Math problem set Grade 7 Math

Foxglove Pump Station Flow Rates: A Grade 7 Unit-Rate Set

A grade 7 problem set where students compute unit rates from ratios of fractions for a fictional pump station's flows, then compare and combine the rates.

Grade: Grade 7
Subject: Math
Material: Math problem set
Time: 40 min
Prep: No prep
Materials: Calculator

Standards

7.RP.A.1 Common Core Math

Compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units.

How this material aligns: Problems 1, 4, and 5 divide a fractional amount by a fractional time to get a unit rate, the ratio-of-fractions computation this standard names.

All materials for 7.RP.A.1 →

7.NS.A.3 Common Core Math

Solve real-world and mathematical problems involving the four operations with rational numbers.

How this material aligns: Every problem uses multiplication and division of fractions and decimals to solve a real pumping problem, the four-operation rational-number work this standard requires.

All materials for 7.NS.A.3 →

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This material is in the public domain (CC0) and may be copied, edited, and shared freely, without attribution. Updated 2026-07-23.

Objectives

Students will compute a unit rate by dividing a fractional amount by a fractional time.
Students will combine and compare pump rates expressed in gallons per hour and per minute.

Vocabulary

Flow rate: Flow rate measures how much water moves past a point in a certain amount of time, like gallons per minute through a garden hose. Bigger pipes and stronger pushes give bigger flow rates.

Also: unit rate, complex fraction

Teacher overview

The fictional Foxglove Pump Station runs several pumps, and its meters rarely land on whole numbers. Students compute each pump's rate by dividing a fractional amount of water by a fractional amount of time, which is a unit rate built from a complex fraction. The set then reuses those rates to compare and combine pumps. A common error is flipping the division and finding time per gallon instead of gallons per time, so students should always divide the amount by the time.

Foxglove Pump Station Flow Rates: A Grade 7 Unit-Rate Set — Grade 7 Math

Name: ____________________________ Date: ______________

Problems

All problems below use the fictional Foxglove Pump Station and the 2,400-gallon tank it fills.

Pump A fills 5/8 of the tank in 1/2 hour. Find Pump A's rate in tanks per hour.
The tank holds 2,400 gallons. Use Pump A's rate from Problem 1 to find how many gallons it moves per hour.
Pump B moves 900 gallons in 3/4 hour. Find Pump B's rate in gallons per hour.
A cracked pipe leaks 2/3 gallon in 1/6 minute. Find the leak rate in gallons per minute.
Pump C empties 3/5 of a holding pond in 1 1/2 hours. Find Pump C's rate in ponds per hour.
Pump D fills 3/8 of the 2,400-gallon tank in 1/2 hour. Find Pump D's rate in gallons per hour.
Pumps A and D run together to fill the empty 2,400-gallon tank. Using their gallons-per-hour rates from Problems 2 and 6, find their combined rate and how long the fill takes.
Compare the leak in Problem 4 with Pump B. Convert Pump B's rate from Problem 3 to gallons per minute, then state which moves more water each minute.

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

Answer Key

Answer: 1.25 tanks per hour. Divide the amount by the time: (5/8) ÷ (1/2) = (5/8) × 2 = 10/8 = 5/4 = 1.25 tanks per hour.
Answer: 3,000 gallons per hour. 1.25 tanks per hour × 2,400 gallons per tank = 3,000 gallons per hour.
Answer: 1,200 gallons per hour. (900 gallons) ÷ (3/4 hour) = 900 × 4/3 = 3,600/3 = 1,200 gallons per hour.
Answer: 4 gpm. (2/3 gallon) ÷ (1/6 minute) = (2/3) × 6 = 12/3 = 4 gallons per minute.
Answer: 0.4 pond per hour. (3/5) ÷ (1 1/2) = (3/5) ÷ (3/2) = (3/5) × (2/3) = 6/15 = 2/5 = 0.4 pond per hour.
Answer: 1,800 gallons per hour. First, 3/8 of 2,400 = 900 gallons in 1/2 hour. Then 900 ÷ (1/2) = 900 × 2 = 1,800 gallons per hour.
Answer: 0.5 hour. Combined rate = 3,000 + 1,800 = 4,800 gallons per hour. Time = 2,400 ÷ 4,800 = 0.5 hour, which is 30 minutes.
Answer: Pump B, at 20 gpm. Pump B's 1,200 gallons per hour ÷ 60 minutes = 20 gallons per minute. Since 20 is greater than the leak's 4 gallons per minute, Pump B moves more water each minute.

Differentiation

Support

Give students a printed reminder that a unit rate divides the amount by the time, with one worked complex fraction shown step by step. Provide the fact that 3/8 of 2,400 is 900 for Problem 6 so they can focus on the division.

Extension

Ask students to find a single pump rate, in gallons per hour, that would fill the 2,400-gallon tank in exactly 40 minutes, then express that rate as a complex fraction of gallons over a fractional hour.

ELL supports

Provide a labeled diagram showing "amount of water" over "amount of time" as a fraction, with an arrow to the words "unit rate." Allow students to show the division of fractions with numbers and the diagram rather than full sentences.