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Worksheet Grade 7 Math

Pethwick's Corrosion Control Results: A Grade 7 Percent Worksheet

A grade 7 worksheet where students compute percent decrease, percent of a budget, and a cost per account from a fictional water district's lead tap results.

Objectives

  1. Students will compute a percent decrease using the original value as the base.
  2. Students will find a whole amount when a part and its percent are known.
  3. Students will express a combined cost as a percent of a stated budget.

Vocabulary

Corrosion control
Treatment that keeps water from slowly eating away at the inside of pipes. It helps stop metals like lead from getting into drinking water.
Action level
A trigger number in lead and copper testing. If too many homes test above this level, the utility must take steps like improving treatment and telling the public.

Also: percent decrease, percentage point, treatment budget

Teacher overview

Lead reaches tap water from the pipe and the solder between the main and the faucet, so a utility with lead in its plumbing adds a corrosion-control chemical that lays a protective coating inside the pipe. The federal action level is 15 parts per billion, and it drops to 10 parts per billion in 2027, so a district that clears the current level may still have to show further progress. This worksheet follows a fictional district through one such program and asks for percents of a sample, a percent decrease, and a share of a budget. The error to correct is dividing a change by the new value instead of the original, and the related shortcut of subtracting 5 percent from 20 percent and reporting 15 percent as the percent decrease.

Pethwick's Corrosion Control Results: A Grade 7 Percent Worksheet — Grade 7 Math
Name: ____________________________ Date: ______________

Student Worksheet

The Pethwick Water District is a fictional utility. Before it started corrosion control, it sampled 60 taps. Of those, 12 came back at or above the action level of 15 parts per billion. After a year of treatment, the district sampled the same 60 taps and 3 came back at or above that level. Over the same year the district's 90th-percentile result fell from 18 parts per billion to 9 parts per billion.

  1. What percent of the 60 sampled taps were at or above the action level before treatment?
  2. What percent of the 60 sampled taps were at or above the action level after treatment?
  3. By what percent did the number of taps at or above the action level decrease?
  4. By what percent did the 90th-percentile result decrease?
  5. The action level drops to 10 parts per billion in 2027. How many parts per billion below that new level is the district's result of 9?
  6. Pethwick spent $46,000 on the corrosion-control chemical last year. That was 8 percent of its treatment budget. What is the treatment budget?
  7. The supplier raises the chemical price by 6 percent for next year. What will the chemical cost?
  8. Pethwick serves 4,800 accounts. What will next year's chemical cost be for each account? Round to the nearest cent.
  9. Suppose Pethwick stopped treating and the rate from item 1 held across all 4,800 accounts. How many accounts would be at or above the action level?
  10. Pethwick also spends $8,740 a year to collect and test tap samples. Testing plus next year's chemical cost is what percent of the treatment budget?
  11. A resident reads item 3 and says lead dropped by that same percent in every home. Explain why the data does not show that.

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: 20 percent.
    Divide 12 by 60 to get 0.20, then multiply by 100.
  2. Answer: 5 percent.
    Divide 3 by 60 to get 0.05, then multiply by 100.
  3. Answer: 75 percent.
    The count fell from 12 to 3, a change of 9. Divide the change by the original count: 9 divided by 12 equals 0.75, or 75 percent.
  4. Answer: 50 percent.
    The result fell from 18 to 9, a change of 9. Divide 9 by 18 to get 0.50, or 50 percent.
  5. Answer: 1 part per billion.
    Subtract the district result from the 2027 level: 10 minus 9 equals 1.
  6. Answer: $575,000.
    The chemical cost is 8 percent of the budget, so divide: 46,000 divided by 0.08 equals 575,000.
  7. Answer: $48,760.
    A 6 percent increase multiplies the cost by 1.06. Then 46,000 times 1.06 equals 48,760.
  8. Answer: $10.16.
    Divide the new cost by the accounts: 48,760 divided by 4,800 equals 10.158, which rounds to 10.16.
  9. Answer: 960 accounts.
    Take 20 percent of 4,800. Multiply 0.20 by 4,800 to get 960.
  10. Answer: 10 percent.
    Add the two costs: 48,760 plus 8,740 equals 57,500. Then divide by the budget: 57,500 divided by 575,000 equals 0.10, or 10 percent.
  11. Answer: The 75 percent applies to the count of taps, not to the lead level in any one home.
    Nine fewer taps landed at or above the action level, which says nothing about how far any single tap fell. One home's result could have dropped a little and still crossed below the line, while another could have dropped a great deal and stayed above it.

Differentiation

Support

Give these students a percent-change organizer with three labeled boxes for the original value, the change, and the quotient, and let them fill the boxes before dividing. Assign items 1 through 5 and item 9, and allow a calculator throughout.

Extension

Ask these students to find what percent increase in the chemical price would push the chemical alone to 10 percent of the same treatment budget, and to state how much more per account that would cost. Have them show both calculations.

ELL supports

Pre-teach "action level," "sample," "decrease," and "budget" with a bar that shrinks from 12 blocks to 3 blocks beside a circle divided into ten parts. Provide the frame "The count fell from ___ to ___, so the change is ___ and the percent decrease is ___" for items 3 and 4.