**Problem 1 :**

The sum of a number and its reciprocal is 65/8. Find the number

**Solution :**

Let x be the required number and 1/x will be its reciprocal

Sum of a number and its reciprocal = 65/8

x + (1/x) = 65/8

(x^{2} + 1)/x = 65/8

8(x^{2} + 1) = 65 x

8x^{2} - 65x + 8 = 0

By factoring the quadratic equation, we get

x-8 = 0 x = 8 |
8x-1 = 0 x = 1/8 |

So, the required number is 8.

**Problem 2 :**

The difference of the squares of two positive numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.

**Solution :**

Let x and y be two required numbers.

x^{2} – y^{2} = 45 ------- (1)

y^{2} = 4x ------- (2)

By applying (2) in (1), we get

x^{2} – 4x = 45

x^{2 }– 4x - 45 = 0

(x – 9) (x + 5) = 0

x-9 = 0 x = 9 |
x+5 = 0 x = -5 |

y^{2} = 4(9)

y^{2} = 36

y = √36

y = ± 6

Here positive 6 only admissible.

Therefore the required numbers are 9 and 6.

**Problem 3 :**

A farmer wishes to start a 100 sq.m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences the sides of the rectangular garden letting his house compound wall acts the fourth side fence. Find the dimension of the garden.

**Solution :**

Let x and y be the dimension of the vegetable garden.

Area of rectangle = Length ⋅ width

x ⋅ y = 100

x = 100/y

Quantity of barbed wire for fencing = 30 m

Using 30 m barbed wire, he is covering only three sides of the house. Compound wall is the fourth side.

x+x+y = 30

2x + y = 30

2(100/y) + y = 30

(200/y) + y = 30

(200 + y^{2})/y = 30

200 + y^{2} = 30y

y^{2 }- 30y + 200 = 0

(y – 10) (y – 20) = 0

y-10 = 0 y = 10 |
y-20 = 0 y = 20 |

By applying the values of y in x = 100/y, we get the values of x.

If y = 10 x = 100/10 x = 10 |
If y = 20 x = 100/20 x = 5 |

So,

length of garden = 10 m and

width of the garden = 5 m.

**Problem 4 :**

A rectangular field is 20 m long and 14 m wide. There is the path of equal width all around it having an area of 111 sq.m. Find the width of the path on the outside.

**Solution :**

Length = 20 m, width = 14 m

Let x be the uniform width all around the path

Length of the rectangular field including path

= 20+x+x

= 20+2x

Width of the rectangular field including path

= 14+x+x

= 14+2x

Area of path

= Area of rectangular field including path - Area of rectangular field

111 = (20 + 2x) (14 + 2x) - (20 x 14)

111 = 280 + 40 x + 28 x + 4 x² – 280

111 = 68x + 4x^{2}

4x^{2} + 68 x - 111 = 0

4x^{2} + 68 x - 111 = 0

2x+37 = 0 2x = -37 x = -37/2 |
2x-3 = 0 2x = 3 x = 3/2 x = 1.5 |

Therefore width of the path = 1.5 m.

**Problem 5 :**

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

**Solution :**

Let x be the usual speed of the train

Let T_{1} be the time taken to cover the distance 90 km in the speed x km/hr

Let T_{2} be the time taken to cover the distance 90 km in the speed (x+15) km/hr

Time = Distance/Speed

T_{1} = 90/x

T_{2} = 90/(x + 15)

By using the given condition

T_{1} - T_{2} = 30/60

(90/x) - (90/(x + 15)) = 1/2

Taking 90 commonly from two fractions

90 [(1/x) – 1/(x+15)] = 1/2

(x + 15 – x) / x(x + 15) = 1/180

15/(x^{2} + 15x) = 1/180

15 (180) = 1(x^{2} + 15x)

x^{2}+15x = 2700

x^{2}+15x–2700 = 0

(x - 45) (x + 60) = 0

x-45 = 0 x = 45 |
x+60 = 0 x = -60 |

Therefore speed of the train is 45 km/hr.

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