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6 th Grade Test Prep

6 th Grade Test Prep. Algebra. 5.A.2 Translate simple expressions into algebraic expressions. Eighteen less than twice a number. 2t - 18. A number squared, plus nineteen. n² + 19. The product of eleven and a number, divided by sixty. 11 × k ÷ 60.

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6 th Grade Test Prep

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  1. 6th Grade Test Prep Algebra

  2. 5.A.2 Translate simple expressions into algebraic expressions Eighteen less than twice a number 2t - 18 A number squared, plus nineteen n² + 19 The product of eleven and a number, divided by sixty 11 × k ÷ 60 One hundred more than the quotient of a number and two hundredths n ÷ 0.02 + 100

  3. 6.A.1 6.A.3Translate two-step verbal equations and expressions Joanna receives $8 per hour when she works at the coffee shop. Last week she earned $256. Write an equation that can be used to find the number of hours Joanna worked last week. Let h = Hours Joanna worked 8h = 256 Write an equation to represent the following. The sum of 7 and a number, divided by 15, is equal to 33. Let n = the number (7 + n) ÷ 15 = 33

  4. 6.A.2Substitute and solve 5.A.3 Evaluate 8a – 2b when a = 7 and b = 4 Evaluate x² + y³ when x = 3 and y = 2 How do you evaluate 6c + d² when c = 9 and d = 5

  5. 5.A.4 5.A.5 One step equations Solve for t: 5t = 25 Step 1: Use the inverse operation (division) to ‘move’ the 5. 5t ÷ 5 = 25 ÷ 5 Step 2: Simplify t = 5 Step 3: Check your work by substituting and solving. t = 5 5t = 25 5(5) = 25 25 = 25

  6. 6.A.4Solve two-step equations Solve for t: 5 + 4t = 25 Step 1: Start by identifying like terms. Use the inverse operation (subtraction) to ‘move’ the 5. 5 + (-5) + 4t = 25 + (-5) 4t = 20 Step 2: To isolate the variable, use the inverse operation (division), to ‘move’ the 4. 4t ÷ 4 = 20 ÷ 4 t = 5 Step 3: Check your work by substituting and solving. t = 5 5 + 4(5) = 25 5 + 20 = 25

  7. 6.A.4Solve two-step equations Solve for n: - 19 = 7 Step 1: Start with the term that does NOT include the variable. Use the inverse operation (subtraction) to ‘move’ the 19. n/8 – 19 (+ 19) = 7 + 19 n/8 = 26 Step 2: To isolate the variable, use the inverse operation (multiplication), to ‘move’ the 8. n/8 (* 8) = 26 (* 8) n = 208 Step 3: Check your work by substituting and solving. n = 208 208 / 8 – 19 = 7 √

  8. 6.A.5Solve simple proportions within context A car uses 9 gallons of gasoline for a 162-mile drive. How many gallons of gasoline will the same car use in a 216-mile drive? Step 1: Write the ratio given in the problem. The ratio is “162 miles uses 9 gallons.” As a fraction, the ratio miles/ gallons is 162/9 Step 2: Represent the unknown by a variable. g = gallons of gasoline for a 216-mile drive Step 3: Write the ratio and fraction for the unknown variable. The ratio is “216 miles uses ‘g’ gallons.” As a fraction, the ratio miles/ gallons is 216 / g Step 4: Write the ratios as proportions. Cross-multiply to solve. 162g = 1944 g = 12 The car will use 12 gallons of gasoline for a 216-mile drive .

  9. 6.A.6Evaluate formulas for given input values Julia’s sister drove 224 miles in 4 hours. If Julia’s sister drove at a constant speed, how fast did she drive? Use the formula d = r t. d = 224 miles t = 4 hours r = ? d = r t 224 = r 4 224 ÷ 4 = r 56 miles per hour = r Terry just got a $1,200 three-year loan. The loan has a 7% interest rate. How much interest will Terry pay on the loan? Use the formula I = P r t. P = $1,200 r = 7% t = 3 years I = ? I = P r t I = (1200)(.07)(3) I = 252 Terry will $252 as interest over the 3 years.

  10. 6.A.6Evaluate formulas for given input values Which temperature is warmer 34°C or 95°F? Use the formula °C = 5/9 * (°F – 32) °C = 34 °F = 95 °C = 5/9 * (°F – 32) °C = 5/9 * (95 – 32) °C = 5/9 * (63) °C = 35° Thus 95°F = 35°C, which is warmer than 34°C.

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