Comparison Problem Solutions
9
1. A pipe that measured 59” was cut into two pieces. One piece was 15” shorter than the
other. Determine the lengths of the longer and shorter pieces of pipe.
2 Unknown quantities : Long piece of pipe, and Short piece of pipe.
Long : 𝒙
Total length : 59”
(Labels & Expressions)
Short : 𝒙– 𝟏𝟓
Equation : (𝒙) + (𝒙 − 𝟏𝟓) = 𝟓𝟗
(Unsimplified Math Equation)
2𝑥 − 15 = 59
+15 +15
2𝑥 = 74
(Solution to Equation)
2𝑥
2
=
74
2

𝒙 = 𝟑𝟕
Long pipe : 𝒙 = 𝟑𝟕"
(Solution to Problem)
Short pipe : 𝒙– 𝟏𝟓 = 37 − 15 = 𝟐𝟐"
Check : 𝟑𝟕 + 𝟐𝟐 = 𝟓𝟗
2. 35 ribbons were awarded in a midwestern music competition. The number of blue
ribbons awarded was 3 less than the number of white ribbons. The number of red ribbons
was 2 more than the number of white ribbons. Determine the number of ribbons awarded of
each color.
Number of ribbons : 35
3 Unknown quantities : Number of blue ribbons, # white ribbons, # red ribbons .
White : 𝒙
(Labels & Expressions)
Blue : 𝒙– 𝟑
Red : 𝒙 + 𝟐
Equation : (𝒙) + (𝒙 − 𝟑) + (𝒙 + 𝟐) = 𝟑𝟓
(Unsimplified Math Equation)
3𝑥 − 1 = 35
+1 +1
3𝑥 = 36
(Solution to Equation)
3𝑥
3
=
36
3

𝒙 = 𝟏𝟐
White : 𝒙
= 𝟏𝟐 white ribbons
Blue : 𝒙– 𝟑 = 12 − 3
=𝟗
Red : 𝒙 + 𝟐 = 12 + 2
= 𝟏𝟒 red ribbons
(Solution to Problem)
blue ribbons
1
Check : 𝟏𝟐 + 𝟗 + 𝟏𝟒 = 𝟑𝟓
Comparison Problem Solutions
3. The top running speed of a cheetah is double the top running speed of a jackal. The top
running speed of an elk is 11 miles per hour faster than that of a jackal. Though not actually
possible, if each of the animals ran at top speed for an hour, they could run a combined
distance of 167 miles. Determine the top running speed of each animal.
Combined speed : 167 mph
3 Unknown quantities : Cheetah speed, Jackal speed, Elk Speed .
Cheetah : 𝟐𝒙
Jackal
: 𝒙
Elk
: 𝒙 + 𝟏𝟏
(Labels & Expressions)
Equation : (𝟐𝒙) + (𝒙) + (𝒙 + 𝟏𝟏) = 𝟏𝟔𝟕
(Unsimplified Math Equation)
4𝑥 + 11 = 167
−11 −11
4𝑥 = 156
(Solution to Equation)
4𝑥
156
=
4

4
𝒙 = 𝟑𝟗
Cheetah : 𝟐𝒙
Jackal
: 𝒙
Elk
: 𝒙 + 𝟏𝟏
= 2(39)
= 𝟕𝟖 mph
= 𝟑𝟗 mph
= 39 + 11 = 𝟓𝟎 mph
(Solution to Problem)
Check : 𝟕𝟖 + 𝟑𝟗 + 𝟓𝟎 = 𝟏𝟔𝟕
4. A particular quadrilateral (four-sided figure) has a perimeter of 51”. The longest side (A)
is twice the length of the shortest side (C). Side B is three-halves the length of side C. Side
D is 7” longer than side C. Determine the dimensions of the quadrilateral (lengths of sides).
4 Unknown quantities : Lengths of Side A, Side B, Side C, and Side D . Perimeter : 51”
Side A
Side B
Side C
Side D
: 𝟐𝒙
𝟑
: 𝟐𝒙
: 𝒙
: 𝒙+𝟕
(Labels & Expressions)
𝟑
Equation : (𝟐𝒙) + (𝟐 𝒙) + (𝒙) + (𝒙 + 𝟕) = 𝟓𝟏
4𝑥 + 2 𝑥 + 7 = 51
3

8
11

11

𝑥=
2
2
11
𝑥 + 7 = 51
∙
11
2
𝑥 = 44 ∙
Side A
Side B
Side C
Side D
2
11
2
3
𝑥 + 2 𝑥 + 7 = 51
2
(Unsimplified Math Equation)


𝑥 + 7 − 7 = 51 − 7
44
11
∙2= 4∙2
: 𝟐𝒙 = 2(8)
= 𝟏𝟔 in.
𝟑
3
: 𝟐 𝒙 = 2 ∙ 8 = 3 ∙ 4 = 𝟏𝟐 in.
: 𝒙
= 𝟖 in.
: 𝒙+𝟕 =8+7
= 𝟏𝟓 in.
(Solution to Equation)

11
2
𝑥 = 44
𝒙=𝟖
(Solution to Problem)
Check : 𝟏𝟔 + 𝟏𝟐 + 𝟖 + 𝟏𝟓 = 𝟓𝟏
2
Comparison Problem Solutions
5. The length of a rectangular garden is 40 yards longer than double the width. The perimeter
of the garden is 410 yards. Determine length and width of the garden. (Be careful to use
only one variable. Length and width should be written as labels of unknowns).
2 Unknown quantities : Length and width of the garden . Perimeter : 410 yd .
Length
: 𝟐𝒙 + 𝟒𝟎
Width
: 𝒙
(Labels & Expressions)
𝑷 = 𝟐 × 𝑳𝒆𝒏𝒈𝒕𝒉 + 𝟐 × 𝑾𝒊𝒅𝒕𝒉
(Formula for Perimeter of a rectangle)
Equation : 𝟐 ∙ (𝟐𝒙 + 𝟒𝟎) + 𝟐 ∙ (𝒙) = 𝟒𝟏𝟎
(Unsimplified Math Equation)

4𝑥 + 80 + 2𝑥 = 410
6𝑥 + 80 − 80 = 410 − 80
6𝑥 + 80 = 410



6𝑥 = 330
Length
: 𝟐𝒙 + 𝟒𝟎 = 2(55) + 40 = 110 + 40 = 𝟏𝟓𝟎 yd.
Width
: 𝒙
= 𝟓𝟓 yd.
(Solution to Equation)
6𝑥
6
=
330
6

𝒙 = 𝟓𝟓
(Solution to Problem)
Check : 2 ∙ 150 + 2 ∙ 55 = 300 + 110 = 410
6. There are 3 times as many people in Fred’s psychology class as in his history class. There
are 13 more people in his algebra class than in his history class. The combined enrollment
of the three classes is 88 students. Determine the number of students in each class. (Note
that it would not make sense if one got a fraction for an answer.)
3 Unknown quantities : Enrollments in Psychology, History, and Algebra. Combined 𝟖 .
Psychology : 𝟑𝒙
(Labels & Expressions)
History
: 𝒙
Algebra
: 𝒙 + 𝟏𝟑
Equation : (𝟑𝒙) + (𝒙) + (𝒙 + 𝟏𝟑) = 𝟖𝟖

5𝑥 + 13 = 88
5𝑥 = 75

History
: 𝒙
Psychology : 𝟑𝒙
Algebra
(Unsimplified Math Equation)
5𝑥 + 13 − 13 = 88 − 13
5𝑥
5
=
75
5

(Solution to Equation)
𝒙 = 𝟏𝟓
= 𝟏𝟓 students
= 3(15)

(Solution to Problem)
= 𝟒𝟓 students
: 𝒙 + 𝟏𝟑 = 15 + 13 = 𝟐𝟖 students
3
Check : 𝟏𝟓 + 𝟒𝟓 + 𝟐𝟖 = 𝟖𝟖
Comparison Problem Solutions
7. A triangular support of a roof has a perimeter of 71 meters. The first side is two-thirds the
length of the second side. The third side is 9 meters shorter than the second side.
Determine the length of the three sides.
3 Unknown quantities : Lengths of : Side 1, Side 2, and Side 3. Combined length : 71 m
𝟐
𝒙
Side 1
:
Side 2
: 𝒙
Side 3
: 𝒙−𝟗
𝟑
(Labels & Expressions)
𝟐
Equation : (𝟑 𝒙) + (𝒙) + (𝒙 − 𝟗) = 𝟕𝟏
2
3
8
3
3
𝑥 + 3 𝑥 + 3 𝑥 − 9 = 71

8
𝑥 − 9 + 9 = 71 + 9

8

𝒙 = 𝟑𝟎
𝑥=
80
8
3
(Unsimplified Math Equation)
∙ 3 = 10 ∙ 3
𝟐
𝒙
Side 1
:
Side 2
: 𝒙
Side 3
: 𝒙−𝟗
𝟑
3
3
𝑥 − 9 = 71 
(Solution to Equation)
3 8

𝑥 = 80
3
∙ 𝑥 = 80 ∙ 8
8 3
2
= 3 ∙ 30 = 2 ∙ 10 = 𝟐𝟎 m

(Solution to Problem)
= 𝟑𝟎 m
= 30 − 9
= 𝟐𝟏 m
Check : 𝟐𝟎 + 𝟑𝟎 + 𝟐𝟏 = 𝟕𝟏
8. Laurie earns $2800 more per year than Don. Together they earn $72,600 per year.
Determine each of their incomes.
2 Unknown quantities : Laurie’s income, Don’s income. Combined income : $72,600 .
Laurie : 𝒙 + 𝟐𝟖𝟎𝟎
Don
(Labels & Expressions)
: 𝒙
Eq : (𝒙 + 𝟐𝟖𝟎𝟎) + (𝒙) = 𝟕𝟐, 𝟔𝟎𝟎
(Unsimplified Math Equation)
2𝑥 + 2800 = 72,600

2𝑥 + 2800 − 2800 = 72,600 − 2800 (Solution to Equation)


2𝑥
2𝑥 = 69,800
2
=
69,800
2

Laurie : 𝒙 + 𝟐𝟖𝟎𝟎 = $34,900 + $2800 = $𝟑𝟕, 𝟕𝟎𝟎
Don
: 𝒙
= $𝟑𝟒, 𝟗𝟎𝟎
𝒙 = 𝟑𝟒, 𝟗𝟎𝟎
(Solution to Problem)
Check : 𝟑𝟕, 𝟕𝟎𝟎 + 𝟑𝟒, 𝟗𝟎𝟎 = 𝟕𝟐, 𝟔𝟎𝟎
4
Comparison Problem Solutions
9. In a triangle, the first angle is 5 degrees more than the second angle. The third angle is 13
degrees less than twice the second angle. Determine each of the three angles. (Note we do not
need to give the combined measure of angles).
3 Unknown quantities : 1st Angle , 2nd Angle , and 3rd Angle . Total of Angles : 180°
1st Angle : 𝒙 + 𝟓
(Labels & Expressions)
2nd Angle : 𝒙
3rd Angle : 𝟐𝒙 − 𝟏𝟑
Equation : (𝒙 + 𝟓) + (𝒙) + (𝟐𝒙 − 𝟏𝟑) = 𝟏𝟖𝟎

4𝑥 − 8 = 180

4𝑥 = 188
(Unsimplified Math Equation)
4𝑥 − 8 + 8 = 180 + 8
4𝑥

4
=
1st Angle : 𝒙 + 𝟓
188
(Solution to Equation)

4
𝒙 = 𝟒𝟕
= 47 + 5 = 𝟓𝟐°
2nd Angle : 𝒙
(Solution to Problem)
= 𝟒𝟕°
3rd Angle : 𝟐𝒙 − 𝟏𝟑 = 2(47) − 13 = 94 − 13 = 𝟖𝟏°
Check : 𝟓𝟐 + 𝟒𝟕 + 𝟖𝟏 = 𝟏𝟖𝟎
10. Helen is two-thirds the age of Janice. Lilly is four years younger than Helen. Their
ages add up to 38 years. Determine the age of each.
3 Unknown quantities : Helen’s age, Janice’s age, and Lilly’s age. Total : 38 yrs
𝟐
Helen
:
Janice
: 𝒙
Lilly
:
𝟑
𝟐
𝟑
𝒙
(Labels & Expressions)
𝒙−𝟒
𝟐
𝟐
Equation : (𝟑 𝒙) + (𝒙) + (𝟑 𝒙 − 𝟒) = 𝟑𝟖
2
3
3
2
𝑥 + 3 𝑥 + 3 𝑥 − 4 = 38
7

𝑥=
3
42
7
𝑥 − 4 + 4 = 38 + 4

∙3= 6∙3
𝟐
Helen
:
Janice
: 𝒙
Lilly
:
𝟑
𝟐
𝟑
𝒙
2

7

7
3
3
(Unsimplified Math Equation)
𝑥 − 4 = 38
𝑥 = 42
(Solution to Equation)

3 7
3
∙ 𝑥 = 42 ∙ 7
7 3

𝒙 = 𝟏𝟖
= 3 (18) = 2 ∙
18
3
= 2 ∙ 6 = 𝟏𝟐 yrs
(Solution to Problem)
= 𝟏𝟖 yrs
2
𝒙 − 𝟒 = 3 (18) − 4 = 12 − 4
5
= 𝟖 yrs
Check : 𝟏𝟐 + 𝟏𝟖 + 𝟖 = 𝟑𝟖