**Q1 : **The product of two consecutive numbers is 72. Find the numbers ?

**Solution:**

Let two consecutive numbers be x and x + 1.

Product of two consecutive numbers = 72.

x ( x +1) = 72.

x^{2} + x - 72 = 0

x^{2} + 9x - 8x - 72 = 0

x(x + 9) - 8(x + 9) = 0

(x + 9)(x - 8) = 0

x + 9 = 0 or x - 8 = 0

x = -9 or x = 8

If x = -9 then x + 1 = -9 + 1 = -8

If x = 8 or x + 1 = 8 + 1 = 9

Therefore the two consecutive numbers are -9, -8 or 8, 9.

**Q2:** The product of two consecutive even numbers is 24 . What are the numbers?

**Solution :** Let two consecutive even numbers be x and x + 2 .

Their product is 24.

x (x + 2) = 24

x^{2 }+ 2 x - 24 = 0

x^{2} + 6 x - 4x - 24 = 0

x(x + 6 ) - 4( x + 6) = 0

(x - 4 )( x + 6) = 0

x - 4 = 0 or x + 6 = 0

x = 4 or x = -6

If x = -6 then x + 2 = -6 +2 = -4

If x = 4 then x + 2 = 4 + 2 = 6.

Therefore the required numbers are -6 , -4 or 4, 6.

**Q3 :** The product of two consecutive odd numbers is 35. Find them .

**Solution :** Let the two consecutive odd numbers are x , x+ 2.

Their product = 35.

x(x+2) = 35.

x^{2} + 2x - 35 =0.

x^{2} + 7x - 5x - 35 = 0

x(x + 7) - 5 ( x + 7) =0

(x + 7 )( x - 5) = 0

x + 7 = 0 or x - 5 = 0

x = - 7 or x = 5

If x = -7 then x + 2 = -7 + 2 = -5.

If x = 5 then x + 2 = 5 + 2 = 7.

Therefore required numbers are -7, -5 or 5, 7.

**Q4: **The difference of two numbers is 3 and their product is 18. Find them.

**Solution :** Let two numbers be x and y .

Given x - y = 3...........(1)

xy = 18

y = 18/x ...................(2)

Substituting (2) in (1) we have,

x - 18/x = 3

x^{2} - 18 = 3x

x^{2} - 3x - 18 = 0

x^{2} -6x + 3x - 18 = 0

x ( x - 6) + 3 ( x - 6 ) = 0

( x - 6 )(x + 3) = 0

x - 6 = 0 or x + 3 = 0

x = 6 or x = -3

If x = 6 then y = 18/6 = 3.

If x = -3 then y = 18 /-3 = -6

Therefore required numbers are 3, 6 or -6, -3.

** ****Q5:** The sum of the squares of two consecutive even numbers is 20 . Find them.

**Solution:** Let x and x + 2 are two consecutive even numbers.

From given x^{2} + (x + 2)^{2} = 20

x^{2} + x^{2} + 4x + 4 = 20

2x^{2} + 4x + 4 - 20 = 0

2x^{2 } + 4x -16 =0

x^{2} +2x - 8 = 0

x^{2} +4x - 2x - 8 = 0

x(x + 4) -2( x + 4) = 0

(x + 4)(x - 2) =0

x = -4 or 2

If x = -4 then x + 2 = -4 +2 = -2

If x = 2 then x +2 = 2 + 2 = 4

Therefore required numbers are -4 , -2 or 2, 4 .

**Q6:** Find the number which is less than its square by 132.

** Solution : **Let the number be x.

From given

x = x^{2} - 132

x^{2} - x -132 =0

x^{2} -12x +11x -132 = 0

x( x - 12 ) + 11 ( x - 12) = 0

(x - 12)(x + 11) = 0

x - 12 =0 or x + 11 = 0

x = 12 or -11

**Q8:** The length of a hall exceeds its breadth by 8 m and its perimeter is 56m . Then find the dimensions of the hall.

**Solution :** Let the breadth of the hall be (b) = x m .

Then length(l) = ( x + 8 )m

Perimeter of a rectangular hall (P) = 2 (l + b)

P = 56 m

2 ( x + 8 + x ) = 56

2x +8 = 56/2

2(x + 4) = 28

x + 4 = 28/2

x = 14 - 4

x = 10

Length = x + 8 = 10 + 8 = 18 m, and breadth = x = 10 m.

Therefore the dimensions of rectangular hall is 18m and 10m .