**Problem 1 :**

Find greatest common factor of the following terms

20x^{3}, 36x^{6}

**Solution :**

20 = 2^{2 }⋅ 5

36 = 2^{2 }⋅ 3^{2}

Greatest common factor = 2^{2 }⋅ 3^{2 }⋅ 5

= 180

So, the greatest common factor is 180x^{6}.

**Problem 2 :**

Simplify the following

**Solution :**

p^{2} - 1 = (p+1) (p-1)

= [(p+1) (p-1)/p]⋅ [p^{3}/(p-1)] ⋅ [1/(p+1)]

= p^{2}

**Problem 3 :**

Find the square root of 9801 by factor method.

**Solution :**

So, the square root of 9801 is 99.

**Problem 4 :**

Solve 6x^{2} + x - 1 = 0

**Solution :**

6x^{2} + x - 1 = 0

(3x - 1) (2x + 1) = 0

3x - 1 = 0 3x = 1 x = 1/3 |
2x + 1 = 0 2x = -1 x = -1/2 |

So the solution is {-1/2, 1/3}.

**Problem 5 :**

The product of two consecutive odd number is 323. Find them.

**Solution :**

Let x and x + 2 are two consecutive odd numbers.

Product of two consecutive odd numbers = 323

x (x + 2) = 323

x^{2} + 2x = 323

x^{2} + 2x - 323 = 0

(x - 17)(x + 19)

x = 17 and x = -19

So, the two odd numbers are 17 and 19.

**Problem 6 :**

Find two consecutive even integers whose product is 224.

**Solution :**

Let x and x + 2 are two consecutive even integers.

Product of even integers = 224

x(x + 2) = 224

x^{2} + 2x = 224

x^{2} + 2x - 224 = 0

(x + 16) (x - 14) = 0

x = -16 and x = 14

So, two consecutive even numbers are 14 and 16.

**Problem 7 :**

The length of the hall is 3 m more than its width. The numerical value of its area is equal to the numerical value of its perimeter. Find the length and width of the hall.

**Solution :**

Let x be the width of the hall

length = x + 3

Area of the hall = Perimeter of the hall

x(x+3) = 2(x+x+3)

x^{2} + 3x = 2(2x+3)

x^{2} + 3x - 4x - 6 = 0

x^{2} - 1x - 6 = 0

(x - 3)(x + 2) = 0

x = 3 and x = -2

x + 3 = 6

So, the width and length of the rectangle are 3 and 6 m.

**Problem 8 :**

Find the area of the sector of a circle with radius 4 cm and of angle 30°. Also, find the area of the corresponding major sector approximately (use π = 3.14).

**Solution :**

Area of major sector = πr^{2} - (θ/360)πr^{2}

= π4^{2} - (30/360)π4^{2}

= π4^{2}(1 - 1/12)

= π4^{2}(11/12)

= (176/12)(3.14)

= 46.05 cm^{2}

**Problem 9 :**

OACB is a quadrant of a circle with center O and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region.

**Solution :**

Area of shaded region = Area of quadrant - Area of triangle ODB

= πr^{2 }- (1/2) ⋅ base ⋅ height

= (22/7)(3.5)^{2 }- (1/2) ⋅ 3.5 ⋅ 2

= 38.5 - 3.5

= 35 cm^{2}

**Problem 10 :**

The wheels of a car are of diameter 80 cm each. Find how many complete revolutions does each wheel make in 10 minutes when the car is traveling at a speed of 66 km per hour.

**Solution :**

Speed of the car = 66 km/hr

1000 m = 1 km

100 cm = 1 m

66 km = 6600000 cm

66 km/hr = 6600000/60

= 110000 cm/min

Distance covered = Time (Speed)

= 10(110000)

= 1100000

Radius of the wheel = 40 cm

Number of revolutions

= Distance covered / Distance covered by 4 wheels

Distance covered by 4 wheels = 2πr

= 1100000 / [2 (3.14) ⋅ 40]

= 1100000/251.2

= 4379

So, each wheel has to revolve 4379 times.

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