MATH 114 - QUANTITATIVE
REASONING - Ratios and Proportions
Question Bank - Set 4
Liberty University
Question 1
Question
Solve the proportion 3x−2
5x+8 =7
9for x.
Solution
Step 1: Cross multiply to get rid of the fractions.
(3x−2) ·9=7·(5x+ 8)
27x−18 = 35x+ 56
Step 2: Move all terms involving xto one side and constants to the other
side. 27x−35x= 56 + 18
−8x= 74
Step 3: Divide by −8 to solve for x.
x=74
−8=−37
4
Thus, the solution to the proportion is x=−37
4.
Question 2
Question
In a chemistry lab, a solution is made by mixing 3 parts of a 20
Solution
Step 1: Calculate the amount of acid in the first solution.
Let xbe the total volume of the 20
Step 2: Calculate the amount of acid in the second solution.
Let ybe the total volume of the 10
Step 3: Set up an equation based on the given mixture.
Since there are 3 parts of the 20
3(0.20x) : 5(0.10y) = c
Step 4: Solve for the concentration of the resulting solution.
We have:
3(0.20x) : 5(0.10y) = c
0.60x: 0.50y=c
1.2x:y=c
Thus, the concentration of the resulting solution is 1.2 times the amount
of acid in the 20
Question 3
Question
If xis directly proportional to yand yis inversely proportional to z, and x= 8
when y= 4 and z= 2, find xwhen y= 6 and z= 3.
Solution
Step 1: Write the proportionality relationships as equations. Let’s denote the
proportionality constants as k1and k2. The given relationships can be written
as: - x=k1y-y=k2
z
Step 2: Find the values of k1and k2using the given values when x= 8,
y= 4, and z= 2. Substitute the values into the equations: - 8 = k1×4
⇒k1= 2 - 4 = k2
2⇒k2= 8
Step 3: Use the found values of k1and k2to find xwhen y= 6 and z= 3.
Substitute k1= 2, k2= 8, y= 6, and z= 3 into the equations: - x= 2 ×6
⇒x= 12
Question 4
Question
A recipe calls for a mixture of ethanol and water in a ratio of 7:3. If you have
450 mL of the mixture, how many milliliters of water are in the mixture?
2
Solution
Step 1: Determine the total number of parts in the ratio. Let xrepresent the
constant of proportionality. The ratio of ethanol to water is 7 : 3, which means
the total ratio is 7 + 3 = 10 parts. Therefore, 450 mL = 10x.
Step 2: Solve for x. Divide both sides by 10 to solve for x:
x=450
10
x= 45
Step 3: Calculate the amount of water in the mixture. Since water is in a
ratio of 3 parts to 10 total parts, we need to find the amount of water represented
by 3 parts:
Water (mL) = 3 ×45
Water (mL) = 135
Therefore, there are 135 milliliters of water in the mixture.
Question 5
Question
Solve for xin the following proportion: 3x−4
2=x+5
4
Solution
To solve the proportion 3x−4
2=x+5
4, we can cross multiply to eliminate the
fractions. Then we can solve for x.
Step 1: Cross multiply to get rid of the fractions.
4(3x−4) = 2(x+ 5)
Step 2: Expand and simplify both sides of the equation.
12x−16 = 2x+ 10
Step 3: Move all terms involving xto one side of the equation.
12x−2x= 10 + 16
10x= 26
Step 4: Solve for xby dividing both sides by 10.
x=26
10
Step 5: Simplify the fraction.
x= 2.6
Therefore, the solution to the proportion is x= 2.6.
3
Question 6
Question
If 3 apples and 4 bananas cost
$
4.80, and 2 apples and 5 bananas cost
$
4.25,
what is the cost of one apple and one banana?
Solution
Let abe the cost of one apple and bbe the cost of one banana.
Step 1: Set up a system of equations based on the given information:
(3a+ 4b= 4.80
2a+ 5b= 4.25
Step 2: Solve the system of equations by elimination or substitution.
Multiplying the first equation by 2 and the second equation by 3, we get:
(6a+ 8b= 9.60
6a+ 15b= 12.75
Step 3: Subtract the equations to eliminate a:
7b= 3.15
Step 4: Solve for b:
b=3.15
7= 0.45
Step 5: Substitute b= 0.45 back into the first equation to find a:
3a+ 4(0.45) = 4.80
Step 6: Solve for a:
3a+ 1.80 = 4.80
3a= 3
a= 1
Therefore, the cost of one apple is
$
1 and the cost of one banana is
$
0.45.
Question 7
Question
Simplify the expression 3x2y3
5yz and express the result in simplest form.
4
Solution
Step 1: Rewrite the expression by canceling out common factors between the
numerator and the denominator.
3x2y3
5yz =3x·xy3
5·y·z
Step 2: Simplify the expression further by canceling out common factors.
=3x·xy3
5·y·z
Step 3: Perform the multiplication in the numerator and the denominator.
=3x2
5z
Step 4: Thus, the simplified expression is 3x2
5z.
Question 8
Question
If x:y= 3 : 5 and y:z= 4 : 7, find the ratio x:z.
Solution
To find the ratio x:z, we first need to find values for x,y, and z.
Step 1: Set up equations using the given ratios: Let x= 3a,y= 5a, and
y= 4b,z= 7bfor some constants aand b.
Step 2: Rewrite the given ratios in terms of x,y, and z:
x
y=3a
5a=3
5and y
z=5a
7b=4
7
Step 3: Solve for a:
3
5=5a
7b=⇒21a= 25a=⇒a=3
5b
Step 4: Calculate the ratio x:z:
x
z=3a
7b=3·3
5b
7b=9
35
Step 5: Therefore, the ratio x:zis 9 : 35 .
5
Question 9
Question
A recipe for a cake requires 3 cups of sugar for every 5 cups of flour. If you want
to make a cake using 15 cups of flour, how many cups of sugar should you use?
Solution
Step 1: Determine the ratio of sugar to flour in the recipe: Let xrepresent the
number of cups of sugar needed for 5 cups of flour. The ratio of sugar to flour
is 3
5=x
5. Cross multiplying gives 5x= 3 ·5 and solving for xyields x= 3 cups
of sugar for every 5 cups of flour.
Step 2: Find the amount of sugar needed for 15 cups of flour: Since the ratio
of sugar to flour is constant, we can set up a proportion to find the amount of
sugar needed for 15 cups of flour. Let yrepresent the number of cups of sugar
needed for 15 cups of flour. The proportion to solve is 3
5=y
15 .
Step 3: Solve the proportion for the unknown variable: Cross multiplying
gives 5y= 3 ·15. Solving for ygives y= 9.
Therefore, you should use 9 cups of sugar when making a cake using 15 cups
of flour to maintain the same sugar to flour ratio as in the original recipe.
Question 10
Question
Solve the proportion 2
x=3
5for x.
Solution
Step 1: Cross multiply to eliminate the fractions.
5·2 = x·3
10 = 3x
Step 2: Divide both sides by 3 to solve for x.
10
3=3x
3
10
3=x
Step 3: Simplify the fraction.
x=10
3
6
Question 11
Question
Sue and Jane are planning a road trip. Sue drove 270 miles in 4 hours while
Jane drove 405 miles in 6 hours. Assuming they both maintained a constant
speed, what is the ratio of Sue’s speed to Jane’s speed?
Solution
Step 1: Calculate Sue’s speed in miles per hour (mph). Step 2: Calculate Jane’s
speed in miles per hour (mph). Step 3: Find the ratio of Sue’s speed to Jane’s
speed.
Step 1: To find Sue’s speed, we divide the distance she drove by the time
it took her. Sue’s speed = 270 miles
4 hours = 67.5 mph.
Step 2: To find Jane’s speed, we divide the distance she drove by the time
it took her. Jane’s speed = 405 miles
6 hours = 67.5 mph.
Step 3: The ratio of Sue’s speed to Jane’s speed is given by Sue’s speed
Jane’s speed .
Ratio of speeds = 67.5
67.5= 1.
Therefore, the ratio of Sue’s speed to Jane’s speed is 1.
Question 12
Question
Sara has a mixture of nuts that contains 20
Solution
Let’s denote the weight of the new mixture as xkilograms. We can set up the
following proportion to solve for x:
0.20 ×8+2
8+2 = 0.40
Step 1: Simplify the left side of the proportion.
1.6+2
10 = 0.40
Step 2: Add the weights on the numerator.
3.6
10 = 0.40
Step 3: Multiply both sides by 10 to solve for x.
x= 10 ×0.40 = 4
Thus, the weight of the new mixture will be 4 kilograms .
7
Question 13
Question
Simplify the following expression and express the answer as a ratio of two inte-
gers: 3(2x+4)
4(3x−5) .
Solution
Step 1: Simplify the expression by multiplying out the terms in the numerator
and the denominator. 3(2x+ 4)
4(3x−5) =6x+ 12
12x−20
Step 2: Factor out a common factor from the numerator and the denomina-
tor. 6x+ 12
12x−20 =6(x+ 2)
4(3x−5)
Step 3: Simplify further by canceling out the common factor of 2 in the
numerator and the denominator.
6(x+ 2)
4(3x−5) =3(x+ 2)
2(3x−5)
Therefore, the simplified expression can be written as 3(x+2)
2(3x−5) .
Question 14
Question
If 15 men can build a wall in 10 days, how many days will it take for 20 men to
build the same wall?
Solution
Let’s denote the number of days it will take for 20 men to build the wall as x.
Step 1: Find the man-days required to build the wall with 15 men in 10
days. The man-days required can be calculated as the product of the number
of men and the number of days:
15 men ×10 days = 150 man-days
Step 2: Use the man-days to find the number of days for 20 men to build
the wall. Since the amount of work is the same, we can set up a proportion
using the man-days:
15 men ×10 days
150 man-days =20 men ×xdays
20xman-days
8
Step 3: Solve for x.150
150 =20x
20
1 = x
Step 4: State the answer. It will take 20 men 10 days to build the wall.
Question 15
Question
If 4 pounds of apples cost
$
6, how many pounds of apples can be purchased for
$
15?
Solution
Step 1: Determine the cost per pound of apples.
Let xrepresent the number of pounds of apples that can be purchased for
$
15.
We can set up a proportion to find the cost per pound:
6
4=15
x
Step 2: Solve for x.
Cross-multiply to solve for x:
6x= 4 ×15
6x= 60
x=60
6
x= 10
Step 3: Answer
Therefore, 10 pounds of apples can be purchased for
$
15.
Question 16
Question
A certain chemical compound is made up of three elements: carbon, hydrogen,
and oxygen. The ratio of the number of atoms of carbon to hydrogen to oxygen
in the compound is 2 : 3 : 1. If the compound contains a total of 108 atoms,
how many atoms of each element are present in the compound?
9
Solution
Step 1: Let’s denote the number of atoms of carbon, hydrogen, and oxygen in
the compound as 2x, 3x, and x, respectively, where xis a positive integer.
Step 2: Since the compound contains a total of 108 atoms, we can set up an
equation based on the ratio of the atoms:
2x+ 3x+x= 108
Step 3: Simplifying the equation, we get:
6x= 108
Step 4: Solving for x:
x=108
6= 18
Step 5: Now, we can find the number of atoms of carbon, hydrogen, and
oxygen: - atoms of carbon: 2x= 2 ×18 = 36 - atoms of hydrogen: 3x=
3×18 = 54 - atoms of oxygen: x= 18
Step 6: Therefore, there are 36 atoms of carbon, 54 atoms of hydrogen, and
18 atoms of oxygen in the compound.
Question 17
Question
If xand yare in a ratio of 3 : 5 and yand zare in a ratio of 2 : 7, what is the
ratio of xto z?
Solution
Let’s set up the given ratios and work towards finding the ratio of xto z.
Step 1: Express xand yin terms of a common variable. Let x= 3kand
y= 5k, where kis a constant.
Step 2: Express yand zin terms of a common variable. Since yand zare
in a ratio of 2 : 7, let y= 2mand z= 7m, where mis another constant.
Step 3: Find the ratio of xto z. Substitute the expressions for xand zinto
the ratio we’re trying to find:
x
z=3k
7m=3
7·k
m
Therefore, the ratio of xto zis 3 : 7 .
Question 18
Question
Solve for x:2x+5
3x−7=3x+8
4x−9.
10
Solution
Step 1: Cross multiply to get rid of the fractions.
2x+ 5
3x−7=3x+ 8
4x−9⇒(2x+ 5)(4x−9) = (3x+ 8)(3x−7)
Step 2: Expand both sides of the equation.
8x2−18x+ 20x−45 = 9x2+ 24x−21x−56
Step 3: Simplify and combine like terms.
8x2+ 2x−45 = 9x2+ 3x−56
Step 4: Move all terms to one side of the equation to set it equal to zero.
0 = x2+x−11
Step 5: Factor the quadratic equation.
0=(x+ 4)(x−3)
Step 6: Set each factor equal to zero and solve for x.
x+ 4 = 0 ⇒x=−4 or x−3=0⇒x= 3
Therefore, the solutions to the equation are x=−4,3 .
Question 19
Question
Solve for x:3x+ 1
2x−1=x+ 2
4.
Solution
Step 1: Cross multiply to eliminate the fractions.
(3x+ 1) ·4=(x+ 2) ·(2x−1)
12x+ 4 = 2x2−x+ 4
Step 2: Rearrange the equation to set it equal to zero.
0 = 2x2−x+ 4 −12x−4
0 = 2x2−13x
Step 3: Factor out xfrom the equation.
0 = x(2x−13)
11
Step 4: Set each factor equal to zero and solve for x.
x= 0
2x−13 = 0
2x= 13
x=13
2
Thus, the solutions for xare x= 0 and x=13
2.
Question 20
Question
A flagpole casts a shadow that is 20 meters long. At the same time, a 5-meter-
tall statue standing next to the flagpole casts a shadow that is 4 meters long.
If the flagpole and the statue are standing on level ground, what is the height
of the flagpole?
Solution
Step 1: Let hbe the height of the flagpole. Since the flagpole is standing
vertically, we can set up the ratio of the height of the flagpole to the length of
its shadow as follows: h
20 =5
4
Step 2: Now, cross multiply to solve for h:
4h= 5 ×20
Step 3: Simplify the right side of the equation:
4h= 100
Step 4: Divide both sides by 4 to solve for h:
h=100
4= 25
Therefore, the height of the flagpole is 25 meters.
Question 21
Question
A group of students decided to share a box of 100 chocolates in the ratio 3:5:7.
How many chocolates should the student who gets the smallest share give to
the student who gets the largest share so that they all have an equal number of
chocolates?
12
Solution
Step 1: Find the total number of parts the chocolates are divided into. The
chocolates are divided into 3 + 5 + 7 = 15 parts.
Step 2: Find the number of chocolates in each part. Each part is equal to
100
15 =20
3chocolates.
Step 3: Calculate the number of chocolates each student should have. The
student who gets the smallest share originally receives 3 parts, which is 3×20
3=
20 chocolates. The student who gets the largest share originally receives 7 parts,
which is 7 ×20
3=140
3chocolates.
Step 4: Calculate the number of chocolates needed to make their shares
equal. The difference in the number of chocolates between the student who gets
the smallest share and the student who gets the largest share is 140
3−20 = 80
3
chocolates.
Step 5: Divide the extra chocolates equally between the students. To make
their shares equal, the student who gets the smallest share should give 80
3∇ · 2 =
40
3chocolates to the student who gets the largest share.
Question 22
Question
Solve the following proportion for x:
2
x=x+ 4
6
Solution
To solve this proportion for x, we will first cross multiply to eliminate the
fractions.
Step 1: Cross multiply to get rid of the fractions.
2·6 = x·(x+ 4)
12 = x2+ 4x
Step 2: Rearrange the equation to set it equal to zero.
x2+ 4x−12 = 0
Step 3: Solve the quadratic equation using the quadratic formula x=
−b±√b2−4ac
2a, where a= 1, b= 4, and c=−12.
x=−4±p42−4(1)(−12)
2(1)
x=−4±√16 + 48
2
13
x=−4±√64
2
x=−4±8
2
Step 4: Solve for xby finding the two potential solutions.
x1=−4+8
2= 2
x2=−4−8
2=−6
Therefore, the solutions to the proportion are x= 2 or x=−6.
Question 23
Question
Given that a
b=3
4and b
c=5
6, find the value of a
c.
Solution
Step 1: First, we need to find a common term between the two given ratios. To
do this, we can find the value of b.
Step 2: To find the value of b, we can set up a proportion using the given
ratios: a
b=3
4and b
c=5
6.
Step 3: Cross-multiplying, we get 4a= 3bfrom the first ratio and 6b= 5c
from the second ratio.
Step 4: Solving for bin the first equation gives us b=4a
3.
Step 5: Substituting b=4a
3into the second equation gives 6 4a
3= 5c.
Step 6: Simplifying, we get 8a= 5c.
Step 7: Therefore, a
c=8
5, so the value of a
cis 8
5.
Question 24
Question
Solve the following proportion for x:
2x+ 1
3x−4=5
7
14
Solution
Step 1: Cross multiply to eliminate the fractions. Step 2: Simplify the equation
by expanding both sides. Step 3: Move all the terms to one side of the equation
to solve for x. Step 4: Solve for xby isolating the variable. Step 5: Check
the solution by substituting the value of xback into the original proportion to
ensure it satisfies the equation.
Step 1: Cross multiply to eliminate the fractions.
(2x+ 1) ×7=5×(3x−4)
Step 2: Simplify the equation by expanding both sides.
14x+ 7 = 15x−20
Step 3: Move all the terms to one side of the equation to solve for x.
14x+ 7 = 15x−20
14x−15x=−20 −7
−x=−27
Step 4: Solve for xby isolating the variable.
x=−27
−1
x= 27
Step 5: Check the solution by substituting the value of xback into the
original proportion.
2(27) + 1
3(27) −4=55
79 =5
7
Therefore, the solution to the proportion is x= 27.
Question 25
Question
Solve the following proportion: 3x−2
7=4x+ 5
11 .
Solution
Step 1: Cross multiply to eliminate the fractions.
11(3x−2) = 7(4x+ 5)
33x−22 = 28x+ 35
15
Step 2: Rearrange the equation to isolate xon one side.
33x−28x= 35 + 22
5x= 57
Step 3: Solve for xby dividing both sides by 5.
x=57
5
x= 11.4
Therefore, the solution to the proportion is x= 11.4.
Question 26
Question
If x+3
2x−5=5
7, find the value of x.
Solution
Step 1: Cross-multiply to get rid of the fractions:
7(x+ 3) = 5(2x−5)
7x+ 21 = 10x−25
Step 2: Rearrange the equation to solve for x:
7x+ 21 = 10x−25
21 + 25 = 10x−7x
46 = 3x
Step 3: Solve for x:
3x= 46
x=46
3
x=46
3×1
1
x=46
3
Hence, the value of xis 46
3.
Question 27
Question
In a class of 50 students, the ratio of male students to female students is 3:2. If
8 more male students join the class, what will be the new ratio of male students
to female students?
16
Solution
Step 1: Determine the number of male and female students in the class.
Let the number of male students be 3xand the number of female students be
2x, where xis a constant.
Given that 3x+ 2x= 50 (total number of students in the class)
Solving for x: 5x= 50
x= 10
So, there are 3(10) = 30 male students and 2(10) = 20 female students in the
class.
Step 2: Find the new ratio after 8 more male students join the class.
After 8 more male students join the class, the total number of male students
will be 30 + 8 = 38. The number of female students remains the same at 20.
Therefore, the new ratio of male students to female students is 38 : 20.
To simplify the ratio, divide both sides by 2: 38∇ · 2 : 20∇ · 2
So, the new ratio is 19 : 10.
Question 28
Question
Solve the proportion: 3
x+2 =5
2x−1.
Solution
Step 1: Cross multiply to eliminate the fractions:
3(2x−1) = 5(x+ 2)
6x−3 = 5x+ 10
Step 2: Simplify the equation:
6x−3=5x+ 10
6x−5x= 10 + 3
x= 13
Step 3: Check the solution: Substitute x= 13 back into the original propor-
tion: 3
13 + 2 =5
2(13) −1
3
15 =5
26
1
5=5
26
Therefore, the solution x= 13 satisfies the proportion.
17
Question 29
Question
Solve the following proportion: 3x
1−x=1
2.
Solution
Step 1: Cross multiply to eliminate the fractions.
3x(2) = 1(1 −x)
6x= 1 −x
Step 2: Simplify the equation by adding xto both sides.
6x+x= 1
7x= 1
Step 3: Solve for xby dividing both sides by 7.
x=1
7
Step 4: Check the solution by substituting x=1
7back into the original
proportion.
31
7
1−1
7=1
2
3
7·7
6=1
2
1 = 1 ✓
Therefore, the solution to the proportion is x=1
7.
Question 30
Question
Solve the proportion: 3x
2=7
5.
Solution
Step 1: Cross multiply to solve the proportion.
3x·5=2·7
15x= 14
18
Step 2: Divide both sides by 15 to solve for x.
x=14
15
x=14
15
Therefore, the solution to the proportion is x=14
15 .
Question 31
Question
A group of students are planning a trip to a conference. If 15 students decide
to go on the trip, the cost per student would be
$
300. However, if 25 students
decide to go on the trip, the cost per student would be
$
250. What is the total
cost of the trip and how much would each student pay if 20 students decide to
go on the trip?
Solution
Step 1: Let xbe the total cost of the trip. We can set up two equations based
on the given information:
(15x= 4500
25x= 6250
Step 2: Solve the system of equations to find the total cost of the trip:
(x=4500
15 = 300
x=6250
25 = 250
Step 3: Since both equations give different values for x, we can see that
the scenario where 15 students decide to go on the trip does not align with the
scenario where 25 students decide to go. This implies that the cost per student
is not directly proportional to the number of students.
Step 4: To find the total cost and cost per student if 20 students decide to
go on the trip, we can calculate the average cost per student based on the two
scenarios: 300 + 250
2=550
2= 275
Thus, if 20 students decide to go on the trip, the total cost would be 20 ×
275 = $5500 and each student would pay
$
275.
19
Question 32
Question
A university is planning to build a new library with an estimated cost of
$
10
million. The university wants to fund 40
Solution
Step 1: Calculate the amount the university still needs to raise after receiving
the donation.
Amount still needed = Total cost −Donation
Amount still needed = $10,000,000 −$2,000,000
Amount still needed = $8,000,000
Step 2: Calculate the amount to be financed through a bond issuance. Since
the university wants to fund 40
Amount to be financed through bond issuance = 60% ×Amount still needed
Amount to be financed through bond issuance = 60% ×$8,000,000
Amount to be financed through bond issuance = 0.60 ×$8,000,000
Amount to be financed through bond issuance = $4,800,000
Therefore, the university should plan to raise
$
4.8 million through a bond
issuance to cover the remaining cost of the new library.
Question 33
Question
Solve for xin the following proportion: 3x+5
2x−1=7
4.
Solution
Step 1: Cross multiply to eliminate the denominators:
(3x+ 5) ·4=7·(2x−1)
12x+ 20 = 14x−7
Step 2: Rearrange the equation to group like terms:
12x+ 20 = 14x−7
12x−14x=−7−20
−2x=−27
20
Step 3: Solve for xby dividing both sides by −2:
x=−27
−2
x= 13.5
Therefore, the solution to the proportion is x= 13.5.
Question 34
Question
If x:y= 4 : 7 and y:z= 5 : 9, find the ratio x:z.
Solution
To find the ratio x:z, we need to first find a common value for yin both ratios.
Step 1: Find a common value for y. Given:
x:y= 4 : 7 and y:z= 5 : 9
Let’s assume that y=k. Then, from the first ratio x:y= 4 : 7, we have:
x
y=4
7
x
k=4
7
x=4
7k
Similarly, from the second ratio y:z= 5 : 9, we have:
y
z=5
9
k
z=5
9
z=9
5k
Step 2: Find the ratio x:z. Substitute the expressions for xand zin terms
of kinto the ratio x:z:
x:z=4
7k:9
5k
x:z=4
7·5
9
x:z=20
63
Therefore, the ratio x:zis 20 : 63 .
21
Solution
Step 1: Calculate the amount of acid in the first solution.
Let xbe the total volume of the 20
Step 2: Calculate the amount of acid in the second solution.
Let ybe the total volume of the 10
Step 3: Set up an equation based on the given mixture.
Since there are 3 parts of the 20
3(0.20x) : 5(0.10y) = c
Step 4: Solve for the concentration of the resulting solution.
We have:
3(0.20x) : 5(0.10y) = c
0.60x: 0.50y=c
1.2x:y=c
Thus, the concentration of the resulting solution is 1.2 times the amount
of acid in the 20
Question 3
Question
If xis directly proportional to yand yis inversely proportional to z, and x= 8
when y= 4 and z= 2, find xwhen y= 6 and z= 3.
Solution
Step 1: Write the proportionality relationships as equations. Let’s denote the
proportionality constants as k1and k2. The given relationships can be written
as: - x=k1y-y=k2
z
Step 2: Find the values of k1and k2using the given values when x= 8,
y= 4, and z= 2. Substitute the values into the equations: - 8 = k1×4
⇒k1= 2 - 4 = k2
2⇒k2= 8
Step 3: Use the found values of k1and k2to find xwhen y= 6 and z= 3.
Substitute k1= 2, k2= 8, y= 6, and z= 3 into the equations: - x= 2 ×6
⇒x= 12
Question 4
Question
A recipe calls for a mixture of ethanol and water in a ratio of 7:3. If you have
450 mL of the mixture, how many milliliters of water are in the mixture?
2
Solution
Step 1: Determine the total number of parts in the ratio. Let xrepresent the
constant of proportionality. The ratio of ethanol to water is 7 : 3, which means
the total ratio is 7 + 3 = 10 parts. Therefore, 450 mL = 10x.
Step 2: Solve for x. Divide both sides by 10 to solve for x:
x=450
10
x= 45
Step 3: Calculate the amount of water in the mixture. Since water is in a
ratio of 3 parts to 10 total parts, we need to find the amount of water represented
by 3 parts:
Water (mL) = 3 ×45
Water (mL) = 135
Therefore, there are 135 milliliters of water in the mixture.
Question 5
Question
Solve for xin the following proportion: 3x−4
2=x+5
4
Solution
To solve the proportion 3x−4
2=x+5
4, we can cross multiply to eliminate the
fractions. Then we can solve for x.
Step 1: Cross multiply to get rid of the fractions.
4(3x−4) = 2(x+ 5)
Step 2: Expand and simplify both sides of the equation.
12x−16 = 2x+ 10
Step 3: Move all terms involving xto one side of the equation.
12x−2x= 10 + 16
10x= 26
Step 4: Solve for xby dividing both sides by 10.
x=26
10
Step 5: Simplify the fraction.
x= 2.6
Therefore, the solution to the proportion is x= 2.6.
3
Question 6
Question
If 3 apples and 4 bananas cost
$
4.80, and 2 apples and 5 bananas cost
$
4.25,
what is the cost of one apple and one banana?
Solution
Let abe the cost of one apple and bbe the cost of one banana.
Step 1: Set up a system of equations based on the given information:
(3a+ 4b= 4.80
2a+ 5b= 4.25
Step 2: Solve the system of equations by elimination or substitution.
Multiplying the first equation by 2 and the second equation by 3, we get:
(6a+ 8b= 9.60
6a+ 15b= 12.75
Step 3: Subtract the equations to eliminate a:
7b= 3.15
Step 4: Solve for b:
b=3.15
7= 0.45
Step 5: Substitute b= 0.45 back into the first equation to find a:
3a+ 4(0.45) = 4.80
Step 6: Solve for a:
3a+ 1.80 = 4.80
3a= 3
a= 1
Therefore, the cost of one apple is
$
1 and the cost of one banana is
$
0.45.
Question 7
Question
Simplify the expression 3x2y3
5yz and express the result in simplest form.
4
Solution
Step 1: Rewrite the expression by canceling out common factors between the
numerator and the denominator.
3x2y3
5yz =3x·xy3
5·y·z
Step 2: Simplify the expression further by canceling out common factors.
=3x·xy3
5·y·z
Step 3: Perform the multiplication in the numerator and the denominator.
=3x2
5z
Step 4: Thus, the simplified expression is 3x2
5z.
Question 8
Question
If x:y= 3 : 5 and y:z= 4 : 7, find the ratio x:z.
Solution
To find the ratio x:z, we first need to find values for x,y, and z.
Step 1: Set up equations using the given ratios: Let x= 3a,y= 5a, and
y= 4b,z= 7bfor some constants aand b.
Step 2: Rewrite the given ratios in terms of x,y, and z:
x
y=3a
5a=3
5and y
z=5a
7b=4
7
Step 3: Solve for a:
3
5=5a
7b=⇒21a= 25a=⇒a=3
5b
Step 4: Calculate the ratio x:z:
x
z=3a
7b=3·3
5b
7b=9
35
Step 5: Therefore, the ratio x:zis 9 : 35 .
5
Question 9
Question
A recipe for a cake requires 3 cups of sugar for every 5 cups of flour. If you want
to make a cake using 15 cups of flour, how many cups of sugar should you use?
Solution
Step 1: Determine the ratio of sugar to flour in the recipe: Let xrepresent the
number of cups of sugar needed for 5 cups of flour. The ratio of sugar to flour
is 3
5=x
5. Cross multiplying gives 5x= 3 ·5 and solving for xyields x= 3 cups
of sugar for every 5 cups of flour.
Step 2: Find the amount of sugar needed for 15 cups of flour: Since the ratio
of sugar to flour is constant, we can set up a proportion to find the amount of
sugar needed for 15 cups of flour. Let yrepresent the number of cups of sugar
needed for 15 cups of flour. The proportion to solve is 3
5=y
15 .
Step 3: Solve the proportion for the unknown variable: Cross multiplying
gives 5y= 3 ·15. Solving for ygives y= 9.
Therefore, you should use 9 cups of sugar when making a cake using 15 cups
of flour to maintain the same sugar to flour ratio as in the original recipe.
Question 10
Question
Solve the proportion 2
x=3
5for x.
Solution
Step 1: Cross multiply to eliminate the fractions.
5·2 = x·3
10 = 3x
Step 2: Divide both sides by 3 to solve for x.
10
3=3x
3
10
3=x
Step 3: Simplify the fraction.
x=10
3
6
Question 11
Question
Sue and Jane are planning a road trip. Sue drove 270 miles in 4 hours while
Jane drove 405 miles in 6 hours. Assuming they both maintained a constant
speed, what is the ratio of Sue’s speed to Jane’s speed?
Solution
Step 1: Calculate Sue’s speed in miles per hour (mph). Step 2: Calculate Jane’s
speed in miles per hour (mph). Step 3: Find the ratio of Sue’s speed to Jane’s
speed.
Step 1: To find Sue’s speed, we divide the distance she drove by the time
it took her. Sue’s speed = 270 miles
4 hours = 67.5 mph.
Step 2: To find Jane’s speed, we divide the distance she drove by the time
it took her. Jane’s speed = 405 miles
6 hours = 67.5 mph.
Step 3: The ratio of Sue’s speed to Jane’s speed is given by Sue’s speed
Jane’s speed .
Ratio of speeds = 67.5
67.5= 1.
Therefore, the ratio of Sue’s speed to Jane’s speed is 1.
Question 12
Question
Sara has a mixture of nuts that contains 20
Solution
Let’s denote the weight of the new mixture as xkilograms. We can set up the
following proportion to solve for x:
0.20 ×8+2
8+2 = 0.40
Step 1: Simplify the left side of the proportion.
1.6+2
10 = 0.40
Step 2: Add the weights on the numerator.
3.6
10 = 0.40
Step 3: Multiply both sides by 10 to solve for x.
x= 10 ×0.40 = 4
Thus, the weight of the new mixture will be 4 kilograms .
7
Question 13
Question
Simplify the following expression and express the answer as a ratio of two inte-
gers: 3(2x+4)
4(3x−5) .
Solution
Step 1: Simplify the expression by multiplying out the terms in the numerator
and the denominator. 3(2x+ 4)
4(3x−5) =6x+ 12
12x−20
Step 2: Factor out a common factor from the numerator and the denomina-
tor. 6x+ 12
12x−20 =6(x+ 2)
4(3x−5)
Step 3: Simplify further by canceling out the common factor of 2 in the
numerator and the denominator.
6(x+ 2)
4(3x−5) =3(x+ 2)
2(3x−5)
Therefore, the simplified expression can be written as 3(x+2)
2(3x−5) .
Question 14
Question
If 15 men can build a wall in 10 days, how many days will it take for 20 men to
build the same wall?
Solution
Let’s denote the number of days it will take for 20 men to build the wall as x.
Step 1: Find the man-days required to build the wall with 15 men in 10
days. The man-days required can be calculated as the product of the number
of men and the number of days:
15 men ×10 days = 150 man-days
Step 2: Use the man-days to find the number of days for 20 men to build
the wall. Since the amount of work is the same, we can set up a proportion
using the man-days:
15 men ×10 days
150 man-days =20 men ×xdays
20xman-days
8
Step 3: Solve for x.150
150 =20x
20
1 = x
Step 4: State the answer. It will take 20 men 10 days to build the wall.
Question 15
Question
If 4 pounds of apples cost
$
6, how many pounds of apples can be purchased for
$
15?
Solution
Step 1: Determine the cost per pound of apples.
Let xrepresent the number of pounds of apples that can be purchased for
$
15.
We can set up a proportion to find the cost per pound:
6
4=15
x
Step 2: Solve for x.
Cross-multiply to solve for x:
6x= 4 ×15
6x= 60
x=60
6
x= 10
Step 3: Answer
Therefore, 10 pounds of apples can be purchased for
$
15.
Question 16
Question
A certain chemical compound is made up of three elements: carbon, hydrogen,
and oxygen. The ratio of the number of atoms of carbon to hydrogen to oxygen
in the compound is 2 : 3 : 1. If the compound contains a total of 108 atoms,
how many atoms of each element are present in the compound?
9
Solution
Step 1: Let’s denote the number of atoms of carbon, hydrogen, and oxygen in
the compound as 2x, 3x, and x, respectively, where xis a positive integer.
Step 2: Since the compound contains a total of 108 atoms, we can set up an
equation based on the ratio of the atoms:
2x+ 3x+x= 108
Step 3: Simplifying the equation, we get:
6x= 108
Step 4: Solving for x:
x=108
6= 18
Step 5: Now, we can find the number of atoms of carbon, hydrogen, and
oxygen: - atoms of carbon: 2x= 2 ×18 = 36 - atoms of hydrogen: 3x=
3×18 = 54 - atoms of oxygen: x= 18
Step 6: Therefore, there are 36 atoms of carbon, 54 atoms of hydrogen, and
18 atoms of oxygen in the compound.
Question 17
Question
If xand yare in a ratio of 3 : 5 and yand zare in a ratio of 2 : 7, what is the
ratio of xto z?
Solution
Let’s set up the given ratios and work towards finding the ratio of xto z.
Step 1: Express xand yin terms of a common variable. Let x= 3kand
y= 5k, where kis a constant.
Step 2: Express yand zin terms of a common variable. Since yand zare
in a ratio of 2 : 7, let y= 2mand z= 7m, where mis another constant.
Step 3: Find the ratio of xto z. Substitute the expressions for xand zinto
the ratio we’re trying to find:
x
z=3k
7m=3
7·k
m
Therefore, the ratio of xto zis 3 : 7 .
Question 18
Question
Solve for x:2x+5
3x−7=3x+8
4x−9.
10
Solution
Step 1: Cross multiply to get rid of the fractions.
2x+ 5
3x−7=3x+ 8
4x−9⇒(2x+ 5)(4x−9) = (3x+ 8)(3x−7)
Step 2: Expand both sides of the equation.
8x2−18x+ 20x−45 = 9x2+ 24x−21x−56
Step 3: Simplify and combine like terms.
8x2+ 2x−45 = 9x2+ 3x−56
Step 4: Move all terms to one side of the equation to set it equal to zero.
0 = x2+x−11
Step 5: Factor the quadratic equation.
0=(x+ 4)(x−3)
Step 6: Set each factor equal to zero and solve for x.
x+ 4 = 0 ⇒x=−4 or x−3=0⇒x= 3
Therefore, the solutions to the equation are x=−4,3 .
Question 19
Question
Solve for x:3x+ 1
2x−1=x+ 2
4.
Solution
Step 1: Cross multiply to eliminate the fractions.
(3x+ 1) ·4=(x+ 2) ·(2x−1)
12x+ 4 = 2x2−x+ 4
Step 2: Rearrange the equation to set it equal to zero.
0 = 2x2−x+ 4 −12x−4
0 = 2x2−13x
Step 3: Factor out xfrom the equation.
0 = x(2x−13)
11
Step 4: Set each factor equal to zero and solve for x.
x= 0
2x−13 = 0
2x= 13
x=13
2
Thus, the solutions for xare x= 0 and x=13
2.
Question 20
Question
A flagpole casts a shadow that is 20 meters long. At the same time, a 5-meter-
tall statue standing next to the flagpole casts a shadow that is 4 meters long.
If the flagpole and the statue are standing on level ground, what is the height
of the flagpole?
Solution
Step 1: Let hbe the height of the flagpole. Since the flagpole is standing
vertically, we can set up the ratio of the height of the flagpole to the length of
its shadow as follows: h
20 =5
4
Step 2: Now, cross multiply to solve for h:
4h= 5 ×20
Step 3: Simplify the right side of the equation:
4h= 100
Step 4: Divide both sides by 4 to solve for h:
h=100
4= 25
Therefore, the height of the flagpole is 25 meters.
Question 21
Question
A group of students decided to share a box of 100 chocolates in the ratio 3:5:7.
How many chocolates should the student who gets the smallest share give to
the student who gets the largest share so that they all have an equal number of
chocolates?
12
Solution
Step 1: Find the total number of parts the chocolates are divided into. The
chocolates are divided into 3 + 5 + 7 = 15 parts.
Step 2: Find the number of chocolates in each part. Each part is equal to
100
15 =20
3chocolates.
Step 3: Calculate the number of chocolates each student should have. The
student who gets the smallest share originally receives 3 parts, which is 3×20
3=
20 chocolates. The student who gets the largest share originally receives 7 parts,
which is 7 ×20
3=140
3chocolates.
Step 4: Calculate the number of chocolates needed to make their shares
equal. The difference in the number of chocolates between the student who gets
the smallest share and the student who gets the largest share is 140
3−20 = 80
3
chocolates.
Step 5: Divide the extra chocolates equally between the students. To make
their shares equal, the student who gets the smallest share should give 80
3∇ · 2 =
40
3chocolates to the student who gets the largest share.
Question 22
Question
Solve the following proportion for x:
2
x=x+ 4
6
Solution
To solve this proportion for x, we will first cross multiply to eliminate the
fractions.
Step 1: Cross multiply to get rid of the fractions.
2·6 = x·(x+ 4)
12 = x2+ 4x
Step 2: Rearrange the equation to set it equal to zero.
x2+ 4x−12 = 0
Step 3: Solve the quadratic equation using the quadratic formula x=
−b±√b2−4ac
2a, where a= 1, b= 4, and c=−12.
x=−4±p42−4(1)(−12)
2(1)
x=−4±√16 + 48
2
13
x=−4±√64
2
x=−4±8
2
Step 4: Solve for xby finding the two potential solutions.
x1=−4+8
2= 2
x2=−4−8
2=−6
Therefore, the solutions to the proportion are x= 2 or x=−6.
Question 23
Question
Given that a
b=3
4and b
c=5
6, find the value of a
c.
Solution
Step 1: First, we need to find a common term between the two given ratios. To
do this, we can find the value of b.
Step 2: To find the value of b, we can set up a proportion using the given
ratios: a
b=3
4and b
c=5
6.
Step 3: Cross-multiplying, we get 4a= 3bfrom the first ratio and 6b= 5c
from the second ratio.
Step 4: Solving for bin the first equation gives us b=4a
3.
Step 5: Substituting b=4a
3into the second equation gives 6 4a
3= 5c.
Step 6: Simplifying, we get 8a= 5c.
Step 7: Therefore, a
c=8
5, so the value of a
cis 8
5.
Question 24
Question
Solve the following proportion for x:
2x+ 1
3x−4=5
7
14
Solution
Step 1: Cross multiply to eliminate the fractions. Step 2: Simplify the equation
by expanding both sides. Step 3: Move all the terms to one side of the equation
to solve for x. Step 4: Solve for xby isolating the variable. Step 5: Check
the solution by substituting the value of xback into the original proportion to
ensure it satisfies the equation.
Step 1: Cross multiply to eliminate the fractions.
(2x+ 1) ×7=5×(3x−4)
Step 2: Simplify the equation by expanding both sides.
14x+ 7 = 15x−20
Step 3: Move all the terms to one side of the equation to solve for x.
14x+ 7 = 15x−20
14x−15x=−20 −7
−x=−27
Step 4: Solve for xby isolating the variable.
x=−27
−1
x= 27
Step 5: Check the solution by substituting the value of xback into the
original proportion.
2(27) + 1
3(27) −4=55
79 =5
7
Therefore, the solution to the proportion is x= 27.
Question 25
Question
Solve the following proportion: 3x−2
7=4x+ 5
11 .
Solution
Step 1: Cross multiply to eliminate the fractions.
11(3x−2) = 7(4x+ 5)
33x−22 = 28x+ 35
15
Step 2: Rearrange the equation to isolate xon one side.
33x−28x= 35 + 22
5x= 57
Step 3: Solve for xby dividing both sides by 5.
x=57
5
x= 11.4
Therefore, the solution to the proportion is x= 11.4.
Question 26
Question
If x+3
2x−5=5
7, find the value of x.
Solution
Step 1: Cross-multiply to get rid of the fractions:
7(x+ 3) = 5(2x−5)
7x+ 21 = 10x−25
Step 2: Rearrange the equation to solve for x:
7x+ 21 = 10x−25
21 + 25 = 10x−7x
46 = 3x
Step 3: Solve for x:
3x= 46
x=46
3
x=46
3×1
1
x=46
3
Hence, the value of xis 46
3.
Question 27
Question
In a class of 50 students, the ratio of male students to female students is 3:2. If
8 more male students join the class, what will be the new ratio of male students
to female students?
16
Solution
Step 1: Determine the number of male and female students in the class.
Let the number of male students be 3xand the number of female students be
2x, where xis a constant.
Given that 3x+ 2x= 50 (total number of students in the class)
Solving for x: 5x= 50
x= 10
So, there are 3(10) = 30 male students and 2(10) = 20 female students in the
class.
Step 2: Find the new ratio after 8 more male students join the class.
After 8 more male students join the class, the total number of male students
will be 30 + 8 = 38. The number of female students remains the same at 20.
Therefore, the new ratio of male students to female students is 38 : 20.
To simplify the ratio, divide both sides by 2: 38∇ · 2 : 20∇ · 2
So, the new ratio is 19 : 10.
Question 28
Question
Solve the proportion: 3
x+2 =5
2x−1.
Solution
Step 1: Cross multiply to eliminate the fractions:
3(2x−1) = 5(x+ 2)
6x−3 = 5x+ 10
Step 2: Simplify the equation:
6x−3=5x+ 10
6x−5x= 10 + 3
x= 13
Step 3: Check the solution: Substitute x= 13 back into the original propor-
tion: 3
13 + 2 =5
2(13) −1
3
15 =5
26
1
5=5
26
Therefore, the solution x= 13 satisfies the proportion.
17
Question 29
Question
Solve the following proportion: 3x
1−x=1
2.
Solution
Step 1: Cross multiply to eliminate the fractions.
3x(2) = 1(1 −x)
6x= 1 −x
Step 2: Simplify the equation by adding xto both sides.
6x+x= 1
7x= 1
Step 3: Solve for xby dividing both sides by 7.
x=1
7
Step 4: Check the solution by substituting x=1
7back into the original
proportion.
31
7
1−1
7=1
2
3
7·7
6=1
2
1 = 1 ✓
Therefore, the solution to the proportion is x=1
7.
Question 30
Question
Solve the proportion: 3x
2=7
5.
Solution
Step 1: Cross multiply to solve the proportion.
3x·5=2·7
15x= 14
18
Step 2: Divide both sides by 15 to solve for x.
x=14
15
x=14
15
Therefore, the solution to the proportion is x=14
15 .
Question 31
Question
A group of students are planning a trip to a conference. If 15 students decide
to go on the trip, the cost per student would be
$
300. However, if 25 students
decide to go on the trip, the cost per student would be
$
250. What is the total
cost of the trip and how much would each student pay if 20 students decide to
go on the trip?
Solution
Step 1: Let xbe the total cost of the trip. We can set up two equations based
on the given information:
(15x= 4500
25x= 6250
Step 2: Solve the system of equations to find the total cost of the trip:
(x=4500
15 = 300
x=6250
25 = 250
Step 3: Since both equations give different values for x, we can see that
the scenario where 15 students decide to go on the trip does not align with the
scenario where 25 students decide to go. This implies that the cost per student
is not directly proportional to the number of students.
Step 4: To find the total cost and cost per student if 20 students decide to
go on the trip, we can calculate the average cost per student based on the two
scenarios: 300 + 250
2=550
2= 275
Thus, if 20 students decide to go on the trip, the total cost would be 20 ×
275 = $5500 and each student would pay
$
275.
19
Question 32
Question
A university is planning to build a new library with an estimated cost of
$
10
million. The university wants to fund 40
Solution
Step 1: Calculate the amount the university still needs to raise after receiving
the donation.
Amount still needed = Total cost −Donation
Amount still needed = $10,000,000 −$2,000,000
Amount still needed = $8,000,000
Step 2: Calculate the amount to be financed through a bond issuance. Since
the university wants to fund 40
Amount to be financed through bond issuance = 60% ×Amount still needed
Amount to be financed through bond issuance = 60% ×$8,000,000
Amount to be financed through bond issuance = 0.60 ×$8,000,000
Amount to be financed through bond issuance = $4,800,000
Therefore, the university should plan to raise
$
4.8 million through a bond
issuance to cover the remaining cost of the new library.
Question 33
Question
Solve for xin the following proportion: 3x+5
2x−1=7
4.
Solution
Step 1: Cross multiply to eliminate the denominators:
(3x+ 5) ·4=7·(2x−1)
12x+ 20 = 14x−7
Step 2: Rearrange the equation to group like terms:
12x+ 20 = 14x−7
12x−14x=−7−20
−2x=−27
20
Step 3: Solve for xby dividing both sides by −2:
x=−27
−2
x= 13.5
Therefore, the solution to the proportion is x= 13.5.
Question 34
Question
If x:y= 4 : 7 and y:z= 5 : 9, find the ratio x:z.
Solution
To find the ratio x:z, we need to first find a common value for yin both ratios.
Step 1: Find a common value for y. Given:
x:y= 4 : 7 and y:z= 5 : 9
Let’s assume that y=k. Then, from the first ratio x:y= 4 : 7, we have:
x
y=4
7
x
k=4
7
x=4
7k
Similarly, from the second ratio y:z= 5 : 9, we have:
y
z=5
9
k
z=5
9
z=9
5k
Step 2: Find the ratio x:z. Substitute the expressions for xand zin terms
of kinto the ratio x:z:
x:z=4
7k:9
5k
x:z=4
7·5
9
x:z=20
63
Therefore, the ratio x:zis 20 : 63 .
21
Solution
Step 1: Calculate the amount of acid in the first solution.
Let xbe the total volume of the 20
Step 2: Calculate the amount of acid in the second solution.
Let ybe the total volume of the 10
Step 3: Set up an equation based on the given mixture.
Since there are 3 parts of the 20
3(0.20x) : 5(0.10y) = c
Step 4: Solve for the concentration of the resulting solution.
We have:
3(0.20x) : 5(0.10y) = c
0.60x: 0.50y=c
1.2x:y=c
Thus, the concentration of the resulting solution is 1.2 times the amount
of acid in the 20
Question 3
Question
If xis directly proportional to yand yis inversely proportional to z, and x= 8
when y= 4 and z= 2, find xwhen y= 6 and z= 3.
Solution
Step 1: Write the proportionality relationships as equations. Let’s denote the
proportionality constants as k1and k2. The given relationships can be written
as: - x=k1y-y=k2
z
Step 2: Find the values of k1and k2using the given values when x= 8,
y= 4, and z= 2. Substitute the values into the equations: - 8 = k1×4
⇒k1= 2 - 4 = k2
2⇒k2= 8
Step 3: Use the found values of k1and k2to find xwhen y= 6 and z= 3.
Substitute k1= 2, k2= 8, y= 6, and z= 3 into the equations: - x= 2 ×6
⇒x= 12
Question 4
Question
A recipe calls for a mixture of ethanol and water in a ratio of 7:3. If you have
450 mL of the mixture, how many milliliters of water are in the mixture?
2
Solution
Step 1: Determine the total number of parts in the ratio. Let xrepresent the
constant of proportionality. The ratio of ethanol to water is 7 : 3, which means
the total ratio is 7 + 3 = 10 parts. Therefore, 450 mL = 10x.
Step 2: Solve for x. Divide both sides by 10 to solve for x:
x=450
10
x= 45
Step 3: Calculate the amount of water in the mixture. Since water is in a
ratio of 3 parts to 10 total parts, we need to find the amount of water represented
by 3 parts:
Water (mL) = 3 ×45
Water (mL) = 135
Therefore, there are 135 milliliters of water in the mixture.
Question 5
Question
Solve for xin the following proportion: 3x−4
2=x+5
4
Solution
To solve the proportion 3x−4
2=x+5
4, we can cross multiply to eliminate the
fractions. Then we can solve for x.
Step 1: Cross multiply to get rid of the fractions.
4(3x−4) = 2(x+ 5)
Step 2: Expand and simplify both sides of the equation.
12x−16 = 2x+ 10
Step 3: Move all terms involving xto one side of the equation.
12x−2x= 10 + 16
10x= 26
Step 4: Solve for xby dividing both sides by 10.
x=26
10
Step 5: Simplify the fraction.
x= 2.6
Therefore, the solution to the proportion is x= 2.6.
3
Question 6
Question
If 3 apples and 4 bananas cost
$
4.80, and 2 apples and 5 bananas cost
$
4.25,
what is the cost of one apple and one banana?
Solution
Let abe the cost of one apple and bbe the cost of one banana.
Step 1: Set up a system of equations based on the given information:
(3a+ 4b= 4.80
2a+ 5b= 4.25
Step 2: Solve the system of equations by elimination or substitution.
Multiplying the first equation by 2 and the second equation by 3, we get:
(6a+ 8b= 9.60
6a+ 15b= 12.75
Step 3: Subtract the equations to eliminate a:
7b= 3.15
Step 4: Solve for b:
b=3.15
7= 0.45
Step 5: Substitute b= 0.45 back into the first equation to find a:
3a+ 4(0.45) = 4.80
Step 6: Solve for a:
3a+ 1.80 = 4.80
3a= 3
a= 1
Therefore, the cost of one apple is
$
1 and the cost of one banana is
$
0.45.
Question 7
Question
Simplify the expression 3x2y3
5yz and express the result in simplest form.
4
Solution
Step 1: Rewrite the expression by canceling out common factors between the
numerator and the denominator.
3x2y3
5yz =3x·xy3
5·y·z
Step 2: Simplify the expression further by canceling out common factors.
=3x·xy3
5·y·z
Step 3: Perform the multiplication in the numerator and the denominator.
=3x2
5z
Step 4: Thus, the simplified expression is 3x2
5z.
Question 8
Question
If x:y= 3 : 5 and y:z= 4 : 7, find the ratio x:z.
Solution
To find the ratio x:z, we first need to find values for x,y, and z.
Step 1: Set up equations using the given ratios: Let x= 3a,y= 5a, and
y= 4b,z= 7bfor some constants aand b.
Step 2: Rewrite the given ratios in terms of x,y, and z:
x
y=3a
5a=3
5and y
z=5a
7b=4
7
Step 3: Solve for a:
3
5=5a
7b=⇒21a= 25a=⇒a=3
5b
Step 4: Calculate the ratio x:z:
x
z=3a
7b=3·3
5b
7b=9
35
Step 5: Therefore, the ratio x:zis 9 : 35 .
5
Question 9
Question
A recipe for a cake requires 3 cups of sugar for every 5 cups of flour. If you want
to make a cake using 15 cups of flour, how many cups of sugar should you use?
Solution
Step 1: Determine the ratio of sugar to flour in the recipe: Let xrepresent the
number of cups of sugar needed for 5 cups of flour. The ratio of sugar to flour
is 3
5=x
5. Cross multiplying gives 5x= 3 ·5 and solving for xyields x= 3 cups
of sugar for every 5 cups of flour.
Step 2: Find the amount of sugar needed for 15 cups of flour: Since the ratio
of sugar to flour is constant, we can set up a proportion to find the amount of
sugar needed for 15 cups of flour. Let yrepresent the number of cups of sugar
needed for 15 cups of flour. The proportion to solve is 3
5=y
15 .
Step 3: Solve the proportion for the unknown variable: Cross multiplying
gives 5y= 3 ·15. Solving for ygives y= 9.
Therefore, you should use 9 cups of sugar when making a cake using 15 cups
of flour to maintain the same sugar to flour ratio as in the original recipe.
Question 10
Question
Solve the proportion 2
x=3
5for x.
Solution
Step 1: Cross multiply to eliminate the fractions.
5·2 = x·3
10 = 3x
Step 2: Divide both sides by 3 to solve for x.
10
3=3x
3
10
3=x
Step 3: Simplify the fraction.
x=10
3
6
Question 11
Question
Sue and Jane are planning a road trip. Sue drove 270 miles in 4 hours while
Jane drove 405 miles in 6 hours. Assuming they both maintained a constant
speed, what is the ratio of Sue’s speed to Jane’s speed?
Solution
Step 1: Calculate Sue’s speed in miles per hour (mph). Step 2: Calculate Jane’s
speed in miles per hour (mph). Step 3: Find the ratio of Sue’s speed to Jane’s
speed.
Step 1: To find Sue’s speed, we divide the distance she drove by the time
it took her. Sue’s speed = 270 miles
4 hours = 67.5 mph.
Step 2: To find Jane’s speed, we divide the distance she drove by the time
it took her. Jane’s speed = 405 miles
6 hours = 67.5 mph.
Step 3: The ratio of Sue’s speed to Jane’s speed is given by Sue’s speed
Jane’s speed .
Ratio of speeds = 67.5
67.5= 1.
Therefore, the ratio of Sue’s speed to Jane’s speed is 1.
Question 12
Question
Sara has a mixture of nuts that contains 20
Solution
Let’s denote the weight of the new mixture as xkilograms. We can set up the
following proportion to solve for x:
0.20 ×8+2
8+2 = 0.40
Step 1: Simplify the left side of the proportion.
1.6+2
10 = 0.40
Step 2: Add the weights on the numerator.
3.6
10 = 0.40
Step 3: Multiply both sides by 10 to solve for x.
x= 10 ×0.40 = 4
Thus, the weight of the new mixture will be 4 kilograms .
7
Question 13
Question
Simplify the following expression and express the answer as a ratio of two inte-
gers: 3(2x+4)
4(3x−5) .
Solution
Step 1: Simplify the expression by multiplying out the terms in the numerator
and the denominator. 3(2x+ 4)
4(3x−5) =6x+ 12
12x−20
Step 2: Factor out a common factor from the numerator and the denomina-
tor. 6x+ 12
12x−20 =6(x+ 2)
4(3x−5)
Step 3: Simplify further by canceling out the common factor of 2 in the
numerator and the denominator.
6(x+ 2)
4(3x−5) =3(x+ 2)
2(3x−5)
Therefore, the simplified expression can be written as 3(x+2)
2(3x−5) .
Question 14
Question
If 15 men can build a wall in 10 days, how many days will it take for 20 men to
build the same wall?
Solution
Let’s denote the number of days it will take for 20 men to build the wall as x.
Step 1: Find the man-days required to build the wall with 15 men in 10
days. The man-days required can be calculated as the product of the number
of men and the number of days:
15 men ×10 days = 150 man-days
Step 2: Use the man-days to find the number of days for 20 men to build
the wall. Since the amount of work is the same, we can set up a proportion
using the man-days:
15 men ×10 days
150 man-days =20 men ×xdays
20xman-days
8
Step 3: Solve for x.150
150 =20x
20
1 = x
Step 4: State the answer. It will take 20 men 10 days to build the wall.
Question 15
Question
If 4 pounds of apples cost
$
6, how many pounds of apples can be purchased for
$
15?
Solution
Step 1: Determine the cost per pound of apples.
Let xrepresent the number of pounds of apples that can be purchased for
$
15.
We can set up a proportion to find the cost per pound:
6
4=15
x
Step 2: Solve for x.
Cross-multiply to solve for x:
6x= 4 ×15
6x= 60
x=60
6
x= 10
Step 3: Answer
Therefore, 10 pounds of apples can be purchased for
$
15.
Question 16
Question
A certain chemical compound is made up of three elements: carbon, hydrogen,
and oxygen. The ratio of the number of atoms of carbon to hydrogen to oxygen
in the compound is 2 : 3 : 1. If the compound contains a total of 108 atoms,
how many atoms of each element are present in the compound?
9
Solution
Step 1: Let’s denote the number of atoms of carbon, hydrogen, and oxygen in
the compound as 2x, 3x, and x, respectively, where xis a positive integer.
Step 2: Since the compound contains a total of 108 atoms, we can set up an
equation based on the ratio of the atoms:
2x+ 3x+x= 108
Step 3: Simplifying the equation, we get:
6x= 108
Step 4: Solving for x:
x=108
6= 18
Step 5: Now, we can find the number of atoms of carbon, hydrogen, and
oxygen: - atoms of carbon: 2x= 2 ×18 = 36 - atoms of hydrogen: 3x=
3×18 = 54 - atoms of oxygen: x= 18
Step 6: Therefore, there are 36 atoms of carbon, 54 atoms of hydrogen, and
18 atoms of oxygen in the compound.
Question 17
Question
If xand yare in a ratio of 3 : 5 and yand zare in a ratio of 2 : 7, what is the
ratio of xto z?
Solution
Let’s set up the given ratios and work towards finding the ratio of xto z.
Step 1: Express xand yin terms of a common variable. Let x= 3kand
y= 5k, where kis a constant.
Step 2: Express yand zin terms of a common variable. Since yand zare
in a ratio of 2 : 7, let y= 2mand z= 7m, where mis another constant.
Step 3: Find the ratio of xto z. Substitute the expressions for xand zinto
the ratio we’re trying to find:
x
z=3k
7m=3
7·k
m
Therefore, the ratio of xto zis 3 : 7 .
Question 18
Question
Solve for x:2x+5
3x−7=3x+8
4x−9.
10
Solution
Step 1: Cross multiply to get rid of the fractions.
2x+ 5
3x−7=3x+ 8
4x−9⇒(2x+ 5)(4x−9) = (3x+ 8)(3x−7)
Step 2: Expand both sides of the equation.
8x2−18x+ 20x−45 = 9x2+ 24x−21x−56
Step 3: Simplify and combine like terms.
8x2+ 2x−45 = 9x2+ 3x−56
Step 4: Move all terms to one side of the equation to set it equal to zero.
0 = x2+x−11
Step 5: Factor the quadratic equation.
0=(x+ 4)(x−3)
Step 6: Set each factor equal to zero and solve for x.
x+ 4 = 0 ⇒x=−4 or x−3=0⇒x= 3
Therefore, the solutions to the equation are x=−4,3 .
Question 19
Question
Solve for x:3x+ 1
2x−1=x+ 2
4.
Solution
Step 1: Cross multiply to eliminate the fractions.
(3x+ 1) ·4=(x+ 2) ·(2x−1)
12x+ 4 = 2x2−x+ 4
Step 2: Rearrange the equation to set it equal to zero.
0 = 2x2−x+ 4 −12x−4
0 = 2x2−13x
Step 3: Factor out xfrom the equation.
0 = x(2x−13)
11
Step 4: Set each factor equal to zero and solve for x.
x= 0
2x−13 = 0
2x= 13
x=13
2
Thus, the solutions for xare x= 0 and x=13
2.
Question 20
Question
A flagpole casts a shadow that is 20 meters long. At the same time, a 5-meter-
tall statue standing next to the flagpole casts a shadow that is 4 meters long.
If the flagpole and the statue are standing on level ground, what is the height
of the flagpole?
Solution
Step 1: Let hbe the height of the flagpole. Since the flagpole is standing
vertically, we can set up the ratio of the height of the flagpole to the length of
its shadow as follows: h
20 =5
4
Step 2: Now, cross multiply to solve for h:
4h= 5 ×20
Step 3: Simplify the right side of the equation:
4h= 100
Step 4: Divide both sides by 4 to solve for h:
h=100
4= 25
Therefore, the height of the flagpole is 25 meters.
Question 21
Question
A group of students decided to share a box of 100 chocolates in the ratio 3:5:7.
How many chocolates should the student who gets the smallest share give to
the student who gets the largest share so that they all have an equal number of
chocolates?
12
Solution
Step 1: Find the total number of parts the chocolates are divided into. The
chocolates are divided into 3 + 5 + 7 = 15 parts.
Step 2: Find the number of chocolates in each part. Each part is equal to
100
15 =20
3chocolates.
Step 3: Calculate the number of chocolates each student should have. The
student who gets the smallest share originally receives 3 parts, which is 3×20
3=
20 chocolates. The student who gets the largest share originally receives 7 parts,
which is 7 ×20
3=140
3chocolates.
Step 4: Calculate the number of chocolates needed to make their shares
equal. The difference in the number of chocolates between the student who gets
the smallest share and the student who gets the largest share is 140
3−20 = 80
3
chocolates.
Step 5: Divide the extra chocolates equally between the students. To make
their shares equal, the student who gets the smallest share should give 80
3∇ · 2 =
40
3chocolates to the student who gets the largest share.
Question 22
Question
Solve the following proportion for x:
2
x=x+ 4
6
Solution
To solve this proportion for x, we will first cross multiply to eliminate the
fractions.
Step 1: Cross multiply to get rid of the fractions.
2·6 = x·(x+ 4)
12 = x2+ 4x
Step 2: Rearrange the equation to set it equal to zero.
x2+ 4x−12 = 0
Step 3: Solve the quadratic equation using the quadratic formula x=
−b±√b2−4ac
2a, where a= 1, b= 4, and c=−12.
x=−4±p42−4(1)(−12)
2(1)
x=−4±√16 + 48
2
13
x=−4±√64
2
x=−4±8
2
Step 4: Solve for xby finding the two potential solutions.
x1=−4+8
2= 2
x2=−4−8
2=−6
Therefore, the solutions to the proportion are x= 2 or x=−6.
Question 23
Question
Given that a
b=3
4and b
c=5
6, find the value of a
c.
Solution
Step 1: First, we need to find a common term between the two given ratios. To
do this, we can find the value of b.
Step 2: To find the value of b, we can set up a proportion using the given
ratios: a
b=3
4and b
c=5
6.
Step 3: Cross-multiplying, we get 4a= 3bfrom the first ratio and 6b= 5c
from the second ratio.
Step 4: Solving for bin the first equation gives us b=4a
3.
Step 5: Substituting b=4a
3into the second equation gives 6 4a
3= 5c.
Step 6: Simplifying, we get 8a= 5c.
Step 7: Therefore, a
c=8
5, so the value of a
cis 8
5.
Question 24
Question
Solve the following proportion for x:
2x+ 1
3x−4=5
7
14
Solution
Step 1: Cross multiply to eliminate the fractions. Step 2: Simplify the equation
by expanding both sides. Step 3: Move all the terms to one side of the equation
to solve for x. Step 4: Solve for xby isolating the variable. Step 5: Check
the solution by substituting the value of xback into the original proportion to
ensure it satisfies the equation.
Step 1: Cross multiply to eliminate the fractions.
(2x+ 1) ×7=5×(3x−4)
Step 2: Simplify the equation by expanding both sides.
14x+ 7 = 15x−20
Step 3: Move all the terms to one side of the equation to solve for x.
14x+ 7 = 15x−20
14x−15x=−20 −7
−x=−27
Step 4: Solve for xby isolating the variable.
x=−27
−1
x= 27
Step 5: Check the solution by substituting the value of xback into the
original proportion.
2(27) + 1
3(27) −4=55
79 =5
7
Therefore, the solution to the proportion is x= 27.
Question 25
Question
Solve the following proportion: 3x−2
7=4x+ 5
11 .
Solution
Step 1: Cross multiply to eliminate the fractions.
11(3x−2) = 7(4x+ 5)
33x−22 = 28x+ 35
15
Step 2: Rearrange the equation to isolate xon one side.
33x−28x= 35 + 22
5x= 57
Step 3: Solve for xby dividing both sides by 5.
x=57
5
x= 11.4
Therefore, the solution to the proportion is x= 11.4.
Question 26
Question
If x+3
2x−5=5
7, find the value of x.
Solution
Step 1: Cross-multiply to get rid of the fractions:
7(x+ 3) = 5(2x−5)
7x+ 21 = 10x−25
Step 2: Rearrange the equation to solve for x:
7x+ 21 = 10x−25
21 + 25 = 10x−7x
46 = 3x
Step 3: Solve for x:
3x= 46
x=46
3
x=46
3×1
1
x=46
3
Hence, the value of xis 46
3.
Question 27
Question
In a class of 50 students, the ratio of male students to female students is 3:2. If
8 more male students join the class, what will be the new ratio of male students
to female students?
16
Solution
Step 1: Determine the number of male and female students in the class.
Let the number of male students be 3xand the number of female students be
2x, where xis a constant.
Given that 3x+ 2x= 50 (total number of students in the class)
Solving for x: 5x= 50
x= 10
So, there are 3(10) = 30 male students and 2(10) = 20 female students in the
class.
Step 2: Find the new ratio after 8 more male students join the class.
After 8 more male students join the class, the total number of male students
will be 30 + 8 = 38. The number of female students remains the same at 20.
Therefore, the new ratio of male students to female students is 38 : 20.
To simplify the ratio, divide both sides by 2: 38∇ · 2 : 20∇ · 2
So, the new ratio is 19 : 10.
Question 28
Question
Solve the proportion: 3
x+2 =5
2x−1.
Solution
Step 1: Cross multiply to eliminate the fractions:
3(2x−1) = 5(x+ 2)
6x−3 = 5x+ 10
Step 2: Simplify the equation:
6x−3=5x+ 10
6x−5x= 10 + 3
x= 13
Step 3: Check the solution: Substitute x= 13 back into the original propor-
tion: 3
13 + 2 =5
2(13) −1
3
15 =5
26
1
5=5
26
Therefore, the solution x= 13 satisfies the proportion.
17
Question 29
Question
Solve the following proportion: 3x
1−x=1
2.
Solution
Step 1: Cross multiply to eliminate the fractions.
3x(2) = 1(1 −x)
6x= 1 −x
Step 2: Simplify the equation by adding xto both sides.
6x+x= 1
7x= 1
Step 3: Solve for xby dividing both sides by 7.
x=1
7
Step 4: Check the solution by substituting x=1
7back into the original
proportion.
31
7
1−1
7=1
2
3
7·7
6=1
2
1 = 1 ✓
Therefore, the solution to the proportion is x=1
7.
Question 30
Question
Solve the proportion: 3x
2=7
5.
Solution
Step 1: Cross multiply to solve the proportion.
3x·5=2·7
15x= 14
18
Step 2: Divide both sides by 15 to solve for x.
x=14
15
x=14
15
Therefore, the solution to the proportion is x=14
15 .
Question 31
Question
A group of students are planning a trip to a conference. If 15 students decide
to go on the trip, the cost per student would be
$
300. However, if 25 students
decide to go on the trip, the cost per student would be
$
250. What is the total
cost of the trip and how much would each student pay if 20 students decide to
go on the trip?
Solution
Step 1: Let xbe the total cost of the trip. We can set up two equations based
on the given information:
(15x= 4500
25x= 6250
Step 2: Solve the system of equations to find the total cost of the trip:
(x=4500
15 = 300
x=6250
25 = 250
Step 3: Since both equations give different values for x, we can see that
the scenario where 15 students decide to go on the trip does not align with the
scenario where 25 students decide to go. This implies that the cost per student
is not directly proportional to the number of students.
Step 4: To find the total cost and cost per student if 20 students decide to
go on the trip, we can calculate the average cost per student based on the two
scenarios: 300 + 250
2=550
2= 275
Thus, if 20 students decide to go on the trip, the total cost would be 20 ×
275 = $5500 and each student would pay
$
275.
19
Question 32
Question
A university is planning to build a new library with an estimated cost of
$
10
million. The university wants to fund 40
Solution
Step 1: Calculate the amount the university still needs to raise after receiving
the donation.
Amount still needed = Total cost −Donation
Amount still needed = $10,000,000 −$2,000,000
Amount still needed = $8,000,000
Step 2: Calculate the amount to be financed through a bond issuance. Since
the university wants to fund 40
Amount to be financed through bond issuance = 60% ×Amount still needed
Amount to be financed through bond issuance = 60% ×$8,000,000
Amount to be financed through bond issuance = 0.60 ×$8,000,000
Amount to be financed through bond issuance = $4,800,000
Therefore, the university should plan to raise
$
4.8 million through a bond
issuance to cover the remaining cost of the new library.
Question 33
Question
Solve for xin the following proportion: 3x+5
2x−1=7
4.
Solution
Step 1: Cross multiply to eliminate the denominators:
(3x+ 5) ·4=7·(2x−1)
12x+ 20 = 14x−7
Step 2: Rearrange the equation to group like terms:
12x+ 20 = 14x−7
12x−14x=−7−20
−2x=−27
20
Step 3: Solve for xby dividing both sides by −2:
x=−27
−2
x= 13.5
Therefore, the solution to the proportion is x= 13.5.
Question 34
Question
If x:y= 4 : 7 and y:z= 5 : 9, find the ratio x:z.
Solution
To find the ratio x:z, we need to first find a common value for yin both ratios.
Step 1: Find a common value for y. Given:
x:y= 4 : 7 and y:z= 5 : 9
Let’s assume that y=k. Then, from the first ratio x:y= 4 : 7, we have:
x
y=4
7
x
k=4
7
x=4
7k
Similarly, from the second ratio y:z= 5 : 9, we have:
y
z=5
9
k
z=5
9
z=9
5k
Step 2: Find the ratio x:z. Substitute the expressions for xand zin terms
of kinto the ratio x:z:
x:z=4
7k:9
5k
x:z=4
7·5
9
x:z=20
63
Therefore, the ratio x:zis 20 : 63 .
21
Question 35
Question
Simplify the following expression: 3x2+ 6x−9
2x2−5x+ 2.
Solution
Step 1: Factor both the numerator and the denominator.
3x2+ 6x−9 = 3(x2+ 2x−3) = 3(x+ 3)(x−1)
2x2−5x+ 2 = 2(x2−5
2x+ 1) = 2(x−2)(x−1)
Step 2: Simplify the expression by dividing both the numerator and the
denominator by their common factor.
3x2+ 6x−9
2x2−5x+ 2 =3(x+ 3)(x−1)
2(x−2)(x−1)
=3
2·x+ 3
x−2
So, the simplified expression is 3
2·x+ 3
x−2.
22
