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(x – 2) represents that the person whose age is (x – 2) years, is 2 years younger to munnu

(X + 4) represents that the person, Whose age is (x + 4) Years, is 4 years elder to munnu

(3x+7) represents that the person whose age is (3x + 7) years, is elder to munnu and his age is 7 years more than three times of the age of munnu

2n may represent the number of students who like either football or some other game such as cricket Where as 11/2 represents the number of students who like Cricket, out of the total number of students who like football.

In future

After m years from now, sara’s age will be (y + h) years in past n years ago , sara’s age was (y – n) years

(y + 7) represents that the person, whose age is (y + 7) years, is 7 years elder to sara (y – 3) represents that the person, whose age is (y – 3) years, is 3 years younger to sara
(y + 41/2) represents that the person

Whose age is (y + 4(1/2)) years, is 4(1/2) years elder to sara (y – 2(1/2)) represents that the person

Whose age is (y – 2(1/2)) years, is 2(1/2) years younger to sara

(b) 20

Let number be x

half of number is x/2

Sum of number and it’s half is given by

x + x/2 = 30 [Multiplying by 2 on both sides]

2x + x = 30 x 2

3x = 60

x = 60/3 = 20

(c) 4

Product of LCM & HCF of 2 numbers is always product of the numbers. [this is property]

Product of LCM & HCF is given as 24.

∴ Product of the 2 nos. is 24

Given one number is 6.

Let other number be ‘x’

∴ 6 × x = 24

∴ x = 24/6 = 4

Origin (0, 0)

40°

Given angles are in the ratio 2 : 3 : 4

Let the angles be 2x, 3x & 4x

Since sum of the angles of a triangle is 180°,
We get

2x + 3x + 4x = 180

∴ 9x = 180

∴ x = 180/9 = 20°

∴ The angles are 2x = 2 x 20 = 40°

3x = 3 x 20 = 60°

4x = 4 x 20 = 80°

∴ Difference between greatest & smallest angle is

80° – 40° = 40°

The coordinates of the point in third quadrant are always negatives.

False

Coordinate of origin is (0, 0), not (1, 1). 

Hence – False

b = 9

Given equation is a + b = 23

a = 14

14 + b = 23

b = 23 – 14 = 9

b = 9

(i) 6x × 4 

= (6 × 4) (x) 

= 24x

(ii) -3x × 7y 

= (-3 × 7) (x × y) 

= -21xy

(iii) (-2m²) × (-5m)³ 

= -2m² × (-)³ (5³ (m)³) 

= -2m² × (-125m³)

= (-) × (-)(2 × 125)(m² × m³) 

= + 250 m⁵ 

= 250 m⁵

(iv) a³ × (-4a² b) 

= (-4) × (a³ × a²) × (b) 

= -4a⁵b

(v) (2p²q³) × (-9pq²) 

= (+) × (-) × (2 × 9) (p² × p(q³ × q²)) 

= -18p³q⁵

No, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables.

Eg. 2y² + 5y^-1 – 3 is a an algebraic expression. But not a polynomial.

(0, -5) point lies on y – axis.

True

(-9, 0) on x – axis, Y – coordinate is always zero.

∴ (-9, 0) lies on x – axis – True

2x + 6y + 9x – 2y

= 2x + 9x + 6y – 2y 

= (2 + 9)x + (6 – 2)y 

= 11x + 4y

Quantity of oil in the second tin = 3x² + 6x – 5 litres.

Quantity of oil added = x + 7 litres

∴ Total quantity of oil in the second tin

= (3x² + 6x – 5) + (x + 7) litres

= 3x² + (6x + x) + (-5 + 7)

= 3x² + (6 + 1)x + 2

= 3x² + 7x + 2 litres

Quantity of oil sold = x + 6 litres

∴ Quantity of oil left in the second tin = (3x² + 7x + 2) – (x² + 6)(3x² – x² ) + 7x + (2 – 6)

= (3 – 1)x² + 7x + (-4) 

= 2x² + 7x – 4

Quantity of oil left = 2x² + 7x – 4 litres

True

(-10, 20)

x = – 10,

y = 20

∴ (- 10, 20) lies in second quadrant – True

True

Let the number be ‘y’

Sum of number & two times that number is 48

Can be written as y + 2y = 48 – True

Let 3, 4 be M, when we mark, we find that it is a different point and not ‘M’

All points on the line X = 5 will have X – coordinate as 5.

Therefore, any point with X – coordinate as 5 will lie on X = 5 line.

Hence the points (5, – 10) & (5,20) will lie on X = 5

True

The points (1, 1), (2, 2), (3, 3) all satisfy the equation y = x

which is straight line. 

Hence, it is true

4, -4

x = 4 is a line parallel to the y – axis and
y = – 4 is a line parallel to the x – axis. The point of intersection is a point that lies on both lines & which should satisfy both the equations. Therefore, that point is (4, -4).

3 units, 25 units

With reference to given graph,

On the x – axis, 1cm = 3 units

y axis, 1cm = 25 units

Let \(\frac{4x+1}{(x-2)(x+1)}\) = \(\frac{A}{x-2}+\frac{B}{x+1} ...(1)\)

Multiply both sides by (x – 2) (x + 1) we get

4x + 1 = A(x + 1) + B(x – 2) … (2)

Put x = -1 in (2) we get

4(-1) + 1 = A(-1 + 1) + B(-1 – 2)

-4 + 1 = A(0) + B(-3)

-3 = B(-3)

B = \(\frac{-3}{-3}\) = 1

Put x = 2 in (2) we get

4(2) + 1 = A(2 + 1) + B(2 – 2)

8 + 1 = A(3) + B(0)

9 = 3A

A = 3

Using A = 3, B = 1 in (1) we get

\(\frac{4x+1}{(x-2)(x+1)}\) = \(\frac{3}{x-2}+\frac{1}{x+1} \)

Given that (n + 2)! = 60(n – 1)! 

(n + 2) (n + 1) n (n – 1)! = 60(n – 1)! 

Cancelling (n – 1)! we get, 

(n + 2)(n + 1)n = 60 

(n + 2)(n + 1)n = 5 x 4 x 3 

Both sides we consecutive product of integers 

∴ n = 3

Given that each number starts at 67, we need a five-digit number. So we have to fill only one’s place, 10’s place, and 100th place. From 0 to 9 there are 10 digits. In these digits, 6 and 7 should not be used as a repetition of digits is not allowed. Except for these two digits, we have 8 digits. Therefore one’s place can be filled by any of the 8 digits in 8 different ways. Now there are 7 digits are left.

Therefore 10’s place can be filled by any of the 7 digits in 7 different ways. Similarly, 100th place can be filled in 6 different ways. 

By multiplication principle, the number of telephone numbers constructed is 8 x 7 x 6 = 336.

(x + y)² = x² + y² + 2xy

12² = x² + y² + 2 × 14

144 = x² + y² + 28

x² + y² = 144 – 28

x² + y² = 116

(i) 18xy – 12yz 

= (2 × 3 × 3 × y × x) – (2 × 2 × 3 × y × z)

Taking out the common factors 2, 3, y, we get

= 2 × 3 × y (3x – 2z) = 6y (3x- 2z)

(ii) 9x⁵y³ + 6x³y² – 18x²y

= (3 × 3 × x² × x³ × y × y²) + (2 × 3 × x² × x × y × y)

Taking out the common factors 3, x², y, we get

= 3 × x² × y (3x³ y² + 2xy – 6)

= 3x²y (3x³ y² + 2xy – 6)

(iii) x(b – 2c) + y(b – 2c)

Taking out the binomial factor (b – 2c) from each term, we have

= (b – 2c)(x + y)

(iv) (ax + ay) + (bx + by)

Taking at ‘a’ from the first term and ‘b’ from the second term we have

(ax + ay) + (bx + by) = a (x + y) + b (x + y)

Now taking out the binomial factor (x + y) from each term

= (x + y)(a + b)

(v) 2x²(4x – 1) – 4x + 1

Taking out -1 from last two terms

2x² (4x – 1) – 4x + 1 = 2x² (4x – 1) – 1 (4x- 1)

Taking out the binomial factor 4x – 1, we get

= (4x – 1)(2x² – 1)

Speed of the car = (x + 30) km / hr.

Time = (y + 2) hours

Distance = Speed × time 

= (x + 30) (y + 2) 

= x(y + 2) + 30 (y + 2) 

= x (y + 2) + 30 (y + 2)

= (x) (y) + (x) (2) + (30) (y) + (30) (2)

= xy + 2x + 30y + 60

Distance covered = (xy + 2x + 30y + 60) km

The correct option is (B) 28p⁷.

The correct option is (C) 6x²y.

11

Given,

y – 9 = (- 5) + 7

y – 9 = 7 – 5 (re-arranging)

y – 9 = 2

∴ y = 2 + 9 = 11 (by transposition method)

The correct option is (A) 6mn.

(i) 4 terms

(ii) 1 term

(iii) 4 terms

(iv) 3 terms

7

Given,

x + 5 = 12

x = 12 – 5 = 7 (by transposition method)

Value of x is 7.

(i) 81a² – 121b²

= (9a)² – (11b)²

[∵ using a² – b² = (a + b)²]

= (9a + 11b) (9a – 11b)

(ii) x² + 8x + 16 

= x² + 2 × x × 4 + 4²

[∵ using a² + 2ab + b² = (a + b)²]

= (x + 4)² 

= (x + 4)(x + 4)

Length of the rectangle = y + 4

breadth of the rectangle = y – 3

Area of the rectangle = length × breadth

= (y + 4)(y – 3) 

= y² + (4 +(-3))y + (4)(-3)

= y² + y – 12

64 – x²

64 – x² = 8² – x²

This is of the form a² – b²

Comparing with a² – b² we have a = 8, b = x

a² – b² = (a + b)(a – b)

64 – x² = (8 + x)(8 – x)

7m(m – 5) + 1(5 – m)

7m(m – 5) + 1(5 – m) 

= 7m(m – 5) + (-1)(-5 + m)

= 7m(m – 5) – 1 (m – 5)

Taking out the common binomial factor (m – 5) = (m – 5)(7m – 1)

(i) 100 (x + y)² – 81 (a + b)²

= {10 (x + y)}² – {(a (a + b)}²

= {10 (x + y) + 9 (a + b)} {10 (x + y) – 9(a + b)}

= (10x + 10y + 9a – 9b)} (10x + 10y – 9a – 9b)

(ii) (x – 1)² – (x – 2)²

= {(x – 1 +(x – 2)} {(x – 1) – (x – 2)}

= (2x – 3) – (x – 1 – x + 2)

= (2x – 3) × 1 = 2x – 3

10x² + 15y²

10x² + 15y² = (2 × 5 × x × x) + (3 × 5 × y × y)

Taking out the common factor 5 we have

10x² + 15y² = 5(2x² + 3y²)

(x + 4)³

Comparing (x + 4)³ with (a + b)³, we have a = x and b = 4.

(a + b)³ = a³ + 3a²b + 3ab² + b³

(x + 4)³ = x³ + 3x²(4) + 3(x)(4)² + 4³

= x³ + 12x² + 48x + 64

Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)

∴ Volume of the cuboid = (l × b × h) units³

= (x + 2) (x – 1) (x – 3) units³

We have (x + a) (x + b) (x + c) = x³ + (a + b + c)

x² + (ab + bc+ ca)x + abc

∴ (x + 2)(x – 1) (x – 3) 

= x³ + (2 – 1 – 3)x² + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2)(-1) (-3)

x³ – 2x² + (-2 + 3 – 6)x + 6

Volume = x³ – 2x² – 5x + 6 units³

3y + 6

3y + 6 = 3 × y + 2 × 3

Taking out the common factor 3 from each term we get 3 (y + 2)

∴ 3y + 6 = 3(y + 2)

(i) (2x – 3) (2x + 5)

= (2x)² + (-3 + 5) (2x) + (-3) (5)

[∵ (x + a) (x + b) = x² + (a + b)x + ab]

= 2²x² + 2 × 2x + (-15)

= 4x² + 4x – 15

(ii) (y – 7) (y + 3)

= y² + (-7 + 3)y + (-7) (3)

[∵ (x + a)(x + b) = x² + (a + b)x + ab]

= y² – 4y + (-21) 

= y² – 4y – 21

(iii) 107 × 103

= (100 + 7) × (100+ 3)

= 100² + (7 + 3) × 100 +(7 × 3)

= 10000 + 10 × 100 + 21 

= 10000 + 1000 + 21 

= 11021

Given side of the cube = (x + 1) cm

Volume of the cube = (side)³ cubic units = (x + 1)³ cm³

We have (a + b)³ = (a³ + 3a²b + 3ab² + b³) cm³

(x + 1)³ = (x³ + 3x² (1) + 3x (1)² + 1³) cm³

Volume = (x³ + 3x² + 3x + 1) cm³