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(i) 4b – 3

Three subtracted from four times b.

(ii) 8(m+5)

Eight times the sum of m and 5.

(iii) 7/(8-x)

Quotient on dividing seven by the difference of eight and x(x<8).

(iv) 17(16/w)

Seventeen times quotient of sixteen divided by w.

In the following expressions, write the number of terms and identify numerical and algebraic expressions in them. 

(i) 8p = 1 term – Algebraic expression

(ii) 5c + s – 7 = 3 terms – Algebraic expression

(iii) – 6 = 1 term – Numerical expression

(iv) (2 + 1) – 6 = 2 terms – Numerical expression

(v) 9t + 15 = 2 terms – Algebraic expression

3pq, – 8pq and qp are like terms in these there is no effects on multiplication pq = qp therefore these are like terms where as – 5p, 6p+ 5 and p² + q, are unlike terms.

Correct option is  A) 999900

Correct option is (A) 999900

1010 \(\times\) 990 = (1000+10) (1000-10)

\(=(1000)^2-10^2\)

= 10,00,000 - 100

= 9,99,900

(i) 998² = (1000 − 2)²

= (1000)² − 2(1000)(2) + (2)² 

= 1000000 − 4000 + 4 = 996004

Note : [(a − b)² = a² − 2ab + b² ]

(ii) (5.2)² = (5.0 + 0.2)²

= (5.0)² + 2(5.0) (0.2) + (0.2)²

= 25 + 2 + 0.04 = 27.04

 Note : [(a + b)² = a² + 2ab + b² ]

(iii) 297 × 303 = (300 − 3) × (300 + 3) 

= (300)² − (3)² 

= 90000 − 9 = 89991

Note : [(a + b) (a − b) = a2 − b2] 

By subtracting 2a² − 7ab + 9b² from a² − 3ab + 2b² we get the required expression

Required expression = (a²– 3ab + 2b²) – (2a² – 7ab + 9b²)

= a²– 3ab + 2b² – 2a² + 7ab – 9b²

Collecting positive and negative like terms together, we get

= a² – 2a² – 3ab + 7ab + 2b² – 9b²   

= – a² + 4ab – 7b²

Correct option is D) 2n

2 lines are needed to make a letter V.

\(\therefore\) A pattern of letter V is 2n where n is number of lines.

Correct option is  C) 3n

3 lines are needed to make a letter z.

\(\therefore\) A pattern of letter z is 3n where n is number of lines.

Correct option is D) -4ab - 8b²

(3a² - 8ab - 2b²) - (3a² - 4ab + 6b²)

 = 3a² - 3a² - 8ab - (-4ab) - 2b² - 6b²

= -8ab + 4ab - 8b²

= -4ab - 8b²

= (2a + 8b + 10) – (-3a + 7b + 16)

= 2a + 8b+ 10 + 3a – 7b – 16

= 2a + 3a + 8b – 7b + 10 – 16

(re-arranging the terms)

= 5a + b – 6

∴ (5a + b – 6) 

should be subtracted from 2a + 8b +10 to get -3a + 7b + 16.

The table can be completed as follows.

Given 3x + 2y – {x – (2y – 3)}

First, we have to remove the parentheses. Then, we have to remove the braces.

Then we get,

= 3x + 2y – {x – 2y + 3}

= 3x + 2y – x + 2y – 3

On simplifying, we get

= 2x + 4y – 3

Given – 2(x² – y² + xy) – 3(x² +y² – xy)

Since the ‘–’ sign precedes the parentheses, we have to change the sign of each term in the parentheses when we remove them. Therefore, we have

= -2x² + 2y² – 2xy – 3x² – 3y² + 3xy

On rearranging,

= -2x² – 3x² + 2y² – 3y² – 2xy + 3xy

On simplifying, we get

= -5x² – y² + xy

Given x² – [3x + [2x – (x² – 1)] + 2]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= x² – [3x + [2x – (x² – 1)] + 2]

= x² – [3x + [2x – x² + 1] + 2]

= x² – [3x + 2x – x² + 1 + 2]

= x² – [5x – x² + 3]

On simplifying we get

= x² – 5x + x² – 3

= 2x² – 5x – 3

Given 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]

First we have to remove the parentheses, then remove braces, and then the square brackets.

Then we get,

= 5 + [x – {2y – (6x + y – 4) + 2x} – {x – (y – 2)}]

= 5 + [x – {2y – 6x – y + 4 + 2x} – {x – y + 2}]

= 5 + [x – {y – 4x + 4} – {x – y + 2}]

= 5 + [x – y + 4x – 4 – x + y – 2]

= 5 + [4x – 6]

= 5 + 4x – 6

= 4x – 1

Perimeter of the triangle = 6p² – 4p + 9

Two sides are;

Side one = p² – 2p + 1 and

Side two = 3p² – 5p + 3

Let’s take third side be = x

As we know perimeter of a triangle = sum of all the sides

So, we have

6p² – 4p + 9 = {(p² – 2p + 1) + (3p² – 5p + 3) + (x)}

6p² – 4p + 9 = p² – 2p + 1 + 3p² – 5p + 3 + x

6p² – 4p + 9 - p² + 2p – 1 - 3p² + 5p - 3 = x

Let’s make the pairs;

(6p² – p² – 3p²) + (- 4p + 2p + 5p) + (9 – 1 – 3) = x

2p² + 3p + 5 = x

The required side is 2p² + 3p + 5.

Correct option is  C) 6

Correct option is (C) 6

Let \(ax_1+bx_2\) is a binomial and \(cy_1+dy_2+ey_3\) is a trinomial.

Then their product \(=(ax_1+bx_2)\) \((cy_1+dy_2+ey_3)\) \(=acx_1y_1+adx_1y_2+aex_1y_3\) \(+bcx_2y_1+bdx_2y_2+bex_2y_3\)

\(\therefore\) No. of terms in the product = 6.

D) 15x² + 8xy – 12y²

Correct option is (D) 15x² + 8xy – 12y²

(5x + 6y) (3x – 2y) \(=5x\times3x+5x\times-2y+6y\times3x+6y\times-2y\)

\(=(5\times3)x^2+(5\times-2)xy+(6\times3)xy+(6\times-2)y^2\)

\(=15x^2-10xy+18xy-12y^2\)

= \( 15x^2+8xy-12y^2\).

 (5m + 7n)² is in the form of (a + b)².

(a + b)² = a² + 2ab + b²   [a = 5m, b = 7n]

(5m + 7n)² = (5m)² + 2 × 5m × 7n + (7n)²

= (5m × 5m) + 70 mn + 7n × 7n

= 25m² + 70mn + 49n²

i) (9m – 2n)² is in the form of (a – b)².

(a – b)² = a² – 2ab + b²

(9m – 2n)² = (9m)² – 2 × 9m × 2n + (2n)²

= (9m × 9m) – 36mn + (2n × 2n)

= 81m² – 36mn + 4n²

ii) (6pq – 7rs)²

a = 6pq, b = 7rs

(6pq – 7rs)² = (6pq)² – 2 × 6pq × 7rs + (7rs)²

= (6pq × 6pq) – 84pqrs + (7rs × 7rs)

= 36p²q² – 84pqrs + 49r²s²

iii) (5x² – 6y²)² = (5x²)² – 2 × 5x² × 6y² + (6y²)²

= (5x² × 5x²) – 60x²y² + (6y² × 6y²)

= 25x⁴ – 60x²y² + 36y⁴

(6m + 7n) (6m -,7n) is in the form of 

(a + b) (a – b) . (a + b) (a – b) = a² – b²,

here a = 6m, b = 7n

(6m + 7n) (6m – 7n) = (6m)² – (7n)²

= 6m × 6m – 7n × 7n

= 36m² – 49n²

i) (292)² = (300 – 8)²

a = 300, b = 8

∴ (300 – 8)² = (300)² – 2 × 300 × 8 + (8)² 

= (300 × 300) – 4800 + (8 × 8)

= 90,000 – 4800 + 64

= 90,064 – 4800

= 85,264

ii) (897)² = (900 – 3)²

= (900)² – 2 × 900 × 3 + (3)²

= 8,10,000 – 5400 + 9

= 8,10,009 – 5400

= 8,04,609

iii) (794)² = (800 – 6)²

= (800)² – 2 × 800 × 6 + (6)²

= 6,40,000 – 9600 + 36

= 6,40,036 – 9600

= 6,30,436

(d) - 24 abc

a + b + c = 0 ⇒ a + b = -c,  c + a = - b, b + c = – a

∴ (a + b – c)³ + (c + a – b)³ + (b + c – a)³

= (– c – c)³ + (– b – b)³ + (– a– a)³

= (– 2c)³ + (– 2b)³ + (– 2a)³

= – 8c³  –  8b³  – 8a³ 

= – 8 (a³ + b³ + c³)

= – 8 x 3 abc = - 24 abc.

(b) 5

(a + b)² = a² + b² + 2ab = 117 + 108 = 225

⇒ a + b = 15

(a + b)² = a² + b² + 2ab = 117 - 108 = 9

⇒ a - b = 3

∴ \(\frac{a+b}{a-b} = \frac{15}{3}=5.\)

(d) 9x

\(x=3^{\frac13}+3^{\frac{-1}{3}}\) ⇒ \(x-3^{\frac13}=3^{\frac{-1}{3}}\)

⇒ \((x-3^{\frac13})^3\) = \((x-3^{\frac{-1}3})^3\)

⇒ \(x^3 - 3.x.3^{\frac13}(x-3^{\frac13})-(3^{\frac13})^3=3^{-1}\)

⇒ \(x^3 - 3.x.3^{\frac13}.3^{\frac{-1}3}-3 =\frac13\) ⇒ \(x^3-3x=3+\frac13\)

⇒ \(3x^3 - 9x = 10 \) ⇒ \(3x^3 - 10 = 9x \)

(d) 27 abc

\(a^\frac13+b^\frac13+c^\frac13=0,\)

\((a^\frac13)^3+(b^\frac13)^3+(c^\frac13)^3\) = \(3\,a^\frac13+b^\frac13+c^\frac13\)

⇒ a + b + c = \(3(abc)^\frac13\) ⇒ (a + b + c)³ = 27 abc.

(b) 3

Since (a-b) + (b-c) + (c-a) = 0

∴ (a-b)³ + (b-c)³ + (c-a)³ = 3(a-b)(b-c)(c-a)

⇒ \(\frac{(a-b)^3+(b-c)^3+(c-a)^3}{(a-b)(b-c)(c-a)}\) = \(\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}\) = 3.

(b) 1 + x¹⁶

(1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)(1 - x)

= (1 + x)(1 - x)(1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)

= (1 - x²)(1 + x²)(1 + x⁴)(1 + x⁸)

= (1 - x⁴)(1 + x⁴)(1 + x⁸)

= (1 - x⁸)(1 + x⁸) = (1 + x¹⁶)

(c) 671

(a+ b + c)² = a² + b² + c² = 11² -2 x 20 = 121 - 40 = 81

Also, we know that

(a+ b + c) = (a² + b² + c² - ab - bc -ca) = a³ + b³ + c³ - 3abc

⇒ a³ + b³ + c³ - 3abc = 11 x (80-20)

⇒ a³ + b³ + c³ - 3abc = 11 x 61 = 671.

(d) (x³ + 4)

Dividend = Divisor × Quotient + Remainder

= (\(x\) + 1) x (\(x^2 - x+1\)) +3

= \(x\) (\(x^2 - x+1\)) +  (\(x^2 - x+1\)) + 3

Now solve.

Let (x + y) = a and (x - y) = b

Then, (x + y)³ + (x - y)³ + 6x(x² - y²) = (x + y)³ + (x - y)³ + 3 x 2\(x\) x (x + y)(x - y)

= (x + y)³ + (x - y)³ - 3 x [(x + y) + (x - y)] x (x + y) + (x - y)

= a³ + b³ + 3ab(a + b)

= (a + b)³ = (x + y + x - y)³ = (2x)³ = 8x³.

Given that (x + 6) (x +6)

But we can write the given expression as (x + 6) (x +6) = (x + 6)²

But we have (a + b)²=a²+2ab+b²

On applying above identity in the given expression we get,

(x + 6)²= x²+2 x (6) + 6²

(x + 6)²= x²+12 x + 36

Given 12x² + 8x³-6x² by -2x²

⇒12x² + 8x³-6x²/ (-2x²)

On dividing polynomial by a monomial we have divide every variables of polynomial

By monomial

On simplifying we get

⇒-6x² -4x + 3

(a) (23x + 54) (y – 2)

Factorised form of 23xy – 46x + 54y – 108 is = 23xy – (2 × 23x) + 54y – (2 × 54)

Take out the common factors,

= 23x (y – 2) + 54 (y – 2)

Again take out the common factor,

= (y – 2) (23x + 54)

(b) – 6m³n²

Like terms are formed from the same variables and the powers of these variables are also the same. But coefficients of like terms need not be the same.

-3y (xy + y²)

= -3xy² – 3y³

According to question:

When x = 4 and y = 5

= - 3xy² – 3y³

= - 3 (4) (5)² – 3 (5)³

= - 300 – 375

= - 675

Correct option is  C) 1

We have m = 3 and n = 2

\(\therefore\) mⁿ - n^m = 3² - 2³ = 9 - 8 = 1

(3x + 2y) (3y – 4x) 

= 3x(3y – 4x) + 2y(3y – 4x)

= 9xy – 12x² + 6y² – 8xy

= xy – 12x² + 6y²

55 × 45 = (50 + 5) (50 – 5) 

Identity (a + b)(a – b) = a² – b² 

Here a = 50,b=5 

(50 + 5)(50 – 5) = (50)² – 5² 

= 2500 – 25 

55 × 45 = 2475

Given that,

x + \(\frac{1}{x}\) = 12

Squaring both sides, we get

(x + \(\frac{1}{x}\) )² = 12²

x² + (\(\frac{1}{x}\))² + 2 × x × \(\frac{1}{x}\) = 144

x² + \(\frac{1}{x^2}\) = 142

Subtract 2 from both sides, we get

x² + \(\frac{1}{x\times x}\) - 2 × x × \(\frac{1}{x}\) = 142 – 2

(x - \(\frac{1}{x}\))² = 140

x - \(\frac{1}{x}\) = \(\sqrt{140}\)

(-3/14)xy⁴ × (7/6)x³y

The coefficient of the product of two monomials is equal to the product of their coefficients.

The variable part in the product of two monomials is equal to the product of the variables in the given monomials.

Then,

= [(-3/14) × (7/6)] × (x × x³) × (y⁴ × y)

= [(-3×7)/ (14×6)] × (x¹⁺³) × (y⁴⁺¹) … [∵ a^m × aⁿ = a^m+n]

= [(-1×1)/ (2×2)] × (x⁴) × (y⁵)

= (-1/4)x⁴y⁵

Correct option is B) -1

We have x = 1, y = 2

\(\therefore\) xy + 3y - 9 = 1 x 2 + 3 x 2 - 9

 = 2 + 6 - 9

 = 8 - 9 = -1

(a – b) (2a + 4b) 

= a(2a + 4b) – b(2a + 4b)

= (a × 2a + a × 4b) – (b × 2a + b × 4b)

= 2a² + 4ab – (2ab + 4b²)

= 2a² + 4ab – 2ab – 4b²

= 2a² + 2ab – 4b²

Given that,

2x + 3y = 14 ............... (1)

2x – 3y = 2 .................. (2)

Now, on squaring both the equation and subtracting (2) from (1), we get,

(2x + 3y)² – (2x – 3y)² = (14)² – (2)²

4x² + 9y² + 12xy – 4x² – 9y² + 12xy = 196 – 4

(The positive and negative terms gets cancelled)

24 xy = 192

xy = 8

Therefore, the value of "xy" is 8.

(a + b) (a – b) = a² – b² 

Here a = 2a,b = 4b 

(2a + 4b)(2a-4b) = (2a)² – (4b)² 

= 4a² – 16b²

(a + b) (a – b) = a² – b² 

Here a = 3x, b = 5 

(3x + 5)(3x – 5) = (3x)² – (5)² 

= 9x² -25

(a + b) (a – b) = a² – b² 

Here a = x,b = 6 

(x + 6)(x – 6) = x² – 6² 

= x² – 36

(198)² = (200 – 2)² 

(a – b)² = a² – 2ab + b² 

Here a = 200, b = 2 

(200 – 2)² = 200² – (2)(200)(2) + 2² 

= 40000 – 800 + 4 

(198)²= 39204

59² = (60 – 1)² (a – b)² = a² – 2ab + b² 

Here a = 60, b = 1 

(60 – 1)² = 60² – (2)(60)(1) +12 

= 3600 – 120 + 1 59² =3481

(9.8)² = (10 – 0.2)² (a – b)² 

= a² – 2ab + b² 

Here a = 10, b=0.2 

(10 – 0.2)² = 10² – (2)(10)(0.2) + (0.2)² 

= 100 – 4 + 0.04 

(9.8)² = 96.04