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Algebra Word Problems Practice Questions Set 2 (page 2)

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Updated on Sep 26, 2011

Answers

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

  1. a.   Using the simple interest formula Interest = principal × rate × time, or I = prt, substitute I = $4,800, p = $12,000, and r = .08 (the interest rate as a decimal); 4,800 = (12,000)(.08)(t). Multiply 12,000 and .08 to get 960, so 4,800 = 960t. Divide both sides by 960 to get 5 = t. Therefore, the time is 5 years.
  2. b.   Using the simple interest formula Interest = principal × rate × time, or I = prt, substitute I = $711, p = $7,900, and t = 3 (36 months is equal to 3 years); 711 = (7,900)(r)(3). Multiply 7,900 and 3 on the right side to get a result of 711 = 23,700r. Divide both sides by 23,700 to get r = .03, which is a decimal equal to 3%.
  3. d.   In the statement, the order of the numbers does not change; however, the grouping of the numbers in parentheses does. Each side, if simplified, results in an answer of 300, even though both sides look different. Changing the grouping in a problem like this is an example of the associative property of multiplication.
  4. c.   Choice a is an example of the associative property of addition, where changing the grouping of the numbers will still result in the same answer. Choice b is an example of the distributive property of multiplication over addition. Choice d is an example of the additive identity, where any number added to zero equals itself. Choice c is an example of the commutative property of addition, where we can change the order of the numbers that are being added and the result is always the same.
  5. b.   In the statement, 3 is being multiplied by the quantity in the parentheses, x + 4. The distributive property allows you to multiply 3 × x and add it to 3 × 4, simplifying to 3x + 12.
  6. c.   Let y = the number. The word product is a key word for multiplication. Therefore the equation is –5y = 35. To solve this, divide each side of the equation by –5; . The variable is now alone: y = –7.
  7. b.   Let x = the number. The opposite of this number is –x. The words subtraction and difference both tell you to subtract, so the equation becomes –x – 10 = 5. To solve this, add 10 to both sides of the equation; –x – 10 + 10 = 5 + 10. Simplify to –x = 15. Divide both sides of the equation by –1. Remember that –x = –1x; . The variable is now alone: x = –15.
  8. b.   Let x = the number. Since sum is a key word for addition, the equation is –4 + x = –48. Add 4 to both sides of the equation; –4 + 4 + x = –48 + 4. The variable is now alone: x = –44.
  9. c.   Let x = the number. Now translate each part of the sentence. Twice a number increased by 11 is 2x + 11; 32 less than 3 times a number is 3x – 32. Set the expressions equal to each other: 2x + 11 = 3x – 32. Subtract 2x from both sides of the equation: 2x – 2x + 11 = 3x - 2x – 32. Simplify: 11 = x – 32. Add 32 to both sides of the equation: 11 + 32 = x – 32 + 32. The variable is now alone: x = 43.
  10. a.   The statement, "If one is added to the difference when 10x subtracted from –18x, the result is 57," translates to the equation –18x – 10x + 1 = 57. Combine like terms on the left side of the equation: –28x + 1 = 57. Subtract 1 from both sides of the equation: –28x + 1 –1 = 57 – 1. Divide each side of the equation by –28: . The variable is now alone: x = –2.
  11. c.   The statement, "If 0.3 is added to 0.2 times the quantity x – 3, the result is 2.5," translates to the equation 0.2(x – 3) + 0.3 = 2.5. Remember to use parentheses for the expression when the words the quantity are used. Use the distributive property on the left side of the equation: 0.2x – 0.6 + 0.3 = 2.5. Combine like terms on the left side of the equation: 0.2x + –0.3 = 2.5. Add 0.3 to both sides of the equation: 0.2x + –0.3 + 0.3 = 2.5 + 0.3. Simplify: 0.2x = 2.8. Divide both sides by 0.2: . The variable is now alone: x = 14.
  12. b.   Let x = the number. The sentence, "If twice the quantity x + 6 is divided by negative seven, the result is 6," translates to . Remember to use parentheses for the expression when the words the quantity are used.
  13. There are different ways to approach solving this problem.

    Method I:

      Multiply both sides of the equation by –7: –7 × = 6 × – 7
      This simplifies to: 2 (x + 6) = –42
      Divide each side of the equation by 2:
      This simplifies to: x + 6 = –21
      Subtract 6 from both sides of the equation: x + 6 – 6 = –21 – 6
      The variable is now alone: x = –27

    Method II:

      Another way to look at the problem is to multiply each side by –7 in the first step to get: 2(x + 6) = –42
      Then use distributive property on the left side: 2x + 12 = –42
      Subtract 12 from both sides of the equation: 2x + 12 –12 = –42 – 12
      Simplify: 2x = –54
      Divide each side by 2:

    The variable is now alone: x = –27

  14. d.   Translating the sentence, "The difference between six times the quantity 6x + 1 minus three times the quantity x – 1 is 108," into symbolic form results in the equation: 6(6x + 1) – 3(x – 1) = 108. Remember to use parentheses for the expression when the words the quantity are used. Perform the distributive property twice on the left side of the equation: 36x + 6 – 3x + 3 = 108. Combine like terms on the left side of the equation: 33x + 9 = 108. Subtract 9 from both sides of the equation: 33x + 9 – 9 = 108 – 9. Simplify: 33x = 99. Divide both sides of the equation by 33: . The variable is now alone: x = 3.
  15. a.   This problem translates to the equation –4 (x + 8) + 6x = 2x + 32. Remember to use parentheses for the expression when the words the quantity are used. Use distributive property on the left side of the equation: –4x – 32 + 6x = 2x + 32. Combine like terms on the left side of the equation: 2x – 32 = 2x + 32. Subtract 2x from both sides of the equation: 2x – 2x – 32 = 2x – 2x + 32. The two sides are not equal. There is no solution: –32 ≠ 32.
  16. c.   Let x = the amount of hours worked so far this week. Therefore, the equation is x + 4 = 10. To solve this equation, subtract 4 from both sides of the equation; x + 4 – 4 = 10 – 4. The variable is now alone: x = 6.
  17. b.   Let x = the number of CDs Kathleen has. Four more than one half the number can be written as + 4. Set this amount equal to 16, which is the number of CDs Michael has. To solve this, subtract 4 from both sides of the equation: + 4 – 4 = 16 – 4. Multiply each side of the equation by 2: = 2 × 12. The variable is now alone: x = 24.
  18. d.   Since the perimeter of the square is x + 4, and a square has four equal sides, we can use the perimeter formula for a square to find the answer to the question: P = 4s where P = perimeter and s = side length of the square. Substituting the information given in the problem, P = x + 4 and s = 24, gives the equation: x = 4 – 4(24). Simplifying yields x + 4 = 96. Subtract 4 from both sides of the equation: x + 4 – 4 = 96 – 4. Simplify: x = 92.
  19. b.   Let x = the width of the rectangle. Let x + 3 = the length of the rectangle, since the length is "3 more than" the width. Perimeter is the distance around the rectangle. The formula is length + width + length + width, P = l + w + l + w, or P = 2l + 2w. Substitute the let statements for l and w and the perimeter (P) equal to 21 into the formula: 21 = 2(x + 3) + 2(x). Use the distributive property on the right side of the equation: 21 = 2x + 6 + 2x. Combine like terms of the right side of the equation: 21 = 4x + 6. Subtract 6 from both sides of the equation: 21 – 6 = 4x + 6 – 6. Simplify: 15 = 4x. Divide both sides of the equation by 4: . The variable is now alone: 3.75 = x.
  20. a.   Two consecutive integers are numbers in order like 4 and 5 or –30 and –29, which are each 1 number apart. Let x = the first consecutive integer. Let x + 1 = the second consecutive integer. Sum is a key word for addition so the equation becomes: (x)+ (x + 1) = 41. Combine like terms on the left side of the equation: 2x + 1 = 41. Subtract 1 from both sides of the equation: 2x + 1 – 1 = 41 – 1. Simplify: 2x = 40. Divide each side of the equation by 2: . The variable is now alone: x = 20. Therefore the larger integer is: x + 1 = 21. The two integers are 20 and 21.
  21. a  . Two consecutive even integers are numbers in order, such as 4 and 6 or –30 and –32, which are each 2 numbers apart. Let x = the first consecutive even integer. Let x + 2 = the second (and larger) consecutive even integer. Sum is a key word for addition so the equation becomes (x) + (x + 2) = 126. Combine like terms on the left side of the equation: 2x + 2 = 126. Subtract 2 from both sides of the equation: 2x + 2 – 2 = 126 – 2; simplify: 2x = 124. Divide each side of the equation by 2: . The variable is now alone: x = 62. Therefore the larger integer is: x + 2 = 64.

More practice problems on algebra word problems can be found at:

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