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(i) (9x + 3y) and (10x – 9y) 

This can be written as (9x + 3y) + (10x – 9y) 

Grouping the like terms, we get 

(9x + 10x) + (3y – 9y) = x(9 + 10) + y(3 – 0) 

= 19x + y(-6) 

= 19x – 6y 

Thus degree of the expression is 1.

(ii) (k² – 25k + 46) and (23 – 2k² + 21k)

This can be written as (k² – 25k + 46) + (23 – 2k² + 21k)

Grouping the like terms, we get

(k² – 2k²) + (-25 k + 21 k) + (46 + 23)

= k²(1 – 2) + k(-25 + 21) + 69 

= – 1k² – 4k + 69

Thus degree of the expression is 2.

(iii) (3m²n + 4pq²) and (5nm² – 2q²p)

This can be written as (3m²n + 4pq²) + (5nm² – 2q²p)

Grouping the like terms, we get

(3m²n + 5m²n) + (4pq² – 2pq²)

= m²n(3 + 5) + pq²(4 – 2) 

= 8m²n + 2pq²

Thus degree of the expression is 3.

(i) 8x + 3x = (8 + 3)x = 11x 

(ii) 7mn + 5mn = (7 + 5)mn = 12mn 

(iii) -9y + 11y + 2y = (-9 + 11 + 2 )y = (2 + 2)y = 4y

(i) (4.9)²

(4.9)² = (5 – 0.1)²

Substituting a = 5 and b = 0.1 in

(a – b)² = a² – 2ab + b², we have

(5 – 0.1)² = 5² – 2(5) (0.1) + (0.1)²

(4.9)² = 25 – 1 + 0.01 

= 24 + 0.01

(4.9)² = 24.01

(ii) (100.1)²

(100.1)² = (100 + 0.1)²

Substituting a = 100 and b = 0.1 in

(a + b)² = a² + 2ab + b², we have

(100 + 0.1)² = (100)² + 2(100) (0.1) + (0.1)²

(100.1)² = 10000 + 20 + 0.01

(100.1)² = 10020.01

(iii) (1.9) x (2.1)

(1.9) x (2.1) = (2 – 0.1) x (2 + 0.1)

Substituting a = 100 and b = 0.1 in

(a – b) (a + b) = a² – b² we have

(2 – 0.1) (2 + 0.1) = 2² – (0.1)²

(1.9) x (2.1) = 4 – 0.01

(1.9) (2.1) = 3.99

(iii) 6

Price of 1 pen = Rs 8 

Price of 1 pack = 10 x 8 = 80 

Number of packs Swetha can buy = x

80x ≤ 500 

8x ≤ 50 

x ≤ \(\frac{50}{8}\) = 6.25 

x is a natural number x = 1, 2, 3, 4, 5, 6

(i) True

(ii) False

(i) 3[(x + y)]

We have (3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)

= 3ab – 2a² – 2b² – 5a² + 7ab – 5b²

= (3ab + 7ab) + (- 2 – 5)a² + (- 2 – 5)b²

= 10 ab + (-7)a² + (-7)b²

= 10ab – 7a² – 7b²

Degree of the expression is 2.

(i) False 

(ii) True 

(iii) True

(i) ab²c³, a²b³c, a³bc² 

ab²c³ = a x b x b x c x c x c 

a²b³c = a x a x b x b x b x c 

a³bc² = a x a x a x b x c x c 

G.C.D. of ab²c³, a²b³c, a³bc² = a x b x c = abc

(ii) 35x⁵y³z⁴, 49x²yz3, 14xy²z² 

35x⁵y³z⁴ = 5 x 7 x⁵ y³ z⁴ 

49x²yz³ = 7 x 7 x² y z³ 

14xy² z² = 2 x 7 x y² z² 

G.C.D. is = 7 x y z²

(iii) 25 ab³ c, 100 a² bc, 125 ab 

25 ab³ c = 5 x 5 ab³ c 

100 a² bc = 2 x 5 x 2 x 5 a² b c 

125 ab = 5 x 5 x 5 ab 

G.C.D is = 5 x 5 ab = 25ab

(iv) 3abc, 5xyz, 7pqr 

3abc = 1 x 3 a bc 

5xyz = 1 x 5 x y z 

7pqr = 1 x 7 p q r 

G.C.D is 1

(ii) 13 

7(3) – 4(2) = 21 – 8 = 13

(i) 1, 100 is a pair of like terms. [∵ 1 = x⁰ and 100 = 100 x⁰]

(ii) -7x, \(\frac{5}{2}\)x is a pair of like terms.

(iii) 4m²p, Amp is a pair of unlike terms.

(iv) 12xz, 12x²z² is a pair of unlike terms.

(i) 24 ab²c² = 2 x 2 x 2 x 3 x a x b x b x c x c

(ii) 36 x³y²z = 2 x 2 x 3 x 3 (x) x (x) x (x) x (x) y (x) y (x) z

(iii) 56 mn²p² = 2 x 2 x 2 x 7 x m x n x n x p x p

1. p² – 2pq + q²

2. x² – 25

3. (x – 2) and (x – 2)

4. 2 x 2 x 2 x 3 x a x b x b x c x c

(i) False

(ii) False 

(iii) True

(i) 1 and 2

5x + 5 ≤ 15

5x ≤ 15 – 5 = 10 

x ≤ \(\frac{10}{2}\) = 2

(v) -4 ≤ x ≤ 2

(i) True

(ii) False

(iii) True

(iv) True

(i) True

(ii) False

(iii) False

(iv) True

\(\frac{1}{3}\) + 6 = 10

(i) False

(ii) False

(iii) True

(iv) False

(i) x² + 5x + 4 = x² + (1 + 4)x + (1 x 4)

Which is of the form x² + (a + b)x + ab

= (x + a) (x + b)

x² + (1 + 4)x + (1 x 4) = (x + 1) (x + 4)

∴ x² + 5x + 4 = (x + 1) (x + 4)

(ii) x² – 5x + 4 = x² + ((-1) + (- 4))x + (-1) (- 4)

Which is of the form x² + (a + b)x + ab

= (x + a) (x + b)

x² + ((-1) + 4))x + ((-1)(-4)) = (x + (-4)) 

= (x – 1) (x – 4)

x² – 5x + 4 = (x – 1) (x – 4)

Let x be the age and y be the height then 

40 ≤ x ≤ 45 and 160 ≤ y ≤ 170

1. x > 25,000, where x is Ramesh’s salary per month. 

2. y ≤ 5, where y is the maximum number of persons the left can carry. 

3. z ≥ 100, where z is the number of days when the exhibition is there.

Number of dots in 1 row = 9

Number of rows = r

Total number of dots in r rows = Number of rows × number of dots in a row

= 9r

Number of dots in 8 rows = 8 × 9 = 72

Number of dots in 10 rows = 10 × 9 = 90

Number of laddus given away = 1

Number of laddus remaining = 5

Total number of laddus = Number of laddus given away + numbers of laddus remaining

= 1 + 5

Number of trees in one row = 4 

Let total number of rows = x 

∴ Total number of trees in x rows = 4 × x = 4x.

(i) Leela’s age in x years. 

∵ Ranu is 5 years younger than Leela. 

∴ Ranu’s age = (x – 5) years. 

(ii) Ranu’s age is P years. 

∵ Leela’s is 5 years older than Ranu. 

∴ Leela’s age = (P + 5) years.

Let Radha’s age be x years

Leele’s age = Radha’s age – 4

= (x – 4) years.

(5x² + 7x – 3) × 4x²

= 4x²(5x² + 7x – 3) Multiplication is commutative

= 4x² (5x² + 4x² (7x) + 4x² (-3)

= (4 × 5)(x² × x²) + (4 × 7)(x² × x) + (4 × -3)(x²)

= 20x⁴ + 28x³ – 12x²

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

56 = 7 × 8 .

∴ y² + (7 + 8)x + 56 = (y + 7)(y + 8)

(i) 4n + 7 ≥ 3n + 10, n is an integer.

4n + 7 – 3n ≥ 3n + 10 – 3n

n(4 – 3) + 7 ≥ 3n + 10 – 3n

n(4 – 3) + 7 ≥ n (3 – 3) + 10

n + 7 ≥ 10

Subtracting 7 on both sides

n + 7 – 7 ≥ 10 – 7

n ≥ 3

Since the solution is an integer and is greater than or equal to 3, the solution will be 3,

4, 5, 6, 7, …

n = 3, 4, 5, 6,7, …

(ii) 6(x + 6) ≥ 5(x – 3), x is a whole number.

6x + 36 ≥ 5x – 15

Subtracting 5x on both sides

6x + 36 – 5x ≥ 5x – 15 – 5x

x (6 – 5) + 36 ≥ x(5 – 5) – 15

x + 36 ≥ -15

Subtracting 36 on both sides

x + 36 – 36 ≥ -15 -36

x ≥ -51

The solution is a whole number and which is greater than or equal to -51

∴ The solution is 0, 1, 2, 3, 4,…

x = 0,1,2, 3,4,…

(iii) -13 ≤ 5x + 2 ≤ 32, x is an integer.

Subtracting throughout by 2

-13 – 2 ≤ 5x + 2 – 2 ≤ 32 – 2

-15 ≤ 5x ≤ 30

Dividing throughout by 5

\(\frac{-15}{5}\) ≤ \(\frac{5x}{5}\) ≤ \(\frac{30}{5}\)

– 3 ≤ x ≤ 6

∴ Since the solution is an integer between -3 and 6 both inclusive, we have the solution

As -3, -2, -1,0, 1,2, 3, 4, 5, 6.

i.e. x = -3, -2, 0, 1, 2, 3,4, 5 and 6.

Answer is (b) n^r 

(i) (2x + 3y + 4z)² 

(a + b + c)² ≡ a² + b² + c² + 2ab + 2bc + 2ca 

∴ (2x + 3y + 4z)² = (2x)² + (3y)² + (4z)² + 2(2x) 

(3y) + 2(3y) (4z) + 2(4z) (2x) 

= 4x² + 9y² + 16z² + 12xy + 24yz + 16zx

(ii) (-p + 2q + 3r)² 

(a + b + c)² ≡ a² + b² + c² + 2ab + 2bc + 2ca 

(-p + 2q + 3r)² = (-p)² + (2q)² + (3r)² + 2(-p) (2q) + 2(2q) (3r) + 2(-p) (3r) 

= p² + 4q² + 9r² – 4pq +12qr – 6rp

(iii) (2p + 3)(2p – 4)(2p – 5) 

(x + a) (x + b) (x + c) ≡ x³ + (a + b + c)x² + (ab + be + ca) x + abc 

(2p + 3)(2p – 4)(2p – 5) = (2p)³ + (3 – 4 – 5) (2p)² + [(3 x – 4)² + (-4 x -5) + (-5 x 3)] 2p + 3 x – 4 x – 5 

= 8p³ + (-6) (4p²) + [-12 + 20 + (-15)] 2p + 60 

= 8p³ – 24p² + (-7)2p + 60 

(2p + 3)(2p – 4)(2p – 5) = 8p³ – 24p² – 14p + 60

(iv) (3a + 1)(3a – 2)(3a + 4) 

(x + a) (x + b) (x + c) ≡ x³ + (a + b + c)x² + (ab + bc + ca)x + abc 

(3a + 1)(3a – 2)(3a + 4) = (3a)³ + (1 – 2 + 4) (3a)² + [1 x (- 2) + (-2 x 4) + 4 x 1] (3a) + 1 x -2 x 4 

= 27a³ + 3 (9a²) + (-2 – 8 + 4)3a – 8 

= 27a³ + 27a² – 8a – 8

(a – b)x² – 6(a + b)x – 9(a – b) = 0 

Here a = a – b; b = – 6(a + b); c = – 9(a – b) 

∆ = b² – 4ac 

= [- 6(a + b)]² – 4(a – b)[-9(a – b)] 

= 36(a + b)² + 36(a – b)(a – b) 

= 36 (a + b)² + 36 (a – b)² 

= 36 [(a + b)² + (a – b)²]

The value is always greater than 0 

∆ = 36 [(a + b)² + (a – b)²] > 0 

∴ The roots are real and unequal.

Let p(x) = x² + px – 4

– 4 is the root of the equation

P(-4) = 0

16 – 4p – 4 = 0

-4p + 12 = 0

-4p = -12

p = 12/4 = 3

The equation x² + px + q = 0 has equal roots

x² + 3 x + q = 0

Here a = 1, b = 3, c = q

Since the roots are real and equal

b² – 4 ac = 0

3² – 4(1)(q) = 0

9 – 4q = 0

9 = 4q

q = 9/4

The value of p = 3 and q = 9/4

(a – b)x² + (b – c)x + (c – a) = 0 

A = (a – b), B = (b – c), C = (c – a) 

Δ = b² – 4ac = 0 

⇒ (b – c)² – 4(a – b)(c – a) 

⇒ b² – 2bc + c² - 4 (ac – bc – a² + ab) 

⇒ b² – 2bc + c² – 4ac + 4bc + 4a² – 4ab = 0 

⇒ 4a² + b² + c² + 2bc – 4ac – 4ab = 0 

⇒- (-2a + b + c)² = 0 [∵ (a + b + c) = a² + b² + c² + 2ab + 2bc + 2ca)] 

⇒ 2a + b + c = 0 

⇒ 2 a = b + c 

∴ a, b, c are in A.P.

Speed = vkm/hr

Distance travelled in 5hrs = 5 × v = 5v km

Total distance between Daspur and Beespur 

= (5v + 20) km

(i) (3m + 5)²

Comparing (3m + 5)² with (a + b)² we have a = 3m and b = 5

(a + b)² = a² + 2 ab + b²

(3m + 5)² = (3m)² + 2 (3m) (5) + 5²

= 3²m² + 30m + 25 

= 9m² + 30m +25

(ii) (5p – 1)²

Comparing (5p – 1)² with (a – b)² we have a = 5p and b = 1

(a – b)² = a² – 2ab + b²

(5p – 1)² = (5p)² – 2 (5p) (1) + 1²

= 5²p² – 10p + 1 = 25p² – 10p + 1

(iii) (2n – 1)(2n + 3)

Comparing (2n – 1) (2n + 3) with (x + a) (x + b) we have a = -1; b = 3

(x + a) (x + b) = x² + (a + b)x + ab

(2n +(- 1)) (2n + 3) 

= (2n)² + (-1 + 3)2n + (-1) (3)

= 2²n² + 2 (2n) – 3 = 4n² + 4n – 3

(iv) 4p² – 25q² = (2p)² – (5q)²

Comparing (2p)² – (5q)² with a² – b² we have a = 2p and b = 5q

(a² – b²) = (a + b)(a – b) 

= (2p + 5q) (2p – 5q)

1 x 18, 2 x 9, 3 x 6, 6 x 3, 9 x 2, 18 x 1 and 1 x 6, 2 x 3, 3 x 2, 6 x 1

xy (k² + 1) + k(x² + y²) = k² xy + xy + kx² + ky²

= (k² xy + kx²) + (ky² + xy)

= kx(ky + x) + y (ky + x)

= (ky + x) (kx + y)

xy (k² – 1) + k(x² – y²) = k² xy – xy + kx² – ky²

= (k²xy + kx²) – xy – ky²

= kx(ky + x) -y (ky + x)

= (ky + x) (kx – y)

L.C.M. = (ky + x) (kx + y) (kx – y)

= (ky + x) (k² x² – y²)

The least common multiple is (ky + x) (k² x² – y²)

Let the three digits numbers be 100a + 10b + c.

100b + 10a + c = 3(100a + 10b + c) + 54 … (1)

100a + 106 + c + 198 = 100c + 106 + a … (2)

(b – a) = 2(b – c) … (3)

(1) ⇒ 100b + 10a + c = 300a + 30b + 3c + 54

⇒ 290a – 70b + 2c = -54

(2) ⇒ 99a – 99c = -198 ⇒ a – c = -2

⇒ a = c – 2

(3) ⇒ a + b – 2c = 0 ⇒ a + b = 2c

⇒ b = 2c – c + 2

⇒ b = c + 2

Substituting a, b in (1)

290(c – 2) – 70 (c + 2) + 2c = -54

290c – 580 – 70c – 140 + 2c = -54

222c = 666 ⇒ c = 3

a = 1, 6 = 5

∴ The number is 153.

Let a natural number be x.

The next number = x + 1

x (x + 1) = 20

x² + x – 20 = 0

(x + 5)(x – 4) = 0

x = -5, 4

∴ x = 4 (∵ x ≠ -5, x is natural number)

The next number = 4 + 1 = 5

Two consecutive numbers are 4, 5

(i) f(x) = 21x² y = 3 x 7x² y 

g(x) = 35xy² = 7 x 5xy² 

G.C.D. = 7xy 

L.C.M. = 7 x 3 x 5 x x² y² = 105x² x y² 

L.C.M x G.C.D = f(x) x g(x) 

105x² y² x 7xy = 21x² y x 35xy² 

735x³ y³ = 735x³ y³ 

Hence verified.

(ii) (x³ – 1)(x + 1) = (x – 1)(x² + x + 1)(x + 1) 

x³ + 1 = (x + 1) (x² – x + 1) 

G.C.D = (x + 1)

L.C.M = (x – 1)(x + 1)(x² + x + 1)(x² – x + 1) 

∴ L.C.M. x G.C.D = f(x) x g(x) 

(x – 1)(x + 1)(x² + x + 1) (x² – x + 1) = (x – 1) 

(x² + x + 1) x (x + 1) (x² – x + 1) 

(x³ – 1)(x + 1)(x³ + 1) = (x³ – 1)(x + 1)(x³ + 1) 

∴ Hence verified.

(iii) f(x) = x² y + xy² = xy(x + y) 

g(x) = x² + xy = x(x + y) 

L.C.M. = x y (x + y) 

G.C.D. = x (x + y) 

To verify: L.C.M. x G.C.D. = xy(x + y) x (x + y) 

= x² y (x + y)² … (1) 

f(x) x g (x) = (x² y + xy²)(x² + xy) 

= x² y (x + y)² … (2) 

∴ L.C.M. x G.C.D = f(x) x g{x). 

Hence verified.

(2) 16x²

x² + 64 = (x²)² + 8² – 2 x (x²) x 8

= (x² – 8)²

2 x (x²) x 8 must be added

i.e, 16x² must be added

Let Omar’s sister helps her mother for x hours.

∴ Omar helps his mother for (x + 1) hours.