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

(x2 + 1)/x  =  65/8

8(x2  + 1)  =  65 x

8x2 - 65x + 8  =  0

By factoring the quadratic equation, we get

 x-8  =  0 x  =  8 8x-1  =  0x  =  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.

x2 – y2  =  45   ------- (1)

y2  =  4x   ------- (2)

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

x2 – 4x  =  45

x– 4x - 45  =  0

(x – 9) (x + 5)  =  0

 x-9  =  0x  =  9 x+5  =  0x  =  -5

y2  =  4(9)

y2  =  36

y  =  36

y  =  ± 6

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

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 + y2)/y  =  30

200 + y2  =  30y

y- 30y + 200  =  0

(y – 10) (y – 20)  =  0

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

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

 If y  =  10x  =  100/10x  =  10 If y  =  20x  =  100/20x  =  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 + 4x2

4x2 + 68 x - 111  =  0

4x2 + 68 x - 111  =  0

 2x+37  =  02x  =  -37x  =  -37/2 2x-3  =  02x  =  3x  =  3/2x  =  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 T1 be the time taken to cover the distance 90 km in the speed x km/hr

Let T2 be the time taken to cover the distance 90 km in the speed (x+15) km/hr

Time  =  Distance/Speed

T1 = 90/x

T2 = 90/(x + 15)

By using the given condition

T1 - T2  =  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/(x2 + 15x)  =  1/180

15 (180)  =  1(x2 + 15x)

x2+15x  =  2700

x2+15x–2700  =  0

(x - 45) (x + 60)  =  0

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

Therefore speed of the train is 45 km/hr.

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