Class 7

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.

Students can Download Maths Chapter 3 Algebra Ex 3.3 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 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.

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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 1 Chapter 5 Geometry Ex 5.6

Miscellaneous Practice Problems

Question 1.
Find the value of x if ∠AOB is a right angle.
Solution:
Given that ∠AOB = 90°

∠AOB = ∠AOC + ∠COB = 90° (Adjacent angles)
3x + 2x = 90°
5x = 90°

Question 2.
In the given figure, find the value of x.
Solution:
Since ∠BOC and ∠AOC are linear pair, their sum = 180°
2x + 23 + 3x – 48 = 180°
5x – 25 = 180°
5x – 25 + 25 = 180° + 25

Question 3.
Find the value of x, y and z.
Solution:

∠DOB and ∠BOC are linear pair
∴ ∠DOB + ∠BOC = 180°
x + 3x + 40 = 180°
4x + 40 = 180°
4x + 40 – 40 = 180° – 40°
4x = 140°

Also ∠BOD and ∠AOC are vertically opposite angles.
∴ ∠BOD = ∠AOC
x = z + 10
35° = z + 10
z + 10 – 10 = 35 – 10
z = 25°
Again ∠AOD and ∠AOC are linear pair.
∴ ∠AOD + ∠AOC = 180°
y + 30 + z + 10 = 180°
y + 30 + 25 + 10 = 180°
y + 65 = 180°
y + 65 – 65 = 180° – 65
y = 115°
∴ x = 35°,
y = 115°,
z = 25°

Question 4.
Two angles are in the ratio 11 : 25. If they are linear pair, find the angles.
Solution:
Given two angles are in the ratio 11 : 25.
Let the angles be 11x and 25x.
They are also linear pair
∴ 11x + 25x = 180°.

∴ The angles 11x = 11 × 5° = 55° and 25x = 25 × 5 = 125°.
∴ The angles are 55° and 125°.

Question 5.
Using the figure, answer the following questions and justify your answer.
(i) Is ∠1 adjacent to ∠2?
(ii) Is ∠AOB adjacent to ∠BOE?
(iii) Does ∠BOC and ∠BOD form a linear pair?
(iv) Are the angles ∠COD and ∠BOD supplementary.
(v) Is ∠3 vertically opposite to ∠1 ?

Solution:
(i) Yes, ∠1 is adjacent to ∠2.
Because they both have the common vertex ‘O’ and the common arm OA . Also their interiors do not overlap.
(ii) No, ∠AOB and ∠BOB are not adjacent angles because they have overlapping interiors.
(iii) No, ∠BOC and ∠BOD does not form a linear pair.
Because ∠BOC itself a straight angle, so the sum of ∠BOC and ∠BOD exceed 180°.
(iv) Yes, the angles ∠COD and ∠BOD are supplementary ∠COD + ∠BOD = 180°, [∵ linear pair of angles]
∴ ∠COD and ∠BOD are supplementary.
(v) No. ∠3 and ∠1 are not formed by intersecting lines. So they are not vertically opposite angles.

Question 6.
In the figure POQ, ROS and TOU are straight lines. Find the x°.

Solution:
Given TOU is a straight line.
∴ The sum of all angles formed at a point on a straight line is 180°
∠TOR + ∠ROP + ∠POV + ∠VOU = 180°.
36° + 47°+ 45° + x° = 180°.
128° + x° = 180°
128° + x° – 128° = 180° – 128°
x = 52°

Question 7.
In the figure AB is parallel to DC. Find the value of ∠1 and ∠2. Justify your answer.

Solution:
Given AB || DC
AB and CD are parallel lines Taking CE as transversal we have.
∠1 = 30°, [∵ alternate interior angles]
Taking DE as transversal
∠2 = 80°.[∵ alternate interior angles]
∠1 = 30° and ∠2 = 80°
Justification:
CDE is a triangle

Question 8.
In the figure AB is parallel to CD. Find x, y and z.

Solution:
Given AB || CD
∴ Z = 42 (∵ Alternate interior angles)
Also y = 42° [vertically opposite angles]
Also x° + 63° + z° = 180°
x° + 63° + 42° = 180°
x° + 105° = 180°
x°+ 105° – 105° = 180° – 105°
x° = 75°
∴ x = 75°;
y = 42°;
z = 42°

Question 9.
Draw two parallel lines and a transversal. Mark two alternate interior angles G and H. If they are supplementary, what is the measure of each angle?
Solution:
l and m are parallel lines and n is the transversal.
∠G and ∠H are alternate interior angles.
∠G = ∠H …… (1)
Given ∠G and ∠G are Suplementary

Question 10.
A plumber must install pipe 2 parallel to pipe 1. He knows that ∠1 is 53. What is the measure of ∠2?

Solution:
Given ∠1 = 53°

Clearly ∠1 and ∠2 are interior angles on the same side of the transversal and so they are supplementary.
∠1 + ∠2 = 180°
53° + ∠2 = 180°
53° + ∠2 – 53° = 180° – 53°
∠2 = 127°

Challenge Problems

Question 11.
Find the value of y.

Solution:
Cleary POQ is a straight line”
Sum of all angles formed at a point on a straight line is 180°
∴ ∠POT + ∠TOS + ∠SOR + ∠ROQ = 180°
60° + (3y – 20°) + y° + (y + 10°) = 180°
60° + 3y – 20° + y° + y° + 10° = 180°
5y + 50° = 180°
5y + 50° – 50° = 180° – 50
5y = 130°
\(y=\frac{130^{\circ}}{5}\)
y = 26°

Question 12.
Find the value of z.

Solution:
The sum of angles at a point is 360°.
∴ ∠QOP + ∠PON + ∠NOM + ∠MOQ = 360°
3z + (2z – 5) + (z + 10) + (4z – 25) = 360°
3z + 2z + z + 4z — 5 +10 — 25 = 360°
10z – 20° = 360°
10z – 20° + 20 = 360°+ 2
10z = 380°
\(z=\frac{380^{\circ}}{10}\)
z = 38°

Question 13.
Find the value of x and y if RS is parallel to PQ.

Solution:
Given RS || PQ
Considering the transversal RU, we have y = 25° (corresponding angles)
Considering ST as transversal

Question 14.
Two parallel lines are cut by a transversal. For each pair of interior angles on the same side of the transversal, if one angle exceeds the twice of the other angle by 48°. Find the angles.
Solution:
Let the two parallel lines be m and n and l be the transversal
Let one of the interior angles on the same side of the transversal be x°
Then the other will be 2x + 48.
We know that they are supplementary.

Question 15.
In the figure, the lines GH and IJ are parallel. If ∠1 = 108° and ∠2 = 123°, find the value of x, y and z.
Solution:
Given GH || IZ
∠1 = 108°
∠2 = 123°
∠1 + ∠KGH = 180 [linear pair]
108° + ∠KGH = 180°

108° + ∠KGH – 108° = 180° – 108°
∠KGH = 72°
∠KGH = x° (corresponding angles if KG is a transversal)
∴ x° = 72°
Similarly
∠2 + ∠GHK = 180° (∵ linear pair)
123° + ∠GHK = 180°
123° + ∠GHK – 123° = 180° – 123°
∠GHK = 57°
Again ∠GHK = y° (corresponding angles if KH is a transversal)
y = 57°
x° +y° + z° = 180° (sum of three angles of a triangle is 180°)
72° + 57° + z° = 180°
129° + z° = 180°
129° + z° – 129° = 180° – 129°
z = 51°
x = 72°,
y = 57°,
z = 51°

Question 16.
In the parking lot shown, the lines that mark the width of each space are parallel. If
∠1 = (2x – 3y)°; ∠2 = (x + 39)° find x° and y°.

Solution:
From the picture
∠2 + 65° = 180° [Sum of interior angles on the same side of a transversal]
x + 39° + 65° = 180°
x + 104° = 180°
x + 104° – 104° = 180° – 104°
x = 76°
Also from the picture
∠1 = 65° [alternate exterior angles]
2x – 3y = 65°
2 (76) – 3y = 65°
152° – 3y = 65°
152° – 3y – 152° = 65 – 152°
-3y = -87

Question 17.
Draw two parallel lines and a transversal. Mark two corresponding angles A and B. If ∠A = 4x, and ∠B = 3x + 7, find the value of x. Explain.
Solution:
Let m and n are two parallel lines and l is the transversal.
A and B are corresponding angles.
We know that corresponding angles are equals,

Question 18.
In the figure AB in parallel to CD. Find x°, y° and z°.
Solution:

Given AB||CD
Then AD and BC are transversals.
x = 48°, alternate interior angles; AD is transversal y = 60°, alternate interior angles; BC is transversal
∠AEB + 48° + y° = 180°, (sum of angles of a triangle is 180°)

Question 19.
Two parallel lines are cut by transversal. If one angle of a pair of corresponding angles can be represented by 42° less than three times the other. Find the corresponding angles.
Solution:
We know that the corresponding angles are equal.
Let one of the corresponding angles be x.
Then the other will be 3x – 42°.

Question 20.
In the given figure, ∠8 = 107°, what is these sum of the angles ∠2 and ∠4.
Solution:
Given ∠8 = 107°
∠2 = 107°
[∵ ∠8 and ∠2 are alternate exterior angles, ∵ ∠8 = ∠2]

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 1 Chapter 1 Number System Intext Questions

(Try These Textbook Page No. 1)

Question 1.
Write the following integers in ascending order: -5,0,2,4, -6,10, -10
Solution:
Plotting the points on the number line, we get

The numbers are placed in an increasing order from left to right.
∴ Ascending order: -10 < -6 < -5 < 0 < 2 < 4 < 10

Question 2.
If the integers -15, 12, -17, 5, -1, -5, 6 are marked on the number line then the integer on the extreme left is _____ .
Solution:
The least number will be on the extreme left.
∴ -17 will be on the extreme left.

Question 3.
Complete the following pattern:
50, ___ 30, 20, _, 0, -10, _, _, -40, _, ___.
Solution:
The difference between the consecutive number is 10.
50, 40, 30,20, 10, 0, -10, -20, -30, -40, -50, -60

Question 4.
Compare the given numbers and write “<”, “>” or in the boxes.

Solution:
(a)  A positive number is greater than a negative number.
(b) 1000,0 is less than all positive integers.
(c)

Question 5.
Write the given integers in descending order, -27, 19, 0, 12, -4, -22, 47, 3, -9, -35.
Solution:
Separating positive and the negative integers, we get -27, -4, -22, -9, -35
Arranging the numbers in descending order -4 > -9 > -22 > -27 > -35
The positive numbers are 19,12,47, 3
Arranging in descending order, we get 47 > 19 > 12 > 3
0 stands in the middle.
∴ Descending order: 47 > 19 > 12 > 3 > 0 > -4 > -9 > -22 > -27 > -35

(Try This Text Book Page No. 3)

Question 1.
Find the value of the following using the number line activity.
(i) (-4) + (+3)
(ii) (-4) + (-3)
(iii) (+4) + (-3)
Solution:
(i) (-4) + (+3)
To find the sum of (-4) and (+3), we start at zero facing positive direction continuing in the same direction and move 4 units backward to represent (-4).
Since the operation is addition we maintain the same direction and move three units forward to represent (+3)
We land at -1
So (-4) + (+3) = -1

(ii) (-4) + (-3)
From zero move 4 steps backward to represent (-4)
From the same direction again move 3 units backward to represent (-3)
We land at -7 So (-4) + (-3) = -7

(iii) (+4) + (-3)
We start at zero facing positive direction and move 4 steps forward to represent (+4) Since the operation is addition we maintain the same direction and move three units backward to represent (-3).
We land at +1.
So (+4) + (-3) = +1

(Properties of Addition Textbook Page No. 6)

Question 1.
Complete the given table and check whether the sum of two integers is an integer or not?
(i) 7 + (-5) = (+2)
(ii) (-6)+ (-13) = (-19)
(iii) 25 + 9 = 34
(iv) (-12) + 4 = -8
(v) 41 + 32 = 73
(vi) (-19) + (-15) = (-34)
(vii) 52 + (-15) = (+37)
(viii)(-7) + 0 = (-7)
(ix) 0 + 12 = 12
(x) 14 + 0 = 14
(xi) (-6) +(-6) = (-12)
(xii) (-27) + 0 = -27
Solution:
The sum of two integers is an integer.

(Try These Textbook Page No. 7)

Question 1.
Fill in the blanks:
(i) 20 + (-11) = -(11)+ 20 [∵ Addition is commutative]
(ii) (-5) + (-8) = (-8) + (-5) [∵ Addition is commutative]
(iii) (-3) +12 =12 + (-3) [∵ Addition is commutative]

Question 2.
Say True or False.
(i) (-11) + (-8) = (-8) + (-11)
(ii) -7 + 2 = 2 + (-7)
(iii) (-33) + 8 = 8 + (-33)
Solution:
(i) True, because addition is commutative for intergers
(ii) True, by commutative property on intergers
(iii) True, by commutative property on intergers

Question 3.
Verify the following.
(i) [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]
(ii) [7 + (-8)] + (-5) = 7 + [(-8) + (-5)]
(iii) [(-11) + 5]+ (-14) = (-11) + [5 + (-14)]
(iv) (-5) + [(-32) +(-2)] = [(-5) + (-32) + (-2)]
Solution:
(i) [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]
[(-2) + (-9)] + 6 = (-11) + 6 = -5
Also (-2) + [(-9) + 6] = (-2) + (-3) = -5
Both the cases the sum is -5.
∴ – [(-2) + (-9)] + 6 = (-2) + [(-9) + 6]

(ii) [7 + (-8)] +(-5) = 7 + [(-8) + (-5)]
Here [7 + (-8)] + (-5) = (-1) + (-5) = -6
Also 7 + [(-8) + (-5)] = 7 + (-13) = 7 – 13 = -6
In both the cases the sum is -6.
∴ [7 + (-8)] + (-5) = 7 + [(-8) + (-5)]

(iii) [(-11) + 5] + (-14) = (-11) + [5 + (-14)]
Here [(-11) + 5] + (-14) = (-6) + (-14) = (-20)
(-11) + [5 + (-14)] = (-11) +(-9) = (-20)
In both the cases the sum is -20.
∴ [(-11) + 5] + (-14) = (-11) + [5 + (-14)]

(iv) (-5) + [(-32) + (-2)] = [(-5) + (-32)] + (-2)
(-5) + [(-32) + (-2)] = (-5) + (-34) = -39
Also [(-5) + (-32)] + (-2) = (-37) + (-2) = -39
In both the cases the sum is -39.
∴ (-5)+ [(-32) +(-2)] = [(-5)+ (-32)] +(-2)

Question 4.
Find the missing integers:
(i) 0 + (-95) = -95
(ii) -611 + 0 = -611
(iii) ____ + 0 = _____ Any integer; the same integer
(iv) 0 + (-140) = -140

Question 5.
Complete the following:
(i) -603 + 603 = 0
(ii) 9847+ (-9847) = 0
(iii) 1652 + (-1652) = 0
(iv) -777 + 777 = 0
(v) –5281 +5281 = 0

Exercise 1.2

Subtraction of Integers

(Try These Text book Page No. 11)

Question 1.
Do the following by using number line.
(i) (-4) – (+3)
Solution:
We start at zero facing positive direction move 4 units backward to represent (-4). Then turn towards negative side and move 3 units forward.

We reach -7.
∴ (-4) – (+3) = -7.

(ii) (-4) – (-3)
Solution:
We start at zero facing positive direction. Move 4 units backward to represent -4. Then turn towards the negative side and move 3 units backwards.

We reach at-1.
∴ (-4) – (-3) = -1.

Question 2.
Find the values and compare the answers.
(i) (-6) – (-2) and (-6) + 2
Solution:
(-6) – (-2) = -6 + (Additive inverse of-2)
= -6 + (+2) = -4
Also (-6)+ 2 = -4
∴ (-6) – (-2) = (-6) + 2

(ii) 35 – (-7) and 35 + 7.
Solution:
35 – (-7) = 35 + (Additive inverse of -7) = 35 + (+7) = 42
Also 35 + 7 = 42 ; 35 – (-7) =35 + 7

(iii) 26 – (+10) and 26 + (-10)
Solution:
26 – (+10) = 26 + (Additive inverse of +10) = 26 + (-10) = 16
Also 26 + (-10) = 16; 26 – (+10) = 26 + (-10)

Question 3.
Put the suitable symbol <, > or = in the boxes.

Solution:
(i) -10 – 8 = -18 & -10 + 8 = – 2
(ii) (-20) + 10 = -10 & (-20) – (-10) = -10
(iii) -70 – 50 = (-70) + (-50) = -20
(iv) 100 – (+100) = 0 & 100 – (-100) = 100 + (+100) = 200
(v) -50 – 30 = -50 + (-30) = -80 Also -100 + 20 = – 80

(Try These Text book Page No. 14)

Question 1.
Fill in the blanks.
(i) (-7) – (-15) = +8
-7 – (-15) = -7 + (Additive inverse of-15)
= -7 + 15 = +8
(ii) 12 – (-7) = 19          12 – (-7) = 19
(iii) -4 – (-5) = 1

Question 2.
Find the values and compare the answers.
(i) 15 – 12 and 12 – 15
(ii) -21 – 32 and -32 – (-21)
Solution:
(i) 15 – 12 = 3 & 12-15 = 12 +(-15) = -3

(ii) -21 – 32 = (-21) + (-32) = -53
Also -32 – (-21) = (-32) + (+21) = -11 ; -53 < -11

Question 3.
Is associative property true for subtraction of integers. Take any three examples and check.
Solution:
Consider the numbers 1,2 and 3. Now (1 – 2) – 3 = -1 – 3 = -4
Also 1 – (2 – 3) = 1 – (-1) = 1 + 1 = 2
∴ (1 – 2) – ≠ 1 – (2 – 3)
∴ Associative property is not true for subtraction of integers.

Exercise 1.3

Multiplication of Integers

(Try These Textbook Page No. 16)

Question 1.
Find the product of the following
(i) (-20) × (-45) = +900 [As we know the product of two negative integers is positive, the answer is +900.]
(ii) (-9) × (-8) = 72 [ ∵ Product of two negative integers is positive]
(iii) (-30) × 40 × (-1) = (+1200) [Product of two integers with opposite sings is negative integer.
(-30) × 40 × (-1) = (-1200) × (-1) = +1200)]
(iv) (-50) × 2 × (-10) = -1000 [Product of two integers with opposite signs is negative.
(+50) × 2 × (-10) = 100 × (-10) = -1000)]

Question 2.
Complete the following table by multiplying the integers in the corresponding row and column headers.

Solution:
We know that
(i) product of two positive integers is positive
(ii) product of two negative integers is
(ii) product of two negative integers is positive
(iii) product of integers with opposite sign is negative.
∴ The table will be as follows:

Question 3.
Which of the following is incorrect?
(i) (-55) × (-22) × (-33) < 0
(ii) (-1521) × 2511 < 0
(iii) 2512 – 1525 < 0
(iv) (1981) × (+2000) < 0
Solution:
(iii) and (iv) are incorrect because 2512 – 1252 is a positive integer.
Also (+1981) × (+2000) is a positive integer.

(Try These Textbook Page No. 18)

Question 1.
Find the product and check for equality
(i) 18 × (-5) and (-5) × 18
Solution:
Here 18 × (-5) = -90 Also (-5) × 18 = -90
∴ 18 × (-5)= (-5) × 18

(ii) 31 × (-6) and (-6) × 31
Solution:
Here 31 × (-6) = -186 Also (-6) × 31 =-186
∴ 31 × (-6)= (-6) × 31

(iii) 4 × 51 and 51 × 4
Solution:
Here 4 × 51 = 204 Also 51 × 4 = 204
∴ 4 × 51 = 51 × 4

Question 2.
Prove the following.
(i) (-20) × (13 × 4) = [(-20) × 13] × 4
Solution:
LHS = (-20) × (13 × 4) = (-20) × 52 = -1040
RHS = [(-20) × 13] × 4 = (-260) × 4 = -1040
LHS = RHS
∴ (-20) × (13 × 4) = [(-20) × 13] × 4

(ii) [(-50) × (-2)] × (-3) = (-50) × [(-2) × (-3)]
Solution:
LHS = [(-50) × (-2)] × (-3) = 100 × (-3) = -300
RHS = (-50) × [(-2) × (-3)] = (-50) × 6 =-300
LHS = RHS
∴ [(-50) × (-2)] × (-3) = (-50) × [-2) × (-3)]

(iii) [(-4) × (-3)] × (-5) = (-4) × [(-3) × (-5)]
Solution:
LHS = [(-4) × (-3)] × (-5) = 12 × (-5) = -60
RHS = (-4) × [(-3) × (-5)] = (-4) × 15 = -60
LHS = RHS
∴ [(-4) × (-3)] × (-5) = (-4) × [(-3) × (-5)]

(Try These Textbook Page No. 19)

Question 1.
Find the values of the following and check for equality:
(i) (-6) × (4 + (-5)) and ((-6) × 4) + ((-6) × (-5))
Solution:
(-6) × (4 + (-5)) = (-6) × (-1) = 6 .
((-6) × 4) + ((-6) × (-5)) = (-24)+ 30 = 6
Hence (-6) × (4 + (-5)) = ((-6) × 4) + ((-6) × (-5))

(ii) (-3) × [2 + (-8)] and [(-3) × 2] + [(-3) × 8]
Solution:
(-3) × [2 + (-8)] = (-3) × (-6) = 18
Also [(-3) × 2] + [(-3) × 8] = (-6)+ (-24) = -30
(-3) × [2 + (-8)] ≠ [(-3) × 2] + [(-3) × 8]

Question 2.
Prove the following.
(i) [(-5) × (-76)] + [(-5) × 8]
Solution:
LHS = (-5) × [(-76) + 8] = (-5) × (-68)
= +340
RHS = [(-5) × (-76)] + [(-5) × 8]
= +380 + (-40) = +380 – 40
= +340
LHS = RHS
∴ (-5) × [(-76) + 8] = [(-5) × (-76)] + [(-5) × 8]

(ii) (42 × 7) + [42 × (-3)]
Solution:
LHS = 42 × [7 + (-3)]
= 168
RHS = (42 × 7) + [42 × (-3)] = 294 – 126
= 168
LHS = RHS
∴ 42 × [7 + (-3)] = (42 × 7) + [42 × (-3)]

(iii) [(-3) × (-4)] + [(-3) × (-5)]
Solution:
LHS = (-3) × [(-4) + (-5)] = (-3) × (-9)
= +27
RHS = [(-3) × (-4)] + [(-3) × (-5)] = 12 + 15 = 27
LHS = RHS
∴ (-3) × [(-4) + (-5)] = [(-3) × (-4)] + [(-3) × (-5)]

(iv) 103 × 25 = (100 + 3) × 25 = (100 × 25) + (3 × 25)
Solution:
First consider 103 × 25 = 2575
Now (100 + 3) × 25 = 103 × 25 = 2575
Also (100 × 25) + (3 × 25) = 2500 + 75
= 2575
∴ All the three are same. 103 × 25 = (100 + 3) × 25 = (100 × 25) +(3 × 25)

Exercise 1.4

Division of Integers

(Try These Text book Page No. 22)

Question 1.
(i) (-32) ÷ 4 = _____
(ii) (-50) ÷ 50 = ____
(iii) 30 ÷ 15 = ______
(iv) -200 ÷ 10 = _____
(v) -48 ÷ 6 = ______
Solution:
(i) -8
(ii) -1
(iii) 2
(iv) -20
(v) -8