y is 1
(iv) Numerical co-efficient of 1216c is 121
(v) Numerical co-efficient of – x is – 1
(vi) Numerical co-efficient of 9pq is 9
(vii) Numerical co-efficient of 2ab is 2

Question 4.
Write the variables, constants and terms of the following expressions,
(i) 18 + x – y
(ii) 7p – 4q + 5
(iii) 29x + 13y
(iv) b + 2
Solution:

Question 5.
Identify the like terms among the following 7x, 5y, -8x, 12y, 6z, z, -12x, -9y, 11 z
Solution:

Question 6.
If x = 2 andy = 3, then find the value of the following expressions,
(i) 2x – 3y
(ii) x + y
(iii) 4y – x
(iv) x + 1 – y
Solution:
Given x = 2; y = 3.
(i) 2x – 3y = 2 (2) – 3 (3) = 4 – 9
= 4 + (Additive inverse of 9)
= 4 +(-9) = -5
(ii) x + y = 2 + 3 = 5
(iii) 4y – x = 4 (3) – 2 = 12 – 2 = 10
(iv) x + 1 – y = 2 + 1 – 3 = 3 – 3 = 0

Objective Type Questions

Question 1.
An algebraic statement which is equivalent to the verbal statement “Three times the sum of ‘x’ and ‘y’ is
(i) 3 (x + y)
(ii) 3 + x + y
(iii) 3x + y
(iv) 3 + xy
Solution:
(i) 3 [(x + y)]

Question 2.
The numerical co-efficient of -7mn is
(i) 7
(ii) -7
(iii) p
(iv) -p
Solution:
(ii) -7

Question 3.
Choose the pair of like terms
(i) 7p, 7x
(ii) 7r, 7x
(iii) – 4x, 4
(iv) – 4x, 7x
Solution:
(iv) -4x, 7x

Question 4.
The value of 7a – 4b when a = 3, b = 2 is
(i) 21
(ii) 13
(iii) 8
(iv) 32
Solution:
(ii) 13
Hint: 7(3) – 4(2) = 21 – 8 = 13

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 5 Geometry Intext Questions

Exercise 5.1

Recap
Try These (Text book Page No. 83)

Question 1.
Complete the following statements.
(i) A Line is a straight path that goes on endlessly in two directions.
(ii) A Line segment is a line with two end points.
(iii) A Ray is a straight path that begins at a point and goes on and extends endlessly the other direction.
(iv) The lines which intersect at right angles are Perpendicular lines.
(v) The lines which intersect each other at a point are called Intersecting lines.
(vi) The lines that never intersect are called Parallel lines.

Question 2.
Use a ruler or straightedge to draw each figure.

Question 3.
Look at the figure and answer the following questions.
(i) Which line is parallel to AB.
(ii) Name a line which intersect CD.
(iii) Name the lines which are perpendicular to GH
(iv) How many lines are parallel to IJ
(v) Will EF intersect AB? Explain.

Solution:
\(\overleftrightarrow { GH } \) is parallel to \(\overleftrightarrow { AB } \)
(ii) \(\overleftrightarrow { IJ } \) and \(\overleftrightarrow { KL } \) intersect \(\overleftrightarrow { CD } \)
(iii) \(\overleftrightarrow { IJ } \) and \(\overleftrightarrow { KL } \) are perpendicular to \(\overleftrightarrow { GH } \)
(iv) Only one line \(\overleftrightarrow { KL } \) is parallel to \(\overleftrightarrow { IJ } \)
(v) Yes, \(\overleftrightarrow { EF } \) will intersect \(\overleftrightarrow { AB } \) at some point.

Try These (Text Book Page No. 85)

Choose the correct answer

Question 1.
A straight angle measures
(a) 45°
(b) 90°
(c) 180°
(d) 100°
Solution:
(c) 180°
Solution:
No, they are not adjacent pairs.

Question 2.
An angle with measure 128° is called ___ angle.
(a) a straight
(b) an obtuse
(c) an acute
(d) Right
Solution:
(b) an obtuse

Question 3.
The corner of the A4 paper has
(a) An acute angle
(b) A right angle
(c) Straight
(d) An obtuse angle
Solution:
(b) a right angle

Question 4.
If a perpendicular line is bisecting the given line, you would have two
(a) right angles
(b) obtuse angles
(c) acute angles
(d) reflex angles
Solution:
(a) right angle

Question 5.
An angle that measure 0° is called
(a) right angle
(b) obtuse angle
(c) acute angle
(d) Zero angle.
Solution:
(d) Zero angle

Try this (Text Book Page No. 86)

Question 1.
In each of the following figures, observe the pair of angles that are marked as ∠1 and ∠2. Do you think that they are adjacent pairs? Justify your answer.

Solution:
No, they are not adjacent pairs.
In (i) and (ii) angles ∠1 and ∠2 have no common vertex.
In (iii) the interiors of ∠1 and ∠2 overlaps.
∴ they are not adjacent angles.

Try these (Text book Page No. 87)

Question 1.
Few real life examples depicting adjacent angles are shown below.

Can you give three more examples of adjacent angles seen in real life?
Solution:

(i) Angles between leaf veins. [ ∠1 and ∠2],
(ii) Angles between adjacent pages of a book, when it is open [ ∠1 and ∠2 ].
(iii) Adjacent angles of scissors [ ∠1 and ∠2 ]

Question 2.
Observe the six angles marked in the picture shown. Write any four pairs of adjacent angles and that are not.
Solution:
Four pairs of adjacent angles are
1. ∠A and ∠B
2. ∠B and ∠C
3. ∠C and ∠D
4. ∠D and ∠E

Four pairs of non adjacent angles are.
1. ∠A and ∠C
2. ∠C and ∠F
3. ∠E and ∠D
4. ∠A and ∠F

Question 3.
Identify the common arm, common vertex of the adjacent angles and shade the interior with two colours in each of the following figures.

Solution:

(ii)

Solution:

Question 4.
Name the adjacent angles in each of the following figure.

Solution:
(i) ∠BAC and ∠CAD are adjacent angles.
(ii) ∠XWY and ∠YWZ are adjacent angles.

Try These (Text Book Page No. 88)

Question 1.
Observe the following pictures and find the other angles of the linear pair.

Solution:
(i) Given one angle 84°
∴ Other angle of the linear pair is 180° – 84° = 96°

(ii) One angle is given as 86°
Other angle of linear pair is 180° – 86° = 94°

(iii) Given one angle = 159°
Other angle of the linear pair = 180° – 159° = 21°

Try this (Text book Page No. 88)

Question 1.
Observe the figure. There are two angles namely ∠PQR = 150° and ∠QPS = 30° Is all this pair of supplementary angles a linear pair? Discuss

Solution:
Given ∠PQR =150°
∠QPS = 30°
They are supplementary angles,
But they are not adjacent angles as they don’t have common vertex or common arm.
∴ They are not a linear pair.

Try this (Text book Page No. 90)

Question 1.
What would happen to the angles if we add 3 or 4 or 5 rays on a line as given below?

Solution:
New adjacent angles are formed.
The new angles become smaller in measure. But their sum is 180° as it is a linear angle.

Try this (Text book Page No. 90)

Question 1.
Can you justify the statement
∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF + ∠FOA = 360°?

Solution:
We know that the sum of angles at a point is 360°
∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF + ∠FOA = 360° as they are the sum of angles at the point ‘O’

Try These (Text book Page No. 91)

Question 1.
Four real life examples of vertically opposite angles are given below.

Solution:
(i) The four angles made in the scissors where the opposite angles are always equal.
(ii) The point where two roads intersect each other.
(iii) Rail road crossing signs.
(iv) An hourglass.

Question 2.
In the given figure two lines \(\overleftrightarrow { AB } \) and \(\overleftrightarrow { CD } \) intersect at ‘O’. Observe the pair of angles and complete the following table. One is done for you.

Solution:

Question 3.
Name the two pairs of vertically opposite angles

Solution:
∠PTS and ∠QTR are vertically opposite angles.
∠PTR and ∠QTS are vertically opposite angles.

Question 4.
Find the value of x° in the figure given below.

Solution:
Lines l and m intersect at a point and making a pair of vertically opposite angles x° and 150°.
We know that vertically opposite angles are equal.
x = 150°

Exercise 5.2

Try this (Text book Page No. 93)

Question 1.
For a given set of lines, it is possible to draw more than one transversal.

Solution:
Yes, it is possible to draw more than one transversal for a given set of lines. l and m are given set of lines. n and p are transversal

Try these (Text Book Page No 94)

Question 1.
Draw as many possible transversals in the given figures.

Solution:

(i) a, b, c are transversal to l, m and n.
(ii) a, b, c are transversal to l, m, n and p. More transversals can be drawn.

Question 2.
Draw a line which is not the transversal to the above figures.
Solution:

Question 3.
How many transversals can you draw for the following two lines

Solution:
Infinite number of transversals can be drawn.

a, b, c, d, e, f, g are transversal to m and n.

Try these (Text book Page No. 96)

Question 1.
Four real life examples for transversal of parallel lines are given below.

Give four more examples for transversal of parallel lines seen in your surroundings.
Solution:
Some examples of parallel lines in our surroundings
(i) Zebra crossing on the road.
(ii) Railway tracks with sleepers.
(iii) Steps
(iv) Parallel bars in men’s gymnastics

Question 2.
Find the value of x.

Solution:
(i) We know that if two parallel lines are cut by a transversal, each pair of corresponding angles are equal.
∴ x = 125°

(ii) m and n are parallel lines and l is a transversal x° and 48° are corresponding angles.
∴ x = 48°

(iii) m and n are parallel lines and 7’ is the transversal.
∴ Corresponding angles are equal.
∴ x° = 138°

Try these (Text book Page No. 98)

Question 1.
Find the value of x°.

(i) m and n are parallel lines. ‘l’ is a transversal.
When two parallel lines are cut by a transversal each pair of alternate interior angles are equal.
∴ x° = 127°

(ii) m and n are parallel lines and l is the transversal.
When two parallel lines are cut by a transversal each pair of alternate exterior angles are equal.
∴ x° = 46°

Try These (Text Book Page No. 99)

Question 1.
Find the values of x.

Solution:
(i) m and n are parallel lines and l is the transversal.
When two parallel lines are cut by a transversal, each pair of interior angles that lie on the same side of the transversal are supplementary

(ii) m and n are parallel line and l is the transversal.
When two parallel lines are cut by a transversal, each pair of exterior angles that lie on the same side of the transversal are supplementary.
the same side of the transversal are supplementary.
∴ x° + 132° = 180°
x° = 180° – 132°
= 48°
∴ x = 48°

Exercise 5.3

Question 1.
What will happen If the radius of the arc is less than half of AB?
Solution:

If the radius of the arc is less than half of AB, then both the arcs will not cut at a point
and we can’t draw perpendicular bisector.