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Basic Algebra Concepts: Lesson 2.2

Basic Algebra Concepts: Lesson 1

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:

  1. Define the Variable:
    Let xx represent the number of pounds of peanuts to be mixed.

  2. Fill the Chart with Known and Unknown Quantities:

ComponentUnit Value ($ per pound)Amount (Pounds)Value ($)
Peanuts0.700.70 xx 0.70x0.70x 
Cashews1.201.20 3030 (1.20)(30)(1.20)(30) 
Mixture1.001.00 x+30x + 30 (1.00)(x+30)(1.00)(x + 30) 
  1. 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)(0.70)(x) + (1.20)(30) = (1.00)(x + 30) 

  1. Simplify the Equation:

    0.70x+36=x+300.70x + 36 = x + 30 

    Subtract 0.70x0.70x  from both sides:

    36=0.30x+3036 = 0.30x + 30 

    Subtract 30 from both sides:

    6=0.30x6 = 0.30x 

    Divide both sides by 0.300.30 :

    x=20x = 20 
  1. Answer:
    x=20x = 20  pounds of peanuts.

Final Mixture Check:

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

  1. Define the Variable: Let xx represent the number of pints of the 30% acid solution to be added.

  2. Fill in the Chart:

ComponentUnit Value (% Acid)Amount (Pints)Value (Pints in Mixture)
30% Acid Solution0.300.30 xx 0.30x0.30x 
40% Acid Solution0.400.40 55 0.40⋅5=20.40 \cdot 5 = 2 
Mixture0.360.36 x+5x + 5 0.36⋅(x+5)0.36 \cdot (x + 5) 
  1. 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)0.30x + 0.40(5) = 0.36(x + 5) 

  1. Solve for xx : Expand and simplify the equation:

    0.30x+2=0.36x+1.80.30x + 2 = 0.36x + 1.8 

    Subtract 0.30x0.30x  and 1.81.8  from both sides:

    0.2=0.06x0.2 = 0.06x 

    Divide by 0.060.06 :

    x=0.20.06=313x = \frac{0.2}{0.06} = 3 \frac{1}{3}
  1. Answer: The amount of 30% acid solution to be added is 3133 \frac{1}{3}  pints.

Final Mixture Check:

  • Acid Content from 30% Solution: 0.30⋅3.333=1.00.30 \cdot 3.333 = 1.0 
  • Acid Content from 40% Solution: 0.40⋅5=2.00.40 \cdot 5 = 2.0 
  • Total Acid Content: 1.0+2.0=3.01.0 + 2.0 = 3.0 
  • Mixture Volume: 3.333+5=8.3333.333 + 5 = 8.333 
  • Final Acid Percentage: 3.08.333≈0.36\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:

  1. Define the Variables:

    • Let xx  represent the number of reserved seats sold.
    • The number of general admission tickets sold will be 500−x500 – x .

  2. Fill in the Chart:

ComponentUnit Price ($)Number of Tickets SoldTotal Value ($)
Reserved Seats4.00xx 4x4x 
General Admission3.00500−x500 – x 3(500−x)3(500 – x) 
Total—500500 1,7601,760 
  1. 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)=1,7604x + 3(500 – x) = 1,760 

  1. Solve the Equation: Expand and simplify:

    4x+1,500−3x=1,7604x + 1,500 – 3x = 1,760 

    Combine like terms:

    x+1,500=1,760x + 1,500 = 1,760 

    Subtract 1,500 from both sides:

    x=260x = 260 
  1. Answer: The number of reserved seats sold is 260.

Final Check:

  • Revenue from Reserved Seats: 4⋅260=1,0404 \cdot 260 = 1,040 
  • Revenue from General Admission: 3⋅(500−260)=3⋅240=7203 \cdot (500 – 260) = 3 \cdot 240 = 720 
  • Total Revenue: 1,040+720=1,7601,040 + 720 = 1,760 

 

Practice Problems with Solutions

  1. 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 xx  represent the amount of silver in grams, and 30−x30 – x the amount of gold.

    • Silver Value: 4x4x 
    • Gold Value: 24(30−x)24(30 – x) 
    • Mixture Value: 12(30)12(30) 

    Equation:

    4x+24(30−x)=12(30)4x + 24(30 – x) = 12(30) 

    Solve:

    4x+720−24x=3604x + 720 – 24x = 360 
    −20x+720=360-20x + 720 = 360 
    −20x=−360-20x = -360 
    x=18 gramsx = 18 \, \text{grams} 

    Answer: 18 grams of silver.

  1. 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 xx represent the amount of pure acid to add.

    • Current Acid: 0.25(200)=500.25(200) = 50 
    • Total Solution: 200+x200 + x 
    • Acid in New Solution: 50+x50 + x 
    • New Solution Concentration: 0.4(200+x)0.4(200 + x) 

    Equation:

    50+x=0.4(200+x)50 + x = 0.4(200 + x) 

    Solve:

    50+x=80+0.4x50 + x = 80 + 0.4x 
    0.6x=300.6x = 30 
    x=50 gramsx = 50 \, \text{grams} 

    Answer: Add 50 grams of pure acid.

  1. 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 xx  represent the gallons of water added.

    • Salt: 0.8(30)=24 gallons0.8(30) = 24 \, \text{gallons} 
    • New Solution Volume: 30+x30 + x 
    • Salt Concentration: 0.6(30+x)0.6(30 + x) 

    Equation:

    24=0.6(30+x)24 = 0.6(30 + x) 

    Solve:

    24=18+0.6x24 = 18 + 0.6x 
    6=0.6x6 = 0.6x 
    x=10 gallonsx = 10 \, \text{gallons} 

    Answer: Add 10 gallons of water.

  1. 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 xx represent the quarts of grape juice.

    • Grape Juice Value: 1.20x1.20x 
    • Apple Juice Value: 0.90(3)=2.70.90(3) = 2.7 
    • Mixture Value: 1.00(x+3)1.00(x + 3) 

    Equation:

    1.20x+2.7=1.00(x+3)1.20x + 2.7 = 1.00(x + 3) 

    Solve:

    1.20x+2.7=x+31.20x + 2.7 = x + 3 
    0.20x=0.30.20x = 0.3 
    x=1.5 quartsx = 1.5 \, \text{quarts} 

    Answer: Add 1.5 quarts of grape juice.

  1. 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 xx  represent the number of balcony tickets, and 540−x540 – x  the number of orchestra tickets.

    • Balcony Tickets Value: 15x15x 
    • Orchestra Tickets Value: 20(540−x)20(540 – x) 

    Equation:

    15x+20(540−x)=9,75015x + 20(540 – x) = 9,750 

    Solve:

    15x+10,800−20x=9,75015x + 10,800 – 20x = 9,750 
    −5x=−1,050-5x = -1,050 
    x=210 ticketsx = 210 \, \text{tickets} 

    Answer: 210 balcony tickets were sold.

  1. 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 xx  represent the number of nickels, and 34−x34 – x the number of dimes.

    • Nickels Value: 0.05x0.05x 
    • Dimes Value: 0.10(34−x)0.10(34 – x) 

    Equation:

    0.05x+0.10(34−x)=2.200.05x + 0.10(34 – x) = 2.20 

    Solve:

    0.05x+3.4−0.10x=2.200.05x + 3.4 – 0.10x = 2.20 
    −0.05x=−1.20-0.05x = -1.20 
    x=24 nickelsx = 24 \, \text{nickels} 

    Answer: The boy has 24 nickels.

Lesson 2-3: Work Problems

The key concept in work problems is that if xx equals the amount of time it takes to complete a specific job, then 1x\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:

14 of the lawn per hour.\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:

Combined Rate=1x1+1x2\text{Combined Rate} = \frac{1}{x_1} + \frac{1}{x_2}

where x1x_1 and x2x_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:

18+16 pools per hour.\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 xx  represent the total time it takes for both pipes to fill the pool together. The work done by each pipe in xx  hours is:

  • Pipe 1: x8 pools\frac{x}{8} \, \text{pools} 
  • Pipe 2: x6 pools 

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

x8+x6=1\frac{x}{8} + \frac{x}{6} = 1 

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

24⋅x8+24⋅x6=24⋅124 \cdot \frac{x}{8} + 24 \cdot \frac{x}{6} = 24 \cdot 1 
3x+4x=243x + 4x = 24 
7x=247x = 24 
x=247=3 37 hours.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: x8+x6=3\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:

  1. Let xx  = the number of minutes it would take for John and Jim working together to polish the car.

  2. Rates:

    • John’s rate: 145\frac{1}{45} of the job per minute.
    • Jim’s rate: 130\frac{1}{30}  of the job per minute.

  3. In xx minutes:

    • John completes x45\frac{x}{45}  of the job.
    • Jim completes x30\frac{x}{30} of the job.

  4. Together, their total work equals 1 job:

    x45+x30=1\frac{x}{45} + \frac{x}{30} = 1 

  5. Eliminate fractions: Multiply through by the least common multiple (LCM) of 45 and 30, which is 90:

    90⋅x45+90⋅x30=90⋅190 \cdot \frac{x}{45} + 90 \cdot \frac{x}{30} = 90 \cdot 1 
    2x+3x=902x + 3x = 90 

  6. Combine and solve:

    5x=905x = 90 
    x=18 minutes.x = 18 \, \text{minutes.} 

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

Practice Problems:

  1. 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 112\frac{1}{12}. Multiply this rate by 8 hours to find the fraction of the pool filled.

  1. Question: If Brutus can eat 16 hamburgers in 1 hour, how many can he eat in 3 hours?

    Hint: Multiply his rate (16 hamburgers/hour16 \, \text{hamburgers/hour} ) by 3 hours.

  1. 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 1128\frac{1}{128} . Multiply this by 40 to find the fraction of the ditch dug.

  1. 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 (12 jobs/hour\frac{1}{2} \, \text{jobs/hour}  and 11.5 jobs/hour\frac{1}{1.5} \, \text{jobs/hour}) and add them. Set the total rate equal to 1x\frac{1}{x}, where xx  is the time taken.

  1. 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 (14+13\frac{1}{4} + \frac{1}{3} ) and set it equal to 1x\frac{1}{x}, where xx  is the time taken.

  1. 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 (14.5+14+13.5\frac{1}{4.5} + \frac{1}{4} + \frac{1}{3.5} ) and set it equal to 1x\frac{1}{x}, where xx  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:

  1. Integers:

    • Positive and negative whole numbers, including zero.
    • Example: {…,−3,−2,−1,0,1,2,3,… }\{ \dots, -3, -2, -1, 0, 1, 2, 3, \dots \}Consecutive Integers:
    • Integers that differ by 1.
    • Example: 7,8,9,107, 8, 9, 10 or represented algebraically as x,x+1,x+2x, x+1, x+2 , where xx  is any integer.

  3. Consecutive Odd Integers:

    • Odd integers that differ by 2.
    • Example: 15,17,19,2115, 17, 19, 21 or represented algebraically as x,x+2,x+4x, x+2, x+4 , where xx  is any odd integer.

  4. Consecutive Even Integers:

    • Even integers that differ by 2.
    • Example: 0,2,4,6,80, 2, 4, 6, 8  or represented algebraically as x,x+2,x+4x, x+2, x+4 , where xx  is any even integer.

Steps to Solving Word Problems:

  1. Define Your Variable:

    • Identify what the problem is asking and represent it with a variable (e.g., xx ).

  2. Set Up Your Equation:

    • Translate the problem into an algebraic equation using the relationships between numbers.

  3. Solve the Equation:

    • Perform operations to isolate the variable and find the solution.

  4. 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:

  1. Define your variable:

    • Let the integers be x,x+1,x+2x, x+1, x+2 .

  2. Set up the equation:

    • The sum of the smallest (xx ) and the largest (x+2x+2 ) equals 28: x+(x+2)=28x + (x+2) = 28 

  3. Solve the equation:

    2x+2=282x + 2 = 28 
    2x=262x = 26 
    x=13x = 13 

  4. Find all integers:

    • The integers are 13,14,1513, 14, 15 

  5. Check your answer:

    • Smallest + Largest = 13+15=2813 + 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:

  1. Define the variable:
    Let xx  represent the number.

  2. Set up the equation:

    6x−2=5x+26x – 2 = 5x + 2 

  3. Solve the equation:
    Subtract 5x5x  from both sides and add 2 to both sides:

    x=4x = 4 

  4. Check your answer:
    Substituting x=4x = 4 :

    6(4)−2=5(4)+2or24−2=20+26(4) – 2 = 5(4) + 2 \quad \text{or} \quad 24 – 2 = 20 + 2 

    Both sides equal 2 , so the solution is correct.

Answer: x=4x = 4 .

Example II:

Question: Find three consecutive integers whose sum is 84.

Solution:

  1. Define the variable:
    Let xx  be the first integer, x+1x+1  the second, and x+2x+2  the third.

  2. Set up the equation:

    x+(x+1)+(x+2)=84x + (x+1) + (x+2) = 84 

  3. Solve the equation:

    3x+3=84⇒3x=81⇒x=273x + 3 = 84 \quad \Rightarrow \quad 3x = 81 \quad \Rightarrow \quad x = 27 

  4. Find the integers:
    The integers are 27,28, and 2927, 28, \text{ and } 29 

Answer: 27,28,2927, 28, 29 

Example III:

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

Solution:

  1. Define the variables:
    Let xx  be the first even integer, x+2x+2 the second, and x+4x+4  the third.

  2. Set up the equation:

    x+2(x+4)=134x + 2(x+4) = 134 

  3. Solve the equation:
    Expand and simplify:

    x+2x+8=134⇒3x+8=134⇒3x=126⇒x=42x + 2x + 8 = 134 \quad \Rightarrow \quad 3x + 8 = 134 \quad \Rightarrow \quad 3x = 126 \quad \Rightarrow \quad x = 42 

  4. Find the integers:
    The integers are 42,44, and 4642, 44, \text{ and } 46 

Answer: 42,44,4642, 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:

  1. Define the variables:
    Let xx  be the first odd integer, x+2x+2 the second, x+4x+4  the third, and x+6x+6 the fourth.

  2. Set up the equation:

    3(x)+2(x+4)=2[(x+2)+(x+6)]+113(x) + 2(x+4) = 2[(x+2) + (x+6)] + 11 

  3. Simplify and solve the equation:
    Expand:

    3x+2x+8=2(2x+8)+113x + 2x + 8 = 2(2x + 8) + 11 

    Simplify:

    5x+8=4x+16+115x + 8 = 4x + 16 + 11 

    Combine terms:

    5x+8=4x+27⇒x=195x + 8 = 4x + 27 \quad \Rightarrow \quad x = 19 

  4. Find the integers:
    The integers are 19,21,23, and 2519, 21, 23, \text{ and } 25 

Answer: 19,21,23,2519, 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+5x, x+1, x+2, x+3, x+4, x+5 
  • Sum: x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)=513x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 513 Simplify: 6x+15=5136x + 15 = 513 
    6x=498⇒x=836x = 498 \quad \Rightarrow \quad x = 83 
  • The integers are 83,84,85,86,87,8883, 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+8x, x+2, x+4, x+6, x+8 
  • Sum: x+(x+2)+(x+4)+(x+6)+(x+8)=225x + (x+2) + (x+4) + (x+6) + (x+8) = 225 Simplify: 5x+20=2255x + 20 = 225  Solve: 5x=205⇒x=415x = 205 \quad \Rightarrow \quad x = 41 
  • The integers are 41,43,45,47,4941, 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 xx . The next even integer is x+1x+1 .
  • Equation: 4x=3(x+1)−144x = 3(x+1) – 14 Simplify: 4x=3x+3−14⇒x=−114x = 3x + 3 – 14 \quad \Rightarrow \quad x = -11 
  • The integers are −11-11  (odd) and −10-10  (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+4x, x+1, x+2, x+3, x+4 
  • Largest: x+4x+4 , Smallest: xx .
    Equation: x+4=2xx+4 = 2x Solve: x=4x = 4 
  • The integers are 4,5,6,7,84, 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+6x, x+2, x+4, x+6 .
  • Average: x+(x+2)+(x+4)+(x+6)4=16\frac{x + (x+2) + (x+4) + (x+6)}{4} = 16 Simplify: 4x+124=16⇒x+3=16⇒x=13\frac{4x + 12}{4} = 16 \quad \Rightarrow \quad x + 3 = 16 \quad \Rightarrow \quad x = 13 
  • The integers are 13,15,17,1913, 15, 17, 19 . The largest is 1919.

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+2x, x+1, x+2 .
  • Equation: x+(x+1)+(x+2)9=7\frac{x + (x+1) + (x+2)}{9} = 7  Simplify: 3x+39=7⇒3x+3=63⇒3x=60⇒x=20\frac{3x + 3}{9} = 7 \quad \Rightarrow \quad 3x + 3 = 63 \quad \Rightarrow \quad 3x = 60 \quad \Rightarrow \quad x = 20 
  • The integers are 20,21,2220, 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+2x, x+1, x+2 .
  • Equation: x3+x+13+x+23=84\frac{x}{3} + \frac{x+1}{3} + \frac{x+2}{3} = 84 Simplify: 3x+33=84⇒x+1=84⇒x=83\frac{3x+3}{3} = 84 \quad \Rightarrow \quad x+1 = 84 \quad \Rightarrow \quad x = 83 
  • The integers are 83,84,8583, 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 xx .
    First: 2x2x , Third: 2(x−5)2(x-5) .
    Equation: 2x+x+2(x−5)=652x + x + 2(x-5) = 65 Simplify: 5x−10=65⇒5x=75⇒x=155x – 10 = 65 \quad \Rightarrow \quad 5x = 75 \quad \Rightarrow \quad x = 15 
  • The numbers are 30,15,2030, 15, 20 . The largest is 3030 .

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 xx .
    Wins: x+32x + 32 .
    Equation: x+(x+32)=150x + (x+32) = 150 Simplify: 2x+32=150⇒2x=118⇒x=592x + 32 = 150 \quad \Rightarrow \quad 2x = 118 \quad \Rightarrow \quad x = 59 
  • Losses: 5959 .

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 xx .
    Larger: x+12x+12 .
    Equation: x+2(x+12)=93x + 2(x+12) = 93  Simplify: x+2x+24=93⇒3x=69⇒x=23x + 2x + 24 = 93 \quad \Rightarrow \quad 3x = 69 \quad \Rightarrow \quad x = 23 
  • The numbers are 2323 and 3535 .

AGE PROBLEMS – LESSON 2-5

Key Concept:

  • If a person is currently xx  years old:
    • 8 years ago: x−8x – 8 
    • In 10 years: x+10x + 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 xx  = Sally’s age now.
      • Father’s age = 2x2x.
    2. Chart of ages:
      • Sally now: xx , in 2 years: x+2x + 2 
      • Father now: 2x2x , in 2 years: 2x+22x + 2 
    3. Set up the equation: (x+2)+(2x+2)=67(x + 2) + (2x + 2) = 67 
    4. Solve the equation: 3x+4=67→3x=63→x=213x + 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(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 xx  = Sue’s age now.
      • Katie’s age = x+2x + 2 .
    2. Chart of ages:
      • Sue now: xx , in 8 years: x+8x + 8 
      • Katie now: x+2x + 2 , in 8 years: x+10x + 10 
    3. Set up the equation: (x+8)+(x+10)=48(x + 8) + (x + 10) = 48 
    4. Solve the equation: 2x+18=48→2x=30→x=152x + 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(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 xx  = Daughter’s current age.
      • Father’s age = x+30x + 30 .
    2. Chart of ages:
      • Daughter now: xx , in 9 years: x+9x + 9 
      • Father now: x+30x + 30 , in 9 years: x+39x + 39 
    3. Set up the equation: 12(x+30)=x+9\frac{1}{2}(x + 30) = x + 9 
    4. Solve the equation: x+30=2(x+9)→x+30=2x+18→30=x+18→x=12x + 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: 12(42)=12+9\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:

  1. Let xx = Roger’s current age.
  2. If he were 32 years older, his age would be x+32x + 32 
  3. According to the problem: x+32=3xx + 32 = 3x 
  4. Solve the equation: 32=3x−x→32=2x→x=1632 = 3x – x \quad \rightarrow \quad 32 = 2x \quad \rightarrow \quad x = 16 
  5. 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:

  1. Let xx  = Sam’s age.
  2. Brad’s age is x+12x + 12 
  3. If Brad were 8 years older, his age would be (x+12)+8=x+20(x + 12) + 8 = x + 20 
  4. According to the problem: x+20=2xx + 20 = 2x 
  5. Solve the equation: 20=2x−x→x=2020 = 2x – x \quad \rightarrow \quad x = 20 
  6. 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:

  1. Let xx = Krista’s age.
  2. Barrie’s age is x+2x + 2 
  3. In 15 years, Barrie’s age will be (x+2)+15=x+17(x + 2) + 15 = x + 17 
  4. According to the problem: x+17=2x+2x + 17 = 2x + 2 
  5. Solve the equation: x+17=2x+2→17−2=x→x=15x + 17 = 2x + 2 \quad \rightarrow \quad 17 – 2 = x \quad \rightarrow \quad x = 15 
  6. Barrie’s current age is x+2=15+2=17x + 2 = 15 + 2 = 17 
  7. In 6 years, Barrie’s age will be 17+6=2317 + 6 = 23 
  8. 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:

  1. Let xx = Marcus’s age.
  2. John’s age is x+10x + 10 
  3. John’s age 5 years from now: (x+10)+5=x+15(x + 10) + 5 = x + 15 
  4. Marcus’s age 5 years ago: x−5x – 5 
  5. According to the problem: 3(x+15)=5(x−5)3(x + 15) = 5(x – 5) 
  6. Solve the equation: 3x+45=5x−25→45+25=5x−3x→70=2x→x=353x + 45 = 5x – 25 \quad \rightarrow \quad 45 + 25 = 5x – 3x \quad \rightarrow \quad 70 = 2x \quad \rightarrow \quad x = 35 
  7. John’s current age is x+10=35+10=45x + 10 = 35 + 10 = 45 
  8. 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:

  1. Let xx  = Danielle’s current age.
  2. Kristen’s age in 1 year: 4(x+1)4(x + 1) 
  3. In 10 years: 4(x+1)+104(x + 1) + 10
  4. According to the problem: 4(x+1)+10=2(x+11)4(x + 1) + 10 = 2(x + 11) 
  5. Solve the equation: 4x+4+10=2x+22→4x−2x=22−14→2x=8→x=44x + 4 + 10 = 2x + 22 \quad \rightarrow \quad 4x – 2x = 22 – 14 \quad \rightarrow \quad 2x = 8 \quad \rightarrow \quad x = 4 
  6. Kristen’s current age is 4(x+1)=4(4+1)=204(x + 1) = 4(4 + 1) = 20 
  7. 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:

  1. Let xx = Daughter’s current age.
  2. Woman’s current age: x+30x + 30 
  3. According to the problem: x+30=3x+6x + 30 = 3x + 6 
  4. Solve the equation: x+30=3x+6→30−6=3x−x→24=2x→x=12x + 30 = 3x + 6 \quad \rightarrow \quad 30 – 6 = 3x – x \quad \rightarrow \quad 24 = 2x \quad \rightarrow \quad x = 12 
  5. Daughter’s age in 5 years: x+5=12+5=17x + 5 = 12 + 5 = 17 
  6. 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:

  1. Let xx = Joe’s current age.
  2. Ralph’s current age: 2x+32x + 3 
  3. Ralph’s age 3 years ago: (2x+3)−3=2x(2x + 3) – 3 = 2x 
  4. Joe’s age 3 years ago: x−3x – 3 
  5. According to the problem: 2x=4(x−3)2x = 4(x – 3) 
  6. Solve the equation: 2x=4x−12→12=4x−2x→2x=12→x=62x = 4x – 12 \quad \rightarrow \quad 12 = 4x – 2x \quad \rightarrow \quad 2x = 12 \quad \rightarrow \quad x = 6 
  7. Joe’s age in 10 years: x+10=6+10=16x + 10 = 6 + 10 = 16 
  8. 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:

  1. Let xx = Number of years ago.
  2. Father’s age xx years ago: 40−x40 – x 
  3. Lisa’s age xx years ago: 15−x15 – x 
  4. According to the problem: 40−x=6(15−x)40 – x = 6(15 – x) 
  5. Solve the equation: 40−x=90−6x→40−90=−6x+x→−50=−5x→x=1040 – x = 90 – 6x \quad \rightarrow \quad 40 – 90 = -6x + x \quad \rightarrow \quad -50 = -5x \quad \rightarrow \quad x = 10 
  6. 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:

  1. Let xx = Roger’s current age.
  2. If he were 32 years older, his age would be x+32x + 32 
  3. According to the problem: x+32=3xx + 32 = 3x 
  4. Solve the equation: 32=3x−x→32=2x→x=1632 = 3x – x \quad \rightarrow \quad 32 = 2x \quad \rightarrow \quad x = 16 
  5. 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:

  1. Let xx = Sam’s age.
  2. Brad’s age is x+12x + 12 
  3. If Brad were 8 years older, his age would be (x+12)+8=x+20(x + 12) + 8 = x + 20 
  4. According to the problem: x+20=2xx + 20 = 2x 
  5. Solve the equation: 20=2x−x→x=2020 = 2x – x \quad \rightarrow \quad x = 20 
  6. 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:

  1. Let xx = Krista’s age.
  2. Barrie’s age is x+2x + 2 
  3. In 15 years, Barrie’s age will be (x+2)+15=x+17(x + 2) + 15 = x + 17 
  4. According to the problem: x+17=2x+2x + 17 = 2x + 2 
  5. Solve the equation: x+17=2x+2→17−2=x→x=15x + 17 = 2x + 2 \quad \rightarrow \quad 17 – 2 = x \quad \rightarrow \quad x = 15 
  6. Barrie’s current age is x+2=15+2=17x + 2 = 15 + 2 = 17 
  7. In 6 years, Barrie’s age will be 17+6=2317 + 6 = 23 
  8. 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:

  1. Let xx = Marcus’s age.
  2. John’s age is x+10x + 10 
  3. John’s age 5 years from now: (x+10)+5=x+15(x + 10) + 5 = x + 15 
  4. Marcus’s age 5 years ago: x−5x – 5 
  5. According to the problem: 3(x+15)=5(x−5)3(x + 15) = 5(x – 5) 
  6. Solve the equation: 3x+45=5x−25→45+25=5x−3x→70=2x→x=353x + 45 = 5x – 25 \quad \rightarrow \quad 45 + 25 = 5x – 3x \quad \rightarrow \quad 70 = 2x \quad \rightarrow \quad x = 35 
  7. John’s current age is x+10=35+10=45x + 10 = 35 + 10 = 45 
  8. 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:

  1. Let xx = Danielle’s current age.
  2. Kristen’s age in 1 year: 4(x+1)4(x + 1) 
  3. In 10 years: 4(x+1)+104(x + 1) + 10 
  4. According to the problem: 4(x+1)+10=2(x+11)4(x + 1) + 10 = 2(x + 11) 
  5. Solve the equation: 4x+4+10=2x+22→4x−2x=22−14→2x=8→x=44x + 4 + 10 = 2x + 22 \quad \rightarrow \quad 4x – 2x = 22 – 14 \quad \rightarrow \quad 2x = 8 \quad \rightarrow \quad x = 4 
  6. Kristen’s current age is 4(x+1)=4(4+1)=204(x + 1) = 4(4 + 1) = 20 
  7. 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:

  1. Let xx = Daughter’s current age.
  2. Woman’s current age: x+30x + 30 
  3. According to the problem: x+30=3x+6x + 30 = 3x + 6 
  4. Solve the equation: x+30=3x+6→30−6=3x−x→24=2x→x=12x + 30 = 3x + 6 \quad \rightarrow \quad 30 – 6 = 3x – x \quad \rightarrow \quad 24 = 2x \quad \rightarrow \quad x = 12 
  5. Daughter’s age in 5 years: x+5=12+5=17x + 5 = 12 + 5 = 17 
  6. 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:

  1. Let xx = Joe’s current age.
  2. Ralph’s current age: 2x+32x + 3 
  3. Ralph’s age 3 years ago: (2x+3)−3=2x(2x + 3) – 3 = 2x 
  4. Joe’s age 3 years ago: x−3x – 3 
  5. According to the problem: 2x=4(x−3)2x = 4(x – 3) 
  6. Solve the equation: 2x=4x−12→12=4x−2x→2x=12→x=62x = 4x – 12 \quad \rightarrow \quad 12 = 4x – 2x \quad \rightarrow \quad 2x = 12 \quad \rightarrow \quad x = 6 
  7. Joe’s age in 10 years: x+10=6+10=16x + 10 = 6 + 10 = 16 
  8. 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:

  1. Let xx = Number of years ago.
  2. Father’s age xx years ago: 40−x40 – x 
  3. Lisa’s age xx years ago: 15−x15 – x 
  4. According to the problem: 40−x=6(15−x)40 – x = 6(15 – x) 
  5. Solve the equation: 40−x=90−6x→40−90=−6x+x→−50=−5x→x=1040 – x = 90 – 6x \quad \rightarrow \quad 40 – 90 = -6x + x \quad \rightarrow \quad -50 = -5x \quad \rightarrow \quad x = 10 
  6. Answer: 10 years ago.

Multiple Choice Problems

  1. If a=9×23a = 9 \times 23 and b=9×124b = 9 \times 124 , what is the value of b−ab – a ?
    • (A) 901
    • (B) 903
    • (C) 906
    • (D) 909
    • (E) 911
  1. 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
  1. 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
  1. 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
  1. 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
  1. Let ∗*  represent one of the four basic arithmetic operations such that, for any nonzero real number rr :
    • r∗0=rr^* 0 = r and r∗r=0r^* r = 0 
      Which arithmetic operation(s) does the symbol ∗* represent?
    • (A) + only
    • (B) −–  only
    • (C) + and −– 
    • (D) × only
    • (E) +, −– , or ×
  1. 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
  1. 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
  1. Let #\# represent one of the four basic arithmetic operations such that, for any nonzero real numbers rr and ss:
    • r#0=rr \# 0 = r  and r#s=s#rr \# s = s \# r .
      Which arithmetic operation(s) does the symbol #\# represent?
    • (A) + only
    • (B) × only
    • (C) −–  only
    • (D) −– and ×
    • (E) + and ÷÷ 
  1. If w=(6)(6)(6)w = (6)(6)(6) , x=(5)(6)(7)x = (5)(6)(7) , and y=(4)(6)(8)y = (4)(6)(8) , which inequality statement is true?
    • (A) x<y<wx < y < w 
    • (B) w<x<yw < x < y 
    • (C) y<w<xy < w < x 
    • (D) y<x<wy < x < w 
    • (E) w<y<xw < y < x 
  1. If xx  and yy  are positive integers, 2x+y<292x + y < 29 , and y>4y > 4 , what is the greatest possible value of x−yx – y ?
    • (A) 5
    • (B) 6
    • (C) 7
    • (D) 8
    • (E) 9

Grid-In Questions

  1. If 24×42=16x2^4 \times 4^2 = 16^x , then x=?x = ? 

  2. If 3(x−1)=93^{(x-1)} = 9  4(y+2)=644^{(y+2)} = 64 , what is the value of xy\frac{x}{y}?

  3. If p+p+pp⋅p⋅p=12\frac{p + p + p}{p \cdot p \cdot p} = 12  and p>0p > 0 , what is the value of pp ?

Divisibility & Factors (Multiple Choice)

  1. If the sum of the factors of 18 is SS and the sum of the prime numbers less than 18 is PP , then PP  exceeds SS  by what number?
    • (A) 19
    • (B) 17
    • (C) 15
    • (D) 13
    • (E) 11
  1. When a whole number NN  is divided by 5, the quotient is 13 and the remainder is 4. What is the value of NN ?
    • (A) 55
    • (B) 59
    • (C) 65
    • (D) 69
    • (E) 79
  1. If pp  is divisible by 3 and qq  is divisible by 4, then pqpq  must be divisible by each of the following EXCEPT:
    • (A) 3
    • (B) 4
    • (C) 6
    • (D) 9
    • (E) 12
  1. Which number has the most factors?
    • (A) 12
    • (B) 18
    • (C) 25
    • (D) 70
    • (E) 100
  1. Which number is divisible by 2 and by 3?
    • (A) 112
    • (B) 4308
    • (C) 6122
    • (D) 23,451
    • (E) 701,456
  1. All numbers that are divisible by both 3 and 10 are also divisible by:
    • (A) 4
    • (B) 9
    • (C) 12
    • (D) 15
    • (E) 20
  1. If xx represents any even number and yy represents any odd number, which of the following numbers is even?
    • (A) y+2y + 2 
    • (B) x−1 
    • (C) (x+1)(y−1)(x + 1)(y – 1) 
    • (D) y(y+2)y(y + 2) 
    • (E) x+yx + y 
  1. For how many different positive integers pp is 105p\frac{105}{p}  also an integer?
    • (A) Five
    • (B) Six
    • (C) Seven
    • (D) Eight
    • (E) Nine
  1. If nn is an odd integer, which expression always represents an odd integer?
    • (A) (2n−1)2(2n – 1)^2 
    • (B) n2+2n+1n^2 + 2n + 1 
    • (C) (n−1)2(n – 1)^2 
    • (D) n+12\frac{n + 1}{2}
    • (E) 3n+13^n + 1 
  1. If k−1k – 1 is a multiple of 4, what is the next larger multiple of 4?
    • (A) k+1k + 1 
    • (B) 4k4k 
    • (C) k−5k – 5 
    • (D) k+3k + 3 
    • (E) 4(k−1)4(k – 1) 
  1. After mm marbles are put into nn jars, each jar contains the same number of marbles, with two marbles remaining. In terms of mm and nn, how many marbles were put into each jar?
    • (A) mn+2\frac{m}{n} + 2 
    • (B) mn−2\frac{m}{n} – 2 
    • (C) m+2n\frac{m + 2}{n}
    • (D) m−2n\frac{m – 2}{n}
    • (E) mnn+2\frac{mn}{n + 2}
  1. When pp  is divided by 4, the remainder is 3, and when pp  is divided by 3, the remainder is 0. What is a possible value of pp ?
    • (A) 8
    • (B) 11
    • (C) 15
    • (D) 18
    • (E) 21
  1. If nn  is an integer and n2+5n^2 + 5  is an odd integer, then which statement(s) must be true?
    I. n2−1n^2 – 1 is even.
    II. nn  is even.
    III. 5n5n  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
  1. 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
  1. If nn  is any even integer, what is the remainder when (n+1)2(n + 1)^2  is divided by 4?
    • (A) 0
    • (B) 1
    • (C) 2
    • (D) 3
    • (E) 4
  1. 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

  1. When aa  is divided by 7, the remainder is 5; when bb  is divided by 7, the remainder is 4. What is the remainder when a+ba + b  is divided by 7?

  2. How many integers from −3000-3000  to 30003000 , inclusive, are divisible by 3?

  3. When a positive integer kk  is divided by 6, the remainder is 1. What is the remainder when 5k5k  is divided by 3?

  4. What is the smallest positive integer pp for which 22p+12^{2p+1}  is not a prime number?

Lesson 3-4: Multiple Choice

  1. If 2b=−32b = -3 , what is the value of 1−4b1 – 4b ?
    • (A) -7
    • (B) -5
    • (C) 5
    • (D) 6
    • (E) 7
  1. p2(2−5)+(−p)2=?p^2(2 – 5) + (-p)^2 = ? 
    • (A) −4p2-4p^2 
    • (B) −p2 
    • (C) −2p2-2p^2 
    • (D) 2p22p^2 
    • (E) 4p24p^2 
  1. If n+5n + 5  is an odd integer, then nn could be which of the following?
    • (A) 1
    • (B) -1
    • (C) -2
    • (D) -3
    • (E) -7
  1. 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
  1. For which value of kk  is the value of k(k+2)(k+1)k(k + 2)(k + 1)  negative?
    • (A) -2
    • (B) -1
    • (C) 0
    • (D) 2
    • (E) 3
  1. Which of the following statements must be true when a<0a < 0  and b>0b > 0 ?
    • I. a+b>0a + b > 0 
    • II. b−a>0 
    • III. a⋅ab>0a \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
  1. (−2)3+(−3)2=?(-2)^3 + (-3)^2 = ? 
    • (A) -12
    • (B) -1
    • (C) 0
    • (D) 1
    • (E) 2

Questions 9 and 10 Refer to the Diagram Above

  1. Which of the following statements must be true?
    • I. c2<cc^2 < c 
    • II. a2>ca^2 > c 
    • III. b<1bb < \frac{1}{b}
    • (A) I only
    • (B) I and II only
    • (C) I and III only
    • (D) I, II, and III
    • (E) None
  1. Which of the following statements must be true?
    • I. ad>bad > b 
    • II. ab>adab > ad 
    • III. 1a>1b\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
  1. If (y−3)2=16(y – 3)^2 = 16 , what is the smallest possible value of y2y^2 ?
    • (A) -4
    • (B) 1
    • (C) 7
    • (D) 16
    • (E) 49
  1. If XX  represents the sum of the 10 greatest negative integers and YY  represents the sum of the 10 least positive integers, which of the following must be true?
    • I. X+Y<0X + Y < 0 
    • II. Y−X=2YY – X = 2Y 
    • III. X2=Y2 
    • (A) None
    • (B) I only
    • (C) III only
    • (D) I and II only
    • (E) II and III only
  1. If a2b3c>0a^2 b^3 c > 0 , which of the following statements must be true?
    • I. bc>0bc > 0 
    • II. ac>0 
    • III. ab>0ab > 0 
    • (A) I only
    • (B) I and II only
    • (C) I and III only
    • (D) II and III only
    • (E) I, II, and III
  1. If a2b3c5a^2 b^3 c^5  is negative, which product is always negative?
    • (A) bcbc 
    • (B) b2cb^2 c 
    • (C) acac 
    • (D) abab 
    • (E) bc2bc^2 
  1. If a≠0a \neq 0 , which of the following statements must be true?
    • I. (−a)2=a2−2a2(-a)^2 = a^2 – 2a^2 
    • II. a−b=−(b−a)a – b = -(b – a) 
    • III. a>−aa > -a 
    • (A) I only
    • (B) II only
    • (C) III only
    • (D) I and III only
    • (E) II and III only

Grid-In Questions

  1. If p2=16p^2 = 16  and q2=36q^2 = 36 , what is the largest possible value of q−pq – p ?

  2. If −4≤x≤2-4 \leq x \leq 2  and y=1−x2y = 1 – x^2 , what number is obtained when the smallest possible value of yy is subtracted from the largest possible value of yy ?

Fractions & Decimals: Lesson 3-5

Multiple Choice

  1. What part of an hour elapses from 4:56 P.M. to 5:32 P.M.?
    • (A) 14\frac{1}{4}
    • (B) 12\frac{1}{2}
    • (C) 35\frac{3}{5}
    • (D) 23\frac{2}{3}
    • (E) 34\frac{3}{4}
  1. If each of the fractions 3k,4k,5k\frac{3}{k}, \frac{4}{k}, \frac{5}{k}  is in lowest terms, which of the following could be the value of kk ?
    • (A) 48
    • (B) 49
    • (C) 50
    • (D) 51
    • (E) 52
  1. Which number has the greatest value?
    • (A) 0.2093
    • (B) 0.2908
    • (C) 0.2893
    • (D) 0.2938
    • (E) 0.2909
  1. 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) 115\frac{1}{15}
    • (B) 114\frac{1}{14}
    • (C) 113\frac{1}{13}
    • (D) 451\frac{4}{51}
    • (E) 112\frac{1}{12}
  1. 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
  1. In which arrangement are the fractions listed from least to greatest?
    • (A) 919,12,815\frac{9}{19}, \frac{1}{2}, \frac{8}{15}
    • (B) 12,815,919\frac{1}{2}, \frac{8}{15}, \frac{9}{19}
    • (C) 919,815,12\frac{9}{19}, \frac{8}{15}, \frac{1}{2}
    • (D) 12,919,815\frac{1}{2}, \frac{9}{19}, \frac{8}{15}
    • (E) 815,12,919\frac{8}{15}, \frac{1}{2}, \frac{9}{19}
  1. After the formula V=43πr3V = \frac{4}{3} \pi r^3 has been evaluated for some positive value of rr , the formula is again evaluated using one-half of the original value of rr . The new value of VV  is what fractional part of the original value of VV ?
    • (A) 116\frac{1}{16}
    • (B) 19\frac{1}{9}
    • (C) 18\frac{1}{8}
    • (D) 14\frac{1}{4}
    • (E) 12\frac{1}{2}
  1. Each inch on ruler AA  is marked in equal 18\frac{1}{8}-inch units, and each inch on ruler BB  is marked in equal 112\frac{1}{12}-inch units. When ruler AA  is used, a side of a triangle measures 12 of the 18\frac{1}{8} -inch units. When ruler BB  is used, how many 112\frac{1}{12}-inch units will the same side measure?
    • (A) 8
    • (B) 12
    • (C) 18
    • (D) 20
    • (E) 24
  1. Evaluate 60+2+48+350060 + 2 + \frac{4}{8} + \frac{3}{500} 
    • (A) 60.256
    • (B) 62.43
    • (C) 62.506
    • (D) 62.53
    • (E) 62.560
  1. If N×712=712×314N \times \frac{7}{12} = \frac{7}{12} \times \frac{3}{14} , then 1N=?\frac{1}{N} = ? 
    • (A) 8
    • (B) 143\frac{14}{3}
    • (C) 127\frac{12}{7}
    • (D) 311\frac{3}{11}
    • (E) 16\frac{1}{6}
  1. If n=2.5×1025n = 2.5 \times 10^{25} , then n=?\sqrt{n} = ? 
    • (A) 0.5×1050.5 \times 10^5 
    • (B) 0.5×10110.5 \times 10^{11} 
    • (C) 5×10245 \times 10^{\sqrt{24}}
    • (D) 5×1055 \times 10^5 
    • (E) 5×10125 \times 10^{12} 
  1. If y=x−2x+3y = \frac{x – 2}{x + 3}, then xx  cannot equal which of the following numbers?
    • (A) -3
    • (B) -2
    • (C) 0
    • (D) 2
    • (E) 3
  1. 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
  1. 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
  1. In the repeating decimal 0.31752=0.3175231752…0.31752 = 0.3175231752 \dots, the set of digits 31752 repeats endlessly. Which digit is in the 968th968^\text{th}  place to the right of the decimal point?
    • (A) 1
    • (B) 2
    • (C) 3
    • (D) 5
    • (E) 7
  1. If 4Δ/6Δ+517=14\Delta / 6\Delta + \frac{5}{17} = 1 , what digit does Δ\Delta represent?
  1. 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?
  1. Four lemons cost $0.68. At the same rate, 1 pound of lemons costs $1.19. How many lemons typically weigh 1 pound?
  1. If the charge for a taxi ride is $2.50 for the first 12\frac{1}{2} mile and $0.75 for each additional 18\frac{1}{8} mile, how many miles did the taxi travel for a ride that cost $10.75?
  1. 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

  1. If y=x+42y = \sqrt{\frac{x+4}{2}} , what is the value of yy when x=12x = \frac{1}{2} ?

  2. What fraction of 109\frac{10}{9}  is 56\frac{5}{6} ?

  3. If 25⋅xy=425 \cdot \frac{x}{y} = 4 , what is the value of 100⋅yx100 \cdot \frac{y}{x} ?

  4. If P=(1+12)(1+13)(1+14)…(1+117)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 (1+1k)\left(1 + \frac{1}{k}\right) , what is the value of PP ?

Fraction Word Problems

Multiple Choice

  1. If 38\frac{3}{8} of a number is 6, what is 78\frac{7}{8} of the same number?
    • (A) 8
    • (B) 12
    • (C) 14
    • (D) 16
    • (E) 24
  1. After Claire has read the first 58\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
  1. 49\frac{4}{9} of 27 is 65\frac{6}{5} of what number?
    • (A) 5
    • (B) 10
    • (C) 12
    • (D) 15
    • (E) 20
  1. If 35\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
  1. What is the sum of all two-digit whole numbers in which one digit is 34\frac{3}{4} of the other digit?
    • (A) 77
    • (B) 102
    • (C) 129
    • (D) 154
    • (E) 231
  1. Boris has DD dollars. If he lends 14\frac{1}{4} of this amount of money and then spends 13\frac{1}{3} of the money that he has left, how many dollars, in terms of DD , does Boris now have?
    • (A) D2\frac{D}{2}
    • (B) D3\frac{D}{3}
    • (C) D4\frac{D}{4}
    • (D) 1112D\frac{11}{12}D 
    • (E) 34D−13\frac{3}{4}D – \frac{1}{3}
  1. The value obtained by increasing aa by 15\frac{1}{5} of its value is numerically equal to the value obtained by decreasing bb by 12\frac{1}{2} of its value. Which equation expresses this fact?
    • (A) 1.2a=0.5b1.2a = 0.5b 
    • (B) 0.2a=0.5b0.2a = 0.5b 
    • (C) 0.8a=1.5b0.8a = 1.5b 
    • (D) 1.2a=1.5b1.2a = 1.5b 
    • (E) a0.2=b−0.5\frac{a}{0.2} = b – 0.5 
  1. How many times must a jogger run around a circular 14\frac{1}{4}mile track in order to have run 3123 \frac{1}{2}  miles?
    • (A) 9
    • (B) 12
    • (C) 14
    • (D) 15
    • (E) 16
  1. A chocolate bar that weighs 916\frac{9}{16} of a pound is cut into seven equal parts. How much do three parts weigh?
    • (A) 21122\frac{21}{122} pound
    • (B) 27112\frac{27}{112} pound
    • (C) 1663\frac{16}{63} pound
    • (D) 4763\frac{47}{63} pound
    • (E) 85112\frac{85}{112} pound
  1. At a high school basketball game, 35\frac{3}{5} of the students who attended were seniors; 13\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
  1. Of the 75 people in a room, 25\frac{2}{5} are college graduates. If 49\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
  1. If 23\frac{2}{3} of 34\frac{3}{4} of a number is 24, what is 14\frac{1}{4} of the same number?
    • (A) 8
    • (B) 12
    • (C) 16
    • (D) 20
    • (E) 24
  1. A man paints 34\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) 14\frac{1}{4} day
    • (B) 12\frac{1}{2} day
    • (C) 23\frac{2}{3} day
    • (D) 1 day
    • (E) 43\frac{4}{3} days
  1. A water tank is 35\frac{3}{5} full. After 12 gallons are poured out, the tank is 13\frac{1}{3}  full. How many gallons of water does the tank hold?
    • (A) 25
    • (B) 32
    • (C) 35
    • (D) 42
    • (E) 45
  1. In a school election, Susan received 23\frac{2}{3} of the ballots cast, Mary received 15\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
  1. After Arlene pumps gas into her car, the gas gauge moves from exactly 18\frac{1}{8}  full to exactly 78\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
  1. 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 110\frac{1}{10} of its original price, and stock B has decreased by 110\frac{1}{10}  of its original price. What fraction of the new price of stock B is the new price of stock A?
    • (A) 210\frac{2}{10}
    • (B) 911\frac{9}{11}
    • (C) 910\frac{9}{10}
    • (D) 1110\frac{11}{10}
    • (E) 119\frac{11}{9}
  1. After 34\frac{3}{4} of the people in a room leave, three people enter. If the remaining people are 13\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

  1. What is 20% of 23\frac{2}{3} of 15?
    • (A) 1
    • (B) 2
    • (C) 3
    • (D) 4
    • (E) 5
  1. Which expression is equivalent to 25%\frac{2}{5}\%?
    • (A) 0.40
    • (B) 0.04
    • (C) 0.004
    • (D) 0.0004
    • (E) 0.00004
  1. The fraction 0.25+0.150.50\frac{0.25 + 0.15}{0.50} is equivalent to what percent?
    • (A) 20%
    • (B) 25%
    • (C) 40%
    • (D) 60%
    • (E) 80%
  1. 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%
  1. 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) 1212%12 \frac{1}{2}\% 
    • (B) 1427%14 \frac{2}{7}\% 
    • (C) 21%
    • (D) 25%
    • (E) 2847%28 \frac{4}{7}\% 
  1. Which expression is equivalent to 0.01%?
    • (A) 1100\sqrt{\frac{1}{100}}
    • (B) 0.0110\frac{0.01}{10}
    • (C) 1100\frac{1}{100}
    • (D) 1100110\frac{\frac{1}{100}}{\frac{1}{10}}
    • (E) 110\frac{1}{10}
  1. If the result of increasing aa  by 300% of aa  is bb , then aa  is what percent of bb ?
    • (A) 20%
    • (B) 25%
    • (C) 3313%33 \frac{1}{3}\% 
    • (D) 40%
    • (E) 6623%66 \frac{2}{3}\% 
  1. 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
  1. 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
  1. 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
  1. What is 20% of 25% of 45\frac{4}{5} ?
    • (A) 0.0025
    • (B) 0.004
    • (C) 0.005
    • (D) 0.04
    • (E) 0.05
  1. 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
  1. 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%
  1. 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%
  1. If aa is 40% of bb , then bb  exceeds aa by what percent of aa ?
    • (A) 60%
    • (B) 100%
    • (C) 140%
    • (D) 150%
    • (E) 250%
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