Chapter 7

Class 8 Mathematics · 125 questions · 122 with answers

Solved examples

example-1Multiple choice

Which is the like term as 24a2bc?

(a)13 × 8a × 2b × c × a
(b)8 × 3 × a × b × c
(c)3 × 8 × a × b × c × c
(d)3 × 8 × a × b × b × c

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Solution : The correct answer is (a).

example-2Multiple choice

Which of the following is an identity?

(a)(p + q)2 = p2 + q2
(b)p2 – q2 = (p – q)2
(c)p2 – q2 = p2 + 2pq – q2
(d)(p + q)2 = p2 + 2pq + q2

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Solution : The correct answer is (d).

example-3Multiple choice

The irreducible factorisation of 3a3 + 6a is

(a)3a (a2 + 2)
(b)3 (a3 + 2)
(c)a (3a2 + 6)
(d)3 × a × a × a + 2 × 3 × a

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Solution : The correct answer is (a).

example-4Short answer

a ( b + c) = ab + ac is (a)commutative property(b) distributive property (c) associative property(d) closure property

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Solution : The correct answer is (b). In examples 5 and 6, fill in the blanks to make the statements true.

example-5Fill in the blanks

The representation of an expression as the product of its factors is called __________.

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Solution : Factorisation.

example-6Fill in the blanks

(x + a) (x + b) = x2 + (a + b)x + ________.

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Solution : ab. In examples 7 to 9, state whether the statements are true (T) or false (F).

example-7Short answer

An identity is true for all values of its variables.

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Solution : True.

example-8Short answer

Common factor of x2y and – xy2 is xy.

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Solution : True.

example-9Short answer

(3x + 3x2) ÷ 3x = 3x2

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Solution : False.

example-10Multiple choice

Simplify

(i)– pqr (p2 + q2 + r 2)
(ii)(px + qy) (ax – by)

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Solution : (i) – pqr (p2 + q2 + r 2) = – (pqr) × p2 – (pqr) × q2 – (pqr) × r 2 = – p3qr – pq3r – pqr3 (ii) (px + qy) (ax – by) = px (ax – by) + qy (ax – by) = apx2 – pbxy + aqxy – qby2 You might think that algebra is a topic found only in textbooks, but you can find algebra all around you – in some of the strongest places. Did you know there is a relationship between the speed at which ants crawl and the air temperature? If you were to find some ants outside and time them as they crawled, you could actually estimate the temperature. Here is the algebraic equation that describes this relationship. Celsius temperature ↓ t = 15s + 3 ↑ ant speed in centimetres per seconds There are many ordinary and extraordinary places where you will encounter algebra. What do you think is the speed of a typical ant?

example-11Multiple choice

Find the expansion of the following using suitable identity. 4 x y 4 x 3y

(i)(3x + 7y) (3x – 7y)
(ii)+ + 5 4 5 4

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Solution : (i) (3x + 7y) (3x – 7y) Since (a + b) (a – b) = a2 – b2, therefore (3x + 7y) (3x – 7y) = (3x)2 – (7y)2 = 9x2 – 49y2 4 x y 4 x 3y (ii) + + 5 4 5 4 Since (x + a) (x + b) = x2 + (a + b)x + ab, therefore 4x y 4 x 3y + + 5 4 5 4 4x y 3y 4 x y 3y = 5 + 4 + 4 × 5 + 4 × 4 4x y 3y = Here, x = , a = and b = 5 4 4 16 x 2 4y 4x 3y 2 = + × + 25 4 5 16 16 x 2 4 xy 3y 2 = + + 25 5 16

example-12Multiple choice

Factorise the following.

(i)21x2y3 + 27x3y2
(ii)a3 – 4a2 + 12 – 3a y2
(iii)4x – 20x + 25
(iv)–9 (v) x4 – 256

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Solution : (i) 21x2y3 + 27x3y2 =3×7× x × x × y× y× y + 3 × 3 × 3 × x × x × x × y× y = 3 × x × x × y × y (7y + 9x) (Using ab + ac = a (b + c)) = 3x2y2 (7y + 9x) (ii) a3 – 4a2 + 12 – 3a = a2 (a – 4) – 3a + 12 = a2 (a – 4) – 3 (a – 4) = (a – 4) (a2 – 3) (iii) 4x2 – 20x + 25 = (2x)2 – 2 × 2x × 5 + (5)2 = (2x – 5)2 (Since a2 – 2ab + b2 = (a – b)2 ) = (2x – 5) (2x – 5) y2 (iv) –9 y 2 = – (3) If there are two numbers you don’t know, that’s not a problem. You can use two different variables, one for each unknown number. In Words Numbers An equation involving variables can be true for all values of the variable The sum of a and b a+b – for example, y + y = 2y (this kind of equation is usually called an The product of v and w v × w, or vw identity). p is subtracted from 9q qq – p Or it can be true for only particular values of the variable – for example, You can use expressions with two q–p (or more) 2y + 3 = 11, which is true only if y = 4. variables to represent situations with more Finding the values that make an than one unknown quantity. equation true is called solving the equation. y y = + 3 – 3 (Since a2 – b2 = (a + b) (a – b)) 3 3 (v) x4 – 256 = (x2)2 – (16)2 = (x2 + 16) (x2 – 16) (using a2 – b2 = (a + b) (a – b)) = (x2 + 16) (x2 – 42) = (x2 + 16) (x + 4) (x – 4) (using a2 – b2 = (a + b) (a – b))

example-13Multiple choice

Evaluate using suitable identities.

(i)(48)2
(ii)18 12 – 192
(iii)497 × 505
(iv)2.07 × 1.93

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Solution : (i) (48)2 = (50 – 2)2 Since (a – b)2 = a2 – 2ab + b2 , therefore (50 – 2)2 = (50)2 – 2 × 50 × 2 + (2)2 = 2500 – 200 + 4 = 2504 – 200 = 2304 (ii) 1812 – 192 = (181 – 19) (181 + 19) [using a2 – b2 = (a – b) (a + b)] = 162 × 200 = 32400 (iii) 497 × 505 = (500 – 3) (500 + 5) = 5002 + (–3 + 5) × 500 + (–3) (5) [using (x + a) (x + b) = x2 + (a + b) x + ab] = 250000 + 1000 – 15 = 250985 (iv) 2.07 × 1.93 = (2 + 0.07) (2 – 0.07) = 22 – (0.07)2 = 3.9951

example-14Short answer

Verify that (3x + 5y)2 – 30xy = 9x2 + 25y2

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Solution : L.H.S= (3x + 5y)2 – 30xy = (3x)2 + 2 × 3x × 5y + (5y)2 – 30xy [Since (a + b)2 = a2 + 2ab + b2] = 9x2 + 30xy + 25y2 – 30xy = 9x2 + 25y2 = R.H.S Hence, verified.

example-15Short answer

Verify that (11pq + 4q)2 – (11pq – 4q)2 = 176pq2

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Solution : L.H.S. (11pq + 4q)2 – (11pq – 4q)2 = (11pq + 4q + 11pq – 4q) × (11pq + 4q – 11pq + 4q) [using a2 – b2 = (a – b) (a + b), here a = 11pq + 4q and b = 11 pq – 4q] = (22pq) (8q) = 176 pq2 R.H.S. Hence Verified To convert a Celsius temperature to a Fahrenheit temperature, find nine-fifths of F = C + 32 the Celsius temperature and then add 32. 5 While the statement on the left may be easier to read and understand at first, the statement on the right has several advantages. It is shorter and easier to write, it shows clearly how the quantities – Celsius temperature and Fahrenheit temperature – are related, and it allows you to try different Celsius temperatures and compute their Fahrenheit equivalents.

example-16Short answer

The area of a rectangle is x2 + 12xy + 27y2 and its length is (x + 9y). Find the breadth of the rectangle. Area

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Solution : Breadth = Length x 2 + 12xy + 27y 2 = ( x + 9y ) x 2 + 9xy + 3xy + 27y 2 = ( x + 9y ) x ( x + 9y ) + 3y ( x + 9y ) = x + 9y ( x + 9y ) ( x + 3y ) = ( x + 9y ) = (x + 3y)

example-17Short answer

Divide 15 (y + 3) (y2 – 16) by 5 (y2 – y – 12).

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Solution : Factorising 15 (y + 3) (y2 – 16), we get 5 × 3 × (y + 3) (y – 4) (y + 4) On factorising 5 (y2 – y – 12), we get 5 (y2 – 4y + 3y – 12) = 5 [y (y – 4) + 3 (y – 4)] = 5 (y – 4) (y + 3) Therefore, on dividing the first expression by the second 15(y + 3) (y 2 –16) expression, we get 5(y 2 – y –12) 5 × 3 × (y + 3)(y − 4)(y + 4) = 5 × ( y − 4)(y + 3) = 3 (y + 4) 2 1 1

example-18Short answer

By using suitable identity, evaluate x + 2 , if x + = 5 . x x

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Solution : Given that x + =5 1 So, x + = 25 x 2 2 1 1 1 Now, x + = x2 + 2 × x × + [Using identity x x x (a + b)2 = a2 + 2ab + b2, with a = x and b = ] 1 = x2 + 2 + 2 x 1 = x2 + 2 + 2 x 1 1 Since x + = 25, therefore x2 + 2 + 2 = 25 x x or x2 + = 25 – 2 = 23 x2 382 – 222

example-19Short answer

Find the value of , using a suitable identity.

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Solution : Since a2 – b2 = (a + b) (a – b), therefore 382 – 222 = (38 – 22) (38 + 22) = 16 × 60 382 – 222 16 × 60 So, = 16 16 = 60

example-20Short answer

Find the value of x, if 10000x = (9982)2 – (18)2

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Solution : R.H.S. = (9982)2 – (18)2 = (9982 + 18) (9982 – 18) [Since a2 – b2 = (a + b) (a – b)] = (10000) × (9964) L.H.S. = (10000) × x Comparing L.H.S. and R.H.S., we get 10000x = 10000 × 9964 10000×9964 or x= = 9964 10000 2 5

Questions

Q1Short answer

Can you find the reciprocal of × ? 11 55

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(b)

Q2Long answer

Can you compare the ratio of this reciprocal with the earlier one? Find each side of a figure given below, if its area is 64 cm2. Understand and Explore the problem • What information is given in the question? AB = BC = DC = AD, and ∠A = ∠B = ∠C = ∠D = 90° Hence ABCD is a square. • What are you trying to find? The value of one of the sides of the square ABCD. • Is there any information that is not needed? No. Make a Plan • In a square all sides are equal, therefore, square of a side gives the area. Solve (Side)2 = Area ⇒ (x + 2)2 = 64 ⇒ (x + 2)2 = 82 ⇒ x+2 =8 ⇒ x=8–2 ∴ x=6 ∴ Side = x + 2 = 6 + 2 = 8 cm Revise • The above answer is verified by squaring the side and comparing the result with the given area. ∴ (Side)2 = 82 = 64 = given area. To become familiar with some of the vocabulary terms in the chapter, consider the following: 1. The word equivalent contains the same root as the word equal. What do you think equivalent expressions are? 2. The word simplify means make less complicated. What do you think it means to simplify an expression?

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(b)

Q3Short answer

The adjective like means alike. What do you suppose like terms are?

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(b)

Q4Multiple choice

A system is a group of related objects. What do you think a system of equations is? In questions 1 to 33, there are four options out of which one is correct. Write the correct answer. 1. The product of a monomial and a binomial is a

(a)monomial
(b)binomial
(c)trinomial
(d)none of these 2. In a polynomial, the exponents of the variables are always (a) integers (b) positive integers (c) non-negative integers (d) non-positive integers 3. Which of the following is correct? (a) (a – b)2 = a2 + 2ab – b2 (b) (a – b)2 = a2 – 2ab + b2 (c) (a – b)2 = a2 – b2 (d) (a + b)2 = a2 + 2ab – b2 4. The sum of –7pq and 2pq is (a) –9pq (b) 9pq (c) 5pq (d) – 5pq

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(d) none of these 2. In a polynomial, the exponents of the variables are always (a) integers (b) positive integers (c) non-negative integers (d) non-positive integers 3. Which of the following is correct? (a) (a – b)2 = a2 + 2ab – b2 (b) (a – b)2 = a2 – 2ab + b2 (c) (a – b)2 = a2 – b2 (d) (a + b)2 = a2 + 2ab – b2 4. The sum of –7pq and 2pq is (a) –9pq (b) 9pq (c) 5pq (d) – 5pq

Q5Multiple choice

If we subtract –3x2y2 from x2y2, then we get

(a)– 4x2y2
(b)– 2x2y2
(c)2x 2y2
(d)4x 2y2

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(d) 4x 2y2

Q6Multiple choice

Like term as 4m 3n 2 is

(a)4m 2n2
(b)– 6m 3n 2
(c)6pm3n2
(d)4m3n

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(b) – 6m 3n 2

Q7Multiple choice

Which of the following is a binomial?

(a)7 × a + a
(b)6a2 + 7b + 2c
(c)4a × 3b × 2c
(d)6 (a2 + b)

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(d) 6 (a2 + b)

Q8Multiple choice

Sum of a – b + ab, b + c – bc and c – a – ac is

(a)2c + ab – ac – bc
(b)2c – ab – ac – bc
(c)2c + ab + ac + bc
(d)2c – ab + ac + bc

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(a) 2c + ab – ac – bc

Q9Multiple choice

Product of the following monomials 4p, – 7q3, –7pq is

(a)196 p2q4
(b)196 pq4
(c)– 196 p2q4
(d)196 p2q3

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(a) 196 p2q4

Q10Multiple choice

Area of a rectangle with length 4ab and breadth 6b2 is

(a)24a2b 2
(b)24ab3
(c)24ab2
(d)24ab

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(b) 24ab3

Q11Multiple choice

Volume of a rectangular box (cuboid) with length = 2ab, breadth = 3ac and height = 2ac is

This question refers to a figure in the original PDF.

(a)12a3bc2
(b)12a3bc
(c)12a2bc
(d)2ab +3ac + 2ac The five figures form a pattern. 1. Copy and complete the table to find the perimeter of each figure. Each side of each individual square is 1 unit. Figure 1 2 3 4 5 Perimeter 2. Without drawing a picture, describe what the sixth figure will look like and predict its perimeter. 3. If you continue this pattern, what will be the perimeter of the 35th figure? 4. Explain how the perimeter of each figure is related to its figure number. 5. Using the variables n for the figure number and P for the perimeter, write an equation for the relationship in Question 4.

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(a) 12a3bc2

Q12Multiple choice

Product of 6a2 – 7b + 5ab and 2ab is

(a)12a3b – 14ab2 + 10ab
(b)12a3b – 14ab2 + 10a2b2
(c)6a2 – 7b + 7ab
(d)12a2b – 7ab2 + 10ab

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(b) 12a3b – 14ab2 + 10a2b2

Q13Multiple choice

Square of 3x – 4y is

(a)9x2 – 16y2
(b)6x2 – 8y2
(c)9x2 + 16y2 + 24xy
(d)9x2 + 16y2 – 24xy

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(d) 9x2 + 16y2 – 24xy

Q14Multiple choice

Which of the following are like terms?

(a)5xyz2, – 3xy2z
(b)– 5xyz2, 7xyz2
(c)5xyz2, 5x2yz
(d)5xyz2, x 2y2z2 −y

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(b) – 5xyz2, 7xyz2

Q15Multiple choice

Coefficient of y in the term is −1 1

(a)– 1
(b)– 3
(c)
(d)3 3

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(c)

Q16Multiple choice

a2 – b2 is equal to

(a)(a – b)2
(b)(a – b) (a – b)
(c)(a + b) (a – b)
(d)(a + b) (a + b)

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Q17Multiple choice

Common factor of 17abc, 34ab2, 51a2b is

(a)17abc
(b)17ab
(c)17ac
(d)17a2b 2c

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(b) 17ab

Q18Multiple choice

Square of 9x – 7xy is

(a)81x2 + 49x2y2
(b)81x2 – 49x2y2
(c)81x2 + 49x2y2 –126x2y
(d)81x2 + 49x2y2 – 63x2y

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Q19Multiple choice

Factorised form of 23xy – 46x + 54y – 108 is

(a)(23x + 54) (y – 2)
(b)(23x + 54y) (y – 2)
(c)(23xy + 54y) (– 46x – 108)
(d)(23x + 54) (y + 2)

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(a) (23x + 54) (y – 2)

Q20Multiple choice

Factorised form of r 2 – 10r + 21 is

(a)(r – 1) (r – 4)
(b)(r – 7) (r – 3)
(c)(r – 7) (r + 3)
(d)(r + 7) (r + 3)

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(b) (r – 7) (r – 3)

Q21Multiple choice

Factorised form of p2 – 17p – 38 is

(a)(p – 19) (p + 2)
(b)(p – 19) (p – 2)
(c)(p + 19) (p + 2)
(d)(p + 19) (p – 2)

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(a) (p – 19) (p + 2)

Q22Multiple choice

On dividing 57p2qr by 114pq, we get 1 3 1

(a)pr
(b)pr
(c)pr
(d)2pr 4 4 2

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Q23Multiple choice

On dividing p (4p2 – 16) by 4p (p – 2), we get

(a)2p + 4
(b)2p – 4
(c)p + 2
(d)p – 2

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Q24Multiple choice

The common factor of 3ab and 2cd is

(a)1
(b)– 1
(c)a
(d)c

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(a) 1

Q25Multiple choice

An irreducible factor of 24x2y2 is

(a)x 2
(b)y 2
(c)x
(d)24x

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Q26Multiple choice

Number of factors of (a + b)2 is

(a)4
(b)3
(c)2
(d)1

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Q27Multiple choice

The factorised form of 3x – 24 is

(a)3x × 24
(b)3 (x – 8)
(c)24 (x – 3)
(d)3(x – 12)

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(b) 3 (x – 8)

Q28Multiple choice

The factors of x – 4 are

(a)(x – 2), (x – 2)
(b)(x + 2), (x – 2)
(c)(x + 2), (x + 2)
(d)(x – 4), (x – 4)

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(b) (x + 2), (x – 2)

Q29Multiple choice

The value of (– 27x2y) ÷ (– 9xy) is

(a)3xy
(b)– 3xy
(c)– 3x
(d)3x

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(d) 3x

Q30Multiple choice

The value of (2x2 + 4) ÷ 2 is

(a)2x2 + 2
(b)x2 + 2
(c)x2 + 4
(d)2x2 + 4

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(b) x2 + 2

Q31Multiple choice

The value of (3x3 +9x2 + 27x ) ÷ 3x is

(a)x2 +9 + 27x
(b)3x3 +3x2 + 27x
(c)3x3 +9x2 + 9
(d)x2 +3x + 9

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(d) x2 +3x + 9

Q32Multiple choice

The value of (a + b)2 + (a – b)2 is

(a)2a + 2b
(b)2a – 2b
(c)2a2 + 2b2
(d)2a2 – 2b2

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Q33Multiple choice

The value of (a + b)2 – (a – b)2 is

(a)4ab
(b)– 4ab
(c)2a2 + 2b2
(d)2a2 – 2b2 In questions 34 to 58, fill in the blanks to make the statements true:

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(a) 4ab

Q34Short answer

The product of two terms with like signs is a term.

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positive

Q35Short answer

The product of two terms with unlike signs is a term.

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negative

Q36Fill in the blanks

a (b + c) = ax ____ × ax _____.

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ab + ac

Q37Fill in the blanks

(a – b) _________ = a2 – 2ab + b2

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(a – b)2

Q38Fill in the blanks

a2 – b2 = (a + b ) __________.

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(a + b) (a – b)

Q39Fill in the blanks

(a – b)2 + ____________ = a2 – b2

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2ab – 2b2

Q40Fill in the blanks

(a + b)2 – 2ab = ___________ + ____________

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a2 + b2

Q41Fill in the blanks

(x + a) (x + b) = x2 + (a + b) x + ________.

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Q42Fill in the blanks

The product of two polynomials is a ________.

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polynomial

Q43Fill in the blanks

Common factor of ax2 + bx is __________.

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Q44Fill in the blanks

Factorised form of 18mn + 10mnp is ________.

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2m (9 + 5p)

Q45Fill in the blanks

Factorised form of 4y2 – 12y + 9 is ________.

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(2y – 3) (2y – 3)

Q46Fill in the blanks

38x3y2z ÷ 19xy2 is equal to _________.

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2x 2z

Q47Fill in the blanks

Volume of a rectangular box with length 2x, breadth 3y and height 4z is _________.

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24 xyz

Q48Fill in the blanks

672 – 372 = (67 – 37) × ________ = _________.

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(67 + 37)

Q49Fill in the blanks

1032 – 1022 = ________ × (103 – 102) = _________.

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205

Q50Fill in the blanks

Area of a rectangular plot with sides 4x2 and 3y2 is __________.

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12 x2y2

Q51Fill in the blanks

Volume of a rectangular box with l = b = h = 2x is _________.

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8x 3

Q52Fill in the blanks

The coefficient in – 37abc is __________.

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– 37

Q53Fill in the blanks

Number of terms in the expression a2 + bc × d is ________.

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Q54Fill in the blanks

The sum of areas of two squares with sides 4a and 4b is _______.

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16 (a2 + b2)

Q55Fill in the blanks

The common factor method of factorisation for a polynomial is based on ___________ property.

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distributive law

Q56Fill in the blanks

The side of the square of area 9y2 is __________. 3x + 3

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Q57Fill in the blanks

On simplification = _________

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x + 1

Q58Fill in the blanks

The factorisation of 2x + 4y is __________. In questions 59 to 80, state whether the statements are True (T) or False (F):

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x + 2y

Q59Short answer

(a + b)2 = a2 + b2

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False

Q60Short answer

(a – b)2 = a2 – b2

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False

Q61Short answer

(a + b) (a – b) = a2 – b2

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True

Q62Short answer

The product of two negative terms is a negative term.

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False

Q63Short answer

The product of one negative and one positive term is a negative term.

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True

Q64Short answer

The coefficient of the term – 6x2y2 is – 6.

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True

Q65Short answer

p2q + q2r + r 2q is a binomial.

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False

Q66Short answer

The factors of a2 – 2ab + b2 are (a + b) and (a + b).

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False

Q67Short answer

h is a factor of 2π (h + r). n2 n 1

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False

Q68Short answer

Some of the factors of + are , n and (n + 1). 2 2 2

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True

Q69Short answer

An equation is true for all values of its variables.

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False

Q70Short answer

x2 + (a + b)x + ab = (a + b) (x + ab)

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False

Q71Short answer

Common factor of 11pq2, 121p2q3, 1331p2q is 11p2q2.

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False

Q72Short answer

Common factor of 12a2b 2 + 4ab2 – 32 is 4.

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True

Q73Short answer

Factorisation of – 3a2 + 3ab + 3ac is 3a (–a – b – c).

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False

Q74Short answer

Factorised form of p2 + 30p + 216 is (p + 18) (p – 12).

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False

Q75Short answer

The difference of the squares of two consecutive numbers is their sum.

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True

Q76Short answer

abc + bca + cab is a monomial. p 3

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True

Q77Short answer

On dividing by , the quotient is 9. 3 p

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False

Q78Short answer

The value of p for 512 – 49 2 = 100p is 2.

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True

Q79Short answer

(9x – 51) ÷ 9 is x – 51.

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False

Q80Short answer

The value of (a + 1) (a – 1) (a2 + 1) is a4 – 1.

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True

Q81Multiple choice

Add:

(i)7a2bc, – 3abc2, 3a2bc, 2abc2
(ii)9ax, + 3by – cz, – 5by + ax + 3cz
(iii)xy2z2 + 3x2y2z – 4x2yz2, – 9x2y2z + 3xy2z2 + x2yz2
(iv)5x2 – 3xy + 4y2 – 9, 7y2 + 5xy – 2x2 + 13 (v) 2p4 – 3p3 + p2 – 5p +7, –3p4 – 7p3 – 3p2 – p – 12 (vi) 3a (a – b + c), 2b (a – b + c) (vii) 3a (2b + 5c), 3c (2a + 2b)

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(i) 7a2bc, – 3abc2, 3a2bc, 2abc2

(ii) 9ax, + 3by – cz, – 5by + ax + 3cz

(iii) xy2z2 + 3x2y2z – 4x2yz2, – 9x2y2z + 3xy2z2 + x2yz2

(iv) 5x2 – 3xy + 4y2 – 9, 7y2 + 5xy – 2x2 + 13 (v) 2p4 – 3p3 + p2 – 5p +7, –3p4 – 7p3 – 3p2 – p – 12 (vi) 3a (a – b + c), 2b (a – b + c) (vii) 3a (2b + 5c), 3c (2a + 2b)

Q82Multiple choice

Subtract :

(i)5a2b2c 2 from – 7a2b2c 2
(ii)6x2 – 4xy + 5y2 from 8y2 + 6xy – 3x2
(iii)2ab2c 2 + 4a2b2c – 5a2bc2 from –10a2b2c + 4ab2c 2 + 2a2bc2
(iv)3t 4 – 4t 3 + 2t 2 – 6t + 6 from – 4t 4 + 8t 3 – 4t 2 – 2t + 11 (v) 2ab + 5bc – 7ac from 5ab – 2bc – 2ac + 10abc (vi) 7p (3q + 7p) from 8p (2p – 7q) (vii) –3p2 + 3pq + 3px from 3p (– p – a – r)

Show answer

(i) 5a2b2c 2 from – 7a2b2c 2

(ii) 6x2 – 4xy + 5y2 from 8y2 + 6xy – 3x2

(iii) 2ab2c 2 + 4a2b2c – 5a2bc2 from –10a2b2c + 4ab2c 2 + 2a2bc2

(iv) 3t 4 – 4t 3 + 2t 2 – 6t + 6 from – 4t 4 + 8t 3 – 4t 2 – 2t + 11 (v) 2ab + 5bc – 7ac from 5ab – 2bc – 2ac + 10abc (vi) 7p (3q + 7p) from 8p (2p – 7q) (vii) –3p2 + 3pq + 3px from 3p (– p – a – r)

Q83Multiple choice

Multiply the following:

(i)– 7pq2r 3, – 13p3q2r
(ii)3x 2y2z2, 17xyz
(iii)15xy2, 17yz2
(iv)–5a2bc, 11ab, 13abc2 (v) –3x2y, (5y – xy) (vi) abc, (bc + ca) (vii) 7pqr, (p – q + r) (viii) x2y2z2, (xy – yz + zx) (ix) (p + 6), (q – 7) (x) 6mn, 0mn (xi) a, a5, a6 (xii) –7st, –1, – 13st2 (xiii) b 3, 3b2, 7ab5 100 3 (xiv) – rs; r 3s 2 9 4 (xv) (a2 – b2), (a2 + b2) (xvi) (ab + c), (ab + c) (xvii) (pq – 2r), (pq – 2r) 3 4 2 3 (xviii) x – y , x + y 4 3 3 2 3 2 2 2 (xix) p + q , (2p2 –3q2) 2 3 (xx) (x2 – 5x + 6), (2x + 7) (xxi) (3x2 + 4x – 8), (2x2 – 4x + 3) (xxii) (2x – 2y – 3), (x + y + 5)

Show answer

(i) – 7pq2r 3, – 13p3q2r

(ii) 3x 2y2z2, 17xyz

(iii) 15xy2, 17yz2

(iv) –5a2bc, 11ab, 13abc2 (v) –3x2y, (5y – xy) (vi) abc, (bc + ca) (vii) 7pqr, (p – q + r) (viii) x2y2z2, (xy – yz + zx) (ix) (p + 6), (q – 7) (x) 6mn, 0mn (xi) a, a5, a6 (xii) –7st, –1, – 13st2 (xiii) b 3, 3b2, 7ab5 100 3 (xiv) – rs; r 3s 2 9 4 (xv) (a2 – b2), (a2 + b2) (xvi) (ab + c), (ab + c) (xvii) (pq – 2r), (pq – 2r) 3 4 2 3 (xviii) x – y , x + y 4 3 3 2 3 2 2 2 (xix) p + q , (2p2 –3q2) 2 3 (xx) (x2 – 5x + 6), (2x + 7) (xxi) (3x2 + 4x – 8), (2x2 – 4x + 3) (xxii) (2x – 2y – 3), (x + y + 5)

Q84Multiple choice

Simplify

(i)(3x + 2y)2 + (3x – 2y)2
(ii)(3x + 2y)2 – (3x – 2y)2 7 9
(iii)a + b – ab 9 7 3 4
(iv)x − y + 2xy 4 3 (v) (1.5p + 1.2q)2 – (1.5p – 1.2q)2 (vi) (2.5m + 1.5q)2 + (2.5m – 1.5q)2 (vii) (x2 – 4) + (x2 + 4) + 16 (viii) (ab – c)2 + 2abc (ix) (a – b) (a2 + b2 + ab) – (a + b) (a2 + b2 – ab) (x) (b2 – 49) (b + 7) + 343 (xi) (4.5a + 1.5b)2 + (4.5b + 1.5a)2 (xii) (pq – qr)2 + 4pq2r (xiii) (s2t + tq2)2 – (2stq)2

Show answer

(i) (3x + 2y)2 + (3x – 2y)2

(ii) (3x + 2y)2 – (3x – 2y)2 7 9

(iii) a + b – ab 9 7 3 4

(iv) x − y + 2xy 4 3 (v) (1.5p + 1.2q)2 – (1.5p – 1.2q)2 (vi) (2.5m + 1.5q)2 + (2.5m – 1.5q)2 (vii) (x2 – 4) + (x2 + 4) + 16 (viii) (ab – c)2 + 2abc (ix) (a – b) (a2 + b2 + ab) – (a + b) (a2 + b2 – ab) (x) (b2 – 49) (b + 7) + 343 (xi) (4.5a + 1.5b)2 + (4.5b + 1.5a)2 (xii) (pq – qr)2 + 4pq2r (xiii) (s2t + tq2)2 – (2stq)2

Q85Multiple choice

Expand the following, using suitable identities.

(i)(xy + yz)2
(ii)(x2y – xy2)2 4 5
(iii)a + b 5 4 2 3
(iv)x – y 3 2 4 5 (v) p + q 5 3 (vi) (x + 3) (x + 7) (vii) (2x + 9) (2x – 7) 4 x y 4 x 3y (viii) + + 5 4 5 4 2x 2 2x 2a (ix) – + 3 3 3 3 (x) (2x – 5y) (2x – 5y) 2a b 2a b (xi) + − 3 3 3 3 (xii) (x2 + y2) (x2 – y2) (xiii) (a2 + b2)2 (xiv) (7x + 5)2 (xv) (0.9p – 0.5q)2 (xvi) x2y2 = (xy)2

Show answer

(i) (xy + yz)2

(ii) (x2y – xy2)2 4 5

(iii) a + b 5 4 2 3

(iv) x – y 3 2 4 5 (v) p + q 5 3 (vi) (x + 3) (x + 7) (vii) (2x + 9) (2x – 7) 4 x y 4 x 3y (viii) + + 5 4 5 4 2x 2 2x 2a (ix) – + 3 3 3 3 (x) (2x – 5y) (2x – 5y) 2a b 2a b (xi) + − 3 3 3 3 (xii) (x2 + y2) (x2 – y2) (xiii) (a2 + b2)2 (xiv) (7x + 5)2 (xv) (0.9p – 0.5q)2 (xvi) x2y2 = (xy)2

Q86Multiple choice

Using suitable identities, evaluate the following.

(i)(52)2
(ii)(49)2
(iii)(103)2
(iv)(98)2 (v) (1005)2 (vi) (995)2 (vii) 47 × 53 (viii) 52 × 53 (ix) 105 × 95 (x) 104 × 97 (xi) 101 × 103 (xii) 98 × 103 (xiii) (9.9)2 (xiv) 9.8 × 10.2 (xv) 10.1 × 10.2 (xvi) (35.4)2 – (14.6)2 (xvii) (69.3)2 – (30.7)2 (xviii) (9.7)2 – (0.3)2 (xix) (132)2 – (68)2 (xx) (339)2 – (161)2 (xxi) (729)2 – (271)2

Show answer

(i) (52)2

(ii) (49)2

(iii) (103)2

(iv) (98)2 (v) (1005)2 (vi) (995)2 (vii) 47 × 53 (viii) 52 × 53 (ix) 105 × 95 (x) 104 × 97 (xi) 101 × 103 (xii) 98 × 103 (xiii) (9.9)2 (xiv) 9.8 × 10.2 (xv) 10.1 × 10.2 (xvi) (35.4)2 – (14.6)2 (xvii) (69.3)2 – (30.7)2 (xviii) (9.7)2 – (0.3)2 (xix) (132)2 – (68)2 (xx) (339)2 – (161)2 (xxi) (729)2 – (271)2

Q87Multiple choice

Write the greatest common factor in each of the following terms.

(i)– 18a2, 108a
(ii)3x2y, 18xy2, – 6xy
(iii)2xy, –y2, 2x2y
(iv)l2m 2n, lm2n2, l2mn2 (v) 21pqr, –7p2q2r 2, 49p2qr (vi) qrxy, pryz, rxyz (vii) 3x3y2z, –6xy3z2, 12x2yz3 (viii) 63p2a2r 2s, – 9pq2r 2s2, 15p2qr2s2, – 60p2a2rs2 (ix) 13x2y, 169xy (x) 11x2, 12y2

Show answer

(i) – 18a2, 108a

(ii) 3x2y, 18xy2, – 6xy

(iii) 2xy, –y2, 2x2y

(iv) l2m 2n, lm2n2, l2mn2 (v) 21pqr, –7p2q2r 2, 49p2qr (vi) qrxy, pryz, rxyz (vii) 3x3y2z, –6xy3z2, 12x2yz3 (viii) 63p2a2r 2s, – 9pq2r 2s2, 15p2qr2s2, – 60p2a2rs2 (ix) 13x2y, 169xy (x) 11x2, 12y2

Q88Multiple choice

Factorise the following expressions.

(i)6ab + 12bc
(ii)–xy – ay
(iii)ax3 – bx2 + cx
(iv)l 2m 2n – lm 2n 2– l 2mn 2 (v) 3pqr –6p 2q 2r 2 – 15r 2 (vi) x 3y2 + x 2y3 – xy4 + xy (vii) 4xy2 – 10x 2y + 16x 2y2 + 2xy (viii) 2a3 – 3a2b + 5ab 2 – ab (ix) 63p 2q 2r 2s – 9pq 2r 2s 2 + 15p 2qr 2s 2 – 60p 2q 2rs 2 (x) 24x 2yz 3 – 6xy 3z 2 + 15x 2y 2z – 5xyz (xi) a3 + a2 + a + 1 (xii) lx + my + mx + ly (xiii) a3x – x4 + a2x2 – ax3 (xiv) 2x2 – 2y + 4xy – x (xv) y2 + 8zx – 2xy – 4yz (xvi) ax2y – bxyz – ax2z + bxy2 (xvii) a2b + a2c + ab + ac + b2c + c2b (xviii) 2ax2 + 4axy + 3bx2 + 2ay2 + 6bxy + 3by2

Show answer

(i) 6ab + 12bc

(ii) –xy – ay

(iii) ax3 – bx2 + cx

(iv) l 2m 2n – lm 2n 2– l 2mn 2 (v) 3pqr –6p 2q 2r 2 – 15r 2 (vi) x 3y2 + x 2y3 – xy4 + xy (vii) 4xy2 – 10x 2y + 16x 2y2 + 2xy (viii) 2a3 – 3a2b + 5ab 2 – ab (ix) 63p 2q 2r 2s – 9pq 2r 2s 2 + 15p 2qr 2s 2 – 60p 2q 2rs 2 (x) 24x 2yz 3 – 6xy 3z 2 + 15x 2y 2z – 5xyz (xi) a3 + a2 + a + 1 (xii) lx + my + mx + ly (xiii) a3x – x4 + a2x2 – ax3 (xiv) 2x2 – 2y + 4xy – x (xv) y2 + 8zx – 2xy – 4yz (xvi) ax2y – bxyz – ax2z + bxy2 (xvii) a2b + a2c + ab + ac + b2c + c2b (xviii) 2ax2 + 4axy + 3bx2 + 2ay2 + 6bxy + 3by2

Q89Multiple choice

Factorise the following, using the identity a2 + 2ab ab + b 2 = (a a + b )2

(i)x2 + 6x + 9
(ii)x2 + 12x + 36
(iii)x2 + 14x + 49
(iv)x2 + 2x + 1 (v) 4x2 + 4x + 1 (vi) a2x2 + 2ax + 1 (vii) a2x2 + 2abx + b2 (viii) a2x2 + 2abxy + b2y2 (ix) 4x2 + 12x + 9 (x) 16x2 + 40x +25 (xi) 9x2 + 24x + 16 (xii) 9x2 + 30x + 25 (xiii) 2x3 + 24x2 + 72x (xiv) a2x3 + 2abx2 + b2x 4 3 2 x2 (xv) 4x + 12x + 9x (xvi) + 2x + 4 y2 (xvii) 9x2 + 2xy + ab + b 2 = (a

Show answer

(i) x2 + 6x + 9

(ii) x2 + 12x + 36

(iii) x2 + 14x + 49

(iv) x2 + 2x + 1 (v) 4x2 + 4x + 1 (vi) a2x2 + 2ax + 1 (vii) a2x2 + 2abx + b2 (viii) a2x2 + 2abxy + b2y2 (ix) 4x2 + 12x + 9 (x) 16x2 + 40x +25 (xi) 9x2 + 24x + 16 (xii) 9x2 + 30x + 25 (xiii) 2x3 + 24x2 + 72x (xiv) a2x3 + 2abx2 + b2x 4 3 2 x2 (xv) 4x + 12x + 9x (xvi) + 2x + 4 y2 (xvii) 9x2 + 2xy + ab + b 2 = (a

Q90Multiple choice

Factorise the following, using the identity a2 – 2ab a – b )2.

(i)x2 – 8x + 16
(ii)x2 – 10x + 25
(iii)y2 – 14y + 49
(iv)p2 – 2p + 1 (v) 4a2 – 4ab + b2 (vi) p2y2 – 2py + 1 (vii) a2y2 – 2aby + b2 (viii) 9x2 – 12x + 4 x2 (ix) 4y2 – 12y + 9 (x) – 2x + 4 4x 2 (xi) a2y3 – 2aby2 + b2y (xii) 9y2 – 4xy +

Show answer

(i) x2 – 8x + 16

(ii) x2 – 10x + 25

(iii) y2 – 14y + 49

(iv) p2 – 2p + 1 (v) 4a2 – 4ab + b2 (vi) p2y2 – 2py + 1 (vii) a2y2 – 2aby + b2 (viii) 9x2 – 12x + 4 x2 (ix) 4y2 – 12y + 9 (x) – 2x + 4 4x 2 (xi) a2y3 – 2aby2 + b2y (xii) 9y2 – 4xy +

Q91Multiple choice

Factorise the following.

(i)x2 + 15x + 26
(ii)x2 + 9x + 20
(iii)y2 + 18x + 65
(iv)p2 + 14p + 13 (v) y2 + 4y – 21 (vi) y2 – 2y – 15 (vii) 18 + 11x + x2 (viii) x2 – 10x + 21 (ix) x2 = 17x + 60 (x) x2 + 4x – 77 (xi) y2 + 7y + 12 (xii) p2 – 13p – 30 (xiii) a2 – 16p – 80

Show answer

(i) x2 + 15x + 26

(ii) x2 + 9x + 20

(iii) y2 + 18x + 65

(iv) p2 + 14p + 13 (v) y2 + 4y – 21 (vi) y2 – 2y – 15 (vii) 18 + 11x + x2 (viii) x2 – 10x + 21 (ix) x2 = 17x + 60 (x) x2 + 4x – 77 (xi) y2 + 7y + 12 (xii) p2 – 13p – 30 (xiii) a2 – 16p – 80

Q92Multiple choice

Factorise the following using the identity a2 – b 2 = (a a+b a – b ). b) (a

(i)x2 – 9
(ii)4x2 – 25y2
(iii)4x2 – 49y2
(iv)3a2b3 – 27a4b (v) 28ay2 – 175ax2 (vi) 9x2 – 1 x 2 y2 (vii) 25ax2 – 25a (viii) − 9 25 2p 2 (ix) – 32q 2 (x) 49x2 – 36y2 y x2 (xi) y3 – (xii) – 625 9 25 x2 y2 4x 2 9y 2 (xiii) − (xiv) − 8 18 9 16 x 3y xy 3 (xv) − (xvi) 1331x3y – 11y3x 9 16 1 2 2 16 2 2 (xvii) a b – b c (xviii) a4 – (a – b)4 36 49 (xix) x4 – 1 (xx) y4 – 625 (xxi) p5 – 16p (xxii) 16x4 – 81 (xxiii) x4 – y4 (xxiv) y4 – 81 (xxv) 16x4 – 625y4 (xxvi) (a – b)2 – (b – c)2 (xxvii) (x + y)4 – (x – y)4 (xxviii) x4 – y4 + x2 – y2 3 2 y2 (xxix) 8a – 2a (xxx) x – (xxxi) 9x2 – (3y + z)2

Show answer

(i) x2 – 9

(ii) 4x2 – 25y2

(iii) 4x2 – 49y2

(iv) 3a2b3 – 27a4b (v) 28ay2 – 175ax2 (vi) 9x2 – 1 x 2 y2 (vii) 25ax2 – 25a (viii) − 9 25 2p 2 (ix) – 32q 2 (x) 49x2 – 36y2 y x2 (xi) y3 – (xii) – 625 9 25 x2 y2 4x 2 9y 2 (xiii) − (xiv) − 8 18 9 16 x 3y xy 3 (xv) − (xvi) 1331x3y – 11y3x 9 16 1 2 2 16 2 2 (xvii) a b – b c (xviii) a4 – (a – b)4 36 49 (xix) x4 – 1 (xx) y4 – 625 (xxi) p5 – 16p (xxii) 16x4 – 81 (xxiii) x4 – y4 (xxiv) y4 – 81 (xxv) 16x4 – 625y4 (xxvi) (a – b)2 – (b – c)2 (xxvii) (x + y)4 – (x – y)4 (xxviii) x4 – y4 + x2 – y2 3 2 y2 (xxix) 8a – 2a (xxx) x – (xxxi) 9x2 – (3y + z)2

Q93Multiple choice

The following expressions are the areas of rectangles. Find the possible lengths and breadths of these rectangles.

(i)x2 – 6x + 8
(ii)x2 – 3x + 2
(iii)x2 – 7x + 10
(iv)x2 + 19x – 20 (v) x2 + 9x + 20

Show answer

(i) x2 – 6x + 8

(ii) x2 – 3x + 2

(iii) x2 – 7x + 10

(iv) x2 + 19x – 20 (v) x2 + 9x + 20

Q94Multiple choice

Carry out the following divisions:

(i)51x3y2z ÷ 17xyz
(ii)76x3yz3 ÷ 19x2y2
(iii)17ab2c 3 ÷ (–abc2)
(iv)–121p3q3r 3 ÷ (–11xy2 z3)

Show answer

(i) 51x3y2z ÷ 17xyz

(ii) 76x3yz3 ÷ 19x2y2

(iii) 17ab2c 3 ÷ (–abc2)

(iv) –121p3q3r 3 ÷ (–11xy2 z3)

Q95Multiple choice

Perform the following divisions:

(i)(3pqr – 6p2q2r 2) ÷ 3pq
(ii)(ax3 – bx2 + cx) ÷ (– dx)
(iii)(x3y3 + x2y3 – xy4 + xy) ÷ xy
(iv)(– qrxy + pryz – rxyz) ÷ (– xyz)

Show answer

(i) (3pqr – 6p2q2r 2) ÷ 3pq

(ii) (ax3 – bx2 + cx) ÷ (– dx)

(iii) (x3y3 + x2y3 – xy4 + xy) ÷ xy

(iv) (– qrxy + pryz – rxyz) ÷ (– xyz)

Q96Multiple choice

Factorise the expressions and divide them as directed:

(i)(x2 – 22x + 117) ÷ (x – 13)
(ii)(x3 + x2 – 132x) ÷ x (x – 11)
(iii)(2x3 – 12x2 + 16x) ÷ (x – 2) (x – 4)
(iv)(9x2 – 4) ÷ (3x + 2) (v) (3x2 – 48) ÷ (x – 4) (vi) (x4 – 16) ÷ x3 + 2x2 + 4x + 8 (vii) (3x4 – 1875) ÷ (3x2 – 75)

Show answer

(i) (x2 – 22x + 117) ÷ (x – 13)

(ii) (x3 + x2 – 132x) ÷ x (x – 11)

(iii) (2x3 – 12x2 + 16x) ÷ (x – 2) (x – 4)

(iv) (9x2 – 4) ÷ (3x + 2) (v) (3x2 – 48) ÷ (x – 4) (vi) (x4 – 16) ÷ x3 + 2x2 + 4x + 8 (vii) (3x4 – 1875) ÷ (3x2 – 75)

Q97Short answer

The area of a square is given by 4x2 + 12xy + 9y2. Find the side of the square.

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2x + 3y

Q98Short answer

The area of a square is 9x2 + 24xy + 16y2. Find the side of the square.

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3x + 4y

Q99Short answer

The area of a rectangle is x2 + 7x + 12. If its breadth is (x + 3), then find its length.

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x + 8

Q100Short answer

The curved surface area of a cylinder is 2π (y2 – 7y + 12) and its radius is (y – 3). Find the height of the cylinder (C.S.A. of cylinder = 2πrh).

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y – 4

Q101Short answer

The area of a circle is given by the expression πx2 + 6πx + 9π. Find the radius of the circle.

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x + 3

Q102Short answer

The sum of first n natural numbers is given by the expression n2 n + . Factorise this expression. 2 2

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n (n + 1)

Q103Short answer

The sum of (x + 5) observations is x4 – 625. Find the mean of the observations.

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(x2 + 25) (x – 5)

Q104Short answer

The height of a triangle is x4 + y4 and its base is 14xy. Find the area of the triangle.

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7xy (x4 + y4)

Q105Short answer

The cost of a chocolate is Rs (x + y) and Rohit bought (x + y) chocolates. Find the total amount paid by him in terms of x. If x = 10, find the amount paid by him.

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Rs x2 + 8x + 16; Rs 196

Q106Short answer

The base of a parallelogram is (2x + 3 units) and the corresponding height is (2x – 3 units). Find the area of the parallelogram in terms of x. What will be the area of parallelogram of x = 30 units?

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4x2 – 9 sq. units; 391 sq. units

Q107Short answer

The radius of a circle is 7ab – 7bc – 14ac. Find the circumference of 22 the circle. π = 7

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44 (ab – b (–2ac))

Q108Short answer

If p + q = 12 and pq = 22, then find p2 + q2.

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100

Q109Short answer

If a + b = 25 and a2 + b2 = 225, then find ab.

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200

Q110Short answer

If x – y = 13 and xy = 28, then find x2 + y2.

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225

Q111Short answer

If m – n = 16 and m2 + n2 = 400, then find mn.

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Q112Short answer

If a2 + b2 = 74 and ab = 35, then find a + b.

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Q113Multiple choice

Verify the following:

(i)(ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab) = 0
(ii)(a + b + c) (a2 + b2 + c 2 – ab – bc – ca) = a3 + b3+ c 3 – 3abc
(iii)(p – q) (p2 + pq + q2) = p3 – q3
(iv)(m + n) (m 2 – mn + n2) = m 3 + n3 (v) (a + b) (a + b) (a + b) = a3 + 3a2b + 3ab2 + b3 (vi) (a – b) (a – b) (a – b) = a3 – 3a2b + 3ab2 – b3 (vii) (a2 – b2) (a2 + b2) + (b2 – c 2) (b2 + c 2) + (c 2 – a2) + (c 2 + a2) = 0 (viii) (5x + 8)2 – 160x = (5x – 8)2 (ix) (7p – 13q)2 + 364pq = (7p + 13q)2 2 2 3p 7 3 7 (x) + − p+ =2 7 6p 7 6p

Q114Multiple choice

Find the value of a, if

(i)8a = 352 – 272
(ii)9a = 762 – 672
(iii)pqa = (3p + q)2 – (3p – q)2
(iv)pq2a = (4pq + 3q)2 – (4pq – 3q)2

Show answer

(i) 8a = 352 – 272

(ii) 9a = 762 – 672

(iii) pqa = (3p + q)2 – (3p – q)2

(iv) pq2a = (4pq + 3q)2 – (4pq – 3q)2

Q115Short answer

What should be added to 4c (– a + b + c) to obtain 3a (a + b + c) – 2b (a – b + c)?

Show answer

3a2 + ab + 7ac + 2b2 – 6bc – 4c 2

Q116Short answer

Subtract b (b2 + b – 7) + 5 from 3b2 – 8 and find the value of expression obtained for b = – 3. 1 1

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–b3 + 2b2 + 7b – 8; 16

Q117Short answer

If x − = 7 then find the value of x 2 + 2 . x x 2 1 3

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51 1 1

Q118Short answer

Factorise x + 2 + 2 − 3x − . x x

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x + x + – 3

Q119Short answer

Factorise p4 + q4 + p2q2.

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(p2 + q2 – pq) (p2 + q2 + pq) x x

Q120Multiple choice

Find the value of 6.25 × 6.25 –1.75 ×1.75

(i)4.5 198 ×198 –102 ×102
(ii)

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(i) 4.5 198 ×198 –102 ×102

(ii)

Q121Short answer

The product of two expressions is x5 + x3 + x. If one of them is x2 + x + 1, find the other.

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x (x2 – x + 1)

Q122Short answer

Find the length of the side of the given square if area of the square is 625 square units and then find the value of x.

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Side = 25 units; x = 5 124. 10x (2x + 1) sq. units

Q123Multiple choice

Take suitable number of cards given in the adjoining diagram [G(x × x) representing x2, R (x × 1) representing x and Y (1 ×1) representing 1] to factorise the following expressions, by arranging the cards in the form of rectangles:

(i)2x2 + 6x + 4
(ii)x2 + 4x + 4. Factorise 2x2 + 6x + 4 by using the figure. Calculate the area of figure.

Q124Short answer

The figure shows the dimensions of a wall having a window and a door of a room. Write an algebraic expression for the area of the wall to be painted.

Q125Multiple choice

Match the expressions of column I with that of column II: Column I Column II (1) (21x + 13y)2

(a)441x2 – 169y2 (2) (21x – 13y)2
(b)441x2 + 169y2 + 546 xy (3) (21x – 13y) (21x + 13y)
(c)441x2 + 169y2 – 546xy
(d)441x2 – 169y2 + 546xy 1. Algebraic Tiles (i) Cut the following tiles from a graph sheet. Now, colour the tiles as per the colour code. Arrange these algebraic tiles to form a square. Find the length of the side of the square so formed. Also find the area of the square. Using the above result factorise x2 + 4x + 4. (ii) 2. Find the length of the side of the rectangle so formed. Also find the area of the rectangle. Using the above result factorise x2 + 5x + 4. Now choose and cut more algebraic tiles from the graph sheet. Create your own colour code and colour the tiles. Arrange them to form square/rectangle. Find the area of the figure so formed using it to factorise a) x2 + 4x + 3 b) x2 + 9x + 18 3. Build a square garden. Divide the square garden into four rectangular flower beds in such a way that each flower bed is as long as one side of the square. The perimeter of each flower bed is 40 m. (a) Draw a diagram to represent the above information. (b) Mention the expression for perimeter of the entire garden. Crossword Number Puzzle Solve the given crossword and then fill up the given boxes. Clues are given below for across as well as downward filling. Also for across and down clues clue number is written at the corner of boxes. Answers of clues have to be filled in their respective boxes. Down 1. A polynomial with two terms. 2. An expression containing one or more terms with non-zero coefficient (with variables having non-negative exponents). 3. To find the value of a mathematical expression. 4. A _________ is formed by the product of variables and constants. 5. The abbreviation of the greatest no. (or expression) that in a factor of two or more numbers. 6. A polynomial with three terms. Across 7. A polynomial with only one term. 8. An expression of the second degree. 9. Terms can be written as product of its _________. 10. The numbers –3, –2, –1, 0, 1, 2, 3 are known as _______. 11. __________ terms are formed from the same variables and the powers of these variables are the same term. 12. The highest power of a polynomial is called the _________ of the polynomial. Solution 1. Binomial 2. Polynomial 3. Evaluate 4. Term 5. GCF 6. Trinomial 7. Monomial 8. Quadratic 9. Factors 10. Integers 11. Like 12. Degree MATHEMATICS Rough Work

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(a) 441x2 – 169y2 (2) (21x – 13y)2

(b) 441x2 + 169y2 + 546 xy (3) (21x – 13y) (21x + 13y)