Algebraic Expressions Class 7 Extra Questions Maths Chapter 12

 Chapter 12 Algebraic Expressions is considered among tough topics in the Class 7 Maths syllabus. In this chapter, students learn how to form and simplify algebraic expressions using numbers and letters. It helps to understand how variables, constants, and coefficients work together in Mathematics. 

The NCERT Solutions Class 7 Chapter 12 guide step by step through different types of expressions like monomials, binomials, and trinomials. Practicing Worksheets Class 7 Maths and Important Questions Class 7 Maths from this chapter improves  problem-solving skills and prepare for higher classes by building strong foundation on this chapter. By the end of this chapter, students feel confident in adding, subtracting, and identifying terms in algebraic expressions easily.

 What is Algebraic Expression

An Algebraic Expression is defined as mathematical phrase that contains numbers, variables (like x or y), and operations such as addition, subtraction, multiplication, or division. It does not have an equal sign (=).

Example 1: 9x + 7 — here, 9x and 7 are called terms.
Example 2: 5a² - 6a + 9 — this has a variable a and three terms.
Example 3: xy + 8 — the expression has two variables x and y.

Algebraic Expressions Class 7 Extra Questions

In Algebraic Expressions Class 7 Maths Extra Questions, students will practice a wide range of important topics such as terms, coefficients, constants, variables, like and unlike terms, addition and subtraction of algebraic expressions, monomials, binomials, trinomials, and formation of expressions. 

These extra questions go beyond the textbook examples and help to think critically while solving different types of mathematical problems.

Algebraic Expressions Class 7 Extra Questions Maths Chapter 12

Question 1. Identify, in each expression below, the terms which are not constants and give their numerical coefficients.
(i) 5x - 3   (ii) 11 - 2x2   (iii) 2x - 1   (iv) 4x2y + 3xy2 - 5

Question 2. Group the like terms from the list:

-8x²y; 3x; 4y; -3/2x; 2x²y; -y

Solution:

Like terms are those that have the same variables raised to the same powers.

Terms with x²y: -8x²y; 2x²y
Terms with x: 3x; -3/2x
Terms with y: 4y; -y

Final Groups: (-8x²y, 2x²y), (3x, -3/2x), (4y, -y)

Question 3. Identify each pair as like or unlike terms:
(i) -³⁄₂x, y
(ii) -x, 3x
(iii) -¹⁄₂y²x, ³⁄₂xy²
(iv) 1000, -2

Solution:

(i) -3/2x , y - Unlike terms (Different variables)
(ii) -x , 3x - Like terms (Same variable x)
(iii) -1/2y²x , 3/2xy² - Like terms (Same variables x and y²)
(iv) 1000 , -2 - Like terms (Both are constants)

Also Check: Rational Numbers Class 7 Maths Extra Questions

Question 4. Classify each expression as a monomial, binomial, or trinomial:

(i) -6 (ii) -5 + x (iii) 3/2x - y (iv) 6x² + 5x - 3 (v) z² + 2

Note: (v) is a binomial (two terms).

Solution:

(i) -6 is a monomial
(ii) -5 + x is a binomial
(iii) 3/2x - y is a binomial
(iv) 6x² + 5x – 3 is a trinomial
(v) z² + 2 is a binomial

Question 5. Draw or describe a tree diagram for each expression:

(i) -3xy + 10
(ii) x² + y²

Solution:

(i) Expression: -3xy + 10
Tree Diagram Description:
Root → Expression (-3xy + 10)
Term 1: -3xy
Term 2: 10

(ii) Expression: x² + y²
Tree Diagram Description:
Root → Expression (x² + y²)
 ├── Term 1: x²
 └── Term 2: y²

Question 6. Identify the constant term in each expression:

(i) -3 + 3/2x
(ii) 3/2 - 5y + y²
(iii) 3x² + 2y - 1

Solution:

(i) Constant term = -3
(ii) Constant term = 3/2
(iii) Constant term = -1

Question 7.

Add:

(i) 3x²y, -5x²y, -x²y
(ii) a + b – 3, b + 2a – 1

Solution:

(i) 3x²y + (-5x²y) + (-x²y)
= 3x²y – 5x²y – x²y
= (3 – 5 – 1)x²y
= -3x²y

(ii) (a + b – 3) + (b + 2a – 1)
= a + b – 3 + b + 2a – 1
= 3a + 2b – 4

Question 8. Subtract 3x² – x from 5x – x².

Solution:

(5x – x²) – (3x² – x)
= 5x – x² – 3x² + x
= 6x – 4x²

Question 9. Simplify by combining the like terms:

(i) a – (a – b) – b – (b – a)
(ii) x² – 3x + y² – x – 2y²

Solution:

(i) a – (a – b) – b – (b – a)
= a – a + b – b – b + a
= a – b

(ii) x² – 3x + y² – x – 2y²
= x² + y² – 2y² – 3x – x
= x² – y² – 4x

Algebraic Expressions Class 7 Extra Questions Very Short Answer Type

Question 1. Identify in the given expressions, terms which are not constants. Give their numerical coefficients.

(i) 5x – 3
(ii) 11 – 2y²
(iii) 2x – 1
(iv) 4x²y + 3xy² – 5

Solution:

(i) Term not constant: 5x → Coefficient = 5
(ii) Term not constant: -2y² → Coefficient = -2
(iii) Term not constant: 2x → Coefficient = 2
(iv) Terms not constant: 4x²y, 3xy² → Coefficients = 4 and 3

Question 2. Group the like terms together from the following expressions:
-8x²y, 3x, 4y, -3/2x, 2x²y, -y

Solution:

(i) -8x²y, 2x²y (terms with x²y)
(ii) 3x, -3/2x (terms with x)
(iii) 4y, -y (terms with y)

Question 3. Identify the pairs of like and unlike terms:

(i) -3/2x , y
(ii) -x , 3x
(iii) -1/2y²x , 3/2xy²
(iv) 1000 , -2

Solution:

(i) -3/2x , y → Unlike terms
(ii) -x , 3x → Like terms
(iii) -1/2y²x , 3/2xy² → Like terms
(iv) 1000 , -2 → Like terms

Question 4. Classify the following into monomials, binomials, and trinomials:

(i) -6
(ii) -5 + x
(iii) 3/2x – y
(iv) 6x² + 5x – 3
(v) z² + 2

Solution:

(i) -6 → Monomial
(ii) -5 + x → Binomial
(iii) 3/2x – y → Binomial
(iv) 6x² + 5x – 3 → Trinomial
(v) z² + 2 → Binomial

Question 5. Draw the tree diagram for the given expressions:

(i) -3xy + 10
(ii) x² + y²

Solution:

(i) -3xy + 10 has two branches: -3xy and 10.
(ii) x² + y² has two branches: x² and y².

Question 6. Identify the constant terms in the following expressions:

(i) -3 + (3/2)x
(ii) (3/2) – 5y + y²
(iii) 3x² + 2y – 1

Solution:

(i) Constant term = -3
(ii) Constant term = 3/2
(iii) Constant term = -1

Question 7.

Add:

(i) 3x²y, -5x²y, -x²y
(ii) a + b – 3, b + 2a – 1

Solution:

(i) 3x²y + (-5x²y) + (-x²y)
= 3x²y – 5x²y – x²y
= (3 – 5 – 1)x²y
= -3x²y

(ii) (a + b – 3) + (b + 2a – 1)
= a + b – 3 + b + 2a – 1
= 3a + 2b – 4

Question 8.

Subtract 3x² – x from 5x – x².

Solution:

(5x – x²) – (3x² – x)
= 5x – x² – 3x² + x
= 6x – 4x²

Question 9. Simplify by combining the like terms:

(i) a – (a – b) – b – (b – a)
(ii) x² – 3x + y² – x – 2y²

Solution:

(i) a – (a – b) – b – (b – a)
= a – a + b – b – b + a
= a – b

(ii) x² – 3x + y² – x – 2y²
= x² + y² – 2y² – 3x – x
= x² – y² – 4x

Algebraic Expressions Class 7 Extra Questions Short Answer Type

Question 10. Subtract 24xy – 10y – 18x from 30xy + 12y – 14x.
Solution:
(30xy + 12y – 14x) – (24xy – 10y – 18x)
= 30xy + 12y – 14x – 24xy + 10y + 18x
= 30xy – 24xy + 12y + 10y – 14x + 18x
= 6xy + 22y + 4x

Question 11. From the sum of 2x² + 3xy – 5 and 7 + 2xy – x² subtract 3xy + x² – 2.

Solution:

Sum of the given terms is (2x² + 3xy – 5) + (7 + 2xy – x²)

= 2x² + 3xy – 5 + 7 + 2xy – x²

= 2x² – x² + 3xy + 2xy – 5 + 7

= x² + 5xy + 2

Now (x² + 5xy + 2) – (3xy + x² – 2)

= x² + 5xy + 2 – 3xy – x² + 2

= x² – x² + 5xy – 3xy + 2 + 2

= 0 + 2xy + 4

Final Answer: 2xy + 4

Question 12. Subtract 3x² – 5y – 2 from 5y – 3x² + xy and find the value of the result if x = 2, y = -1.

Solution:

(5y – 3x² + xy) – (3x² – 5y – 2)

= 5y – 3x² + xy – 3x² + 5y + 2

= -3x² – 3x² + 5y + 5y + xy + 2

= -6x² + 10y + xy + 2

Putting x = 2 and y = -1, we get

= -6(2)² + 10(-1) + (2)(-1) + 2

= -6 × 4 – 10 – 2 + 2

= -24 – 10 – 2 + 2

Final Answer: -34

Question 13. Simplify the following expressions and then find the numerical values for x = -2.
(i) 3(2x – 4) + x2 + 5
(ii) -2(-3x + 5) – 2(x + 4)
Solution:
(i) 3(2x – 4) + x2 + 5
= 6x – 12 + x2 + 5
= x2 + 6x – 7
Putting x = -2, we get
= (-2)2 + 6(-2) – 7
= 4 – 12 – 7
= 4 – 19
= -15
(ii) -2(-3x + 5) – 2(x + 4)
= 6x – 10 – 2x – 8
= 6x – 2x – 10 – 8
= 4x – 18
Putting x = -2, we get
= 4(-2) – 18
= -8 – 18
= -26

Question 14. Find the value of t if the value of 3x² + 5x – 2t equals 8, when x = -1.

Solution:

3x² + 5x – 2t = 8 at x = -1

⇒ 3(-1)² + 5(-1) – 2t = 8

⇒ 3(1) – 5 – 2t = 8

⇒ 3 – 5 – 2t = 8

⇒ -2 – 2t = 8

⇒ -2t = 8 + 2

⇒ -2t = 10

⇒ t = -5

Hence, the required value of t = -5.

Question 15. Subtract the sum of (-3x³y² + 2x²y³) and (-3x²y³ – 5y⁴) from (x⁴ + x³y² + x²y³ + y⁴).

Solution:

Sum of the given terms = (-3x³y² + 2x²y³) + (-3x²y³ – 5y⁴)

= -3x³y² + 2x²y³ - 3x²y³ - 5y⁴

= -3x³y² - x²y³ - 5y⁴

Now, subtract this sum from (x⁴ + x³y² + x²y³ + y⁴):

(x⁴ + x³y² + x²y³ + y⁴) - (-3x³y² - x²y³ - 5y⁴)

= x⁴ + x³y² + x²y³ + y⁴ + 3x³y² + x²y³ + 5y⁴

= x⁴ + (x³y² + 3x³y²) + (x²y³ + x²y³) + (y⁴ + 5y⁴)

= x⁴ + 4x³y² + 2x²y³ + 6y⁴

Hence, the required expression = x⁴ + 4x³y² + 2x²y³ + 6y⁴.

Also Check: Class 7 Maths Extra Questions Congruence of Triangles

Question 16. What should be subtracted from (2x³ – 3x²y + 2xy² + 3y²) to get (x³ – 2x²y + 3xy² + 4y²)?

Solution:

Let the required expression be A.

(2x³ – 3x²y + 2xy² + 3y²) – A = (x³ – 2x²y + 3xy² + 4y²)

⇒ A = (2x³ – 3x²y + 2xy² + 3y²) – (x³ – 2x²y + 3xy² + 4y²)

= 2x³ – 3x²y + 2xy² + 3y² – x³ + 2x²y – 3xy² – 4y²

= (2x³ – x³) + (-3x²y + 2x²y) + (2xy² – 3xy²) + (3y² – 4y²)

= x³ – x²y – xy² – y²

Hence, the required expression = x³ – x²y – xy² – y².

Question 17. To what expression must (99x³ – 33x² – 13x – 41) be added to make the sum zero?

Solution:

Given expression: 99x³ – 33x² – 13x – 41

To make the sum zero, we must add its negative:

Negative of the given expression = -99x³ + 33x² + 13x + 41

(99x³ – 33x² – 13x – 41) + (-99x³ + 33x² + 13x + 41)

= 99x³ – 33x² – 13x – 41 – 99x³ + 33x² + 13x + 41

= 0

Hence, the required expression = -99x³ + 33x² + 13x + 41.

Also Check: Class 7 Maths Sample Papers

Algebraic Expressions Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 18. If P = 2x² – 5x + 2, Q = 5x² + 6x – 3 and R = 3x² – x – 1. Find the value of 2P – Q + 3R.

Solution:

2P – Q + 3R = 2(2x² – 5x + 2) – (5x² + 6x – 3) + 3(3x² – x – 1)

= 4x² – 10x + 4 – 5x² – 6x + 3 + 9x² – 3x – 3

= 4x² – 5x² + 9x² – 10x – 6x – 3x + 4 + 3 – 3

= 8x² – 19x + 4

Hence, the required expression = 8x² – 19x + 4.

Question 19. If A = -(2x + 3), B = -3(x – 2) and C = -2x + 7. Find the value of k if (A + B + C) = kx.
Solution:
A + B + C = -(2x + 13) – 3(x – 2) + (-2x + 7)
= -2x – 13 – 3x + 6 – 2x + 7
= -2x – 3x – 2x – 13 + 6 + 7
= -7x
Since A + B + C = kx
-7x = kx
Thus, k = -7

Question 20. Find the perimeter of the given figure ABCDEF.

Solution:
Required perimeter of the figure
ABCDEF = AB + BC + CD + DE + EF + FA
= (3x – 2y) + (x + 2y) + (x + 2y) + (3x – 2y) + (x + 2y) + (x + 2y)
= 2(3x – 2y) + 4(x + 2y)
= 6x – 4y + 4x + 8y
= 6x + 4x-4y + 8y
= 10x + 4y
Required expression.

Question 21. Rohan’s mother gave him ₹ 3xy² and his father gave him ₹ 5(xy² + 2). Out of this total money he spent ₹ (10 – 3xy²) on his birthday party. How much money is left with him? [NCERT Exemplar]

Solution:

Money given by Rohan’s mother = ₹ 3xy²

Money given by his father = ₹ 5(xy² + 2)

Total money given to him = ₹ 3xy² + ₹ 5(xy² + 2)

= ₹ [3xy² + 5(xy² + 2)]

= ₹ (3xy² + 5xy² + 10)

= ₹ (8xy² + 10)

Money spent by him = ₹ (10 – 3xy²)

Money left with him = ₹ (8xy² + 10) – ₹ (10 – 3xy²)

= ₹ (8xy² + 10 – 10 + 3xy²)

= ₹ (11xy²)

Hence, the required money left with him = ₹ 11xy².

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