Problem 1 :
The area of a rectangular field is 2000 m^{2} and its perimeter is 180 m. Find the length and width of the field.
Solution :
Let x and y be the length and width of the rectangular field respectively.
Given : Perimeter of the rectangle is 180 m.
2x + 2y = 180
Divide both sides by 2.
x + y = 90
Subtract x from both sides.
y = 90 - x
Given : Area of the rectangle is 2000 m^{2}.
length ⋅ width = 2000
x ⋅ y = 2000
Substitute y = 90 - x.
x(90 - x) = 2000
90x - x^{2} = 2000
x^{2} - 90x + 2000 = 0
(x - 50)(x - 40) = 0
x = 50 or x = 40
If x = 50,
y = 90 - 50
y = 40
If x = 40,
y = 90 - 40
y = 50
(x, y) = (50, 40) or (40, 50)
So, the length and width of the rectangle are 40 m and 50 m respectively.
Problem 2 :
A distributor of apple juice has 5000 bottles in the store that it wishes to distribute in a month. From experience, it is known that demand D is given by D = -2000p^{2} + 2000p + 17000. Find the price per bottle that will result zero inventory.
Solution :
Stock in the store is 5000 bottles.
If the inventory is zero, the demand has to be 5000 bottles.
To know the price for zero inventory, we have to substitute D = 5000 in the quadratic equation
D = -2000p^{2} + 2000p + 17000
5000 = -2000p^{2} + 2000p + 17000
2000p^{2} - 2000p - 12000 = 0
Divide both sides by 2000.
p^{2} - p - 6 = 0
(p - 3)(p + 2) = 0
p - 3 = 0 or p + 2 = 0
p = 3 or p = -2
p = -2 can not be accepted. Because, price can not be negative.
So, price per bottle that will result zero inventory is $3.
Problem 3 :
Two squares have sides p cm and (p + 5) cm. The sum of their squares is 625 sq.cm. Find the sides of the squares.
Solution :
According to the question, we have
p^{2} + (p + 5)^{2} = 625
p^{2} + p^{2} + 10p + 25 = 625
2p^{2} + 10p - 600 = 0
p^{2} + 5p - 300 = 0
(p - 15)(p + 20) = 0
p = 15 or p = -20
p = -20 can not be accepted. Because, the side of the square can not be negative.
If p = 15,
p + 5 = 15 + 5
p + 5 = 20
So, the sides of the square are 15 cm and 20 cm.
Problem 4 :
A ball is thrown upwards from a rooftop which is above from the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball h from the ground at time t seconds is given by, h = -16t^{2} + 64t + 80. How long will the ball take to hit the ground?
Solution :
When the ball hits the ground, height h = 0.
0 = -16t^{2} + 64t + 80
16t^{2} - 64t - 80 = 0
t^{2} - 4t - 5 = 0
(t - 5)(t + 1) = 0
t = 5 or t = -1
t = -1 can not be accepted. Because time can never be a negative value.
So, the ball will take 5 seconds to hit the ground.
Problem 5 :
A picture has a height that is 4/3 its width. It is to be enlarged to have an area of 192 square inches. What will be the dimensions of the enlargement?
Solution :
After enlargement, let x be the width of picture.
Then, the height of the picture is
= 4x/3
(From the given information, it is very clear that the picture is in the shape of rectangle. So, the given height can be considered as length of the rectangle)
Given : Area after enlargement is 192 sq. inches.
length ⋅ width = 192
(4x/3) ⋅ x = 192
4x^{2}/3 = 192
Multiply both sides by 3/4.
x^{2 }= 144
Take square root on both sides.
x = 12
So, the width is 12 inches.
Then, the height is
= 4(12)/3
= 16 inches
So, the dimensions of the picture are 12 inches and 16 inches.
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