1) Find three consecutive integers such that twice the first minus the third is 3 less
than the second.
Discussion:
Let x represent the smallest of the three integers.
Then the other two numbers may be represented as x + 1 and x + 2.
Now we can represent 3 less than the second as (x + 1) – 3.
We can also represent twice the first minus the third as 2x – (x + 2).
We are told that 3 less than the second is the first minus the third which is 2x – (x
+ 2)
We have two expressions for 3 less than the second so by the Transitive Law (of
equality) they must be equal.
This yields the equation (x + 1) – 3 = 2x – (x + 2).
This is the mathematical model and the question is answered by solving the
equation
(x + 1) – 3 = 2x – (x + 2)
x–2=x–2
x=x
This equation is clearly an identity.
Which means it is true for all real numbers.
Therefore we conclude that for ANY three consecutive integers, twice the first
minus the third is equal to three less than the second.
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2) One number is 5 times another number. If the sum of the two numbers is 270,
what are the numbers?
Discussion:
Let x represent the smallest of the two numbers.
Then the other number is 5x.
The sum of the two numbers is x + 5x
The sum of the two numbers is 270.
We now have two expressions for the sum of the two numbers.
By the Transitive Law (of equality) they must be equal.
This yields the equation x + 5x = 270.
This is the mathematical model and the question is answered by solving the
equation
x + 5x = 270
6x = 270
x = 45.
Therefore we conclude the smaller of the two numbers is 45 and the larger is 5(45)
= 225.
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3) The sides of a triangle are x, 5x and 6x – 3. If
the perimeter of the triangle is 483 ft. what are
the lengths of the sides of the triangle?
Discussion:
The perimeter of the triangle is x + 5x + (6x – 3).
The perimeter of the triangle is 483.
We now have two expressions for the perimeter of the triangle.
By the Transitive Law (of equality) they must be equal.
This yields the equation x + 5x + (6x – 3) = 483.
This is the mathematical model and the question is answered by solving the
equation
x + 5x + (6x – 3) = 483
12x = 486
x = 40.5
5x = 202.5
6x – 3 = 240
Therefore we conclude the lengths of the sides of the triangle are
40.5 ft., 202.5 ft., and 240 ft.
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4)
A cone has radius 3 feet and volume 36 cubic feet.
What is its height?
Discussion:
The relevant formula here is the formula V  1 r 2h for the
3
volume of a cone.
Substitute the given information to obtain 36  1 (3)2 h
3
which we must solve for h.
36  1 (3)2 h
3
36  3h
36  h
3
h  12
Therefore we conclude that the cone is 12 feet high.
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5) A can manufacturer has a contract to make cylindrical cans
with a radius of 3 inches and a volume of 15 cubic
inches. What should be the height of the cans?
Discussion: The relevant formula is the formula V  r 2h for
the volume of a cylinder.
Substitute the known quantities to obtain
15    3 h
h  15  5
9 3
2
Therefore we conclude the height must be 5 inches.
3
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6) The station fire in CA has burned more than 226 square miles. How many
acres have burned?
Discussion: As soon as we realize that there are 640 acres per square mile, we
recognize this problem as nothing but a conversion of units. We need to convert
square miles to acres.
If 1 square mile is 640 acres, then 226 square miles is (226)(640) = 144,640 acres.
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7) A koi pond measuring 5 feet by 7 feet is to be
surrounded by a concrete sidewalk of uniform
width. The total area of pool and sidewalk is
required to be 99 square feet. What should be
the width of the sidewalk?
Discussion: In this discussion, “structure” will refer
to the combined pond and sidewalk.
Let x be the width of the sidewalk.
Then the length of the structure is 2x + 7 and the
width of the structure is 2x + 5.
The area of the structure is (2x + 7)(2x + 5)
The area of the structure is 99.
We now have two expressions for the area of the
structure.
By the Transitive Law (of equality) they must be
equal.
This yields the equation (2x + 7)(2x + 5) = 99.
This is the mathematical model and the question is answered by solving the
equation
(2x + 7)(2x + 5) = 99
4x2 + 24x + 35 = 99
4x2 + 24x – 64 = 0
x2 + 6x – 16 = 0
(x + 8)(x – 2) = 0
By the Zero Factor Property
x + 8 = 0 or x – 2 = 0
x = – 8 or x = 2
Because x represents a distance it cannot be negative.
Therefore we conclude that the sidewalk should be 2 feet wide.
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8) A trough is 12 feet long, 3 feet
deep, and 3 feet wide. Find the
depth of water when the
trough contains 70 gallons of
water.
NOTE: 1 US Gallons =
0.13370 Cubic Feet so that
70 gal = 9.3576
Discussion: The relevant formula
is V = mwh (where m is the
length)
Let x represent the depth of water.
Then the volume of water in the trough is V = (12)(3)x cu. ft.
The volume of water in the trough is 9.3576 cu. ft.
We now have two expressions for the volume of water.
By the Transitive Law (of equality) they must be equal.
This yields the equation (12)(3)x = 9.3576.
This is the mathematical model and the question is answered by solving the
equation
(12)(3)x = 9.3576
x = 0.2599 ft.
Therefore we conclude the water is 0.2599 ft. or 3.12 inches deep.
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9) The diameter of a cylindrical propane gas
tank with flat ends is 6 feet. The total
volume of the tank is 603.2 cubic feet.
Find the length of the tank.
Discussion: The relevant formula is V  r 2h
Substitute the known quantities into the
formula to obtain 603.2 = (3.14)(6)2h .
Solve this equation for h.
h
603.2  5.34 ft.
(3.14)(36)
Therefore we conclude the tank is 5.34 feet long.
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10) The diameter of a cylindrical propane gas tank with
hemispherical ends is 6 feet. The total volume of
the tank is 360  cubic feet. Find the length of the
cylindrical part of the tank.
Discussion: The volume of the tank is the sum of the
volume of the cylindrical part and the two hemispherical ends.
The sum of the volume of the two hemispherical ends is equal to the volume of a
sphere with the same radius.
The two relevant formulas are the volume of a sphere V  4 r 3 and the volume of
3
2
a cylinder V  r h .
The formula for the volume of the tank is V  4 r 3  r 2h
3
A bit of factoring will make this formula simpler.

V  4 r 3  r 2h  r 2 4 r  h
3
3


So the formula we want to work with is V  r 2 4 r  h
3

Substitute known quantities into this formula to obtain
360  (3)2  4 (3)  h 
3

This is the mathematical model and the question will be answered by solving this
equation for h.
360  (3)2  4 (3)  h 
3

360  4  h

9 
40  4  h
h  36
Therefore we conclude that the cylindrical part of the tank is 36 feet long
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11) A rectangular parcel of land is 50 ft. wide. The
length of a diagonal between opposite corners is
10 ft. more than the length of the parcel. What
is the length of the parcel?
Discussion: The relevant formula is a2 + b2 = c2
from the Pythagorean Theorem.
Let x the length of the parcel.
Then the length of the diagonal is x + 10
Substitute into the formula to obtain (50)2 + x2 = (x+ 10)2
This is the model and the problem is answered by solving the equation
(50)2 + x2 = (x+ 10)2
2500 + x2 = x2 + 20x + 100
20x = 2400
x = 120
Therefore we conclude the parcel is 120 feet long.
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12) An item originally priced at $55 is marked 15% off for a sale. What is the sale
price?
Discussion: The relevant formula is A = PB where P is the percent (marked with
%), B is the base amount and A is the percentage.
Let x represent the sale price.
The sale price is the original price minus the discount (the percentage).
The sale price can be written algebraically as 55 – (55)(0.15)
We now have two expressions for the sale price, therefore according to the
Transitive Property of equality those two expressions are equal. This yields the
equation x = 55 – (55)(0.15) from which we can compute
x = 46.75
Therefore we conclude that the sale price is $46.75
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13) In a trapezoid the lengths of one of its two bases is one more than twice that of
the other. The length of the altitude is 2. If its area is 19, find the lengths of the
bases.
Discussion: The relevant formula is A  1  B  b  h for the area of a trapezoid.
2
Let b represent the length of the smaller base.
Then the length of the other base B is 2b + 1
The area of the trapezoid is 19
The area of the trapezoid is 1  (2b  1)  b   2    3b  1
2
We have two expressions for the area of the trapezoid. By the Transitive property
of equality, these two expressions must be equal. This observation yields the
equation 3b + 1 = 19 which is the model for this question.
3b + 1 = 19 implies b = 6 and then it follows that B = 13
Therefore we conclude that the bases of the trapezoid are 6 and 13.
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14) John has $20,000 to invest. He invests part of his money at an annual interest
rate of 6%, the rest at 9% annual rate. The return on these two investments over
one year is $1,440. How much did he invest at each rate?
Discussion: The relevant formula is A = PB where P is the percent (marked with
%), B is the base amount and A is the percentage.
Let x represent the amount invested at 6%.
Then 20000 – x is the amount invested at 9%.
The interest earned at 6% on the amount x is 0.06x.
The interest earned at 9% on the amount (20000 – x) is 0.09(20000 – x).
The total interest earned at the two rates is 0.06x + 0.09(20000 – x)
The total interest earned is 1440.
We have two expressions for the total interest earned. By the Transitive Property
of equality, these two expressions must be equal. This yields the equation
0.06x + 0.09(20000 – x) = 1440 which is the mathematical model for this question
and whose solution will answer the question.
0.06x + 0.09(20000 – x) = 1440
6x + 9(20000 – x) = 144000
6x + 180000 – 9x = 144000
–3x = –36000
x = 12000
We therefore conclude that John invested $12,000 at 6% and $8,000 at 9%.
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15) Suppose one painter Joe can paint the entire house in
twelve hours, and the second painter Fred takes eight
hours. How long would it take the two painters
working together to paint the house?
Discussion:
The key to these kinds of problems is to concentrate on
how much can be done in one unit of time.
Let x be the amount of time it takes Joe and Fred, working
together, to paint the house.
Joe can paint the house in 12 hours, so he paints 1 th of the house in one hour.
12
Fred can paint the house in 8 hours, so he paints 1 th of the house in one hour.
8
Working together they paint 1  1  2  3  5 th of the house in one hour.
12
8
24
24
24
Working together they paint 5 th of the house in one hour.
24
Working together they paint the house in x hours, so they paint 1 th of the house
x
in one hour.
The mathematical model for this work problem is the rational equation in
one variable 1  5 .
x
24
Ordinary methods now may be used to solve this equation.
1
5

multiply both sides of the equation by 24x
x 24
24 = 5x
multiply both sides of the equation by 1
5
x  24
5
Therefore we conclude that if Joe and Fred work together, they can paint the house
in 4 and 4 hours which is 4 hrs. and 48 min.
5
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16) The perimeter of a rectangle is 104 in. Three times the width is four less than
twice the length. What are it dimensions?
Discussion:
Let x represent the width of the rectangle.
Let y represent the length of the rectangle.
The perimeter of a rectangle is 2x + 2y.
The perimeter of the rectangle is 104.
We now have two expressions for the perimeter of the rectangle.
By the Transitive Law (of equality) they must be equal.
This yields the equation 2x + 2y = 104.
This equation contains two variables so we cannot solve for either of them.
However we are also told that three times the width is four less than twice the
length.
We can express that statement algebraically as 3x = 2y – 4.
By adding 4 to both sides we obtain the equation 2y = 3x + 4.
We now have the system of equations
 2x  2y  104
which can be solved by substitution

 2y  3x  4
x  20
 2x  2y  104
2x  (3x  4)  104
5x  4  104

x  20









2y  3x  4
 2y  3x  4

 2y  3x  4
2y  3(20)  4
 y  32
Therefore we conclude the rectangle is 32 inches long and 20 inches wide.
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