Algebra Word Problems
Basic algebra problems ask you to solve equations in which one or more elements are unknown. The unknown quantities are represented by variables, which are letters of the alphabet, such as x or y. The questions in this set of practice problems give you practice in writing algebraic equations and using these expressions to solve problems.
- The perimeter of a rectangle is 104 inches. The width is 6 inches less than 3 times the length. Find the width of the rectangle.
- 13.5 inches
- 37.5 inches
- 14.5 inches
- 15 inches
- The perimeter of a parallelogram is 50 cm. The length of the parallelogram is 5 cm more than the width. Find the length of the parallelogram.
- 15 cm
- 11 cm
- 5 cm
- 10 cm
- Jackie invested money in two different accounts, one of which earned 12% interest per year and another that earned 15% interest per year. The amount invested at 15% was 100 more than twice the amount at 12%. How much was invested at 12% if the total annual interest earned was $855?
- $4,100
- $2,100
- $2,000
- $4,000
- Kevin invested $4,000 in an account that earns 6% interest per year and $x in a different account that earns 8% interest per year. How much is invested at 8% if the total amount of interest earned annually is $405.50?
- $2,075.00
- $4,000.00
- $2,068.75
- $2,075.68
- Megan bought x pounds of coffee that cost $3 per pound and 18 pounds of coffee at $2.50 per pound for the company picnic. Find the total number of pounds of coffee purchased if the average cost per pound of both types together is $2.85.
- 42
- 18
- 63
- 60
- The student council bought two different types of candy for the school fair. They purchased 40 pounds of candy at $2.15 per pound and x pounds at $1.90 per pound. What is the total number of pounds they bought if the total amount of money spent on candy was $162?
- 42
- 40
- 80
- 52
- The manager of a garden store ordered two different kinds of marigold seeds for her display. The first type cost her $1 per packet and the second type cost $1.26 per packet. How many packets of the first type did she purchase if she bought 50 more of the $1.26 packets than the $1 packets and spent a total of $402?
- 150
- 200
- 250
- 100
- Harold used a 3% iodine solution and a 20% iodine solution to make a 95- ounce solution that was 19% iodine. How many ounces of the 3% iodine solution did he use?
- 5
- 80
- 60
- 20
- A chemist mixed a solution that was 34% acid with another solution that was 18% acid to produce a 30-ounce solution that was 28% acid. How much of the 34% acid solution did he use?
- 27
- 11.25
- 18.75
- 28
- Bob is 2 years from being twice as old as Ellen. The sum of twice Bob's age and three times Ellen's age is 66. How old is Ellen?
- 15
- 10
- 18
- 20
- Sam's age is 1 less than twice Shari's age. The sum of their ages is 92. How old is Shari?
- 48
- 32
- 65
- 31
- At the school bookstore, two binders and three pens cost $12.50. Three binders and five pens cost $19.50. What is the total cost of 1 binder and 1 pen?
- $4.50
- $4.00
- $1.50
- $5.50
- Two angles are complementary. The larger angle is 15° more than twice the smaller. Find the measure of the smaller angle.
- 25°
- 65°
- 90°
- 82.5°
- The cost of a student ticket is $1 more than half of an adult ticket. Six adults and four student tickets cost $76. What is the cost of one adult ticket?
- $2.50
- $3.00
- $5.50
- $4.00
- Three shirts and five ties cost $23. Five shirts and one tie cost $20. What is the price of one shirt?
- $3.50
- $2.50
- $6.00
- $3.00
- Noel rode 3x miles on his bike and Jamie rode 5x miles on hers. In terms of x, what is the total number of miles they rode?
- 15x miles
- 15x^{2} miles
- 8x miles
- 8x^{2} miles
- If the areas of two sections of a garden are 6a + 2 and 5a, what is the difference between the areas of the two sections in terms of a?
- a – 2
- 3a + 2
- a + 2
- 11a – 2
- Laura has a rectangular garden whose width is x^{3} and whose length is x^{4}. In terms of x, what is the area of her garden?
- 2x^{7}
- x^{7}
- x^{12}
- 2x^{12}
- Jonestown High School has a soccer field whose dimensions can be expressed as 7y^{2} and 3xy. What is the area of this field in terms of x and y?
- 10xy^{2}
- 10xy^{3}
- 21xy^{3}
- 21xy^{2}
- The area of a parallelogram is x^{8}. If the base is x^{4}, what is the height in terms of x?
- x^{4}
- x^{2}
- x^{12}
- x^{32}
Answers
The following explanations show one way in which each problem can be solved. You may have another method for solving these problems.
- b. Let l = the length of the rectangle and let w = the width of the rectangle. Since the width is 6 inches less than 3 times the length, one equation is w = 3l – 6. The formula for the perimeter of a rectangle is 2l + 2w = 104. Substituting the first equation into the perimeter equation for w results in 2l + 2(3l – 6) = 104. Use the distributive property on the left side of the equation: 2l + 6l – 12 = 104. Combine like terms on the left side of the equation: 8l – 12 = 104; add 12 to both sides of the equation: 8l – 12 + 12 = 104 + 12. Simplify to: 8l = 116. Divide both sides of the equation by 8: . Therefore, the length is l = 14.5 inches and the width is w = 3(14.5) – 6 = 37.5 inches.
- a. Let w = the width of the parallelogram and let l = the length of the parallelogram. Since the length is 5 more than the width, then l = w + 5. The formula for the perimeter of a parallelogram 2l + 2w = 50. Substituting the first equation into the second for l results in 2(w + 5) + 2w = 50. Use the distributive property on the left side of the equation: 2w + 10 + 2w = 50; combine like terms on the left side of the equation: 4w + 10 = 50. Subtract 10 on both sides of the equation: 4w + 10 – 10 = 50 – 10. Simply to: 4w = 40. Divide both sides of the equation by 4: ; w = 10. Therefore, the width is 10 cm and the length is 10 + 5 = 15 cm.
- c. Let x = the amount invested at 12% interest. Let y = the amount invested at 15% interest. Since the amount invested at 15% is 100 more then twice the amount at 12%, then y = 2x + 100. Since the total interest was $855, use the equation 0.12x + 0.15y = 855. You have two equations with two variables. Use the second equation 0.12x + 0.15y = 855 and substitute (2x + 100) for y: 0.12x + 0.15(2x + 100) = 855. Use the distributive property: 0.12x + 0.3x + 15 = 855. Combine like terms: 0.42x + 15 = 855. Subtract 15 from both sides: 0.42x + 15 – 15 = 855 – 15; simplify: 0.42x = 840. Divide both sides by 0.42: . Therefore, x = $2,000, which is the amount invested at 12% interest.
- c. Let x = the amount invested at 8% interest. Since the total interest is $405.50, use the equation 0.06(4,000) + 0.08x = 405.50. Simplify the multiplication: 240 + 0.08x = 405.50. Subtract 240 from both sides: 240 – 240 + 0.8x = 405.50 – 240; simplify: 0.08x = 165.50. Divide both sides by 0.08: . Therefore, x = $2,068.75, which is the amount invested at 8% interest.
- d. Let x = the amount of coffee at $3 per pound. Let y = the total amount of coffee purchased. If there are 18 pounds of coffee at $2.50 per pound, then the total amount of coffee can be expressed as y = x + 18. Use the equation 3x + 2.50(18) = 2.85y since the average cost of the y pounds of coffee is $2.85 per pound. To solve, substitute y = x + 18 into 3x + 2.50(18) = 2.85y. 3x + 2.50(18) = 2.85(x + 18). Multiply on the left side and use the distributive property on the right side: 3x + 45 = 2.85x + 51.30. Subtract 2.85x on both sides: 3x – 2.85x + 45 = 2.85x – 2.85x + 51.30. Simplify: 0.15x + 45 = 51.30. Subtract 45 from both sides: 0.15x + 45 – 45 = 51.30 – 45. Simplify: 0.15x = 6.30. Divide both sides by 0.15: ; so, x = 42 pounds, which is the amount of coffee that costs $3 per pound. Therefore, the total amount of coffee is 42 + 18, which is 60 pounds.
- c. Let x = the amount of candy at $1.90 per pound. Let y = the total number of pounds of candy purchased. If it is known that there are 40 pounds of candy at $2.15 per pound, then the total amount of candy can be expressed as y = x + 40. Use the equation 1.90x + 2.15(40) = $162 since the total amount of money spent was $162. Multiply on the left side: 1.90x + 86 = 162. Subtract 86 from both sides: 1.90x + 86 – 86 = 162 – 86. Simplify: 1.90x = 76. Divide both sides by 1.90: ; so, x = 40 pounds, which is the amount of candy that costs $1.90 per pound. Therefore, the total amount of candy is 40 + 40, which is 80 pounds.
- a. Let x = the amount of marigolds at $1 per packet. Let y = the amount of marigolds at $1.26 per packet. Since there are 50 more packets of the $1.26 seeds than the $1 seeds, y = x + 50. Use the equation 1x + 1.26y = 420 to find the total number of packets of each. By substituting into the second equation, you get 1x + 1.26(x + 50) = 402. Multiply on the left side using the distributive property: 1x + 1.26x + 63 = 402. Combine like terms on the left side: 2.26x + 63 = 402. Subtract 63 from both sides: 2.26x + 63 – 63 = 402 – 63. Simplify: 2.26x = 339. Divide both sides by 2.26: ; so, x = 150 packets, which is the number of packets that costs $1 each.
- a. Let x = the amount of 3% iodine solution. Let y = the amount of 20% iodine solution. Since the total amount of solution was 85 oz., then x + y = 85. The amount of each type of solution added together and set equal to the amount of 19% solution can be expressed in the equation 0.03x + 0.20y = 0.19(85); Use both equations to solve for x. Multiply the second equation by 100 to eliminate the decimal point: 3x + 20y = 19(85). Simplify that equation: 3x + 20y = 1805. Multiply the first equation by –20: –20x + –20y = –1700. Add the two equations to eliminate y: –17x + 0y = –85. Divide both sides of the equation by –17: ; x = 5. The amount of 3% iodine solution is 5 ounces.
- c. Let x = the amount of 34% acid solution. Let y = the amount of 18% iodine solution. Since the total amount of solution was 30 oz., then x + y = 30. The amount of each type of solution added together and set equal to the amount of 28% solution can be expressed in the equation 0.34x + 0.18y = 0.28(30). Use both equations to solve for x. Multiply the second equation by 100 to eliminate the decimal point: 34x + 18y = 28(30); simplify that equation: 34x + 22y = 840. Multiply the first equation by –18: –18x + –18y = –540. Add the two equations to eliminate y: 16x + 0y = 300. Divide both sides of the equation by 16: , x = 18.75. The amount of 34% acid solution is 18.75 ounces.
- b. Let x = Ellen's age and let y = Bob's age. Since Bob is 2 years from being twice as old as Ellen, than y = 2x – 2. The sum of twice Bob's age and three times Ellen's age is 66 and gives a second equation of 2y + 3x = 66. Substituting the first equation for y into the second equation results in 2(2x – 2) + 3x = 66. Use the distributive property on the left side of the equation: 4x – 4 + 3x = 66; combine like terms on the left side of the equation: 7x – 4 = 66. Add 4 to both sides of the equation: 7x – 4 + 4 = 66 + 4. Simplify: 7x = 70. Divide both sides of the equation by 7: . The variable, x, is now alone: x = 10. Therefore, Ellen is 10 years old.
- d. Let x = Shari's age and let y = Sam's age. Since Sam's age is 1 less than twice Shari's age this gives the equation y = 2x – 1. Since the sum of their ages is 104, this gives a second equation of x + y = 92. By substituting the first equation into the second for y, this results in the equation x + 2x – 1 = 92. Combine like terms on the left side of the equation: 3x – 1 = 92. Add 1 to both sides of the equation: 3x – 1 + 1 = 92 + 1. Simplify: 3x = 93. Divide both sides of the equation by 3: . The variable, x, is now alone: x = 31. Therefore, Shari's age is 31.
- d. Let x = the cost of one binder and let y = the cost of one pen. The first statement, "two binders and three pens cost $12.50," translates to the equation 2x + 3y = 12.50. The second statement, "three binders and five pens cost $19.50," translates to the equation:
Therefore, the cost of one pen is $1.50. Since the cost of 2 binders and 3 pens is 12.50, substitute y = 1.50 into the first equation: 3 × $1.50 = $4.50; $12.50 – 4.50 = $8.00; $8.00 ÷ 2= $4.00, so each binder is $4.00. The total cost of 1 binder and 1 pen is $4.00 + $1.50 = $5.50.
- a. Let x = the number of degrees in the smaller angle and let y = the number of degrees in the larger angle. Since the angles are complementary, x + y = 90. In addition, since the larger angle is 15 more than twice the smaller, y = 2x + 15. Substitute the second equation into the first equation for y: x + 2x + 15 = 90. Combine like terms on the left side of the equation: 3x + 15 = 90. Subtract 15 from both sides of the equation: 3x + 15 – 15 = 90 – 15; simplify: 3x = 75. Divide both sides by 3: . The variable, x, is now alone: x = 25. The number of degrees in the smaller angle is 25.
- b. Let x = the cost of a student ticket. Let y = the cost of an adult ticket. The first sentence, "The cost of a student ticket is $1 more than half of an adult ticket," gives the equation x = + 1; the second sentence, "six adults and four student tickets cost $76," gives the equation 6y + 4x = 76. Substitute the first equation into the second for x: 6y + 4( + 1) = 76. Use the distributive property on the left side of the equation: 6y + 2y + 4 = 76. Combine like terms: 8y + 4 = 76. Subtract 4 on both sides of the equation: 8y + 4 – 4 = 76 – 4; simplify: 8y = 72. Divide both sides by 8: . The variable is now alone: y = 9. The cost of one adult ticket is $9.
- a. Let x = the cost of one shirt. Let y = the cost of one tie. The first part of the question, "three shirts and 5 ties cost $23," gives the equation 3x + 5y = 23; the second part of the question, "5 shirts and one tie cost $20," gives the equation 5x + 1y = 20. Multiply the second equation by –5: –25x – 5y = –100. Add the first equation to that result to eliminate y. The combined equation is: –22x = – 77. Divide both sides of the equation by –22: . The variable is now alone: x = 3.50; the cost of one shirt is $3.50.
- c. The terms 3x and 5x are like terms because they have exactly the same variable with the same exponent. Therefore, you just add the coefficients and keep the variable. 3x + 5x = 8x.
- c. Because the question asks for the difference between the areas, you need to subtract the expressions: 6a + 2 – 5a. Subtract like terms: 6a – 5a + 2 = 1a + 2; 1a = a, so the simplified answer is a + 2.
- b. Since the area of a rectangle is A = length times width, multiply (x^{3})(x^{4}). When multiplying like bases, add the exponents: x^{3}+4 = x^{7}.
- c. Since the area of the soccer field would be found by the formula A = length × width, multiply the dimensions together: 7y^{2} × 3xy. Use the commutative property to arrange like variables and the coefficients next to each other: 7 × 3 × x × y^{2} × y. Multiply: remember that y^{2} × y = y^{2} × y^{1} = y^{2} + 1 = y^{3}. The answer is 21xy^{3}.
- a. Since the area of a parallelogram is A = base times height, then the area divided by the base would give you the height; ; when dividing like bases, subtract the exponents; x^{8–4} = x^{4}.
More practice problems on algebra word problems can be found at:
View Full Article
From 501 Math Word Problems. Copyright © 2003 by LearningExpress, LLC. All Rights Reserved.