## Algebraic Expressions Class 7 Extra Questions Maths Chapter 12

**Extra Questions for Class 7 Maths Chapter 12 Algebraic Expressions**

### 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 – 2y^{2}

(iii) 2x – 1

(iv) 4x^{2}y + 3xy^{2} – 5

Solution:

Question 2.

Group the like terms together from the following expressions:

-8x^{2}y, 3x, 4y, \(\frac { -3 }{ 2 }\)x , 2x^{2}y, -y

Solution:

Group of like terms are:

(i) -8x^{2}y, 2x^{2}y

(ii) 3x, \(\frac { -3 }{ 2 }\)x

(iii) 4y, -y

Question 3.

Identify the pairs of like and unlike terms:

(i) \(\frac { -3 }{ 2 }\)x, y

(ii) -x, 3x

(iii) \(\frac { -1 }{ 2 }\)y2x, \(\frac { 3 }{ 2 }\)xy^{2}

(iv) 1000, -2

Solution:

(i) \(\frac { -3 }{ 2 }\)x, y → Unlike Terms

(ii) -x, 3x → Like Terms

(iii) \(\frac { -1 }{ 2 }\)y2x, \(\frac { 3 }{ 2 }\)xy^{2} → Like Terms

(iv) 1000, -2 → Like Terms

Question 4.

Classify the following into monomials, binomial and trinomials.

(i) -6

(ii) -5 + x

(iii) \(\frac { 3 }{ 2 }\)x – y

(iv) 6x^{2} + 5x – 3

(v) z^{2} + 2

Solution:

(i) -6 is monomial

(ii) -5 + x is binomial

(iii) \(\frac { 3 }{ 2 }\)x – y is binomial

(iv) 6x^{2} + 5x – 3 is trinomial

(v) z^{2} + z is binomial

Question 5.

Draw the tree diagram for the given expressions:

(i) -3xy + 10

(ii) x^{2} + y^{2}

Solution:

Question 6.

Identify the constant terms in the following expressions:

(i) -3 + \(\frac { 3 }{ 2 }\)x

(ii) \(\frac { 3 }{ 2 }\) – 5y + y^{2}

(iii) 3x^{2} + 2y – 1

Solution:

(i) Constant term = -3

(ii) Constant term = \(\frac { 3 }{ 2 }\)

(iii) Constant term = -1

Question 7.

Add:

(i) 3x^{2}y, -5x^{2}y, -x^{2}y

(ii) a + b – 3, b + 2a – 1

Solution:

(i) 3x^{2}y, -5x^{2}y, -x^{2}y

= 3x^{2}y + (-5x^{2}y) + (-x^{2}y)

= 3x^{2}y – 5x^{2}y – x^{2}y

= (3 – 5 – 1 )x^{2}y

= -3x^{2}y

(ii) a + b – 3, b + 2a – 1

= (a + b – 3) + (b + 2a – 1)

= a + b – 3 + b + 2a – 1

= a + 2a + b + b – 3 – 1

= 3a + 2b – 4

Question 8.

Subtract 3x^{2} – x from 5x – x^{2}.

Solution:

(5x – x^{2}) – (3x^{2} – x)

= 5x – x^{2} – 3x^{2} + x

= 5x + x – x^{2} – 3x^{2}

= 6x – 4x^{2}

Question 9.

Simplify combining the like terms:

(i) a – (a – b) – b – (b – a)

(ii) x^{2} – 3x + y^{2} – x – 2y^{2}

Solution:

(i) a – (a – b) – b – (b – a)

= a – a + b – b – b + a

= (a – a + a) + (b – b – b)

= a – b

(ii) x^{2} – 3x + y^{2} – x – 2y^{2}

= x^{2} + y^{2} – 2y^{2} – 3x – x

= x^{2} – y^{2} – 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 2x^{2} + 3xy – 5 and 7 + 2xy – x^{2} subtract 3xy + x^{2} – 2.

Solution:

Sum of the given term is (2x^{2} + 3xy – 5) + (7 + 2xy – x^{2})

= 2x^{2} + 3xy – 5 + 7 + 2xy – x^{2}

= 2x^{2} – x^{2} + 3xy + 2xy – 5 + 7

= x^{2} + 5xy + 2

Now (x^{2} + 5xy + 2) – (3xy + x^{2} – 2)

= x^{2} + 5xy + 2 – 3xy – x^{2} + 2

= x^{2} – x^{2} + 5xy – 3xy + 2 + 2

= 0 + 2xy + 4

= 2xy + 4

Question 12.

Subtract 3x^{2} – 5y – 2 from 5y – 3x^{2} + xy and find the value of the result if x = 2, y = -1.

Solution:

(5y – 3x^{2} + xy) – (3x^{2} – 5y – 2)

= 5y – 3x^{2} + xy – 3x^{2} + 5y + 2

= -3x^{2} – 3x^{2} + 5y + 5y + xy + 2

= -6x^{2} + 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

= -34

Question 13.

Simplify the following expressions and then find the numerical values for x = -2.

(i) 3(2x – 4) + x^{2} + 5

(ii) -2(-3x + 5) – 2(x + 4)

Solution:

(i) 3(2x – 4) + x^{2} + 5

= 6x – 12 + x^{2} + 5

= x^{2} + 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 3x^{2} + 5x – 2t equals to 8, when x = -1.

Solution:

3x^{2} + 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 -3x^{3}y^{2} + 2x^{2}y^{3} and -3x^{2}y^{3} – 5y^{4} from x^{4} + x^{3}y^{2} + x^{2}y^{3} + y^{4}.

Solution:

Sum of the given terms:

Required expression

Question 16.

What should be subtracted from 2x^{3} – 3x^{2}y + 2xy^{2} + 3y^{2} to get x^{3} – 2x^{2}y + 3xy^{2} + 4y^{2}? [NCERT Exemplar]

Solution:

We have

Required expression

Question 17.

To what expression must 99x^{3} – 33x^{2} – 13x – 41 be added to make the sum zero? [NCERT Exemplar]

Solution:

Given expression:

99x^{3} – 33x^{2} – 13x – 41

Negative of the above expression is

-99x^{3} + 33x^{2} + 13x + 41

(99x^{3} – 33x^{2} – 13x – 41) + (-99x^{3} + 33x^{2} + 13x + 41)

= 99x^{3} – 33x^{2} – 13x – 41 – 99x^{3} + 33x^{2} + 13x + 41

= 0

Hence, the required expression is -99x^{3} + 33x^{2} + 13x + 41

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

Question 18.

If P = 2x^{2} – 5x + 2, Q = 5x^{2} + 6x – 3 and R = 3x^{2} – x – 1. Find the value of 2P – Q + 3R.

Solution:

2P – Q + 3R = 2(2x^{2} – 5x + 2) – (5x^{2} + 6x – 3) + 3(3x^{2} – x – 1)

= 4x^{2} – 10x + 4 – 5x^{2} – 6x + 3 + 9x^{2} – 3x – 3

= 4x^{2} – 5x^{2} + 9x^{2} – 10x – 6x – 3x + 4 + 3 – 3

= 8x^{2} – 19x + 4

Required expression.

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 ₹ 3xy^{2} and his father gave him ₹ 5(xy^{2} + 2). Out of this total money he spent ₹ (10 – 3xy^{2}) on his birthday party. How much money is left with him? [NCERT Exemplar]

Solution:

Money give by Rohan’s mother = ₹ 3xy^{2}

Money given by his father = ₹ 5(xy^{2} + 2)

Total money given to him = ₹ 3xy^{2} + ₹ 5 (xy^{2} + 2)

= ₹ [3xy^{2} + 5(xy^{2} + 2)]

= ₹ (3xy^{2} + 5xy^{2} + 10)

= ₹ (8xy^{2} + 10).

Money spent by him = ₹ (10 – 3xy)^{2}

Money left with him = ₹ (8xy^{2} + 10) – ₹ (10 – 3xy^{2})

= ₹ (8xy^{2} + 10 – 10 + 3x^{2}y)

= ₹ (11xy^{2})

Hence, the required money = ₹ 11xy^{2}