Problem 1 :
The sum of a positive number and its square is 240. Find the number.
Solution :
Let x be the positive number.
x + x^{2} = 240
x^{2 }+ x - 240 = 0
Solve by factoring.
x^{2 }- 15x + 16x - 240 = 0
x(x^{ }- 15) + 16(x - 15) = 0
(x - 15)(x + 16) = 0
x - 15 = 0 or x + 16 = 0
x = 15 or x = -16
Since x is a positive number, it can not be negative .
So,
x = 15
Therefore, the number is 15.
Problem 2 :
Find two numbers whose sum is 27 and product is 182.
Solution :
Let x be one of the two numbers.
Then the other number is (27 - x).
Product of those numbers = 182
x(27 - x) = 182
27x - x^{2} = 182
x^{2} - 27x + 182 = 0
Solve by factoring.
x^{2} - 14x - 13x + 182 = 0
x(x - 14) - 13(x - 14) = 0
(x - 14)(x - 13) = 0
x - 14 = 0 or x - 13 = 0
x = 14 or x = 13
If x = 14,
= 27 - x
= 27 - 14
= 13
If x = 13,
= 27 - x
= 27 - 13
= 14
Therefore, the two numbers are 13 and 14.
Problem 3 :
Find two consecutive positive integers, sum of whose squares is 365.
Solution :
Let x and x + 1 be the two consecutive integers.
Sum of squares of two consecutive integers = 365
x^{2} + (x + 1)^{2} = 365
x^{2} + (x + 1)(x + 1) = 365
x^{2} + x^{2} + x + x + 1 = 365
2x^{2}^{ }+ 2x + 1 = 365
2x^{2} + 2x - 364 = 0
Divide both sides by 2.
x^{2} + x - 182 = 0
Solve by factoring.
x^{2} - 13x + 14x - 182 = 0
x(x - 13) + 14(x - 13) = 0
(x - 13)(x + 14) = 0
x - 13 = 0 or x + 14 = 0
x = 13 or x = -14
Since x is a positive integer, it can not be negative .
So,
x = 13
x + 1 = 14
Therefore, the two consecutive integers are 13 and 14.
Problem 4 :
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 5 :
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 6 :
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 7 :
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
Multiply both sides by 16.
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 8 :
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
Multiply both sides by 3.
4x^{2} = 576
Divide both sides by 4.
x^{2} = 144
x^{2} = 12^{2}
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.
Problem 9 :
The sum of the number and its positive square root is ⁷⁄₃₆. Find the number.
Solution :
Let x be the required number
The sum of the number and its positive square root is ⁷⁄₃₆.
x + √x = ⁷⁄₃₆
√x = ⁷⁄₃₆ - x
By taking squares on both sides, we get
x = (⁷⁄₃₆ - x)^{2}
x = (⁷⁄₃₆ - x)(⁷⁄₃₆ - x)
x = (⁷⁄₃₆)^{2} - (⁷⁄₃₆)x - (⁷⁄₃₆)x + x^{2}
x = (⁴⁹⁄₁₂₉₆) - 2(⁷⁄₃₆)x + x^{2}
x = (⁴⁹⁄₁₂₉₆) - (⁷⁄₁₈)x + x^{2}
Multiply both sides by 1296.
1296x = 49 - 504x + 1296x^{2}
1296x^{2} - 504x - 1296x + 49 = 0
1296x^{2} - 1800x + 49 = 0
Use quadratic formula and solve the above quadraticc equation.
When x = ⁴⁹⁄₃₆,
The given condition is not satisfied.
When x = ¹⁄₃₆,
The given condition is satisfied.
Therefore, the required number is ¹⁄₃₆.
Problem 10 :
The sum of two numbers is 15 and sum of their reciprocals is ³⁄₁₀. Find the numbers.
Solution :
Let x and y be the two numbers.
Given : The sum of two numbers is 15.
x + y = 15
y = 15 - x ----(1)
Given : The sum of their reciprocals is ³⁄₁₀.
By cross-multiplication,
10(x + y) = 3xy
10x + 10y = 3xy
Substitute y = 15 - x.
10x + 10(15 - x) = 3x(15 - x)
10x + 150 - 10x = 45x - 3x^{2}
3x^{2} - 45x + 150 = 0
Divide both sides by 3.
x^{2} - 15x + 50 = 0
Solve by factoring.
x^{2} - 10x - 5x + 50 = 0
x(x - 10) - 5(x - 10) = 0
(x - 10)(x - 5) = 0
x - 10 = 0 or x - 5 = 0
x = 10 or x = 5
When x = 10, (1)----> y = 15 - 10 y = 5 |
When x = 5, (1)----> y = 15 - 5 y = 10 |
x = 10 and y = 5
or
x = 5 and y = 10
Therefore, the two numbers are 5 and 10.
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