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 square of a positive number is 49. What is the number?
- –7
- 7 or –7
- 7
- The square of a number added to 25 equals 10 times the number. What is the number?
- –5
- 10
- –10
- 5
- The sum of the square of a number and 12 times the number is –27. What is the smaller possible value of this number?
- –3
- –9
- 3
- 9
- The area of a rectangle is 24 square inches. The length of the rectangle is 5 inches more than the width. How many inches is the width?
- 4 in
- 3 in
- 6 in
- 8 in
- The height of a parallelogram measures 5 meters more than its base. If the area of the parallelogram is 36 m^{2}, what is the height in meters?
- 6 m
- 9 m
- 12 m
- 4 m
- Patrick has a rectangular patio whose length is 5 m less than the diagonal and a width that is 7 m less than the diagonal. If the area of his patio is 195 m^{2}, what is the length of the diagonal?
- 10 m
- 8 m
- 16 m
- 20 m
- Samantha owns a rectangular field that has an area of 3,280 square feet. The length of the field is 2 more than twice the width. What is the width of the field?
- 40 ft
- 82 ft
- 41 ft
- 84 ft
- A garden in the shape of a rectangle is surrounded by a walkway of uniform width. The dimensions of the garden only are 35 by 24. The area of the garden and the walkway together is 1,530 square feet. What is the width of the walkway in feet?
- 4 ft
- 5 ft
- 34.5 ft
- 24 ft
- A pool is surrounded by a deck that has the same width all the way around. The total area of the deck only is 400 square feet. The dimensions of thepool are 18 feet by 24 feet. How many feet is the width of the deck?
- 4 ft
- 8 ft
- 24 ft
- 25 ft
- Jessica has a picture in a frame with a total area of 288 in^{2}. The dimension of the picture without the frame is 12 in by 14 in. What is the larger dimension, in inches, of the frame?
- 2 in
- 14 in
- 18 in
- 16 in
- What is the lesser of two consecutive positive integers whose product is 90?
- –9
- 9
- –10
- 10
- What is the greater of two consecutive negative integers whose product is 210?
- –14
- –15
- 14
- 15
- Find the lesser of two consecutive positive even integers whose product is 168.
- 12
- 14
- 10
- 16
- Find the greater of two consecutive positive odd integers whose product is 143.
- 10
- 11
- 12
- 13
- The sum of the squares of two consecutive positive odd integers is 74. What is the value of the smaller integer?
- 3
- 7
- 5
- 11
- If the difference between the squares of two consecutive integers is 15, find the larger integer.
- 8
- 7
- 6
- 9
- The square of one integer is 55 less than the square of the next consecutive integer. Find the lesser integer.
- 23
- 24
- 27
- 28
- A 4-inch by 6-inch photograph is going to be enlarged by increasing each side by the same amount. The new area is 168 square inches. How many inches is each dimension increased?
- 12
- 10
- 8
- 6
- A photographer decides to reduce a picture she took in order to fit it into a certain frame. She needs the picture to be one-third of the area of the original. If the original picture was 4 inches by 6 inches, how many inches is the smaller dimension of the reduced picture if each dimension changes the same amount?
- 2
- 3
- 4
- 5
- A rectangular garden has a width of 20 feet and a length of 24 feet. If each side of the garden is increased by the same amount, how many feet is the new length if the new area is 141 square feet more than the original?
- 23
- 24
- 26
- 27
- Ian can remodel a kitchen in 20 hours and Jack can do the same job in 15 hours. If they work together, how many hours will it take them to remodel the kitchen?
- 5.6
- 8.6
- 7.5
- 12
- Peter can paint a room in an hour and a half and Joe can paint the same room in 2 hours. How many minutes will it take them to paint the room if they do it together? Round answer to nearest minute.
- 51
- 64
- 30
- 210
- Carla can plant a garden in 3 hours and Charles can plant the same garden in 4.5 hours. If they work together, how many hours will it take them to plant the garden?
- 1.5
- 2.1
- 1.8
- 7.5
- If Jim and Jerry work together they can finish a job in 4 hours. If working alone takes Jim 10 hours to finish the job, how many hours would it take Jerry to do the job alone?
- 16
- 5.6
- 6.7
- 6.0
- Bill and Ben can clean the garage together in 6 hours. If it takes Bill 15 hours working alone, how long will it take Ben working alone?
- 11 hours
- 9 hours
- 16 hours
- 10 hours
Answers
The following explanations show one way in which each problem can be solved. You may have another method for solving these problems.
- d. Let x = the number. The sentence, "The square of a positive number is 49," translates to the equation x^{2} = 49. Take the square root of each side to get so x = 7 or –7. Since you are looking for a positive number, the final solution is 7.
- d. Let x = the number. The statement, "The square of a number added to 25 equals 10 times the number," translates to the equation x^{2} + 25 = 10x. Put the equation in standard form ax^{2} + bx + c = 0, and set it equal to zero: x^{2} – 10x + 25 = 0. Factor the left side of the equation: (x – 5)(x – 5) = 0. Set each factor equal to zero and solve: x – 5 = 0 or x – 5 = 0; x = 5 or x = 5. The number is 5.
- b. Let x = the number. The statement, "The sum of the square of a number and 12 times the number is –27," translates to the equation x^{2} + 12x = –27. Put the equation in standard form and set it equal to zero: x^{2} + 12x + 27 = 0. Factor the left side of the equation: (x + 3)(x + 9) = 0. Set each factor equal to zero and solve: x + 3 = 0 or x + 9 = 0; x = –3 or x = –9. The possible values of this number are –3 or –9, the smaller of which is –9.
- b. Let x = the number of inches in the width and let x + 5 = the number of inches in the length. Since area of a rectangle is length times width, the equation for the area of the rectangle is x(x + 5) = 24. Multiply the left side of the equation using the distributive property: x^{2} + 5x = 24. Put the equation in standard form and set it equal to zero: x^{2} + 5x – 24 = 0. Factor the left side of the equation: (x + 8)(x – 3) = 0. Set each factor equal to zero and solve: x + 8 = 0 or x – 3 = 0; x = –8 or x = 3. Reject the solution of –8 because a distance will not be negative. The width is 3 inches.
- b. Let x = the measure of the base and let x + 5 = the measure of the height. Since the area of a parallelogram is base times height, then the equation for the area of the parallelogram is x(x + 5) = 36. Multiply the left side of the equation using the distributive property: x^{2} + 5x = 36; Put the equation in standard form and set it equal to zero: x^{2} + 5x – 36 = 0. Factor the left side of the equation: (x + 9)(x – 4) = 0. Set each factor equal to zero and solve: x + 9 = 0 or x – 4 = 0; x = –9 or x = 4. Reject the solution of –9 because a distance will not be negative. The height is 4 + 5 = 9 meters.
- d. Let x = the length of the diagonal. Therefore, x – 5 = the length of the patio and x –7 = the width of the patio. Since the area is 195 m^{2}, and area is length times the width, the equation is (x – 5)(x – 7) = 195. Use the distributive property to multiply the binomials: x^{2} –5x – 7x + 35 = 195. Combine like terms: x^{2} – 12x + 35 = 195. Subtract 195 from both sides: x^{2} – 12x + 35 – 195= 195 – 195. Simplify: x^{2} – 12x – 160 = 0. Factor the result: (x – 20)(x + 8) = 0. Set each factor equal to 0 and solve: x – 20 = 0 or x + 8 = 0; x = 20 or x = –8. Reject the solution of –8 because a distance will not be negative. The length of the diagonal is 20 m.
- a. Let w = the width of the field and let 2w + 2 = the length of the field (two more than twice the width). Since area is length times width, multiply the two expressions together and set them equal to 3,280: w(2w + 2) = 3,280. Multiply using the distributive property: 2w^{2} + 2w = 3,280. Subtract 3,280 from both sides: 2w^{2} + 2w – 3,280 = 3,280 – 3,280; simplify: 2w^{2} + 2w – 3,280 = 0. Factor the trinomial completely: 2(w^{2} + w – 1640) = 0; 2(w + 41)(w – 40) = 0. Set each factor equal to zero and solve: 2 ≠ 0 or w + 41 = 0 or w – 40 = 0; w = –41 or w = 40. Reject the negative solution because you will not have a negative width. The width is 40 feet.
- b. Let x = the width of the walkway. Since the width of the garden only is 24, the width of the garden and the walkway together is x + x + 24 or 2x + 24. Since the length of the garden only is 35, the length of the garden and the walkway together is x + x + 35 or 2x + 35. Area of a rectangle is length times width, so multiply the expressions together and set the result equal to the total area of 1,530 square feet: (2x + 24)(2x + 35) = 1,530. Multiply the binomials using the distributive property: 4x^{2} + 70x + 48x + 840 = 1,530. Combine like terms: 4x^{2} + 118x + 840 = 1,530. Subtract 1,530 from both sides: 4x^{2} + 118x + 840 – 1,530 = 1,530 – 1,530; simplify: 4x^{2} + 118x – 690 = 0. Factor the trinomial completely: 2(2x^{2} + 59x – 345) = 0; 2(2x + 69)(x – 5) = 0. Set each factor equal to zero and solve: 2 ≠ 0 or 2x + 69 = 0 or x – 5 = 0; x = –34.5 or x = 5. Reject the negative solution because you will not have a negative width. The width is 5 feet.
- a. Let x = the width of the deck. Since the width of the pool only is 18, the width of the pool and the deck is x + x + 18 or 2x + 18. Since the length of the pool only is 24, the length of the pool and the deck together is x + x + 24 or 2x + 24. The total area for the pool and the deck together is 832 square feet, 400 square feet added to 432 square feet for the pool. Area of a rectangle is length times width so multiply the expressions together and set them equal to the total area of 832 square feet: (2x + 18)(2x + 24) = 832. Multiply the binomials using the distributive property: 4x^{2} + 36x + 48x + 432 = 832. Combine like terms: 4x^{2} + 84x + 432 = 832. Subtract 832 from both sides: 4x^{2} + 84x + 432 – 832 = 832 – 832; simplify: 4x^{2} + 84x – 400 = 0. Factor the trinomial completely: 2(2x^{2} + 42x – 200) = 0; 2(2x – 8)(x + 25) = 0. Set each factor equal to zero and solve: 2 ≠ 0 or 2x – 8 = 0 or x + 25 = 0; x = 4 or x = –25. Reject the negative solution because you will not have a negative width. The width is 4 feet.
- c. To solve this problem, find the width of the frame first. Let x = the width of the frame. Since the width of the picture only is 12, the width of the frame and the picture is x + x + 12 or 2x + 12. Since the length of the picture only is 14, the length of the frame and the picture together is x + x + 14 or 2x + 14. The total area for the frame and the picture together is 288 square inches. Area of a rectangle is length times width so multiply the expressions together and set them equal to the total area of 288 square inches: (2x + 12)(2x + 14) = 288. Multiply the binomials using the distributive property: 4x^{2} + 28x + 24x + 168 = 288. Combine like terms: 4x^{2} + 52x + 168 = 288. Subtract 288 from both sides: 4x^{2} + 52x + 168 – 288 = 288 – 288; simplify: 4x^{2} + 52x – 120 = 0. Factor the trinomial completely: 4(x^{2} + 13x – 30) = 0; 4(x – 2)(x + 15) = 0. Set each factor equal to zero and solve: 4 ≠ 0 or x – 2 = 0 or x + 15 = 0; x = 2 or x = –15. Reject the negative solution because you will not have a negative width. The width is 2 feet. Therefore, the larger dimension of the frame is 2(2) + 14 = 4 + 14 = 18 inches.
- b. Let x = the lesser integer and let x + 1 = the greater integer. Since product is a key word for multiplication, the equation is x(x + 1) = 90. Multiply using the distributive property on the left side of the equation: x^{2} + x = 90. Put the equation in standard form and set it equal to zero: x^{2} + x – 90 = 0. Factor the trinomial: (x – 9)(x + 10) = 0. Set each factor equal to zero and solve: x – 9 = 0 or x + 10 = 0; x = 9 or x = –10. Since you are looking for a positive integer, reject the x-value of –10. Therefore, the lesser positive integer would be 9.
- a. Let x = the lesser integer and let x + 1 = the greater integer. Since product is a key word for multiplication, the equation is x(x + 1) = 210. Multiply using the distributive property on the left side of the equation: x^{2} + x = 210. Put the equation in standard form and set it equal to zero: x^{2} + x – 210 = 0. Factor the trinomial: (x – 14)(x + 15) = 0. Set each factor equal to zero and solve: x – 14 = 0 or x + 15 = 0; x = 14 or x = –15. Since you are looking for a negative integer, reject the x-value of 14. Therefore, x = –15 and x + 1 = –14. The greater negative integer is –14.
- a. Let x = the lesser even integer and let x + 2 = the greater even integer. Since product is a key word for multiplication, the equation is x(x + 2) = 168. Multiply using the distributive property on the left side of the equation: x^{2} + 2x = 168. Put the equation in standard form and set it equal to zero: x^{2} + 2x – 168 = 0. Factor the trinomial: (x – 12)(x + 14) = 0. Set each factor equal to zero and solve: x – 12 = 0 or x + 14 = 0; x = 12 or x = –14. Since you are looking for a positive integer, reject the x-value of –14. Therefore, the lesser positive integer would be 12.
- d. Let x = the lesser odd integer and let x + 2 = the greater odd integer. Since product is a key word for multiplication, the equation is x(x + 2) = 143. Multiply using the distributive property on the left side of the equation: x^{2} + 2x = 143. Put the equation in standard form and set it equal to zero: x^{2} + 2x – 143 = 0. Factor the trinomial: (x – 11)(x + 13) = 0. Set each factor equal to zero and solve: x – 11 = 0 or x + 13 = 0; x = 11 or x = –13. Since you are looking for a positive integer, reject the x-value of –13. Therefore, x = 11 and x + 2 = 13. The greater positive odd integer is 13.
- c. Let x = the lesser odd integer and let x + 2 = the greater odd integer. The translation of the sentence, "The sum of the squares of two consecutive odd integers is 74," is the equation x^{2} + (x + 2)^{2} = 74. Multiply (x + 2)^{2} out as (x + 2)(x + 2) using the distributive property: x^{2} + (x^{2} + 2x + 2x + 4) = 74. Combine like terms on the left side of the equation: 2x^{2} + 4x + 4 = 74. Put the equation in standard form by subtracting 74 from both sides, and set it equal to zero: 2x^{2} + 4x – 70 = 0; factor the trinomial completely: 2(x^{2} + 2x – 35) = 0; 2(x – 5)(x + 7) = 0. Set each factor equal to zero and solve: 2 ≠ 0 or x – 5 = 0 or x + 7 = 0; x = 5 or x = –7. Since you are looking for a positive integer, reject the solution of x = –7. Therefore, the smaller positive integer is 5.
- a. Let x = the lesser integer and let x + 1 = the greater integer. The sentence, "the difference between the squares of two consecutive integers is 15," can translate to the equation (x + 1)^{2} – x^{2} = 15. Multiply the binomial (x + 1)2 as (x + 1)(x + 1) using the distributive property: x^{2} + 1x + 1x + 1 – x^{2} = 15. Combine like terms: 2x + 1 = 15; subtract 1 from both sides of the equation: 2x + 1 – 1 = 15 – 1. Divide both sides by 2: . The variable is now alone: x = 7. Therefore, the larger consecutive integer is x + 1 = 8.
- c. Let x = the lesser integer and let x + 1 = the greater integer. The sentence, "The square of one integer is 55 less than the square of the next consecutive integer," can translate to the equation x^{2} = (x + 1)2 –55. Multiply the binomial (x + 1)2 as (x + 1)(x + 1) using the distributive property: x^{2} = x^{2} + 1x + 1x + 1 – 55. Combine like terms: x^{2} = x^{2} + 2x – 54. Subtract x^{2} from both sides of the equation: x^{2} – x^{2} = x^{2} – x^{2} + 2x – 54. Add 54 to both sides of the equation: 0 + 54 = 2x – 54 + 54. Divide both sides by 2: . The variable is now alone: 27 = x. The lesser integer is 27.
- c. Let x = the amount each side is increased. Then, x + 4 = the new width and x + 6 = the new length. Since area is length times width, the formula using the new area is (x + 4)(x + 6) = 168. Multiply using the distributive property on the left side of the equation: x^{2} + 6x + 4x + 24 = 168; combine like terms: x^{2} + 10x + 24 = 168. Subtract 168 from both sides: x^{2} + 10x + 24 – 168 = 168 – 168. Simplify: x^{2} + 10x – 144 = 0. Factor the trinomial: (x – 8)(x + 18) = 0. Set each factor equal to zero and solve: x – 8 = 0 or x + 18 = 0; x = 8 or x = –18. Reject the negative solution because you won't have a negative dimension. The correct solution is 8 inches.
- a. Let x = the amount of reduction. Then 4 – x = the width of the reduced picture and 6 – x = the length of the reduced picture. Since area is length times width, and one-third of the old area of 24 is 8, the equation for the area of the reduced picture would be (4 – x)(6 – x) = 8. Multiply the binomials using the distributive property: 24 – 4x – 6x + x^{2} = 8; combine like terms: 24 – 10x + x^{2} = 8. Subtract 8 from both sides: 24 – 8 – 10x + x^{2} = 8 – 8. Simplify and place in standard form: x^{2} – 10x + 16 = 0. Factor the trinomial into 2 binomials: (x – 2)(x – 8) = 0. Set each factor equal to zero and solve: x – 2 = 0 or x – 8 = 0; x = 2 or x = 8. The solution of 8 is not reasonable because it is greater than the original dimensions of the picture. Accept the solution of x = 2 and the smaller dimension of the reduced picture would be 4 – 2 = 2 inches.
- d. Let x = the amount that each side of the garden is increased. Then, x + 20 = the new width and x + 24 = the new length. Since the area of a rectangle is length times width, then the area of the old garden is 20 × 24 = 480 and the new area is 480 + 141 = 621. The equation using the new area becomes (x + 20)(x + 24) = 621. Multiply using the distributive property on the left side of the equation: x^{2} + 24x + 20x + 480 = 621; combine like terms: x^{2} + 44x + 480 = 621. Subtract 621 from both sides: x^{2} + 44x + 480 – 621 = 621 – 621; simplify: x^{2} + 44x – 141 = 0. Factor the trinomial: (x – 3)(x + 47) = 0. Set each factor equal to zero and solve: x – 3 = 0 or x + 47 = 0; x = 3 or x = – 47. Reject the negative solution because you won't have a negative increase. Thus, each side will be increased by 3 and the new length would be 24 + 3 = 27 feet.
- b. Let x = the number of hours it takes Ian and Jack to remodel the kitchen if they are working together. Since it takes Ian 20 hours if working alone, he will complete of the job in one hour, even when he's working with Jack. Similarly, since it takes Jack 15 hours to remodel a kitchen, he will complete of the job in one hour, even when he's working with Ian. Since it takes x hours for Ian and Jack to complete the job together, it stands to reason that at the end of one hour, their combined effort will have completed of the job. Therefore, Ian's work + Jack's work = combined work and we have the equation: . Multiply through by the least common denominator of 20, 15 and x which is 60x: . Simplify: 3x + 4x = 60. Simplify: 7x = 60. Divide by 7: which is about 8.6 hours.
- a. Let x = the number of hours it takes Peter and Joe to paint a room if they are working together. Since it takes Peter 1.5 hours if working alone, he will complete of the job in one hour, even when he's working with Joe. Similarly, since it takes Joe 2 hours to paint a room working alone, he will complete of the job in one hour, even when working with Peter. Since it takes x hours for Peter and Joe to complete the job together, it stands to reason that at the end of one hour, their combined effort will have completed of the job. Therefore, Peter's work + Joe's work = combined work and we have the equation: + = . Multiply through by the least common denominator of 1.5, 2 and x which is 6x: . Simplify: 4x + 3x = 6. Simplify: 7x = 6. Divide by 7: hours. Change hours into minutes by multiplying by 60 since there are 60 minutes in one hour. divided by 7 equals 51.42 minutes which rounds to 51 minutes.
- c. Let x = the number of hours it takes Carla and Charles to plant a garden if they are working together. Since it takes Carla 3 hours if working alone, she will complete of the job in one hour, even when she's working with Charles. Similarly, since it takes Charles 4.5 hours to plant a garden working alone, he will complete of the job in one hour, even when working with Carla. Since it takes x hours for Carla and Charles to complete the job together, it stands to reason that at the end of one hour, their combined effort will have completed of the job. Therefore, Carla's work + Charles's work = combined work and we have the equation: . Multiply through by the least common denominator of 3, 4.5 and x which is 9x: . Simplify: 3x + 2x = 9. Simplify: 5x = 9. Divide by 5: ; hours which is equal to 1.8 hours.
- c. Let x = the number of hours it will take Jerry to do the job alone. In 1 hour Jim can do of the work, and Jerry can do of the work. As an equation this looks like where represents what part of the job they can complete in one hour together. Multiplying both sides of the equation by the least common denominator, 40x, results in the equation 4x + 40 = 10x. Subtract 4x from both sides of the equation. 4x – 4x + 40 = 10x – 4x. This simplifies to 40 = 6x. Divide each side of the equation by 6; . Therefore, 6.666 = x, and it would take Jerry about 6.7 hours to complete the job alone.
- d. Let x = the number of hours Ben takes to clean the garage by himself. In 1 hour Ben can do of the work and Bill can do of the work. As an equation this looks like , where represents what part they can clean in one hour together. Multiply both sides of the equation by the least common denominator, 30x, to get an equation of 30 + 2x = 5x. Subtract 2x from both sides of the equation; 30 + 2x – 2x = 5x – 2x. This simplifies to 30 = 3x, and dividing both sides by 3 results in a solution of 10 hours.
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.