Chapter 8 – Algebraic Expressions and Operations on them
Practice Set 32
Classify the following algebraic expressions as monomials, binomials, trinomials or polynomials.
(i) 7x
Ans: Monomial
(ii) 5y – 7z
Ans: Binomial
(iii) 3x³ – 5x² – 11
Ans: Trinomial
(iv) 1 – 8a – 7a² – 7a³
Ans: Polynomial
(v) 5m – 3
Ans: Binomial
(vi) a
Ans: Monomial
(vii) 4
Ans: Monomial
(viii) 3y² – 7y + 5
Ans: Trinomial
Practice Set 33
Add.
(i) 9p + 16q ; 13p + 2q
Solution:
(9p + 16q) + (13p + 2q)
= (9p + 13p) + (16q + 2q)
= 22p + 18q
∴ (9p + 16q) + (13p + 2q) = 22p + 18q
(ii) 2a + 6b + 8c; 16a + 13c + 18b
Solution:
(2a + 6b + 8c) + (16a + 13c + 18b)
= (2a + 16a) + (6b + 18b) + (8c + 13c)
= 18a + 24b + 21c
∴ (2a + 6b + 8c) + (16a + 13c + 18b) = 18a + 24b + 21c
(iii) 13x² – 12y² ; 6x² – 8y²
Solution:
(13x² – 12y²) + (6x² – 8y²)
= (13x² + 6x²) + [(– 12y²) + (– 8y²)]
= 19x² + (– 20y²)
= 19x² – 20y²
∴ (13x² – 12y²) + (6x² – 8y²) = 19x² – 20y²
(iv) 17a²b² + 16c ; 28c – 28a²b²
Solution:
(17a²b² + 16c) + (28c – 28a²b²)
= [17a²b² + (– 28a²b²)] + (16c + 28c)
= – 11a²b² + 44c
∴ (17a²b² + 16c) + (28c – 28a²b²) = – 11a²b² + 44c
(v) 3y² – 10y + 16 ; 2y – 7
Solution:
(3y² – 10y + 16) + (2y – 7)
= 3y² + (– 10y + 2y) + (16 – 7)
= 3y² – 8y + 9
∴ (3y² – 10y + 16) + (2y – 7) = 3y² – 8y + 9
(vi) – 3y² + 10y – 16 ; 7y² + 8
Solution:
(– 3y² + 10y – 16) + (7y² + 8)
= (– 3y² + 7y²) + (10y) + (– 16 + 8)
= 4y² + 10y – 8
∴ (– 3y² + 10y – 16) + (7y² + 8) = 4y² + 10y – 8
Practice Set 34
Subtract the second expression from the first.
(i) (4xy – 9z) ; (3xy – 16z)
Solution:
(4xy – 9z) – (3xy – 16z)
= 4xy – 9z – 3xy + 16z
= (4xy – 3xy) + (16z – 9z)
= xy + 7z
∴ (4xy – 9z) – (3xy – 16z) = xy + 7z
(ii) (5x + 4y + 7z) ; (x + 2y + 3z)
Solution:
(5x + 4y + 7z) – (x + 2y + 3z)
= 5x + 4y + 7z – x – 2y – 3z
= (5x – x) + (4y – 2y) + (7z – 3z)
= 4x + 2y + 4z
∴ (5x + 4y + 7z) – (x + 2y + 3z) = 4x + 2y + 4z
(iii) (14x² + 8xy + 3y²) ; (26x² – 8xy – 17y²)
Solution:
(14x² + 8xy + 3y²) – (26x² – 8xy – 17y²)
= 14x² + 8xy + 3y² – 26x² + 8xy + 17y²
= (14x² – 26x²) + (8xy + 8xy) + (3y² + 17y²)
= – 12x² + 16xy + 20y²
∴ (14x² + 8xy + 3y²) – (26x² – 8xy – 17y²) = – 12x² + 16xy + 20y²
(iv) (6x² + 7xy + 16y²);(16x² – 17xy)
Solution:
(6x² + 7xy + 16y²) – (16x² – 17xy)
= 6x² + 7xy + 16y² – 16x² + 17xy
= (6x² – 16x²) + (7xy + 17xy) + 16y²
= – 10x² + 24xy + 16y²
∴ (6x² + 7xy + 16y²) – (16x² – 17xy) = – 10x² + 24xy + 16y²
(v) (4x + 16z) ; (19y – 14z + 16x)
Solution:
(4x + 16z) – (19y – 14z + 16x)
= 4x + 16z – 19y + 14z – 16x
= (4x – 16x) – 19y + (16z + 14z)
= – 12x – 19y + 30z
∴ (4x + 16z) – (19y – 14z + 16x) = – 12x – 19y + 30z
Practice Set 35
1. Multiply.
(i) 16xy × 18xy
Solution:
16xy × 18xy
= 16 × 18 × xy × xy
= 288x²y²
∴ 16xy × 18xy = 288x²y²
(ii) 23xy² × 4yz²
Solution:
23xy² × 4yz²
= 23 × 4 × xy² × yz²
= 92xy³z²
∴ 23xy² × 4yz² = 92xy³z²
(iii) (12a + 17b) × 4c
Solution:
(12a + 17b) × 4c
= 12a × 4c + 17b × 4c
= 48ac + 68bc
∴ (12a + 17b) × 4c = 48ac + 68bc
(iv) (4x + 5y) × (9x + 7y)
Solution:
(4x + 5y) × (9x + 7y)
= 4x × (9x + 7y) + 5y × (9x + 7y)
= (4x × 9x) + (4x × 7y) + (5y × 9x) + (5y × 7y)
= 36x² + 28xy + 45xy + 35y²
= 36x² + 73xy + 35y²
∴ (4x + 5y) × (9x + 7y) = 36x² + 73xy + 35y²
2. A rectangle is (8x + 5) cm long and (5x + 3) cm broad. Find its area.
Solution:
Length of the rectangle = (8x + 5) cm
Breadth of the rectangle = (5x + 3) cm
∴ Area of the rectangle
= length × breadth
= (8x + 5) × (5x + 3)
= 8x × (5x + 3) + 5 × (5x + 3)
= (8x × 5x) + (8x × 3) + (5 × 5x) + (5 × 3)
= 40x² + 24x + 25x + 15
= 40x² + 49x + 15
Ans: The area of the rectangle is (40x² + 49x + 15) sq. cm.
Practice Set 36
1. Simplify (3x – 11y) – (17x + 13y) and choose the right answer.
(i) 7x – 12y
(ii) – 14x – 54y
(iii) – 3 (5x + 4y)
(iv) – 2 (7x + 12y)
Ans: Option (D) : – 2 (7x + 12y)
Solution:
(3x – 11y) – (17x + 13y)
= 3x – 11y – 17x – 13y
= – 14x – 24y
= – 2 × 7x – 2 × 12y
= – 2(7x + 12y)
2. The product of (23 x²y³z) and (– 15 x³yz²) is ______.
(i) – 345 x⁵y⁴z³
(ii) 345 x²y³z⁵
(iii) 145 x³y²z
(iv) 170 x³y²z³
Ans: Option (A) : – 345 x⁵y⁴z³
Solution:
(23 x²y³z) × (– 15 x³yz²)
= 23 × (– 15) × x² × y³ × z × x³ × y × z²
= – 345x⁵y⁴z³
3. Solve the following equations.
(i) 4x + 12 = 92
Solution:
4x + \(\large \frac {1}{2}\) = \(\large \frac {9}{2}\)
∴ 4x = \(\large \frac {9}{2}\) – \(\large \frac {1}{2}\)
∴ 4x = \(\large \frac {9\,–\,1}{2}\)
∴ 4x = \(\large \frac {8}{2}\)
∴ 4x = 4
∴ x = \(\large \frac {4}{4}\)
∴ x = 1
(ii) 10 = 2y + 5
Solution:
10 = 2y + 5
∴ 2y + 5 = 10
∴ 2y = 10 – 5
∴ 2y = 5
∴ y = \(\large \frac {5}{2}\)
(iii) 5m – 4 = 1
Solution:
5m – 4 = 1
∴ 5m = 1 + 4
∴ 5m = 5
∴ m = \(\large \frac {5}{5}\)
∴ m = 1
(iv) 6x – 1 = 3x + 8
Solution:
6x – 1 = 3x + 8
∴ 6x – 1 = 3x + 8
∴ 6x – 3x = 8 + 1
∴ 3x = 9
∴ x = \(\large \frac {9}{3}\)
∴ x = 3
(v) 2 (x – 4) = 4x + 2
Solution:
2 (x – 4) = 4x + 2
∴ 2x – 8 = 4x + 2
∴ 2x – 4x = 2 + 8
∴ – 2x = 10
∴ x = \(\large \frac {10}{–\,5}\)
∴ x = – 2
(vi) 5 (x + 1) = 74
Solution:
5 (x + 1) = 74
∴ 5x + 5 = 74
∴ 5x = 74 – 5
∴ 5x = 69
∴ x = \(\large \frac {69}{5}\)
4. Rakesh’s age is less than Sania’s age by 5 years. The sum of their ages is 27 years. How old are they?
Solution:
Let the age of Rakesh be x years.
∴ Sania’s age = (x + 5) years.
According to the given condition,
x + (x + 5) = 27
∴ 2x + 5 = 27
∴ 2x = 27 – 5
∴ 2x = 22
∴ x = \(\large \frac {22}{11}\)
∴ x = 11
Sania’s age
= x + 5
= 11 + 5
= 16 years
Ans: The ages of Rakesh and Sania are 11 years and 16 years respectively.
5. When 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?
Solution:
Let the number of jambhul trees planted be x.
∴ Number of ashoka trees = x – 60
According to the given condition,
x + x – 60 = 200
∴ 2x = 200 + 60
∴ 2x = 260
∴ x = \(\large \frac {260}{2}\)
∴ x = 130
Ans: 130 jambhul trees were planted.
6. Shubhangi 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?
Solution:
Let the number of 50-rupee notes with Shubhangi be x.
∴ Number of 20 rupee notes = 2x
According to the given condition
x × 50 + 2x × 20 = 2700
∴ 50x + 40x = 2700
∴ 90x = 2700
∴ x = \(\large \frac {2700}{90}\)
∴ x = 30
Ans: Shubhangi has 30 notes of 50 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?
Solution:
Let the runs made by Rohit be x.
∴ Runs made by Virat = 2x
According to the given condition,
x + 2x = 200 – 2
∴ 3x = 198
∴ x = \(\large \frac {198}{3}\)
∴ x = 66
Runs made by Virat
= 2x
= 2 × 66
= 132
Ans: The runs made by Virat and Rohit are 132 and 66 respectively.