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Algebra Word Problems Practice Questions Set 6 (page 2)

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Updated on Sep 26, 2011

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The following explanations show one way in which each problem can be solved. You may have another method for solving these problems.

  1. d.   Let x = the number. The sentence, "The square of a positive number is 49," translates to the equation x2 = 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.
  2. d.   Let x = the number. The statement, "The square of a number added to 25 equals 10 times the number," translates to the equation x2 + 25 = 10x. Put the equation in standard form ax2 + bx + c = 0, and set it equal to zero: x2 – 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.
  3. 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 x2 + 12x = –27. Put the equation in standard form and set it equal to zero: x2 + 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.
  4. 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: x2 + 5x = 24. Put the equation in standard form and set it equal to zero: x2 + 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.
  5. 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: x2 + 5x = 36; Put the equation in standard form and set it equal to zero: x2 + 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.
  6. 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 m2, and area is length times the width, the equation is (x – 5)(x – 7) = 195. Use the distributive property to multiply the binomials: x2 –5x – 7x + 35 = 195. Combine like terms: x2 – 12x + 35 = 195. Subtract 195 from both sides: x2 – 12x + 35 – 195= 195 – 195. Simplify: x2 – 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.
  7. 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: 2w2 + 2w = 3,280. Subtract 3,280 from both sides: 2w2 + 2w – 3,280 = 3,280 – 3,280; simplify: 2w2 + 2w – 3,280 = 0. Factor the trinomial completely: 2(w2 + 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.
  8. 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: 4x2 + 70x + 48x + 840 = 1,530. Combine like terms: 4x2 + 118x + 840 = 1,530. Subtract 1,530 from both sides: 4x2 + 118x + 840 – 1,530 = 1,530 – 1,530; simplify: 4x2 + 118x – 690 = 0. Factor the trinomial completely: 2(2x2 + 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.
  9. 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: 4x2 + 36x + 48x + 432 = 832. Combine like terms: 4x2 + 84x + 432 = 832. Subtract 832 from both sides: 4x2 + 84x + 432 – 832 = 832 – 832; simplify: 4x2 + 84x – 400 = 0. Factor the trinomial completely: 2(2x2 + 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.
  10. 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: 4x2 + 28x + 24x + 168 = 288. Combine like terms: 4x2 + 52x + 168 = 288. Subtract 288 from both sides: 4x2 + 52x + 168 – 288 = 288 – 288; simplify: 4x2 + 52x – 120 = 0. Factor the trinomial completely: 4(x2 + 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.
  11. 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: x2 + x = 90. Put the equation in standard form and set it equal to zero: x2 + 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.
  12. 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: x2 + x = 210. Put the equation in standard form and set it equal to zero: x2 + 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.
  13. 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: x2 + 2x = 168. Put the equation in standard form and set it equal to zero: x2 + 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.
  14. 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: x2 + 2x = 143. Put the equation in standard form and set it equal to zero: x2 + 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.
  15. 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 x2 + (x + 2)2 = 74. Multiply (x + 2)2 out as (x + 2)(x + 2) using the distributive property: x2 + (x2 + 2x + 2x + 4) = 74. Combine like terms on the left side of the equation: 2x2 + 4x + 4 = 74. Put the equation in standard form by subtracting 74 from both sides, and set it equal to zero: 2x2 + 4x – 70 = 0; factor the trinomial completely: 2(x2 + 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.
  16. 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)2x2 = 15. Multiply the binomial (x + 1)2 as (x + 1)(x + 1) using the distributive property: x2 + 1x + 1x + 1 – x2 = 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.
  17. 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 x2 = (x + 1)2 –55. Multiply the binomial (x + 1)2 as (x + 1)(x + 1) using the distributive property: x2 = x2 + 1x + 1x + 1 – 55. Combine like terms: x2 = x2 + 2x – 54. Subtract x2 from both sides of the equation: x2x2 = x2x2 + 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.
  18. 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: x2 + 6x + 4x + 24 = 168; combine like terms: x2 + 10x + 24 = 168. Subtract 168 from both sides: x2 + 10x + 24 – 168 = 168 – 168. Simplify: x2 + 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.
  19. 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 + x2 = 8; combine like terms: 24 – 10x + x2 = 8. Subtract 8 from both sides: 24 – 8 – 10x + x2 = 8 – 8. Simplify and place in standard form: x2 – 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.
  20. 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: x2 + 24x + 20x + 480 = 621; combine like terms: x2 + 44x + 480 = 621. Subtract 621 from both sides: x2 + 44x + 480 – 621 = 621 – 621; simplify: x2 + 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.
  21. 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.
  22. 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.
  23. 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.
  24. 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.
  25. 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:

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