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Number of eggs for the first women be ‘x’ 

Let the selling price of each women be ‘y’ 

Selling price of one egg for the first women = y/(100 - x)

By the given condition 

(100 – x) (y/x) = 15 (for first women) 

y = 15/(100 - x) … (1) 

(x) x (y/(100 - x)) = 20/3 [For second women] 

y = 20(100 - x)/3x … (2) 

From (1) and (2) We get

15/(100 - x) = 20(100 - x)/3x

45x² = 20(100 – x)² 

(100 – x)² = 45x²/20 = (9/4)x² 

∴ 100 – x = √(9/4)x² 

100 – x = 3x/2 

3x = 2(100 – x) 

3x = 200 – 2x 

3x + 2x = 200 ⇒ 5x = 200 

x = 200/5 ⇒ x = 40 

Number of eggs with the first women = 40 

Number of eggs with the second women = (100 – 40) = 60

(i) 15x² + 11x + 2 = 0 comparing with ax² + bx + c = 0. 

Here a = 15, 6 = 11, c = 2. 

Δ = b² – 4ac 

= 11² - 4 x 15 x 2 

= 121 – 120 

= 1 > 1. 

∴ The roots are real and unequal.

(ii) x² – x – 1 = 0, 

Here a = 1, b = -1, c = -1.

Δ = b² – 4ac 

= (-1)² – 4 x 1 x -1 

= 1 + 4 = 5 > 0. 

∴ The roots are real and unequal.

(iii) √2t² – 3t + 3√2 = 0 

Here a = √2, b = -3, c = 3√2

Δ = b² – 4ac 

= (-3)² – 4 x √2 x 3√2 

= 9 – 24 = -15 < 0. 

∴ The roots are not real.

(iv) 9y² – 6√2y + 2 = 0 

a = 9, b = 6√2, c = 2 

Δ = b² – 4ac = 

(6√2)² – 4 x 9 x 2 

= 36 x 2 – 72 

= 72 – 72 = 0 

∴ The roots are real and equal.

(v) 9a² b² x² – 24abcdx + 16c² d² = 0

Δ = b² – 4ac 

= (-24abcd)² – 4 x 9a² b² x 16c² d² 

= 576a² b² c² d² – 576a² b² c² d² = 0 

∴ The roots are real and equal.

(1) straight line

(2) 1

(x + 2)² = (x + 2)(x + 2)

= x = -2, -2 = 1

Let the work done by Pari and Yuvan together be x 

Work done by part = 1/4

Work done by Yuvan = 1/6

By the given condition

1/4 + 1/6 = 1/x ⇒ (3 + 2)/12 = 1/x

5/12 = 1/x

5x = 12 ⇒ x = 12/5 

x = 2(2/5) hours (or) 2 hours 24 minutes

(3) \(-\frac{3}{2}\)

(2) x² – y² 

x⁴ – y⁴ = (x²)² – (y²)² = (x² + y²) (x² – y²)

x² – y² = x² – y² 

G.C.D. is = x² – y²

Answer is (3) 1

Answer is (4) a³ – b³

3x² – 48 = 3(x² – 16) = 3(x² – 4³) = 3(x + 4)(x – 4)

x² – 7x + 12 = x² – 3x – 4x + 12 = x(x – 3) – 4 (x – 3) = (x – 3) (x – 4)

Therefore, GCD = x – 4

y³ – 1 = (y – 1)(y³ + y + 1)

y – 1 = y – 1

Therefore, GCD = y – 1

(i) 7² x 3⁴ = (7 x 7) x (3 x 3 x 3 x 3)

= 49 x 81 

= 3969

(ii) 3² x 2⁴ = (3 x 3) x (2 x 2 x 2 x 2)

= 9 x 16 

= 144

(iii) 5² x 10⁴ = (5 x 5) x (10 x 10 x 10 x 10)

= 25 x 10000 

= 2,50,000

(i) 6³ or 3⁶

6³ = 6 x 6 x 6 

= 36 x 6 

= 216

3⁶ = 3 x 3 x 3 x 3 x 3 x 3 

= 729

729 > 216 gives 3⁶ > 6³

∴ 3⁶ is greater.

(ii) 5³ or 3⁵

5³ = 5 x 5 x 5 

= 125

3⁵ = 3 x 3 x 3 x 3 x 3 

= 243

243 > 125 gives 3⁵ > 5³

∴ 3⁵ is greater.

(iii) 2⁸ or 8²

2⁸ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 

= 256

8² = 8 x 8 

= 64

256 > 64 gives 2⁸ > 8²

∴ 2⁸ is greater.

(i) 5³ or 3⁵ 5³ = 5 x 5 x 5 

= 125

3⁵ = 3 x 3 x 3 x 3 x 3 

= 243

243 > 125 

∴ 3⁵ > 5³

(ii) 2⁸ or 8² 

2⁸ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 

= 256

8² = 8 x 8 

= 64

256 > 64 

∴ 2⁸ > 8²

(iii) 100² or 2¹⁰⁰⁰

We have 100² = 100 x 100 

= 10000

2¹⁰⁰ = (2¹⁰)¹⁰ 

= (2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2)¹⁰

= (1024)¹⁰ 

= [(1024)²]⁵

= (1024 x 1024)⁵ 

= (1048576)⁵

Since 1048576 > 10000

(1048576)⁵ > 10000

i.e., (1048576) > 100²

(2¹⁰)¹⁰ > 100²

2¹⁰⁰ > 100²

(iv) 2¹⁰ or 10²

We have 2¹⁰ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 

= 1024

10² = 10 x 10 

= 100

Since 1024 > 100

2¹⁰ > 10²

p(2√2) = (2√2)² - 2√2(2√2) + 1

= 4 x 2 – 4 x 2 + 1 

= 8 – 8 + 1 = 1

(i) At y = 1, 

f(1) = 6(1) – 3(1)² + 3 = 6 – 3 + 3 = 6 

(ii) At y = -1, 

f(-1) = 6(-1) – 3(-1)² + 3 = -6 – 3 + 3 = -6 

(iii) At y = 0, 

f(0) = 6(0) – 3(0)² + 3 = 0 – 0 + 3 = 3

(2x⁴ + 4x² – 3x + 7) – Q(x) = 3x³ – x² + 2x + 1

Q(x) = (2x⁴ + 4x² – 3x + 7) – 3x³ – x² + 2x + 1

The required polynomial = 2x⁴ + 4x² – 3x + 7 – 3x³ + x² – 2x – 1

= 2x⁴ – 3x³ + 5x² – 5x + 6

(i) x³ – 1

The terms of the given expression are x³, -1

Degree of each of the terms: 3,0

Terms with highest degree: x³.

Therefore, degree of the expression is 3.

(ii) 3x² + 2x + 1

The terms of the given expression are 3x², 2x, 1

Degree of each of the terms: 2, 1, 0

Terms with highest degree: 3x²

Therefore, degree of the expression is 2.

(iii) 3t⁴ – 5st² + 7s²t²

The terms of the given expression are 3t⁴, – 5st², 7s³t²

Degree of each of the terms: 4, 3, 5

Terms with highest degree: 7s²t²

Therefore, degree of the expression is 5.

(iv) 5 – 9y + 15y² – 6y³

The terms of the given expression are 5, – 9y, 15y², – 6y³

Degree of each of the terms: 0, 1, 2, 3

Terms with highest degree: – 6y³

Therefore, degree of the expression is 3.

(v) u⁵ + u⁴v + u³v² + u₂v³ + uv⁴

The terms of the given expression are u⁵, u⁴v, u³v², u²v³, uv⁴

Degree of each of the terms: 5, 5, 5, 5, 5

Terms with highest degree: u⁵, u⁴v, u³v², u²v³, uv⁴

Therefore, degree of the expression is 5.

As the graph intersects x-axis at one Point

.'. The number of zeroes of p(x) is 1.

Let the two consecutive odd numbers be x and x + 2 

∴ Their sum = 200 

x + (x + 2) = 200 

x + x + 2 = 200 

2x + 2 = 200 

2x + 2 – 2 = 200 – 2 [∵ Subtracting 2 from both sides] 

2x = 198

\(\frac{2x}{2}=\frac{198}{2}\) [Dividing both sides by 2] 

x = 99 

The numbers will be 99 and 99 + 2. 

∴ The numbers will be 99 and 101.

Let the distance travelled by taxi be ‘x’ km 

For the first 5 km the charge = Rs 100 

For additional kms the charge = Rs 16(x – 5) 

∴ For x kms the charge = 100 + 16(x – 5) 

Amount paid = Rs 740 

∴ 100 + 16(x – 5) = 740 

100 + 16(x – 5) – 100 = 740- 100 

16(x – 5) = 640

\(\frac{16(x-5)}{16}=\frac{640}{16}\)

x – 5 = 40 

x – 5 + 5 = 45 + 5 

x = 45 

x = 45 km 

∴ Total distance travelled = 45 km

Present weight = weight on last birthday + weight put on after a year

= 40 kg + m kg

= (40 + m) kg

We have given, length of the board = r cm and breadth = t cm.

(i) The length of aluminium strip to frame the board

= Perimeter of the board

= 2(length + breadth)

= 2(r + t) cm

But 10 cm extra strip is required to fix it properly.

∴ Total length of the aluminium strip = 2(r + t) cm + 10 cm.

(ii) Number of nails required to repair 1 board = x.

∴ Number of nails required to repair 15 boards = 15 × x = 15x.

(iii) Area of the cloth required for 1 board = Area of the board

= length × breadth

= r cm × t cm

= (rt) sq cm

Area of the cloth required for 8 boards

= 8 × (rt) sq cm

= 8rt sq cm

But 500 sq cm extra cloth per board is required to cover the edges.

∴ For 8 boards we need 8 × 500 sq cm

= 4000 sq cm extra cloth.

Thus, the total area of the cloth required

= (8rt + 4000) sq cm

(iv) The carpenter charges for 1 board = Rs. x.

∴ The carpenter charges for 23 boards

= Rs. (23 × x) = Rs. 23x

Let the present age of Sunita be x years.
The present age of her mother Geeta = 2(Sunita’s present age) = 2x years.

(i) After 4 years,

Sunita’s age = (x + 4) years

Geeta’s age = (2x + 4) years

(ii) Before 3 years,

Sunita’s age = (x – 3) years

Geeta’s age = (2x – 3) years.

Length = 3 × Breadth – 4

l = (3b – 4) metres

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

= (x + 1) (x² + 1)

x⁴ - 1 = (x²)² – 1

= (x² + 1) (x² – 1)

= (x² + 1) (x + 1) (x – 1)

H.C.F. = (x² + 1)(x + 1)

Let a number be x.

Its reciprocal is 1/x

x - 1/x = 24/5

(x² - 1)/x = 24/5

5x² – 5 - 24x = 0 

⇒ 5x² – 24x – 5 = 0

5x² – 25x + x – 5 = 0

5x(x – 5) + 1 (x – 5) = 0

(5x + 1)(x – 5) = 0

x = -1/5, 5

∴ The number is -1/5 or 5.

Answer is (2) 5

(1) x = 1, y = 2, z = 3

x + y – 3x = – 6 …(1) 

– 7y + 7z = 7 ...(2) 

3z = 9 ...(3) 

From (3) we get 

z = 9/3 = 3 

Substitute the value of z in (2) 

-7y + 7(3) = 7 

-7y = -14 

Substitute the value of y = 2 and z = 3 in (1) 

x + 2 – 3(3) = -6 

x + 2 – 9 = -6 

x = -6 + 7 

x = 1 

The value of x = 1, y = 2 and z = 3

Let the height of Empire State building be y.

∴ Height of Mount Everest = 20y.

Let the numerator of the fraction be x.

∴ The denominator of the fraction can be expressed as x + 1.

∴ The fraction = x/(x+1) .

Multiple of 5 can be expressed as 5n, where n is an integer.

Two consecutive even integers are 2m and 2m + 2, where m is any integer.