Std 7 Mathematics · Chapter 8

8. Algebraic Expressions and Operations on them

Step-by-step solutions to Practice Sets 32 to 36. Maharashtra State Board (Balbharati), Grade Seven Mathematics.

Practice Set 32

1Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.

(i) 7x — Monomial (1 term)
(ii) 5y − 7z — Binomial (2 terms)
(iii) 3x³ − 5x² − 11 — Trinomial (3 terms)
(iv) 1 − 8a − 7a² − 7a³ — Polynomial (4 terms)
(v) 5m − 3 — Binomial (2 terms)
(vi) a — Monomial (1 term)
(vii) 4 — Monomial (1 term)
(viii) 3y² − 7y + 5 — Trinomial (3 terms)

A number alone, like 4, is also a monomial. An expression with more than three terms is called a polynomial.

Practice Set 33

1Add. (i) 9p + 16q ; 13p + 2q (ii) 2a + 6b + 8c ; 16a + 13c + 18b (iii) 13x² − 12y² ; 6x² − 8y² (iv) 17a²b² + 16c ; 28c − 28a²b² (v) 3y² − 10y + 16 ; 2y − 7 (vi) −3y² + 10y − 16 ; 7y² + 8

Add the coefficients of like terms; unlike terms are written as they are.

(i) (9p + 16q) + (13p + 2q)
= (9p + 13p) + (16q + 2q)
= 22p + 18q

(ii) (2a + 6b + 8c) + (16a + 13c + 18b)
= (2a + 16a) + (6b + 18b) + (8c + 13c)
= 18a + 24b + 21c

(iii) (13x² − 12y²) + (6x² − 8y²)
= (13x² + 6x²) + (−12y² − 8y²)
= 19x² − 20y²

(iv) (17a²b² + 16c) + (28c − 28a²b²)
= (17a²b² − 28a²b²) + (16c + 28c)
= −11a²b² + 44c

(v) (3y² − 10y + 16) + (2y − 7)
= 3y² + (−10y + 2y) + (16 − 7)
= 3y² − 8y + 9

(vi) (−3y² + 10y − 16) + (7y² + 8)
= (−3y² + 7y²) + 10y + (−16 + 8)
= 4y² + 10y − 8

Practice Set 34

1Subtract the second expression from the first.

Change the sign of every term of the second expression and add.

(i) (4xy − 9z) − (3xy − 16z)
= 4xy − 9z − 3xy + 16z
= (4xy − 3xy) + (−9z + 16z)
= xy + 7z

(ii) (5x + 4y + 7z) − (x + 2y + 3z)
= 5x + 4y + 7z − x − 2y − 3z
= (5x − x) + (4y − 2y) + (7z − 3z)
= 4x + 2y + 4z

(iii) (14x² + 8xy + 3y²) − (26x² − 8xy − 17y²)
= 14x² + 8xy + 3y² − 26x² + 8xy + 17y²
= (14x² − 26x²) + (8xy + 8xy) + (3y² + 17y²)
= −12x² + 16xy + 20y²

(iv) (6x² + 7xy + 16y²) − (16x² − 17xy)
= 6x² + 7xy + 16y² − 16x² + 17xy
= (6x² − 16x²) + (7xy + 17xy) + 16y²
= −10x² + 24xy + 16y²

(v) (4x + 16z) − (19y − 14z + 16x)
= 4x + 16z − 19y + 14z − 16x
= (4x − 16x) − 19y + (16z + 14z)
= −12x − 19y + 30z

Practice Set 35

1Multiply. (i) 16xy × 18xy (ii) 23xy² × 4yz² (iii) (12a + 17b) × 4c (iv) (4x + 5y) × (9x + 7y)

Multiply the coefficients (with their signs) first, then the variables, adding the indices of the same variable.

(i) 16xy × 18xy
= 16 × 18 × x × x × y × y
= 288x²y²

(ii) 23xy² × 4yz²
= 23 × 4 × x × y² × y × z²
= 92xy³z²

(iii) (12a + 17b) × 4c
= 12a × 4c + 17b × 4c
= 48ac + 68bc

(iv) (4x + 5y) × (9x + 7y)
= 4x(9x + 7y) + 5y(9x + 7y)
= 36x² + 28xy + 45xy + 35y²
= 36x² + 73xy + 35y²

2A rectangle is (8x + 5) cm long and (5x + 3) cm broad. Find its area.

Area of rectangle = length × breadth
= (8x + 5) × (5x + 3)
= 8x(5x + 3) + 5(5x + 3)
= 40x² + 24x + 25x + 15
= 40x² + 49x + 15

The area of the rectangle is (40x² + 49x + 15) cm².

Practice Set 36

1Simplify (3x − 11y) − (17x + 13y) and choose the right answer. (i) 7x − 12y (ii) −14x − 54y (iii) −3(5x + 4y) (iv) −2(7x + 12y)

(3x − 11y) − (17x + 13y)
= 3x − 11y − 17x − 13y
= −14x − 24y
= −2(7x + 12y)

Answer: (iv) −2(7x + 12y)

2The product of (23x²y³z) and (−15x³yz²) is ______ .

(23x²y³z) × (−15x³yz²)
= 23 × (−15) × x² × x³ × y³ × y × z × z²
= −345x⁵y⁴z³

Answer: (i) −345x⁵y⁴z³

3Solve the following equations.

When a term is transposed to the other side of “=”, its sign changes.

(i) 4x + 12 = 92
∴ 4x = 92 − 12 = 82 = 4
∴ x = 1

(ii) 10 = 2y + 5
∴ 2y = 10 − 5 = 5
∴ y = 52 = 2.5

(iii) 5m − 4 = 1
∴ 5m = 1 + 4 = 5
∴ m = 1

(iv) 6x − 1 = 3x + 8
∴ 6x − 3x = 8 + 1
∴ 3x = 9
∴ x = 3

(v) 2(x − 4) = 4x + 2
∴ 2x − 8 = 4x + 2
∴ 2x − 4x = 2 + 8
∴ −2x = 10
∴ x = −5

(vi) 5(x + 1) = 74
∴ 5x + 5 = 74
∴ 5x = 69
∴ x = 695 = 13.8

4Rakesh’s age is less than Sania’s age by 5 years. The sum of their ages is 27 years. How old are they?

Let Sania’s age be x years. Then Rakesh’s age is (x − 5) years.

x + (x − 5) = 27
∴ 2x − 5 = 27
∴ 2x = 32
∴ x = 16

Sania is 16 years old and Rakesh is 16 − 5 = 11 years old.

Check: 16 + 11 = 27 and 16 − 11 = 5.

5When planting a forest, the number of jambhul trees planted was greater than the number of ashoka trees by 60. If there are altogether 200 trees of these two types, how many jambhul trees were planted?

Let the number of ashoka trees be x. Then the number of jambhul trees is x + 60.

x + (x + 60) = 200
∴ 2x + 60 = 200
∴ 2x = 140
∴ x = 70

Jambhul trees = 70 + 60 = 130.

Check: 70 + 130 = 200 and 130 − 70 = 60.

6Shubhangi has twice as many 20-rupee notes as she has 50-rupee notes. Altogether, she has 2700 rupees. How many 50-rupee notes does she have?

Let the number of 50-rupee notes be x. Then the number of 20-rupee notes is 2x.

50 × x + 20 × 2x = 2700
∴ 50x + 40x = 2700
∴ 90x = 2700
∴ x = 30

Shubhangi has 30 fifty-rupee notes (and 60 twenty-rupee notes).

Check: 30 × 50 + 60 × 20 = 1500 + 1200 = 2700 rupees.

7*Virat made twice as many runs as Rohit. The total of their scores is 2 less than a double century. How many runs did each of them make?

A double century is 200 runs, so the total is 200 − 2 = 198 runs. Let Rohit’s runs be x; then Virat’s runs are 2x.

x + 2x = 198
∴ 3x = 198
∴ x = 66

Rohit made 66 runs and Virat made 2 × 66 = 132 runs.

Check: 66 + 132 = 198.

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