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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