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(i) (5 – x)³

Comparing (5 – x)³ with (a – b)³ we have a = 5 and b = x

(a – b)³ = a³ – 3a²b + 3ab² – b³

(5 – x)³ = 5³ – 3 (5)² (x) + 3(5)(x²) – x³

= 125 – 3(25)(x) + 15x² – x³ = 125

(ii) (2x – 4y)³

Comparing (2x – 4y)³ with (a – b)³ we have a = 2x and b = 4y

(a – b)³ = a³ – 3a²b + 3ab³ – b³

(2x – 4y)³ = (2x)³ – 3(2x)² (4y) + 3(2x) (4y)² – (4y)³

= 2³x³ – 3(2²x²) (4y) + 3(2x) (4²y²) – (4³y³)

= 8x³ – 48x²y + 96xy² – 64y³

(iii) (ab – c)³

Comparing (ab – c)³ with (a – b)³ we have a = ab and b = c

(a – b)³ = a³ – 3a²b + 3ab² – b³

(ab – c)³ = (ab)³ – 3 (ab)² c + 3 ab (c)² – c³

= a³b³ – 3(a²b²) c + 3abc² – c³

= a³b³ – 3a²b² c + 3abc² – c³

(iv) (48)³ = (50 – 2)³

Comparing (50 – 2)³ with (a – b)³ we have a = 50 and b = 2

(a – b)³ = a³ – 3a²b + 3ab² – b³

(50 – 2)³ = (50)³ – 3(50)²(2) + 3 (50)(2)² – 2³

= 1,25,000 – 15000 + 600 – 8 = 1,10,000 + 592

= 1,10,592

(v) (97xy)³

= 97³ x³ y³ = (100 – 3)³ x³y³

Comparing (100 – 3)³ with (a – b)³ we have a = 100, b = 3

(a – b)³ = a³ – 3a²b + 3ab² – b³

(100 – 3)³ = (100)³ – 3(100)² (3) + 3 (100)(3)² – 3³

97³ = 10,00,000 – 90000 + 2700 – 27

97³ = 910000 + 2673

97³ = 912673

97x³y³ = 912673x³y³

(p – 2)(p + 1)(p – 4) = (p + (-2))0 +1)(p + (-4))

Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2; b = 1 ; c = -4.

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

= p³ + (-2 + 1 + (-4))p² + ((-2) (1) + (1) (-4) (-4) (-2)p + (-2) (1) (-4)

= p³ + (-5 )p² + (-2 + (-4) + 8)p + 8

= p³ – 5p² + 2p + 8

(i) (2x + 3y)²

= (2x)² + 2 × (2x) × (3y) + (3y)²

[using (a + b)² = a² + 2ab + b²]

= 4x² + 12xy + 9y²

(ii) (2x – 3y)²

= (2x)² – 2(2x) (3y) + (3y)²

[∵ using (a – b)² = a² – 2ab + b²]

= 4x² – 12xy + 9y²

(5y + 1) (5y + 2) (5y + 3)

Comparing (5y + 1) (5y + 2) (5y + 3) with (x + a) (x + b) (x + c) we have x = 5y ; a = 1; b = 2 and c = 3.

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

= (5y)³ + (1 + 2 + 3) (5y)² + [(1) (2) + (2) (3) + (3) (1)] 5y + (1)(2) (3)

= 5³y³ + 6(5²y²) + (2 + 6 + 3)5y + 6

= 125³ + 150y² + 55y + 6

(5 – x) (3 – 2x) (4 – 3x)

= {(5 – x)(3 – 2x)} × (4 – 3x) [∴ Multiplication in association]

= {5 (3 – 2x) -x (3 – 2x)} × (4 – 3x)

= (15 – 10x – 3x + 2x²) × (4 – 3x)

= (2x² – 13x + 15) (4 – 3x)

= 2x² × (4 – 3x) – 13x (4 – 3x) + 15 (4 – 3x)

= 8x³ – 6³ – 52x + 39x² + 60 – 45x

= -6x³ + 47x² – 97x + 60

(3x – 2) (x – 1) (3x + 5)

= {(3x – 2) (x – 1)} × (3x + 5) [∴Multiplication in associative]

= {3x (x – 1) – 2 x – 1)} × (3x + 5)

= (3x² – 3x – 2x + 2) × (3x + 5)

= (3x² – 5x + 2) (3x + 5)

= 3x² × (3x + 5) – 5x (3x + 5) + 2 (3x + 5)

= (9x³ + 15x²) + (-15x² – 25x) + (6x + 10)

= 9x³ + 15x² – 15x² – 25x + 6x + 10

= 9x³ – 19x + 10

(i) (3ab²) × (-2a²b²) 

= (+) × (-) × (3 × 2) × (a × a²) × (b² × b³) 

= -6a³ b⁵

(ii) (4xy) × (5y²x) × (-x²)

= (+) × (+) × (-) × (4 × 5 × 1) × (x × x × x²) × (y × y²)

= -20x⁴y³

(iii) (2m) × (-5n) × (-3p) 

= (+) × (-) × (-) × (2 × 5 × 3) × m × n × p

= + 30mnp 

= 30 mnp

(i) (3ab³c³) × (5a³b²c)

= (3 × 5)(a × a³ × b² × b² × c² × c)

= 15a¹⁺³.b²⁺².c³⁺¹ 

= 15a⁴b⁴c⁴

(ii) 4x²yz by 32 x²yz²

= (4 × 3/2) × (x² × x² × y × y × z × z²)

= -6x²⁺² y¹⁺¹ x¹⁺² 

= -6x⁴y²z³

p kg of potatoes are bought for Rs 70. Cost of 1kg of potatoes (in Rs) is 70/p.

Given, p kg of potatoes are bought for ₹ 70

Then, cost of 1 kg of potato = 70/p

An auto rickshaw charges Rs 10 for the first kilometre then Rs 8 for each such subsequent kilometre. The total charge (in Rs) for d kilometres is 8d + 2.

From the question it is given that,

An auto rickshaw charges ₹ 10 for the first kilometre

Then ₹ 8 for each such subsequent kilometre.

So, The total charge (in Rs) for d kilometres is = 10 + (d – 1)8

= 10 + 8d – 8

= 2 + 8d

If 7x + 4 = 25, then the value of x is 3.

Consider the equation, 7x + 4 = 25

Transposing 4 from left hand side to right hand side it becomes -4,

7x = 25 – 4

7x = 21

x = 21/7

x = 3

The solution of the equation 3x + 7 = –20 is -9.

Consider the equation, 3x + 7 = –20

Transposing 7 from left hand side to right hand side it becomes -7,

3x = – 20 – 7

3x = – 27

x = -27/3

x = – 9

‘x exceeds y by 7’ can be expressed as x = y + 7.

‘8 more than three times the number x’ can be written as 3x + 8.

As per the condition given in the question, three times the number x = 3x

So, 8 more than three times the number x = 3x + 8

The number of days in w weeks is 7w.

We know that, there are 7 days in a week.

Therefore, number of days in w weeks is 7w.

Number of pencils bought for Rs x at the rate of Rs 2 per pencil is x/2.

From the question it is given that, cost of pencil = ₹ x

Amount per pencil = ₹ 2

Therefore, number of pencil bought = ₹ x/2

x metres = x × 100 centimetres

we know that, 1 meter = 100 centimeter.

Therefore, x metres × 100 centimetres = x 100 centimetres

p litres = p× 1000 millilitres

we know that, 1 litre = 1000 millilitres

Therefore, p litres × 1000 millilitres = p1000 milliliters.

True.

If m is a whole number, then 2m denotes a multiple of 2.

r rupees = r100 paise

we know that, 1 rupee = 100 paise

True.

Consider equations x + 1 = 0

So, x = -1

Consider the equation, 2x + 2 = 0

Divide both the side by 2,

Then we get, x + 1 = 0

Therefore, x = – 1

False.

Consider the equation, x + 1 = 0

Then, x = -1

If the present age of Ramandeep is n years, then her age after 7 years will be n + 7.

If I spend f rupees from 100 rupees, the money left with me is 100 – f rupees.

Cost of a pen is 6 times the cost of a pencil.

Khader’s monthly salary increased by Rs. 1000 in the year 2006 than that of 2005.

The price of petrol per litre decreased this month by Rs. 5 than its price last month.

The correct option is (B) unknown quantities.

(A) can take different values

The word ‘variable’ means something that can vary, i.e., change. The value of a variable is not fixed. We use a variable to represent a number and denote it by any letter such as l, m, n, p, x, y, z etc.

(A) can take different values 

(C) x is subtracted from 10 

LHS = (x + 2y)² – (x – 2y)²

= x² + (2 (x) x (x) 2y) + (2y)² – [x² – (2 (x) x (x) 2y) + (2y)²]

= x² + 4xy + 4y² – [x² – 4xy + 2²y²]

= x² + 4xy + 4y² – x² + 4xy – 4y²

= x² – x² + 4xy + 4xy + 4y² – 4y²

= x² (1 – 1) + xy (4 + 4) + y² (4 – 4)

= 0x² + 8xy + 0y² 

= 8xy 

= RHS

∴ (x + 2y)² – (x – 2y)² = 8xy

[∵ (a + b)² = a² + 2ab + b² (a – b)² = a² – 2ab + b²]

False

The number of lines that can be drawn through a point is a variable.

Let the number be x.

Twice of the number = 2x.

Now, 1 more than 2x can be expressed as 2x + 1.

(a) A pattern of letter T as T

From the figure, it can be observed that it will required two matchsticks to make a T, 

Therefore the pattern is 2n

(b) A pattern of letter Z as Z

From the figure, it can be observed that it will require three matchsticks two make a Z there fore, the pattern is 3n.

(c) A pattern of letter U as U

From the figure, it can be observed that it will require there matchsticks to make a U.

 Therefore the pattern 3n

(d) A pattern of letter V as V

From the figure it can be observed that it will required two matchsticks to make, a V, 

Therefore, the pattern is 2n

(e) A pattern of letter E as E

From the figure, it can be observed that it will required five matchsticks to make an E. 

Therefore, the pattern in 5n.

(f) A pattern of letter S as S

From the figure, it can be observed that it will require five matchsticks to make a S therefore, the pattern is 5n

(g) A pattern of letter A as A

From the figure, it can be observed that it will required six matchsticks to make an A, therefore, the pattern is 6n.

If can be observed that in the given matchsticks pattern, the number of matchsticks are 4, 7, 10 and 13. which is 1 more than thrice of the numbers of square in the pattern. Hence, the pattern is 3n +1 Where n is the number of squares.

Let the present temperature be t° C.

Now, 20°C, less than t = (t – 20)°C

It can be observed that in the given matchsticks pattern, the number of matchstick pattern the number are 3, 5, 7 and 9. which is 1 more than twice of the number of ∆ les in the pattern

Hence, the pattern is 2n + 1 Where n is the number of ∆.les

Let the integer be n.

Now, successor of the integer n = n + 1.

Perimeter of a rectangle is twice the sum of its length and breadth.

We have given, side of the equilateral triangle = m.

The perimeter of an equilateral triangle = 3 × side = 3 m

The number of girls enrolled this year is 10 less than 3 times the girls enrolled last year.

Side of equilateral ∆ le = l

Perimeter = l + l + l + 3l

Taking -(5y² + 2y – 6) 

= 5y² + [(-)(+) 2y] + [(-) × (-)6]

= -5y² – 2y + 6

≠ -5y² – 2y + 6

∴ Correct answer is -5y² + 2y – 6 = -(5y² + 2y + 6)