Solutions to Math Problems (1)

Detailed solutions to the maths problems are presented.

Solution to Problem 1:
The circumference of a circle is given by
C = 2 π r, where r is the radius of the circle.
Substitute C by 72 π to obtain the equation
72 π = 2 π r
Simplify and solve for r to obtain
r = 36

Solution to Problem 2:
Let L and W be the length and width of the garden. The statement "the length of a rectangular garden is 2 feet longer than 3 times its width" may be formulated by
L = 2 + 3 W
The formula for the perimeter is given by
P = 2 L + 2 W
Substitute P and L in the above equation by 100 and 2 + 3 W respectively to obtain
100 = 2(2 + 3 W) + 2 W
Solve for W and L
W = 12 and L = 2 + 3 W = 38.
Check that the perimeter of the rectangular garden is 100
P = 2 L + 2 W = 76 + 24 = 100

Solution to Problem 3:
Let x and y be the length and width of the field. The statement "a length 10 feet more than it is width" may be formulated by
x = 10 + y
The formula for the area is given by
A = x y
Substitute A and x in the above equation by 264 and 10 + y respectively to obtain
264 = (10 + y) y
Write the above equation in standard form as follows
y 2 + 10 y - 264 = 0
Solve the above equation for to obtain
y = 12 and y = - 22
Since y is the value of a width it must be positive. The dimensions of the field are given by
y = 12 and x = 10 + y = 22
As an exercise, check that the rectangular field has length 10 feet more than it is width and an area of 264.

Solution to Problem 4:
Let us use profit = revenue - costs to find a formula for the profit P
P = ( x 2 + 100 x ) - ( 240 x + 500 )
Substitute P in the above equation by 10,000 dollars to obtain
10000 = x 2 - 140 x - 500
Write the equation in standard form and solve for x
x 2 - 140 x - 10500 = 0
Solve the above equation for x
x = 194.10 and x = -54.01
The number of units to produce must be positive, so
x = 194.
As an exercise, check that for the above value of x the profit is approximately (because of the rounding) equal to 10,000.

Solution to Problem 5:
Let x be the side of the smaller square and y be the side of the larger square. The statement "A square has a side 5 centimeters shorter than the side of a second square" may be formulated by
x = y - 5
The area of the smaller square is equal to x 2 = (y - 5) 2 and the area of the larger square is equal to y 2
The statement "the area of the larger square is four times the area of the smaller square" may be formulated by
y 2 = 4 (y - 5) 2
Write the above equation so that the right side is equal to 0
y 2 - 4 (y - 5) 2 = 0
The left side is made up of the difference of two squares and can be easily factored as follows
[ y - 2 (y - 5) ] [ y + 2 (y - 5) ] = 0
Solve for y to find
y = 10 and y = 10 / 3.
We now use the equation x = y - 5 to find x
y = 10 and x = 5
For the second solution y = 10 / 3, x is negative and cannot be accepted as the length of the side of a square must be positive.

Solution to Problem 6:
Let the two numbers be a and b and use the sum and product to write two equations with two unknowns
a + b = 26 and a b = 165
Solve the first equation for b
b = 26 - a
Substitute b in the equation a b = 165 by 26 - a
a (26 - a) = 165
Write the above equation in standard form
- a 2 + 26 a - 165 = 0
Solve the above equation for a
a = 11 and a = 15.
Use b = 26 - a to find b
when a = 11 , b = 15 and when a = 15 , b = 11.
The two numbers are 11 and 15.

Solution to Problem 7:
Let x and y be the length and width of the rectangle. Using the formulas for the area and the perimeter, we can write two equations.
15 = x y and 16 = 2 x + 2 y
Solve the second equation for x
x = 8 - y
Substitute x in the equation 15 = x y by 8 - y to rewrite the equation as
15 = (8 - y) y
Solve for y to find
y = 3 and y = 5
Use x = 8 - y to find x
when y = 3 , x = 5 and when y = 5 , x = 3.
The dimensions of the rectangle are 3 and 5.

As an exercise, check that the perimeter of this rectangle is 16 and its area is 15.

Solution to Problem 8:
Let x and y be the two numbers such that x is larger than y. The statement "the larger number is four less than twice the smaller number" may be formulated by
x = 2y - 4

We use the sum of the two numbers to write a second equation.
x + y = 20
Substitute x by 2y - 4 in x + y = 20 to obtain
2y - 4 + y = 20
Solve for y to find
y = 8 and x = 2y - 4 = 12

Solution to Problem 9:
Let a and b be the two sides of the triangle such that a is longer than b. The statement "the hypotenuse of a right triangle is 2 centimeters more than the longer side of the triangle" may be fomulated by
h = a + 2 or a = h - 2
The statement "the shorter side of the triangle is 7 centimeters less than the longer side" may be formulated .
b = a - 7 or b = (h - 2) - 7 = h - 9
We now use Pythagora's theorem to write a third equation
h 2 = a 2 + b 2
Substitute a by h - 2 and b by h - 9 in the above equation to obtain an equation in one variable only.
h 2 = (h - 2) 2 + (h - 9) 2
Simplify and rewrite the above equation in standard form.
h 2 -22 h + 85 = 0
Solve for h.
h = 5 and h = 17.
Only the solution h = 17 gives a and b positive and it is the length of the hypotenuse of the triangle.
As an exercise, find a and b and see if a, b and h satisfies Pythagora's theorem.

More math problems with detailed solutions in this site.

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