Basic Algebra Concepts: Lesson 2.2

Table of Contents

Mixture Problems Lesson 2-2

Example Problem:

How many pounds of peanuts, costing $0.70 per pound, should be mixed with 30 pounds of cashews, worth $1.20 per pound, to obtain a mixture worth $1.00 per pound?

Solution Steps:

Define the Variable:
Let $x$ represent the number of pounds of peanuts to be mixed.
Fill the Chart with Known and Unknown Quantities:

Component	Unit Value ($ per pound)	Amount (Pounds)	Value ($)
Peanuts	$0.70$	$x$	$0.70x$
Cashews	$1.20$	$30$	$(1.20)(30)$
Mixture	$1.00$	$x + 30$	$(1.00)(x + 30)$

Set Up the Equation:
The total value of peanuts and cashews must equal the total value of the mixture: $(0.70)(x) + (1.20)(30) = (1.00)(x + 30)$

Simplify the Equation:
$0.70x + 36 = x + 30$
Subtract $0.70x$ from both sides:
$36 = 0.30x + 30$
Subtract 30 from both sides:
$6 = 0.30x$
Divide both sides by $0.30$ :
$x = 20$

Answer:
$x = 20$ pounds of peanuts.

Final Mixture Check:

Peanuts’ Value: $(20 \cdot 0.70) = \$14$
Cashews’ Value: $(30 \cdot 1.20) = \$36$
Mixture Value: $(50 \cdot 1.00) = \$50$

Everything checks out! The required amount of peanuts is 20 pounds.

Mixture Problem Example

Problem:

How many pints of a 30% acid solution must be added to 5 pints of a 40% acid solution to produce a solution that is 36% acid?

Solution Steps:

Define the Variable: Let $x$ represent the number of pints of the 30% acid solution to be added.
Fill in the Chart:

Component	Unit Value (% Acid)	Amount (Pints)	Value (Pints in Mixture)
30% Acid Solution	$0.30$	$x$	$0.30x$
40% Acid Solution	$0.40$	$5$	$0.40 \cdot 5 = 2$
Mixture	$0.36$	$x + 5$	$0.36 \cdot (x + 5)$

Set Up the Equation: The total acid content from both solutions equals the acid content in the final mixture: $0.30x + 0.40(5) = 0.36(x + 5)$

Solve for $x$ : Expand and simplify the equation:
$0.30x + 2 = 0.36x + 1.8$
Subtract $0.30x$ and $1.8$ from both sides:
$0.2 = 0.06x$
Divide by $0.06$ :
$x = \frac{0.2}{0.06} = 3 \frac{1}{3}$

Answer: The amount of 30% acid solution to be added is $3 \frac{1}{3}$ pints.

Final Mixture Check:

Acid Content from 30% Solution: $0.30 \cdot 3.333 = 1.0$
Acid Content from 40% Solution: $0.40 \cdot 5 = 2.0$
Total Acid Content: $1.0 + 2.0 = 3.0$
Mixture Volume: $3.333 + 5 = 8.333$
Final Acid Percentage: $\frac{3.0}{8.333} \approx 0.36$

The solution checks out!

Example Problem: Baseball Ticket Sales

Problem:

Tickets to a baseball game cost $4.00 for reserved seats and $3.00 for general admission. In total, 500 tickets were sold for $1,760. How many reserved seats were sold?

Solution Steps:

Define the Variables:
- Let $x$ represent the number of reserved seats sold.
- The number of general admission tickets sold will be $500 – x$ .
Fill in the Chart:

Component	Unit Price ($)	Number of Tickets Sold	Total Value ($)
Reserved Seats	4.00	$x$	$4x$
General Admission	3.00	$500 – x$	$3(500 – x)$
Total	—	$500$	$1,760$

Set Up the Equation: The total value of tickets sold is the sum of the revenue from reserved seats and general admission: $4x + 3(500 – x) = 1,760$

Solve the Equation: Expand and simplify:
$4x + 1,500 – 3x = 1,760$
Combine like terms:
$x + 1,500 = 1,760$
Subtract 1,500 from both sides:
$x = 260$

Answer: The number of reserved seats sold is 260.

Final Check:

Revenue from Reserved Seats: $4 \cdot 260 = 1,040$
Revenue from General Admission: $3 \cdot (500 – 260) = 3 \cdot 240 = 720$
Total Revenue: $1,040 + 720 = 1,760$

Practice Problems with Solutions

Problem: How much silver valued at $4.00 per gram must be mixed with gold valued at $24.00 per gram to obtain 30 grams of an alloy worth $12.00 per gram?
Solution: Let $x$ represent the amount of silver in grams, and $30 – x$ the amount of gold.
- Silver Value: $4x$
- Gold Value: $24(30 – x)$
- Mixture Value: $12(30)$
Equation:
$4x + 24(30 – x) = 12(30)$
Solve:
$4x + 720 – 24x = 360$ $-20x + 720 = 360$ $-20x = -360$ $x = 18 \, \text{grams}$
Answer: 18 grams of silver.

Problem: A 200-gram solution is 25% acid. How much pure acid must be added to produce a solution that is 40% acid?
Solution: Let $x$ represent the amount of pure acid to add.
- Current Acid: $0.25(200) = 50$
- Total Solution: $200 + x$
- Acid in New Solution: $50 + x$
- New Solution Concentration: $0.4(200 + x)$
Equation:
$50 + x = 0.4(200 + x)$
Solve:
$50 + x = 80 + 0.4x$ $0.6x = 30$ $x = 50 \, \text{grams}$
Answer: Add 50 grams of pure acid.

Problem: A 30-gallon solution is 80% salt. How much pure water must be added to produce a solution that is 60% salt?
Solution: Let $x$ represent the gallons of water added.
- Salt: $0.8(30) = 24 \, \text{gallons}$
- New Solution Volume: $30 + x$
- Salt Concentration: $0.6(30 + x)$
Equation:
$24 = 0.6(30 + x)$
Solve:
$24 = 18 + 0.6x$ $6 = 0.6x$ $x = 10 \, \text{gallons}$
Answer: Add 10 gallons of water.

Problem: How many quarts of grape juice worth $1.20 a quart should be mixed with 3 quarts of apple juice worth $0.90 a quart to produce a punch worth $1.00 a quart?
Solution: Let $x$ represent the quarts of grape juice.
- Grape Juice Value: $1.20x$
- Apple Juice Value: $0.90(3) = 2.7$
- Mixture Value: $1.00(x + 3)$
Equation:
$1.20x + 2.7 = 1.00(x + 3)$
Solve:
$1.20x + 2.7 = x + 3$ $0.20x = 0.3$ $x = 1.5 \, \text{quarts}$
Answer: Add 1.5 quarts of grape juice.

Problem: Tickets to a concert cost $15.00 for the balcony and $20.00 for an orchestra seat. If 540 tickets were sold for a total price of $9,750, how many balcony tickets were sold?
Solution: Let $x$ represent the number of balcony tickets, and $540 – x$ the number of orchestra tickets.
- Balcony Tickets Value: $15x$
- Orchestra Tickets Value: $20(540 – x)$
Equation:
$15x + 20(540 – x) = 9,750$
Solve:
$15x + 10,800 – 20x = 9,750$ $-5x = -1,050$ $x = 210 \, \text{tickets}$
Answer: 210 balcony tickets were sold.

Problem: A boy has 34 coins in his pocket consisting of nickels and dimes. If the total value of the coins is $2.20, how many nickels does he have?
Solution: Let $x$ represent the number of nickels, and $34 – x$ the number of dimes.
- Nickels Value: $0.05x$
- Dimes Value: $0.10(34 – x)$
Equation:
$0.05x + 0.10(34 – x) = 2.20$
Solve:
$0.05x + 3.4 – 0.10x = 2.20$ $-0.05x = -1.20$ $x = 24 \, \text{nickels}$
Answer: The boy has 24 nickels.

Lesson 2-3: Work Problems

The key concept in work problems is that if $x$ equals the amount of time it takes to complete a specific job, then $\frac{1}{x}$ is the rate at which the job is done.

Example 1:

If a boy can cut a lawn in 4 hours, his work rate is:

$\frac{1}{4} \, \text{of the lawn per hour.}$

Key Formula for Combined Work:

The total work done is the sum of the component parts. For two entities working together:

$\text{Combined Rate} = \frac{1}{x_1} + \frac{1}{x_2}$

where $x_1$ and $x_2$ are the times it takes for each to complete the job independently.

Example 2:

If one pipe can fill a pool in 8 hours and a second pipe can fill the same pool in 6 hours, their combined work rate is:

$\frac{1}{8} + \frac{1}{6} \, \text{pools per hour.}$

Problem:

How many hours will it take both pipes working together to fill the pool?

Solution: Let $x$ represent the total time it takes for both pipes to fill the pool together. The work done by each pipe in $x$ hours is:

Pipe 1: $\frac{x}{8} \, \text{pools}$
Pipe 2: $\frac{x}{6} pools$

Since the total work equals 1 pool, the equation becomes:

$\frac{x}{8} + \frac{x}{6} = 1$

Multiply through by the least common denominator (LCD = 24) to eliminate the fractions:

$24 \cdot \frac{x}{8} + 24 \cdot \frac{x}{6} = 24 \cdot 1$ $3x + 4x = 24$ $7x = 24$ $x = \frac{24}{7} = 3 \, \frac{3}{7} \, \text{hours.}$

Final Answer:

It will take both pipes working together 3 hours and 25.7 minutes (approximately) to fill the pool.

Notes:

If the problem had asked to fill 3 pools, the equation would be set equal to 3 instead of 1: $\frac{x}{8} + \frac{x}{6} = 3$

Work Problems: Example and Practice Problems

Example Problem:

Question: John can polish his car in 45 minutes, while Jim can polish the same car in 30 minutes. How long would it take to polish the car if both boys worked together?

Solution:

Let $x$ = the number of minutes it would take for John and Jim working together to polish the car.
Rates:
- John’s rate: $\frac{1}{45}$ of the job per minute.
- Jim’s rate: $\frac{1}{30}$ of the job per minute.
In $x$ minutes:
- John completes $\frac{x}{45}$ of the job.
- Jim completes $\frac{x}{30}$ of the job.
Together, their total work equals 1 job:
$\frac{x}{45} + \frac{x}{30} = 1$
Eliminate fractions: Multiply through by the least common multiple (LCM) of 45 and 30, which is 90:
$90 \cdot \frac{x}{45} + 90 \cdot \frac{x}{30} = 90 \cdot 1$ $2x + 3x = 90$
Combine and solve:
$5x = 90$ $x = 18 \, \text{minutes.}$

Final Answer: It would take 18 minutes for John and Jim to polish the car together.

Practice Problems:

Question: If one pipe can fill a pool in 12 hours, how much of the pool is filled after 8 hours?
Hint: The rate is $\frac{1}{12}$ . Multiply this rate by 8 hours to find the fraction of the pool filled.

Question: If Brutus can eat 16 hamburgers in 1 hour, how many can he eat in 3 hours?
Hint: Multiply his rate ( $16 \, \text{hamburgers/hour}$ ) by 3 hours.

Question: One machine can dig a ditch in 128 minutes. How much of the ditch can be dug in 40 minutes?
Hint: The rate is $\frac{1}{128}$ . Multiply this by 40 to find the fraction of the ditch dug.

Question: One machine can process a payroll in 2 hours. A second machine can process the same payroll in 90 minutes. How long would it take to process the payroll if both machines worked together?
Hint: Find the rates for both machines ( $\frac{1}{2} \, \text{jobs/hour}$ and $\frac{1}{1.5} \, \text{jobs/hour}$ ) and add them. Set the total rate equal to $\frac{1}{x}$ , where $x$ is the time taken.

Question: One pipe can drain a pool in 4 hours. A second pipe can drain the same pool in 3 hours. How many hours would it take to drain the pool if both pipes worked together?
Hint: Combine their rates ( $\frac{1}{4} + \frac{1}{3}$ ) and set it equal to $\frac{1}{x}$ , where $x$ is the time taken.

Question: Frank can cut his lawn in 4.5 hours, his wife Mary can cut the same lawn in 4 hours, and their son Tom can cut the lawn in 3.5 hours. How long would it take to cut the lawn if all three worked together?
Hint: Add their rates ( $\frac{1}{4.5} + \frac{1}{4} + \frac{1}{3.5}$ ) and set it equal to $\frac{1}{x}$ , where $x$ is the time taken.

Number Problems: Lesson 2-4

Understanding basic number concepts can help solve word problems effectively. Here’s a breakdown of essential terms and methods:

Key Definitions:

Integers:
- Positive and negative whole numbers, including zero.
- Example: $\{ \dots, -3, -2, -1, 0, 1, 2, 3, \dots \}$ Consecutive Integers:
- Integers that differ by 1.
- Example: $7, 8, 9, 10$ or represented algebraically as $x, x+1, x+2$ , where $x$ is any integer.
Consecutive Odd Integers:
- Odd integers that differ by 2.
- Example: $15, 17, 19, 21$ or represented algebraically as $x, x+2, x+4$ , where $x$ is any odd integer.
Consecutive Even Integers:
- Even integers that differ by 2.
- Example: $0, 2, 4, 6, 8$ or represented algebraically as $x, x+2, x+4$ , where $x$ is any even integer.

Steps to Solving Word Problems:

Define Your Variable:
- Identify what the problem is asking and represent it with a variable (e.g., $x$ ).
Set Up Your Equation:
- Translate the problem into an algebraic equation using the relationships between numbers.
Solve the Equation:
- Perform operations to isolate the variable and find the solution.
Check Your Answer:
- Plug the solution back into the original problem to verify it satisfies all conditions.

Example Problem:

Question: Find three consecutive integers such that the sum of the smallest and the largest is 28.

Solution:

Define your variable:
- Let the integers be $x, x+1, x+2$ .
Set up the equation:
- The sum of the smallest ( $x$ ) and the largest ( $x+2$ ) equals 28: $x + (x+2) = 28$
Solve the equation:
$2x + 2 = 28$ $2x = 26$ $x = 13$
Find all integers:
- The integers are $13, 14, 15$
Check your answer:
- Smallest + Largest = $13 + 15 = 28$ (Correct!).

Examples of Solving Number Problems:

Example I:

Question: Two less than six times a certain number equals five times the number plus 2. What is the number?

Solution:

Define the variable:
Let $x$ represent the number.
Set up the equation:
$6x – 2 = 5x + 2$
Solve the equation:
Subtract $5x$ from both sides and add 2 to both sides:
$x = 4$
Check your answer:
Substituting $x = 4$ :
$6(4) – 2 = 5(4) + 2 \quad \text{or} \quad 24 – 2 = 20 + 2$
Both sides equal 2 , so the solution is correct.

Answer: $x = 4$ .

Example II:

Question: Find three consecutive integers whose sum is 84.

Solution:

Define the variable:
Let $x$ be the first integer, $x+1$ the second, and $x+2$ the third.
Set up the equation:
$x + (x+1) + (x+2) = 84$
Solve the equation:
$3x + 3 = 84 \quad \Rightarrow \quad 3x = 81 \quad \Rightarrow \quad x = 27$
Find the integers:
The integers are $27, 28, \text{ and } 29$

Answer: $27, 28, 29$

Example III:

Question: Find three consecutive even integers such that the sum of the first and twice the third is 134.

Solution:

Define the variables:
Let $x$ be the first even integer, $x+2$ the second, and $x+4$ the third.
Set up the equation:
$x + 2(x+4) = 134$
Solve the equation:
Expand and simplify:
$x + 2x + 8 = 134 \quad \Rightarrow \quad 3x + 8 = 134 \quad \Rightarrow \quad 3x = 126 \quad \Rightarrow \quad x = 42$
Find the integers:
The integers are $42, 44, \text{ and } 46$

Answer: $42, 44, 46$

Example IV:

Question: Find four consecutive odd integers such that the sum of three times the first and twice the third is 11 more than twice the sum of the second and fourth integers.

Solution:

Define the variables:
Let $x$ be the first odd integer, $x+2$ the second, $x+4$ the third, and $x+6$ the fourth.
Set up the equation:
$3(x) + 2(x+4) = 2[(x+2) + (x+6)] + 11$
Simplify and solve the equation:
Expand:
$3x + 2x + 8 = 2(2x + 8) + 11$
Simplify:
$5x + 8 = 4x + 16 + 11$
Combine terms:
$5x + 8 = 4x + 27 \quad \Rightarrow \quad x = 19$
Find the integers:
The integers are $19, 21, 23, \text{ and } 25$

Answer: $19, 21, 23, 25$

Practice Problems with Solutions

1. Find six consecutive integers whose sum is 513.

Let the integers be $x, x+1, x+2, x+3, x+4, x+5$
Sum: $x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 513$ Simplify: $6x + 15 = 513$ $6x = 498 \quad \Rightarrow \quad x = 83$
The integers are $83, 84, 85, 86, 87, 88$

2. Find five consecutive odd integers whose sum is 225.

Let the integers be $x, x+2, x+4, x+6, x+8$
Sum: $x + (x+2) + (x+4) + (x+6) + (x+8) = 225$ Simplify: $5x + 20 = 225$ Solve: $5x = 205 \quad \Rightarrow \quad x = 41$
The integers are $41, 43, 45, 47, 49$

3. Four times one odd integer is 14 less than three times the next even integer. Find the integers.

Let the odd integer be $x$ . The next even integer is $x+1$ .
Equation: $4x = 3(x+1) – 14$ Simplify: $4x = 3x + 3 – 14 \quad \Rightarrow \quad x = -11$
The integers are $-11$ (odd) and $-10$ (even).

4. The largest of five consecutive integers is twice the smallest. Find the smallest integer.

Let the integers be $x, x+1, x+2, x+3, x+4$
Largest: $x+4$ , Smallest: $x$ .
Equation: $x+4 = 2x$ Solve: $x = 4$
The integers are $4, 5, 6, 7, 8$ .

5. The average of four consecutive odd integers is 16. Find the largest integer.

Let the integers be $x, x+2, x+4, x+6$ .
Average: $\frac{x + (x+2) + (x+4) + (x+6)}{4} = 16$ Simplify: $\frac{4x + 12}{4} = 16 \quad \Rightarrow \quad x + 3 = 16 \quad \Rightarrow \quad x = 13$
The integers are $13, 15, 17, 19$ . The largest is $19$ .

6. When the sum of three consecutive integers is divided by 9, the result is 7. Find the integers.

Let the integers be $x, x+1, x+2$ .
Equation: $\frac{x + (x+1) + (x+2)}{9} = 7$ Simplify: $\frac{3x + 3}{9} = 7 \quad \Rightarrow \quad 3x + 3 = 63 \quad \Rightarrow \quad 3x = 60 \quad \Rightarrow \quad x = 20$
The integers are $20, 21, 22$ .

7. If each of three consecutive integers is divided by 3, the sum of the quotients is 84. Find the smallest integer.

Let the integers be $x, x+1, x+2$ .
Equation: $\frac{x}{3} + \frac{x+1}{3} + \frac{x+2}{3} = 84$ Simplify: $\frac{3x+3}{3} = 84 \quad \Rightarrow \quad x+1 = 84 \quad \Rightarrow \quad x = 83$
The integers are $83, 84, 85$ .

8. The sum of three numbers is 65. The first is twice the second, and the third is twice the result obtained by subtracting 5 from the second. Find the largest number.

Let the second number be $x$ .
First: $2x$ , Third: $2(x-5)$ .
Equation: $2x + x + 2(x-5) = 65$ Simplify: $5x – 10 = 65 \quad \Rightarrow \quad 5x = 75 \quad \Rightarrow \quad x = 15$
The numbers are $30, 15, 20$ . The largest is $30$ .

9. A baseball team played 150 games and won 32 more than it lost. How many did it lose?

Let the number of losses be $x$ .
Wins: $x + 32$ .
Equation: $x + (x+32) = 150$ Simplify: $2x + 32 = 150 \quad \Rightarrow \quad 2x = 118 \quad \Rightarrow \quad x = 59$
Losses: $59$ .

10. One number exceeds another number by 12, and the sum of the smaller number and twice the larger number is 93. Find the numbers.

Let the smaller number be $x$ .
Larger: $x+12$ .
Equation: $x + 2(x+12) = 93$ Simplify: $x + 2x + 24 = 93 \quad \Rightarrow \quad 3x = 69 \quad \Rightarrow \quad x = 23$
The numbers are $23$ and $35$ .

AGE PROBLEMS – LESSON 2-5

Key Concept:

If a person is currently $x$ years old:
- 8 years ago: $x – 8$
- In 10 years: $x + 10$

Example I:

Problem: Sally is half her father’s age. In 2 years, the sum of their ages will be 67. How old is each now?
Solution:
1. Define the variables:
  - Let $x$ = Sally’s age now.
  - Father’s age = $2x$ .
2. Chart of ages:
  - Sally now: $x$ , in 2 years: $x + 2$
  - Father now: $2x$ , in 2 years: $2x + 2$
3. Set up the equation: $(x + 2) + (2x + 2) = 67$
4. Solve the equation: $3x + 4 = 67 \quad \rightarrow \quad 3x = 63 \quad \rightarrow \quad x = 21$
5. Answer:
  - Sally is 21 years old, and her father is 42.
6. Verification: $(21 + 2) + (42 + 2) = 67$

Example II:

Problem: Katie is 2 years older than Sue. In 8 years, the sum of their ages will be 48. How old is each now?
Solution:
1. Define the variables:
  - Let $x$ = Sue’s age now.
  - Katie’s age = $x + 2$ .
2. Chart of ages:
  - Sue now: $x$ , in 8 years: $x + 8$
  - Katie now: $x + 2$ , in 8 years: $x + 10$
3. Set up the equation: $(x + 8) + (x + 10) = 48$
4. Solve the equation: $2x + 18 = 48 \quad \rightarrow \quad 2x = 30 \quad \rightarrow \quad x = 15$
5. Answer:
  - Sue is 15 years old, and Katie is 17.
6. Verification: $(15 + 8) + (17 + 8) = 48$

Example III:

Problem: A father is 30 years older than his daughter. Half the father’s current age equals the daughter’s age in 9 years. How old is the daughter?
Solution:
1. Define the variables:
  - Let $x$ = Daughter’s current age.
  - Father’s age = $x + 30$ .
2. Chart of ages:
  - Daughter now: $x$ , in 9 years: $x + 9$
  - Father now: $x + 30$ , in 9 years: $x + 39$
3. Set up the equation: $\frac{1}{2}(x + 30) = x + 9$
4. Solve the equation: $x + 30 = 2(x + 9) \quad \rightarrow \quad x + 30 = 2x + 18 \quad \rightarrow \quad 30 = x + 18 \quad \rightarrow \quad x = 12$
5. Answer:
  - Daughter is 12 years old.
6. Verification: $\frac{1}{2}(42) = 12 + 9$

Problem 1:

Question: If Roger were 32 years older, he would be three times as old as he is now. How old is Roger?

Solution:

Let $x$ = Roger’s current age.
If he were 32 years older, his age would be $x + 32$
According to the problem: $x + 32 = 3x$
Solve the equation: $32 = 3x – x \quad \rightarrow \quad 32 = 2x \quad \rightarrow \quad x = 16$
Answer: Roger is 16 years old.

Problem 2:

Question: Brad is 12 years older than Sam. If Brad were 8 years older than he is now, he would be twice as old as Sam. How old is Sam now?

Solution:

Let $x$ = Sam’s age.
Brad’s age is $x + 12$
If Brad were 8 years older, his age would be $(x + 12) + 8 = x + 20$
According to the problem: $x + 20 = 2x$
Solve the equation: $20 = 2x – x \quad \rightarrow \quad x = 20$
Answer: Sam is 20 years old.

Problem 3:

Question: Barrie is now 2 years older than Krista. In 15 years, Barrie’s age will be 2 years more than twice Krista’s age now. How old will Barrie be in 6 years?

Solution:

Let $x$ = Krista’s age.
Barrie’s age is $x + 2$
In 15 years, Barrie’s age will be $(x + 2) + 15 = x + 17$
According to the problem: $x + 17 = 2x + 2$
Solve the equation: $x + 17 = 2x + 2 \quad \rightarrow \quad 17 – 2 = x \quad \rightarrow \quad x = 15$
Barrie’s current age is $x + 2 = 15 + 2 = 17$
In 6 years, Barrie’s age will be $17 + 6 = 23$
Answer: Barrie will be 23 years old in 6 years.

Problem 4:

Question: John is now 10 years older than Marcus. Three times John’s age 5 years from now will be the same as five times Marcus’s age 5 years ago. How old is John now?

Solution:

Let $x$ = Marcus’s age.
John’s age is $x + 10$
John’s age 5 years from now: $(x + 10) + 5 = x + 15$
Marcus’s age 5 years ago: $x – 5$
According to the problem: $3(x + 15) = 5(x – 5)$
Solve the equation: $3x + 45 = 5x – 25 \quad \rightarrow \quad 45 + 25 = 5x – 3x \quad \rightarrow \quad 70 = 2x \quad \rightarrow \quad x = 35$
John’s current age is $x + 10 = 35 + 10 = 45$
Answer: John is 45 years old.

Problem 5:

Question: In 1 year, Kristen will be four times as old as Danielle. Ten years from then, Kristen will only be twice as old as Danielle. How old is Kristen now?

Solution:

Let $x$ = Danielle’s current age.
Kristen’s age in 1 year: $4(x + 1)$
In 10 years: $4(x + 1) + 10$
According to the problem: $4(x + 1) + 10 = 2(x + 11)$
Solve the equation: $4x + 4 + 10 = 2x + 22 \quad \rightarrow \quad 4x – 2x = 22 – 14 \quad \rightarrow \quad 2x = 8 \quad \rightarrow \quad x = 4$
Kristen’s current age is $4(x + 1) = 4(4 + 1) = 20$
Answer: Kristen is 20 years old.

Problem 6:

Question: A woman was 30 years old when her daughter was born. Her age is now 6 years more than three times her daughter’s age. How old will the daughter be in 5 years?

Solution:

Let $x$ = Daughter’s current age.
Woman’s current age: $x + 30$
According to the problem: $x + 30 = 3x + 6$
Solve the equation: $x + 30 = 3x + 6 \quad \rightarrow \quad 30 – 6 = 3x – x \quad \rightarrow \quad 24 = 2x \quad \rightarrow \quad x = 12$
Daughter’s age in 5 years: $x + 5 = 12 + 5 = 17$
Answer: The daughter will be 17 years old.

Problem 7:

Question: Ralph’s age is currently 3 years more than twice Joe’s age. But 3 years ago, Ralph was four times as old as Joe was then. How old will Joe be in 10 years?

Solution:

Let $x$ = Joe’s current age.
Ralph’s current age: $2x + 3$
Ralph’s age 3 years ago: $(2x + 3) – 3 = 2x$
Joe’s age 3 years ago: $x – 3$
According to the problem: $2x = 4(x – 3)$
Solve the equation: $2x = 4x – 12 \quad \rightarrow \quad 12 = 4x – 2x \quad \rightarrow \quad 2x = 12 \quad \rightarrow \quad x = 6$
Joe’s age in 10 years: $x + 10 = 6 + 10 = 16$
Answer: Joe will be 16 years old in 10 years.

Problem 8:

Question: Lisa is 15 years old, and her father is 40. How many years ago was the father six times as old as Lisa?

Solution:

Let $x$ = Number of years ago.
Father’s age $x$ years ago: $40 – x$
Lisa’s age $x$ years ago: $15 – x$
According to the problem: $40 – x = 6(15 – x)$
Solve the equation: $40 – x = 90 – 6x \quad \rightarrow \quad 40 – 90 = -6x + x \quad \rightarrow \quad -50 = -5x \quad \rightarrow \quad x = 10$
Answer: 10 years ago.

Problem 1:

Question: If Roger were 32 years older, he would be three times as old as he is now. How old is Roger?

Solution:

Let $x$ = Roger’s current age.
If he were 32 years older, his age would be $x + 32$
According to the problem: $x + 32 = 3x$
Solve the equation: $32 = 3x – x \quad \rightarrow \quad 32 = 2x \quad \rightarrow \quad x = 16$
Answer: Roger is 16 years old.

Problem 2:

Question: Brad is 12 years older than Sam. If Brad were 8 years older than he is now, he would be twice as old as Sam. How old is Sam now?

Solution:

Let $x$ = Sam’s age.
Brad’s age is $x + 12$
If Brad were 8 years older, his age would be $(x + 12) + 8 = x + 20$
According to the problem: $x + 20 = 2x$
Solve the equation: $20 = 2x – x \quad \rightarrow \quad x = 20$
Answer: Sam is 20 years old.

Problem 3:

Question: Barrie is now 2 years older than Krista. In 15 years, Barrie’s age will be 2 years more than twice Krista’s age now. How old will Barrie be in 6 years?

Solution:

Let $x$ = Krista’s age.
Barrie’s age is $x + 2$
In 15 years, Barrie’s age will be $(x + 2) + 15 = x + 17$
According to the problem: $x + 17 = 2x + 2$
Solve the equation: $x + 17 = 2x + 2 \quad \rightarrow \quad 17 – 2 = x \quad \rightarrow \quad x = 15$
Barrie’s current age is $x + 2 = 15 + 2 = 17$
In 6 years, Barrie’s age will be $17 + 6 = 23$
Answer: Barrie will be 23 years old in 6 years.

Problem 4:

Question: John is now 10 years older than Marcus. Three times John’s age 5 years from now will be the same as five times Marcus’s age 5 years ago. How old is John now?

Solution:

Let $x$ = Marcus’s age.
John’s age is $x + 10$
John’s age 5 years from now: $(x + 10) + 5 = x + 15$
Marcus’s age 5 years ago: $x – 5$
According to the problem: $3(x + 15) = 5(x – 5)$
Solve the equation: $3x + 45 = 5x – 25 \quad \rightarrow \quad 45 + 25 = 5x – 3x \quad \rightarrow \quad 70 = 2x \quad \rightarrow \quad x = 35$
John’s current age is $x + 10 = 35 + 10 = 45$
Answer: John is 45 years old.

Problem 5:

Question: In 1 year, Kristen will be four times as old as Danielle. Ten years from then, Kristen will only be twice as old as Danielle. How old is Kristen now?

Solution:

Let $x$ = Danielle’s current age.
Kristen’s age in 1 year: $4(x + 1)$
In 10 years: $4(x + 1) + 10$
According to the problem: $4(x + 1) + 10 = 2(x + 11)$
Solve the equation: $4x + 4 + 10 = 2x + 22 \quad \rightarrow \quad 4x – 2x = 22 – 14 \quad \rightarrow \quad 2x = 8 \quad \rightarrow \quad x = 4$
Kristen’s current age is $4(x + 1) = 4(4 + 1) = 20$
Answer: Kristen is 20 years old.

Problem 6:

Question: A woman was 30 years old when her daughter was born. Her age is now 6 years more than three times her daughter’s age. How old will the daughter be in 5 years?

Solution:

Let $x$ = Daughter’s current age.
Woman’s current age: $x + 30$
According to the problem: $x + 30 = 3x + 6$
Solve the equation: $x + 30 = 3x + 6 \quad \rightarrow \quad 30 – 6 = 3x – x \quad \rightarrow \quad 24 = 2x \quad \rightarrow \quad x = 12$
Daughter’s age in 5 years: $x + 5 = 12 + 5 = 17$
Answer: The daughter will be 17 years old.

Problem 7:

Question: Ralph’s age is currently 3 years more than twice Joe’s age. But 3 years ago, Ralph was four times as old as Joe was then. How old will Joe be in 10 years?

Solution:

Let $x$ = Joe’s current age.
Ralph’s current age: $2x + 3$
Ralph’s age 3 years ago: $(2x + 3) – 3 = 2x$
Joe’s age 3 years ago: $x – 3$
According to the problem: $2x = 4(x – 3)$
Solve the equation: $2x = 4x – 12 \quad \rightarrow \quad 12 = 4x – 2x \quad \rightarrow \quad 2x = 12 \quad \rightarrow \quad x = 6$
Joe’s age in 10 years: $x + 10 = 6 + 10 = 16$
Answer: Joe will be 16 years old in 10 years.

Problem 8:

Question: Lisa is 15 years old, and her father is 40. How many years ago was the father six times as old as Lisa?

Solution:

Let $x$ = Number of years ago.
Father’s age $x$ years ago: $40 – x$
Lisa’s age $x$ years ago: $15 – x$
According to the problem: $40 – x = 6(15 – x)$
Solve the equation: $40 – x = 90 – 6x \quad \rightarrow \quad 40 – 90 = -6x + x \quad \rightarrow \quad -50 = -5x \quad \rightarrow \quad x = 10$
Answer: 10 years ago.

Multiple Choice Problems

If $a = 9 \times 23$ and $b = 9 \times 124$ , what is the value of $b – a$ ?
- (A) 901
- (B) 903
- (C) 906
- (D) 909
- (E) 911

By how much does the product of 8 and 25 exceed the product of 15 and 10?
- (A) 25
- (B) 50
- (C) 75
- (D) 100
- (E) 125

How many containers, each holding 16 fluid ounces of milk, are needed to hold 5 quarts of milk?
(1 quart = 32 fluid ounces)
- (A) 6
- (B) 8
- (C) 10
- (D) 12
- (E) 14

If the current odometer reading of a car is 31,983 miles, what is the LEAST number of miles that the car must travel before the odometer displays four digits that are the same?
- (A) 17
- (B) 239
- (C) 350
- (D) 650
- (E) 1350

In store A, a scarf costs $12, and in store B the same scarf is on sale for $8. How many scarves can be bought in store B with the amount of money needed to buy 10 scarves in store A?
- (A) 4
- (B) 12
- (C) 15
- (D) 18
- (E) 21

Let $*$ represent one of the four basic arithmetic operations such that, for any nonzero real number $r$ :
- $r^* 0 = r$ and $r^* r = 0$
  Which arithmetic operation(s) does the symbol $*$ represent?
- (A) + only
- (B) $–$ only
- (C) + and $–$
- (D) × only
- (E) +, $–$ , or ×

Kurt has saved $160 to buy a stereo system that costs $400 including taxes. If he earns $8 an hour after payroll deductions, how many hours will he have to work to save enough money?
- (A) 20
- (B) 24
- (C) 25
- (D) 30
- (E) 40

If the present time is exactly 1:00 PM, what was the time exactly 39 hours ago?
- (A) 4:00 PM
- (B) 4:00 AM
- (C) 9:00 PM
- (D) 9:00 AM
- (E) 10:00 PM

Let $\#$ represent one of the four basic arithmetic operations such that, for any nonzero real numbers $r$ and $s$ :
- $r \# 0 = r$ and $r \# s = s \# r$ .
  Which arithmetic operation(s) does the symbol $\#$ represent?
- (A) + only
- (B) × only
- (C) $–$ only
- (D) $–$ and ×
- (E) + and $÷$

If $w = (6)(6)(6)$ , $x = (5)(6)(7)$ , and $y = (4)(6)(8)$ , which inequality statement is true?
- (A) $x < y < w$
- (B) $w < x < y$
- (C) $y < w < x$
- (D) $y < x < w$
- (E) $w < y < x$

If $x$ and $y$ are positive integers, $2x + y < 29$ , and $y > 4$ , what is the greatest possible value of $x – y$ ?
- (A) 5
- (B) 6
- (C) 7
- (D) 8
- (E) 9

Grid-In Questions

If $2^4 \times 4^2 = 16^x$ , then $x = ?$
If $3^{(x-1)} = 9$ $4^{(y+2)} = 64$ , what is the value of $\frac{x}{y}$ ?
If $\frac{p + p + p}{p \cdot p \cdot p} = 12$ and $p > 0$ , what is the value of $p$ ?

Divisibility & Factors (Multiple Choice)

If the sum of the factors of 18 is $S$ and the sum of the prime numbers less than 18 is $P$ , then $P$ exceeds $S$ by what number?
- (A) 19
- (B) 17
- (C) 15
- (D) 13
- (E) 11

When a whole number $N$ is divided by 5, the quotient is 13 and the remainder is 4. What is the value of $N$ ?
- (A) 55
- (B) 59
- (C) 65
- (D) 69
- (E) 79

If $p$ is divisible by 3 and $q$ is divisible by 4, then $pq$ must be divisible by each of the following EXCEPT:
- (A) 3
- (B) 4
- (C) 6
- (D) 9
- (E) 12

Which number has the most factors?
- (A) 12
- (B) 18
- (C) 25
- (D) 70
- (E) 100

Which number is divisible by 2 and by 3?
- (A) 112
- (B) 4308
- (C) 6122
- (D) 23,451
- (E) 701,456

All numbers that are divisible by both 3 and 10 are also divisible by:
- (A) 4
- (B) 9
- (C) 12
- (D) 15
- (E) 20

If $x$ represents any even number and $y$ represents any odd number, which of the following numbers is even?
- (A) $y + 2$
- (B) $x - 1$
- (C) $(x + 1)(y – 1)$
- (D) $y(y + 2)$
- (E) $x + y$

For how many different positive integers $p$ is $\frac{105}{p}$ also an integer?
- (A) Five
- (B) Six
- (C) Seven
- (D) Eight
- (E) Nine

If $n$ is an odd integer, which expression always represents an odd integer?
- (A) $(2n – 1)^2$
- (B) $n^2 + 2n + 1$
- (C) $(n – 1)^2$
- (D) $\frac{n + 1}{2}$
- (E) $3^n + 1$

If $k – 1$ is a multiple of 4, what is the next larger multiple of 4?
- (A) $k + 1$
- (B) $4k$
- (C) $k – 5$
- (D) $k + 3$
- (E) $4(k – 1)$

After $m$ marbles are put into $n$ jars, each jar contains the same number of marbles, with two marbles remaining. In terms of $m$ and $n$ , how many marbles were put into each jar?
- (A) $\frac{m}{n} + 2$
- (B) $\frac{m}{n} – 2$
- (C) $\frac{m + 2}{n}$
- (D) $\frac{m – 2}{n}$
- (E) $\frac{mn}{n + 2}$

When $p$ is divided by 4, the remainder is 3, and when $p$ is divided by 3, the remainder is 0. What is a possible value of $p$ ?
- (A) 8
- (B) 11
- (C) 15
- (D) 18
- (E) 21

If $n$ is an integer and $n^2 + 5$ is an odd integer, then which statement(s) must be true?
I. $n^2 – 1$ is even.
II. $n$ is even.
III. $5n$ is even.
- (A) I only
- (B) I and II only
- (C) I and III only
- (D) II and III only
- (E) I, II, and III

When the number of people who contribute equally to a gift decreases from four to three, each person must pay an additional $10. What is the cost of the gift?
- (A) $30
- (B) $60
- (C) $90
- (D) $120
- (E) $180

If $n$ is any even integer, what is the remainder when $(n + 1)^2$ is divided by 4?
- (A) 0
- (B) 1
- (C) 2
- (D) 3
- (E) 4

A jar contains between 40 and 50 marbles. If the marbles are taken out of the jar three at a time, two marbles will be left in the jar. If the marbles are taken out of the jar five at a time, four marbles will be left in the jar. How many marbles are in the jar?
- (A) 41
- (B) 43
- (C) 44
- (D) 47
- (E) 49

Grid-In Questions

When $a$ is divided by 7, the remainder is 5; when $b$ is divided by 7, the remainder is 4. What is the remainder when $a + b$ is divided by 7?
How many integers from $-3000$ to $3000$ , inclusive, are divisible by 3?
When a positive integer $k$ is divided by 6, the remainder is 1. What is the remainder when $5k$ is divided by 3?
What is the smallest positive integer $p$ for which $2^{2p+1}$ is not a prime number?

Lesson 3-4: Multiple Choice

If $2b = -3$ , what is the value of $1 – 4b$ ?
- (A) -7
- (B) -5
- (C) 5
- (D) 6
- (E) 7

$p^2(2 – 5) + (-p)^2 = ?$
- (A) $-4p^2$
- (B) $- p^{2}$
- (C) $-2p^2$
- (D) $2p^2$
- (E) $4p^2$

If $n + 5$ is an odd integer, then $n$ could be which of the following?
- (A) 1
- (B) -1
- (C) -2
- (D) -3
- (E) -7

If the product of five numbers is positive, then at most how many of the five numbers could be negative?
- (A) One
- (B) Two
- (C) Three
- (D) Four
- (E) Five

For which value of $k$ is the value of $k(k + 2)(k + 1)$ negative?
- (A) -2
- (B) -1
- (C) 0
- (D) 2
- (E) 3

Which of the following statements must be true when $a < 0$ and $b > 0$ ?
- I. $a + b > 0$
- II. $b - a > 0$
- III. $a \cdot \frac{a}{b} > 0$
- (A) I only
- (B) II only
- (C) I and II only
- (D) II and III only
- (E) I, II, and III

$(-2)^3 + (-3)^2 = ?$
- (A) -12
- (B) -1
- (C) 0
- (D) 1
- (E) 2

Questions 9 and 10 Refer to the Diagram Above

Which of the following statements must be true?
- I. $c^2 < c$
- II. $a^2 > c$
- III. $b < \frac{1}{b}$
- (A) I only
- (B) I and II only
- (C) I and III only
- (D) I, II, and III
- (E) None

Which of the following statements must be true?
- I. $ad > b$
- II. $ab > ad$
- III. $\frac{1}{a} > \frac{1}{b}$
- (A) II only
- (B) I and II only
- (C) II and III only
- (D) I, II, and III
- (E) None

If $(y – 3)^2 = 16$ , what is the smallest possible value of $y^2$ ?
- (A) -4
- (B) 1
- (C) 7
- (D) 16
- (E) 49

If $X$ represents the sum of the 10 greatest negative integers and $Y$ represents the sum of the 10 least positive integers, which of the following must be true?
- I. $X + Y < 0$
- II. $Y – X = 2Y$
- III. $X^{2} = Y^{2}$
- (A) None
- (B) I only
- (C) III only
- (D) I and II only
- (E) II and III only

If $a^2 b^3 c > 0$ , which of the following statements must be true?
- I. $bc > 0$
- II. $a c > 0$
- III. $ab > 0$
- (A) I only
- (B) I and II only
- (C) I and III only
- (D) II and III only
- (E) I, II, and III

If $a^2 b^3 c^5$ is negative, which product is always negative?
- (A) $bc$
- (B) $b^2 c$
- (C) $ac$
- (D) $ab$
- (E) $bc^2$

If $a \neq 0$ , which of the following statements must be true?
- I. $(-a)^2 = a^2 – 2a^2$
- II. $a – b = -(b – a)$
- III. $a > -a$
- (A) I only
- (B) II only
- (C) III only
- (D) I and III only
- (E) II and III only

Grid-In Questions

If $p^2 = 16$ and $q^2 = 36$ , what is the largest possible value of $q – p$ ?
If $-4 \leq x \leq 2$ and $y = 1 – x^2$ , what number is obtained when the smallest possible value of $y$ is subtracted from the largest possible value of $y$ ?

Fractions & Decimals: Lesson 3-5

Multiple Choice

What part of an hour elapses from 4:56 P.M. to 5:32 P.M.?
- (A) $\frac{1}{4}$
- (B) $\frac{1}{2}$
- (C) $\frac{3}{5}$
- (D) $\frac{2}{3}$
- (E) $\frac{3}{4}$

If each of the fractions $\frac{3}{k}, \frac{4}{k}, \frac{5}{k}$ is in lowest terms, which of the following could be the value of $k$ ?
- (A) 48
- (B) 49
- (C) 50
- (D) 51
- (E) 52

Which number has the greatest value?
- (A) 0.2093
- (B) 0.2908
- (C) 0.2893
- (D) 0.2938
- (E) 0.2909

The elapsed time from 11:00 A.M. to 3:00 P.M. on Wednesday is what fraction of the elapsed time from 11:00 A.M. on Wednesday to 3:00 P.M. on Friday?
- (A) $\frac{1}{15}$
- (B) $\frac{1}{14}$
- (C) $\frac{1}{13}$
- (D) $\frac{4}{51}$
- (E) $\frac{1}{12}$

In which number is the digit 3 in the hundredths place?
- (A) 300.000
- (B) 30.000
- (C) 0.300
- (D) 0.030
- (E) 0.003

In which arrangement are the fractions listed from least to greatest?
- (A) $\frac{9}{19}, \frac{1}{2}, \frac{8}{15}$
- (B) $\frac{1}{2}, \frac{8}{15}, \frac{9}{19}$
- (C) $\frac{9}{19}, \frac{8}{15}, \frac{1}{2}$
- (D) $\frac{1}{2}, \frac{9}{19}, \frac{8}{15}$
- (E) $\frac{8}{15}, \frac{1}{2}, \frac{9}{19}$

After the formula $V = \frac{4}{3} \pi r^3$ has been evaluated for some positive value of $r$ , the formula is again evaluated using one-half of the original value of $r$ . The new value of $V$ is what fractional part of the original value of $V$ ?
- (A) $\frac{1}{16}$
- (B) $\frac{1}{9}$
- (C) $\frac{1}{8}$
- (D) $\frac{1}{4}$
- (E) $\frac{1}{2}$

Each inch on ruler $A$ is marked in equal $\frac{1}{8}$ -inch units, and each inch on ruler $B$ is marked in equal $\frac{1}{12}$ -inch units. When ruler $A$ is used, a side of a triangle measures 12 of the $\frac{1}{8}$ -inch units. When ruler $B$ is used, how many $\frac{1}{12}$ -inch units will the same side measure?
- (A) 8
- (B) 12
- (C) 18
- (D) 20
- (E) 24

Evaluate $60 + 2 + \frac{4}{8} + \frac{3}{500}$
- (A) 60.256
- (B) 62.43
- (C) 62.506
- (D) 62.53
- (E) 62.560

If $N \times \frac{7}{12} = \frac{7}{12} \times \frac{3}{14}$ , then $\frac{1}{N} = ?$
- (A) 8
- (B) $\frac{14}{3}$
- (C) $\frac{12}{7}$
- (D) $\frac{3}{11}$
- (E) $\frac{1}{6}$

If $n = 2.5 \times 10^{25}$ , then $\sqrt{n} = ?$
- (A) $0.5 \times 10^5$
- (B) $0.5 \times 10^{11}$
- (C) $5 \times 10^{\sqrt{24}}$
- (D) $5 \times 10^5$
- (E) $5 \times 10^{12}$

If $y = \frac{x – 2}{x + 3}$ , then $x$ cannot equal which of the following numbers?
- (A) -3
- (B) -2
- (C) 0
- (D) 2
- (E) 3

If eight pencils cost $0.42, how many pencils can be purchased with $2.10?
- (A) 16
- (B) 24
- (C) 30
- (D) 36
- (E) 40

A store sells 8-ounce containers of orange juice at $0.69 each and 12-ounce containers of orange juice at $0.95 each. How much money will be saved by purchasing a total of 48 ounces of orange juice in 12-ounce containers rather than 8-ounce containers?
- (A) $0.24
- (B) $0.32
- (C) $0.34
- (D) $0.48
- (E) $0.56

In the repeating decimal $0.31752 = 0.3175231752 \dots$ , the set of digits 31752 repeats endlessly. Which digit is in the $968^\text{th}$ place to the right of the decimal point?
- (A) 1
- (B) 2
- (C) 3
- (D) 5
- (E) 7

If $4\Delta / 6\Delta + \frac{5}{17} = 1$ , what digit does $\Delta$ represent?

On a certain map, 1.5 inches represents a distance of 45 miles. If two points on the map are 0.8 inch apart, how many miles apart are these two points?

Four lemons cost $0.68. At the same rate, 1 pound of lemons costs $1.19. How many lemons typically weigh 1 pound?

If the charge for a taxi ride is $2.50 for the first $\frac{1}{2}$ mile and $0.75 for each additional $\frac{1}{8}$ mile, how many miles did the taxi travel for a ride that cost $10.75?

One cubic foot of a certain metal weighs 8 pounds and costs $4.20 per pound. If 1 cubic foot is equivalent to 1728 cubic inches, what is the cost of 288 cubic inches of the same metal?

Operations with Fractions

Grid-In Questions

If $y = \sqrt{\frac{x+4}{2}}$ , what is the value of $y$ when $x = \frac{1}{2}$ ?
What fraction of $\frac{10}{9}$ is $\frac{5}{6}$ ?
If $25 \cdot \frac{x}{y} = 4$ , what is the value of $100 \cdot \frac{y}{x}$ ?
If $P = \left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right)\left(1 + \frac{1}{4}\right)\dots\left(1 + \frac{1}{17}\right)$ , where each factor is of the form $\left(1 + \frac{1}{k}\right)$ , what is the value of $P$ ?

Fraction Word Problems

Multiple Choice

If $\frac{3}{8}$ of a number is 6, what is $\frac{7}{8}$ of the same number?
- (A) 8
- (B) 12
- (C) 14
- (D) 16
- (E) 24

After Claire has read the first $\frac{5}{8}$ of a book, there are 120 pages left to read. How many pages of the book has Claire read?
- (A) 160
- (B) 200
- (C) 240
- (D) 300
- (E) 320

$\frac{4}{9}$ of 27 is $\frac{6}{5}$ of what number?
- (A) 5
- (B) 10
- (C) 12
- (D) 15
- (E) 20

If $\frac{3}{5}$ of a class that includes 10 girls are boys, how many students are in the class?
- (A) 15
- (B) 20
- (C) 21
- (D) 25
- (E) 30

What is the sum of all two-digit whole numbers in which one digit is $\frac{3}{4}$ of the other digit?
- (A) 77
- (B) 102
- (C) 129
- (D) 154
- (E) 231

Boris has $D$ dollars. If he lends $\frac{1}{4}$ of this amount of money and then spends $\frac{1}{3}$ of the money that he has left, how many dollars, in terms of $D$ , does Boris now have?
- (A) $\frac{D}{2}$
- (B) $\frac{D}{3}$
- (C) $\frac{D}{4}$
- (D) $\frac{11}{12}D$
- (E) $\frac{3}{4}D – \frac{1}{3}$

The value obtained by increasing $a$ by $\frac{1}{5}$ of its value is numerically equal to the value obtained by decreasing $b$ by $\frac{1}{2}$ of its value. Which equation expresses this fact?
- (A) $1.2a = 0.5b$
- (B) $0.2a = 0.5b$
- (C) $0.8a = 1.5b$
- (D) $1.2a = 1.5b$
- (E) $\frac{a}{0.2} = b – 0.5$

How many times must a jogger run around a circular $\frac{1}{4}$ mile track in order to have run $3 \frac{1}{2}$ miles?
- (A) 9
- (B) 12
- (C) 14
- (D) 15
- (E) 16

A chocolate bar that weighs $\frac{9}{16}$ of a pound is cut into seven equal parts. How much do three parts weigh?
- (A) $\frac{21}{122}$ pound
- (B) $\frac{27}{112}$ pound
- (C) $\frac{16}{63}$ pound
- (D) $\frac{47}{63}$ pound
- (E) $\frac{85}{112}$ pound

At a high school basketball game, $\frac{3}{5}$ of the students who attended were seniors; $\frac{1}{3}$ of the other students who attended were juniors, and the remaining 80 students who attended were all sophomores. How many seniors attended this game?
- (A) 120
- (B) 175
- (C) 180
- (D) 210
- (E) 300

Of the 75 people in a room, $\frac{2}{5}$ are college graduates. If $\frac{4}{9}$ of the students who are not college graduates are seniors in high school, how many people in the room are neither college graduates nor high school seniors?
- (A) 15
- (B) 20
- (C) 25
- (D) 36
- (E) 40

If $\frac{2}{3}$ of $\frac{3}{4}$ of a number is 24, what is $\frac{1}{4}$ of the same number?
- (A) 8
- (B) 12
- (C) 16
- (D) 20
- (E) 24

A man paints $\frac{3}{4}$ of a house in 2 days. If he continues to work at the same rate, how much more time will he need to paint the rest of the house?
- (A) $\frac{1}{4}$ day
- (B) $\frac{1}{2}$ day
- (C) $\frac{2}{3}$ day
- (D) 1 day
- (E) $\frac{4}{3}$ days

A water tank is $\frac{3}{5}$ full. After 12 gallons are poured out, the tank is $\frac{1}{3}$ full. How many gallons of water does the tank hold?
- (A) 25
- (B) 32
- (C) 35
- (D) 42
- (E) 45

In a school election, Susan received $\frac{2}{3}$ of the ballots cast, Mary received $\frac{1}{5}$ of the remaining ballots, and Bill received all of the other votes. If Bill received 48 votes, how many votes did Susan receive?
- (A) 75
- (B) 90
- (C) 120
- (D) 150
- (E) 180

After Arlene pumps gas into her car, the gas gauge moves from exactly $\frac{1}{8}$ full to exactly $\frac{7}{8}$ full. If gas costs $1.50 per gallon and Arlene is charged $18.00, what is the capacity, in gallons, of the gas tank?
- (A) 24
- (B) 20
- (C) 18
- (D) 16
- (E) 15

At the beginning of the day, the prices of stocks A and B are the same. At the end of the day, stock A has increased by $\frac{1}{10}$ of its original price, and stock B has decreased by $\frac{1}{10}$ of its original price. What fraction of the new price of stock B is the new price of stock A?
- (A) $\frac{2}{10}$
- (B) $\frac{9}{11}$
- (C) $\frac{9}{10}$
- (D) $\frac{11}{10}$
- (E) $\frac{11}{9}$

After $\frac{3}{4}$ of the people in a room leave, three people enter. If the remaining people are $\frac{1}{3}$ of the original number of people, how many people left the room?
- (A) 9
- (B) 18
- (C) 24
- (D) 27
- (E) 36

Multiple Choice

What is 20% of $\frac{2}{3}$ of 15?
- (A) 1
- (B) 2
- (C) 3
- (D) 4
- (E) 5

Which expression is equivalent to $\frac{2}{5}\%$ ?
- (A) 0.40
- (B) 0.04
- (C) 0.004
- (D) 0.0004
- (E) 0.00004

The fraction $\frac{0.25 + 0.15}{0.50}$ is equivalent to what percent?
- (A) 20%
- (B) 25%
- (C) 40%
- (D) 60%
- (E) 80%

In a movie theater, 480 of the 500 seats were occupied. What percent of the seats were NOT occupied?
- (A) 0.4%
- (B) 2%
- (C) 4%
- (D) 20%
- (E) 40%

After 2 months on a diet, John’s weight dropped from 168 pounds to 147 pounds. By what percent did John’s weight drop?
- (A) $12 \frac{1}{2}\%$
- (B) $14 \frac{2}{7}\%$
- (C) 21%
- (D) 25%
- (E) $28 \frac{4}{7}\%$

Which expression is equivalent to 0.01%?
- (A) $\sqrt{\frac{1}{100}}$
- (B) $\frac{0.01}{10}$
- (C) $\frac{1}{100}$
- (D) $\frac{\frac{1}{100}}{\frac{1}{10}}$
- (E) $\frac{1}{10}$

If the result of increasing $a$ by 300% of $a$ is $b$ , then $a$ is what percent of $b$ ?
- (A) 20%
- (B) 25%
- (C) $33 \frac{1}{3}\%$
- (D) 40%
- (E) $66 \frac{2}{3}\%$

After a 20% increase, the new price of a radio is $78.00. What was the original price of the radio?
- (A) $15.60
- (B) $60.00
- (C) $62.40
- (D) $65.00
- (E) $97.50

After a discount of 15%, the price of a shirt is $51. What was the original price of the shirt?
- (A) $44.35
- (B) $58.65
- (C) $60.00
- (D) $64.00
- (E) $65.00

Three students use a computer for a total of 3 hours. If the first student uses the computer 28% of the total time, and the second student uses the computer 52% of the total time, how many minutes does the third student use the computer?
- (A) 24
- (B) 30
- (C) 36
- (D) 42
- (E) 50

What is 20% of 25% of $\frac{4}{5}$ ?
- (A) 0.0025
- (B) 0.004
- (C) 0.005
- (D) 0.04
- (E) 0.05

In a factory that manufactures light bulbs, 0.04% of the bulbs manufactured are defective. It is expected that there will be one defective light bulb in what number of bulbs that are manufactured?
- (A) 2500
- (B) 1250
- (C) 1000
- (D) 500
- (E) 250

A discount of 25% on the price of a pair of shoes, followed by another discount of 8% on the new price of the shoes, is equivalent to a single discount of what percent of the original price?
- (A) 17%
- (B) 29%
- (C) 31%
- (D) 33%
- (E) 35%

If 8 kilograms of alcohol are added to 17 kilograms of pure water, what percent by weight of the resulting solution is alcohol?
- (A) 8%
- (B) 17%
- (C) 32%
- (D) 68%
- (E) 75%

If $a$ is 40% of $b$ , then $b$ exceeds $a$ by what percent of $a$ ?
- (A) 60%
- (B) 100%
- (C) 140%
- (D) 150%
- (E) 250%

Mixture Problems Lesson 2-2

Example Problem:

Solution Steps:

Final Mixture Check:

Mixture Problem Example

Problem:

Solution Steps:

Final Mixture Check:

Example Problem: Baseball Ticket Sales

Problem:

Solution Steps:

Final Check:

Practice Problems with Solutions

Lesson 2-3: Work Problems

Example 1:

Key Formula for Combined Work:

Example 2:

Problem:

Final Answer:

Notes:

Work Problems: Example and Practice Problems

Example Problem:

Practice Problems:

Number Problems: Lesson 2-4

Key Definitions:

Steps to Solving Word Problems:

Example Problem:

Examples of Solving Number Problems:

Example I:

Example II:

Example III:

Example IV:

Practice Problems with Solutions

1. Find six consecutive integers whose sum is 513.

2. Find five consecutive odd integers whose sum is 225.

3. Four times one odd integer is 14 less than three times the next even integer. Find the integers.

4. The largest of five consecutive integers is twice the smallest. Find the smallest integer.

5. The average of four consecutive odd integers is 16. Find the largest integer.

6. When the sum of three consecutive integers is divided by 9, the result is 7. Find the integers.

7. If each of three consecutive integers is divided by 3, the sum of the quotients is 84. Find the smallest integer.

8. The sum of three numbers is 65. The first is twice the second, and the third is twice the result obtained by subtracting 5 from the second. Find the largest number.

9. A baseball team played 150 games and won 32 more than it lost. How many did it lose?

10. One number exceeds another number by 12, and the sum of the smaller number and twice the larger number is 93. Find the numbers.

AGE PROBLEMS – LESSON 2-5

Problem 1:

Problem 2:

Problem 3:

Problem 4:

Problem 5:

Problem 6:

Problem 7:

Problem 8:

Problem 1:

Problem 2:

Problem 3:

Problem 4:

Problem 5:

Problem 6:

Problem 7:

Problem 8:

Multiple Choice Problems

Grid-In Questions

Divisibility & Factors (Multiple Choice)

Grid-In Questions

Lesson 3-4: Multiple Choice

Questions 9 and 10 Refer to the Diagram Above

Grid-In Questions

Fractions & Decimals: Lesson 3-5

Multiple Choice

Operations with Fractions

Grid-In Questions

Fraction Word Problems

Multiple Choice

Multiple Choice