Samacheer Kalvi Class 7 Maths Term 1 Chapter 3 Algebra Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 1 chapter 3 Algebra. Here students can easily find all the solutions for Algebra Exercise 3.1, 3.3, 3.3 and 3.4. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 3 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Algebra Exercise 3.1 Solutions :
(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 + 14n, the type of the terms are ___ terms.
(iv) The number of terms in the expression 3ab + 4c –9 is ___
(v) The numerical co-efficient of the term −xy is ___
Solution :
(i) → x
(ii) → -6
(iii) → Unlike terms.
(iv) → Three.
(v) → -1
(2) Say True or False.
(i) x + ( -x) = 0
(ii) The co-efficient of ab in the term 15abc is 15.
(iii) 2pq and -7qp are like terms
(iv) When y = -1, the value of the expression 2y − 1 is 3
Solution :
(i) → True
(ii) → False
(iii) → True
(iv) → False
(3) Find the numerical coefficient of each of the following terms : -3yx, 12k, y, 121bc, -x, 9pq, 2ab
Solution :
(i) Term : −3yx
Numerical coefficient = -3
(ii) Term : 12k
Numerical coefficient = 12
(iii) Term : y
Numerical coefficient = 1
(iv) Term : 121bc
Numerical coefficient = 121
(v) Term : -x
Numerical coefficient = -1
(vi) Term : 9pq
Numerical coefficient = 9
(vii) Term : 2ab
Numerical coefficient = 2
(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 :
(i) 18 + x − y
Expression = 18 + x − y
Variable = x, y
Constant = 18
Terms = 18, x, -y
(ii) 7p − 4q + 5
Expression = 7p − 4q + 5
Variable = p, q
Constant = 5
Terms = 7p , -4q , 5
(iii) 29x + 13y
Expression = 29x + 13y
Variable = x, y
Constant = 0
Terms = 29x , 13y
(iv) b + 2
Expression = b + 2
Variable = b
Constant = 2
Terms = b, 2
(5) Identify the like terms among the following : 7x, 5y, −8x, 12y, 6z, z, −12x, −9y, 11z.
Solution :
(a) 7x , -8x, -12x
(b) 5y , 12y and -9y
(c) 6z , z and 11z
Objective Type Question Solutions :
(8) The numerical co-efficient of −7mn is
(i) 7
(ii) −7
(iii) p
(iv) −p
Solution :
Correct Option → (ii)
The numerical co-efficient of −7mn is −7.
(9) Choose the pair of like terms
(i) 7p, 7x
(ii) 7r, 7x
(iii) −4x, 4
(iv) −4x, 7x
Solution :
Correct Option → (iv)
Here, the pair of like term is −4x, 7x
(10) The value of 7a − 4b when a = 3, b = 2 is
(i) 21
(ii) 13
(iii) 8
(iv) 32
Solution :
Correct Option → (ii)
The value of 7a − 4b when a = 3, b = 2 is 13
Algebra Exercise 3.2 Solutions :
(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
(2) Say True or False.
(i) The expressions 8x + 3y and 7x + 2y can not be added.
(ii) If x is a natural number, then x + 1 is its predecessor.
(iii) Sum of a – b + c and -a + b – c is zero.
Solution :
(i) Statement is → False.
(ii) Statement is → False.
(3) Add :
(i) 8x, 3x
(ii) 7mn, 5mn
(iii) -9y, 11y, 2y
Solution :
(i) 8x, 3x
= 8 x+ 3x
= (8+3) x = 11x
(ii) 7mn, 5mn
= 7mn + 5mn
= ( 7 + 5) mn = 12mn
(iii) −9y, 11y, 2y
= -9y + 11 y + 2y
= (-9+11+2) y
= 4y
(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
(5) Find the sum of the following expressions
(i) 7p + 6q, 5p − q, q + 16p
(ii) a + 5b + 7c, 2a + 10b + 9c
(iii) mn + t, 2mn − 2t, − 3t + 3mn
(iv) u + v, u − v, 2u + 5v, 2u − 5v.
(v) 5xyz − 3xy, 3zxy − 5yx
Solution :
(i) 7p + 6q, 5p − q, q + 16p
= 7p + 6q + 5p − q + q + 16p
= (7+5+16)p + (6-1+1)q
= 28p + 6q
(ii) a + 5b + 7c, 2a + 10b + 9c
= a + 5b + 7c + 2a + 10b + 9c
= ( 1+2)a + (5+10)b + (7+9)c
= 3a + 15b + 16c
(iii) mn + t, 2mn − 2t, − 3t + 3mn
= mn + t + 2mn − 2t +( − 3t + 3mn)
= (1+2+3) mn + (1 -2 -3) t
= 6mn – 4t
(iv) u + v, u − v, 2u + 5v, 2u − 5v.
= u + v + u − v + 2u + 5v + 2u − 5v
= (1 + 1 + 2 + 2)u + (1 -1 + 5 -5)v
= 6u
(v) 5xyz − 3xy, 3zxy − 5yx
= 5xyz − 3xy + 3zxy − 5yx
= (5 + 3)xyz – (3 + 5)yx
= 8 xyz – 8xy
(6) Subtract
(i) 13x + 12y − 5 from 27x + 5y − 43
(ii) 3p + 5 from p − 2q + 7
(iii) m + n from 3m − 7n.
(iv) 2y + z from 6z − 5y.
Solution :
(i) 13x + 12y − 5 from 27x + 5y − 43
= 27x + 5y − 43 – (13x + 12y − 5)
= (27 – 13) x + ( 5 -12)y – 43 + 5
= 14x – 7y -38
(ii) 3p + 5 from p − 2q + 7
= p − 2q + 7 – ( 3p + 5)
= (1-3)p -2q + 7 -5
= -2p -2q + 2
(iii) m + n from 3m − 7n.
= 3m − 7n – ( m + n)
= (3-1) m – (7 + 1) n
= 2m – 8n
(iv) 2y + z from 6z − 5y.
= 6z − 5y – ( 2y + z)
= (6 -1)z – ( 5 + 2)y
= 5z – 7y
(7) Simplify :
(i) (x + y − z) + (3x − 5y + 7z) − (14x + 7y − 6z)
(ii) p + p + 2 + p + 3 − p − 4 − p − 5 + p + 10
(iii) n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
Solution :
(i) (x + y − z) + (3x − 5y + 7z) − (14x + 7y − 6z)
= (x + y − z) + (3x − 5y + 7z) − (14x + 7y − 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
= p + p + 2 + p + 3 − p − 4 − p − 5 + p + 10
= (1 + 1 + 1 – 1 – 1 + 1)p + 2 + 3 – 4 – 5 + 10
= 2p + 6
(iii) n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
= n + (m + 1) + (n + 2) + (m + 3) + (n + 4) + (m + 5)
= (1 + 1 + 1)n + (1 + 1 + 1)m + 1 + 2 + 3 + 4 + 5
= 3n + 3m + 15
Objective Type Question Solution :
(8) The addition of 3mn, −5mn, 8mn and −4mn is
(i) mn
(ii) −mn
(iii) 2mn
(iv) 3mn
Solution :
Correct Option → (iii)
The addition of 3mn, −5mn, 8mn and −4mn is 2mn.
(9) When we subtract ‘a’ from ‘– a’, we get
(i) 0
(ii) 2a
(iii) −2a
(iv) −a
Solution :
Correct Option → (iii)
If we subtract ‘a’ from ‘–a’, we get −2a
(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 :
Correct Option → (i)
In an expression, we can add or subtract only Like terms.
Algebra Exercise 3.3 Solutions :
(1) Fill in the blanks.
(i) An expression 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) → Equation
(ii) → 15
(iii) → 6x
(2) Say True or False.
(i) Every algebraic expression is an equation
(ii) The expression 7x + 1 can not 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
(3) Solve :
(i) x + 5 = 8
(ii) p − 3 = 7
(iii) 2x = 30
(iv) m/6 = 5
(v) 7x + 10 = 80
Solution :
(i) x + 5 = 8
x + 5 = 8
x + 5 – 5 = 8 – 5 ….. (Subtract 8 on both sides )
x = 3
(ii) p − 3 = 7
p − 3 = 7
p -3 + 3 = 7 + 3 …. ( add 3 on both sides)
p = 10
(iii) 2x = 30
2x = 30
2x/2 = 30/2 …. (Divide 2 on both sides)
x = 15
(iv) m/6 = 5
m/6 = 5
m/6 * 6 = 5*6
m = 30
(v) 7x + 10 = 80
7x + 10 = 80
7x +10 – 10 = 80 – 10
7x = 70
7x/7 = 70/7
x = 10
(4) What should be added to 3x + 6y to get 5x + 8y?
Solution :
Let A should be added to 3x + 6y
Thus, A + 3x + 6y = 5x + 8y
A + 3x + 6y – ( 3x + 6y )
= 5x + 8y – ( 3x + 6y )
A = (5 -3) x + (8-6) y
A = 2x + 2y
Hence, 2x + 2y should be added to 3x + 6y to get 5x + 8y.
(5) Nine added to thrice a whole number gives 45. Find the number.
Solution :
Let A be a number
According to question
9 + 3A = 45
9 – 9 + 3A = 45 – 9
3 A = 36
A = 36/3
A = 12
Hence this number is 12
(6) Find two consecutive odd numbers whose sum is 200.
Solution :
Let, x and x + 2 be two consecutive odd numbers
Thus, x + x + 2 = 200
2x + 2 = 200
2x + 2 – 2 = 200 – 2
2x = 198
x = 198/2
x = 99
Therefore
x + 2 = 99 + 2 = 101.
Hence, two consecutive odd numbers is 99 and 101.
Objective Type Question Solutions :
(8) The generalization of the number pattern 3, 6, 9, 12, is
(i) n
(ii) 2n
(iii) 3n
(iv) 4n
Solution :
Correct option → (iii)
The generalization of the number pattern 3, 6, 9, 12, is 3n
(9) The solution of 3x + 5 = x + 9 is
(i) 2
(ii) 3
(iii) 5
(iv) 4
Solution :
Correct option → (i)
The solution of 3x + 5 = x + 9 is 2.
(10) The equation y + 1 = 0 is true only when y is
(i) 0
(ii) − 1
(iii) 1
(iv) − 2
Solution :
Correct option → (ii)
The equation y + 1 = 0 is true only when y is − 1
Algebra Exercise 3.4 Solutions :
(1) Subtract −3ab −8 from 3ab + 8. Also, subtract 3ab + 8 from −3ab −8.
Solution :
3ab + 8 – (−3ab − 8)
= (3 + 3)ab + 8 + 8
= 6ab + 16
Also, −3ab − 8 – (3ab + 8)
= (-3 – 3)ab – 8 – 8
= -6ab – 16
(2) Find the perimeter of a triangle whose sides are x + 3y, 2x + y, x − y.
Solution :
Given, sides of triangle are x + 3y, 2x + y, x − y.
Perimeter of a triangle
= x + 3y + 2x + y + x − y
= (1 + 2 + 1) x + (3 + 1 – 1)y
= 4x + 3y
(3) Thrice a number when increased by 5 gives 44. Find the number.
Solution :
Let, the number = x
From question, 3x + 5 = 44
= 3x = 44 – 5
= x = 39/3
= x = 13
Therefore, the number is 13
(5) Six times a number subtracted from 40 gives −8. Find the number.
Solution :
Let, A be a number
According to question,
40 – 6A = -8
40 – 40 – 6A = -8 – 40
-6A = – 48
A = – 48/-6
A = 8
Hence, 8 be a number.
Challenge Problem Solution :
(6) From the sum of 5x + 7y − 12 and 3x − 5y + 2, subtract the sum of 2x − 7y − 1 and −6x + 3y + 9.
Solution :
5x + 7y − 12 + 3x − 5y + 2
= (5+3)x + ( 7-5)y -12 + 2
= 8x + 2y -10 ….. (1)
2x − 7y − 1 + (−6x + 3y + 9)
= (2-6)x +(-7+3)y + (-1+9)z
= -4x -4y +8 … (2)
From 1 and 2 , we have
8x + 2y -10 – (-4x -4y +8)
= (8+4)x +(2+4)y -10 -8
= 12x +6y -18
(7) Find the expression to be added with 5a − 3b + 2c to get a − 4b − 2c?
Solution :
Let A be a expression.
Now, A + (5a − 3b + 2c) = a − 4b − 2c
A = a − 4b − 2c – (5a − 3b + 2c)
A = (1 – 5)a + (-4 + 3)b + (-2 – 2)c
A = -4a – b – 4c
Hence, the required expression is -4a – b – 4c.
(8) What should be subtracted from 2m + 8n + 10 to get −3m + 7n + 16?
Solution :
Let A be a expression.
Now, 2m + 8n + 10 – A = −3m + 7n + 16
A = 2m + 8n + 10 – (−3m + 7n + 16)
A = (2 + 3)m + (8 – 7)n + 10 – 16
A = 5m + n – 6
Hence, the required expression is 5m + n – 6.
(10) Add 2a + b + 3c and a + 1/3b + 2/5c
Solution :
2a + b + 3c + a + 1/3b + 2/5c
= (2 + 1)a + (1 + 1/3)b + (3 + 2/5)c
= 3a + 4/3b + 17/5c
(11) Fill in the blanks :
(i) If the cost of 8 apples is ₹56 then the cost of 12 apples is __.
(ii) If the weight of one fruit box is 3 1/2 kg, then the weight of 6 such boxes is __
(iii) A car travels 60 km with 3 liters of petrol. If the car has to cover the distance of 200 km, it requires __ liters of petrol.
(iv) If the cost of 7 m cloth is ₹294, then the cost of 5 m of cloth is __.
(v) If a machine in a cool drinks factory fills 600 bottles in 5 hrs, then it will fill __ bottles in 3 hours
Solution :
(i) → ₹ 84
(ii) → 21 kg
(iii) → 10 litre
(iv) → Rs 210
(v) → 360 bottles
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