Math problem set Grade 7 Science

Wetherby Reservoir Road-Salt Chloride: A Grade 7 Proportion Set

A grade 7 problem set tracking chloride from winter road salt in a reservoir: percent increase, proportional rates, ppm equivalence, and percent error.

Objectives

  1. Students will compute the percent increase in a chloride reading.
  2. Students will use a proportional rate to predict chloride from a salt amount.
  3. Students will convert milligrams per liter to parts per million and find percent error.

Vocabulary

Parts per million (ppm)
Parts per million tells how many tiny bits of a substance are mixed into a million bits of water. Scientists use it to measure very small amounts of chemicals in water.

Also: chloride, milligrams per liter, percent error

Teacher overview

Salt spread on winter roads washes into lakes and streams, where it raises the level of chloride in the water. This set follows chloride readings at the fictional Wetherby Reservoir from summer into winter and has students work with the change. A common misconception to watch for is treating the raw rise in milligrams per liter as if it were the percent increase; the two are different, and problems 1 and 2 are built to separate them. A second point to watch: in fresh water like the reservoir's, milligrams per liter and parts per million are almost exactly the same size, so no arithmetic is needed to switch between them. This shortcut does not hold for very salty water such as seawater, where a correction is needed.

Answer Key

  1. Answer: 55 mg/L.
    95 mg/L minus 40 mg/L equals 55 mg/L of increase.
  2. Answer: 138 percent.
    The increase is 55 mg/L over the starting 40 mg/L. 55 divided by 40 equals 1.375, which is a 138 percent increase.
  3. Answer: 95 ppm.
    Because 1 mg/L is about 1 ppm in fresh water, 95 mg/L is about 95 ppm.
  4. Answer: 43 percent.
    95 divided by 220 equals about 0.432, so the winter level is about 43 percent of the taste target.
  5. Answer: 99 mg/L.
    The rate is 55 mg/L per 500 pounds, which is 0.11 mg/L per pound. Then 0.11 times 900 pounds equals 99 mg/L.
  6. Answer: 100 pounds.
    At 0.11 mg/L per pound, 11 mg/L divided by 0.11 mg/L per pound equals 100 pounds of salt.
  7. Answer: 6 percent.
    The difference is 95 minus 90, which is 5 mg/L. Then 5 divided by the true value 90 equals about 0.056, or about 6 percent error.
  8. Answer: 47.5 mg/L, yes.
    Half of 95 mg/L is 47.5 mg/L, which is well below the 220 mg/L target.

Differentiation

Support

Give students the percent formulas written out with labels (change divided by starting amount for percent increase; part divided by whole for percent of a target). Provide the salt-to-chloride rate already computed (0.11 milligrams per liter per pound) so students can focus on problems 5 and 6.

Extension

Ask students to find the summer-to-winter percent increase if the winter reading had instead reached the 220 milligrams per liter target, and to explain why the same raw increase can be a larger or smaller percent depending on the starting level.

ELL supports

Pre-teach "chloride," "increase," and "target" with a simple thermometer-style bar that fills from summer to winter. Allow students to label the parts of each percent formula before computing.