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Worksheet Grade 6 Math

Nitrate Levels in Hazel Run: Decimal Operations and Percent Change

A grade 6 worksheet using decimal nitrate readings from a fictional creek to practice decimal operations, percent increase, and percent-of comparisons.

Objectives

  1. Students will add, subtract, multiply, and divide multi-digit decimal water-quality readings.
  2. Students will calculate percent increase between two decimal measurements.
  3. Students will distinguish a percent increase from a percent-of comparison using the same data.

Vocabulary

Nitrate
Nitrate is a form of nitrogen that plants use as food. It gets into water from fertilizer, animal waste, and septic systems, and too much of it is unhealthy for both people and streams.

Also: decimal, percent increase, mean

Teacher overview

Hazel Run (a fictional creek used for this worksheet) flows past a farm field that applies fertilizer each growing season. This worksheet uses a set of fictional decimal nitrate readings, in milligrams per liter, taken from spring through late summer, to give students practice with decimal addition, subtraction, and division, followed by percent increase and percent-of comparisons on the same data.

A common misconception is treating "percent increase" and "what percent one number is of another" as the same calculation. Problem 8 asks students to explain why those two questions, applied to readings from the same data set, produce very different answers.

Nitrate Levels in Hazel Run: Decimal Operations and Percent Change — Grade 6 Math
Name: ____________________________ Date: ______________

Student Worksheet

Hazel Run (fictional) is checked for nitrate all through the growing season. Too much nitrate can feed algae, and algae blooms can lower the oxygen fish need. Use the readings below for all problems.

  1. In spring, Hazel Run's nitrate level measured 2.4 mg/L. By early summer, it measured 3.9 mg/L. Find the increase, in mg/L. Use decimal subtraction.

  2. By late summer, the level rose again to 5.85 mg/L. Find the increase from early summer to late summer.

  3. Add your answers from problems 1 and 2 to find the total increase from spring to late summer. Then check your work. Subtract the spring reading from the late-summer reading. Do the two answers match?

  4. A crew takes four more readings that same week in late summer: 5.7, 6.1, 5.9, and 5.75 mg/L. Find the sum of all four readings.

  5. Find the mean (average) of the four readings in problem 4. Divide your sum by 4. Round to the nearest hundredth.

  6. Find the percent increase in nitrate level from spring (2.4 mg/L) to late summer (5.85 mg/L). Round to the nearest whole percent.

  7. A different stretch of Hazel Run, away from the farm, held steady at 1.2 mg/L all season. What percent of the late-summer reading (5.85 mg/L) is that steady reading? Round to the nearest whole percent.

  8. Look at problems 6 and 7. Why does a percent increase give different information than a percent-of comparison, even though both use the same data? Explain.

  9. Challenge. Suppose the nitrate level keeps rising by the same amount it did from early to late summer (problem 2), for one more measuring period. What would the new reading be? What would the total percent increase from spring be at that point?

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: 1.5 mg/L.
    3.9 − 2.4 = 1.5.

  2. Answer: 1.95 mg/L.
    5.85 − 3.9 = 1.95.

  3. Answer: 3.45 mg/L.
    1.5 + 1.95 = 3.45. Check: 5.85 − 2.4 = 3.45, which matches.

  4. Answer: 23.45 mg/L.
    5.7 + 6.1 + 5.9 + 5.75 = 23.45.

  5. Answer: 5.86 mg/L.
    23.45 ÷ 4 = 5.8625, which rounds to 5.86.

  6. Answer: 144 percent.
    Increase = 3.45. Percent increase = 3.45 ÷ 2.4 = 1.4375, or about 144 percent.

  7. Answer: 21 percent.
    1.2 ÷ 5.85 = 0.2051..., which rounds to 21 percent.

  8. Answer: Percent increase compares a change to the starting value; percent-of compares one full reading to another.
    Percent increase (problem 6) divides the amount of change (3.45) by the spring starting value (2.4). Percent-of (problem 7) divides one complete reading (1.2) by a different complete reading (5.85). The two calculations use different numbers as the "part" and the "whole," so they answer different questions even though both readings appear in the data set.

  9. Answer: 7.80 mg/L, a 225 percent total increase from spring.
    New reading = 5.85 + 1.95 = 7.80 mg/L. Total increase from spring = 7.80 − 2.40 = 5.40. Percent increase = 5.40 ÷ 2.40 = 2.25, or 225 percent.

Differentiation

Support (below grade level)

Provide a decimal place-value chart for problems 1 through 5, with the decimal points already aligned in each pair of numbers, and allow a calculator for problem 5's division. Work problem 6 together as a class model before students attempt problem 7 independently.

Extension (above grade level)

Using the fictional spring reading from problem 1, ask students to find what late-summer reading would make the percent increase from spring equal exactly 200 percent, and to check their answer using the percent-increase formula.

ELL supports

Provide a labeled number line showing the three seasonal readings (spring, early summer, late summer) in order, with the amount of increase marked between each pair. Allow the explanation in problem 8 to be supported with the labeled number line and short phrases rather than full sentences.