Spectrum Math Grade 7 Chapter 4 Lesson 1 Answer Key Unit Rates with Fractions

This handy Spectrum Math Grade 7 Answer Key Chapter 4 Lesson 4.1 Unit Rates with Fractions provides detailed answers for the workbook questions.

Spectrum Math Grade 7 Chapter 4 Lesson 4.1 Unit Rates with Fractions Answers Key

A rate is a special ratio in which two terms are in different units. A unit rate is when one of those terms is expressed as a value of 1. Rates can be calculated with whole numbers or with fractions.
Emily ate \(\frac{1}{4}\) of an ice-cream cone in \(\frac{1}{2}\) of a minute. How long would it take her to eat one ice-cream cone?

1.Set up equivalent ratios using the information from the problem and l to represent the ice cream cone. Let t represent the time.
Spectrum Math Grade 7 Chapter 4 Lesson 1 Answer Key Unit Rates with Fractions 1
\(\frac{1}{4}\) × t = \(\frac{1}{2}\) × l
\(\frac{1}{4}\) × t ÷ \(\frac{1}{4}\) = \(\frac{1}{2}\) × 1 ÷ \(\frac{1}{4}\)
t = 2
2. Use cross multiplication.
3. Isolate the variable.
4. Solve.

Find the unit rate in each problem.

Question 1.
For Bill’s birthday his mom is bringing donuts to school. She has a coupon to get 2\(\frac{1}{2}\) dozen donuts for $8.00. How much would just one dozen donuts cost at this price?
Let c represent the cost of the donuts.
Equivalent ratios: _______________________
One dozen donuts would cost ____________________
Answer: Equivalent ratios: 2\(\frac{1}{2}\) ÷ 8 = \(\frac{1}{c}\)
One dozen donuts would cost $3.20

Bill’s mom has a coupon to get 2\(\frac{1}{2}\) dozen donuts for $8.00
Let c represent the cost of the donuts
2\(\frac{1}{2}\) dozen donuts = 2\(\frac{1}{2}\)  × 12
= \(\frac{5}{2}\) × 12
= 30 donuts
Therefore, Bill’s mom has a coupon to get 30 donuts for $8.00
Equivalent ratios:
2\(\frac{1}{2}\) ÷ 8 = \(\frac{1}{c}\)
\(\frac{30}{8}\) = \(\frac{1}{c}\)
By cross multiplication,
30 × c = 1 × 8
c = \(\frac{8}{30}\)
so, c = 2.6667
the cost of each donut = 2.667, so the cost of One dozen donuts would be 2.6667 × 12 = $3.20

Question 2.
Jake ate 4\(\frac{1}{2}\) pounds of candy in one week. If he ate the same amount of candy every day, how much candy did he eat each day?
Let c represent the amount of candy.
Equivalent ratios: ______________________
He ate ___________________ pounds of candy each day.
Answer: Equivalent ratios: 4\(\frac{1}{2}\) ÷ 7 = \(\frac{1}{c}\)
He ate 1\(\frac{5}{9}\) pounds of candy each day.

Jake ate 4\(\frac{1}{2}\) pounds of candy in one week.
Converting the above mixed fraction into the improper fraction
4\(\frac{1}{2}\) = \(\frac{9}{2}\)
Therefore, Jake ate \(\frac{9}{2}\) pounds of candy in 7 days.
Let c represent the amount of candy.
Equivalent ratios:
4\(\frac{1}{2}\) ÷ 7 = \(\frac{1}{c}\)
\(\frac{9}{2}\) ÷ 7 = \(\frac{1}{c}\)
\(\frac{9}{14}\) = \(\frac{1}{c}\)
By cross multiplication,
9 × c = 1 × 14
c = \(\frac{14}{9}\)
c =1\(\frac{5}{9}\)
Therefore, He ate 1\(\frac{5}{9}\) pounds of candy each day.

Question 3.
A bakery used 6\(\frac{1}{4}\) cups of Hour this morning to make 5 batches of cookies. How much flour went into each batch of cookies?
Let f represent the amount of flour.
Equivalent ratios: ______________
Each batch of cookies used _______________________ cups of flour.
Answer: Equivalent ratios: 6\(\frac{1}{4}\) ÷ 5 = \(\frac{f}{1}\)
Each batch of cookies used 1\(\frac{1}{4}\) cups of flour.

A bakery used 6\(\frac{1}{4}\) cups of Hour this morning to make 5 batches of cookies.
Let f represent the amount of flour.
Equivalent ratios:
6\(\frac{1}{4}\) ÷ 5 = \(\frac{f}{1}\)
\(\frac{25}{4}\) ÷ 5 = \(\frac{f}{1}\)
\(\frac{25}{20}\) = \(\frac{f}{1}\)
By cross multiplication,
20 × f = 1 × 25
f = \(\frac{25}{20}\)
f =  \(\frac{5}{4}\)
f = 1\(\frac{1}{4}\)
Therefore, Each batch of cookies used 1\(\frac{1}{4}\) cups of flour.

Using unit rates can help you compare two items.

Mike’s car can travel 425 miles on 10\(\frac{1}{2}\) gallons of gas. Jason’s car can travel 275 miles on 5\(\frac{4}{5}\) gallons of gas. Which car gets better gas mileage?
Let m represent Mike’s car and j represent Jason’s car.
Equivalent Ratio 1: = \(\frac{425}{10 \frac{1}{2}}\) = \(\frac{m}{l}\) m = 40\(\frac{10}{21}\) miles per gallon
Equivaient Ratio 2: \(\frac{275}{5 \frac{4}{5}}\) = \(\frac{j}{l}\) j = 47\(\frac{12}{29}\) miles per gallon
Jason’s car gets better gas mileage because it can go farther on one gallon of gas.

Calculate unit rates to solve each problem.

Question 1.
Cara can run 3 miles in 27\(\frac{1}{2}\) minutes. Melanie can run 6 miles in 53\(\frac{1}{3}\) minutes. Who can run faster?
Let c represent Cara’s speed and m represent Melanie’s speed.
Equivalent Ratio 1: _____________________
Equivalent Ratio 2: ______________________
_______________________can run faster.
Answer: Equivalent Ratio 1:\(\frac{3}{27 \frac{1}{2}}\) = \(\frac{d}{t}\) c
Equivalent Ratio 2: \(\frac{6}{53 \frac{1}{3}}\) = \(\frac{d}{t}\) m
Melanie can run faster.

Cara can run 3 miles in 27\(\frac{1}{2}\) minutes.
Melanie can run 6 miles in 53\(\frac{1}{3}\) minutes.
Let c represent Cara’s speed and m represent Melanie’s speed.
Equivalent Ratio 1: \(\frac{3}{27 \frac{1}{2}}\) = \(\frac{d}{t}\) c
\(\frac{3}{ \frac{55}{2}}\) = \(\frac{d}{t}\) c
= \(\frac{6}{55}\) = 0.109090
Equivalent Ratio 2: \(\frac{6}{53 \frac{1}{3}}\) = \(\frac{d}{t}\) m
\(\frac{6}{\frac{160}{3}}\) = \(\frac{d}{t}\) m
= \(\frac{18}{160}\) = 0.1125
Therefore, by comparing the above equivalent ratios, Melanie can run faster.

Question 2.
Bob goes to Shop and Save and buys 3\(\frac{1}{3}\) pounds of turkey for $10.50. Sonia goes to Quick Stop and buys 2\(\frac{1}{2}\) pounds of turkey for $6.25. Who got a better deal?
Let b represent Bob’s price and s represent Sonia’s price.
Equivalent Ratio 1: ____________________
Equivalent Ratio 2: _____________________
__________ got a better deal on turkey.

Answer: Equivalent Ratio 1: \(\frac{10.5}{3 \frac{1}{3}}\) = \(\frac{c}{q}\) b
Equivalent Ratio 2: \(\frac{6.25}{2 \frac{1}{2}}\) = \(\frac{c}{q}\) s
Sonia got a better deal on turkey.
Bob goes to Shop and Save and buys 3\(\frac{1}{3}\) pounds of turkey for $10.50.
Sonia goes to Quick Stop and buys 2\(\frac{1}{2}\) pounds of turkey for $6.25.
Let b represent Bob’s price and s represent Sonia’s price.
Equivalent Ratio 1: \(\frac{10.5}{3 \frac{1}{3}}\) = \(\frac{c}{q}\) b
\(\frac{10.5}{ \frac{10}{3}}\) = \(\frac{c}{q}\) b
= \(\frac{31.5}{10}\) = 3.15
Equivalent Ratio 2: \(\frac{6.25}{2 \frac{1}{2}}\) = \(\frac{c}{q}\) s
\(\frac{6.25}{ \frac{5}{2}}\) = \(\frac{c}{q}\) s
= \(\frac{12.5}{5}\) = 2.5
Therefore, by comparing the above equivalent ratios, Sonia got a better deal on turkey.

Question 3.
Thomas went for a long hike and burned 675 calories in 2\(\frac{1}{2}\) hours. Marvin decided to go for a bike ride and burned 1,035 calories in 3\(\frac{1}{4}\) hours. Who burned the most calories per hour?
Let t represent Thomas’s calories burned and m represent Marvin’s calories burned.
Equivalent Ratio 1: ______________________
Equivalent Ratio 2: _____________________
____________ burned the most calories per hour.

Answer: Equivalent Ratio 1: \(\frac{675}{2 \frac{1}{2}}\) = \(\frac{c}{t}\) t
Equivalent Ratio 2: \(\frac{1035}{3 \frac{1}{4}}\) = \(\frac{c}{t}\) m
Marvin  burned the most calories per hour.
Thomas went for a long hike and burned 675 calories in 2\(\frac{1}{2}\) hours.
Marvin decided to go for a bike ride and burned 1,035 calories in 3\(\frac{1}{4}\) hours.
Let t represent Thomas’s calories burned and m represent Marvin’s calories burned.
Equivalent Ratio 1: \(\frac{675}{2 \frac{1}{2}}\) = \(\frac{c}{t}\) t
\(\frac{675}{ \frac{5}{2}}\) = \(\frac{c}{t}\) t
= \(\frac{1350}{5}\) = 270 calories per hour.
Equivalent Ratio 2: \(\frac{1035}{3 \frac{1}{4}}\) = \(\frac{c}{t}\) m
\(\frac{1035}{ \frac{13}{4}}\) = \(\frac{c}{t}\) m
= \(\frac{4140}{13}\) = 318.4615 calories per hour.
Therefore, by comparing the above equivalent ratios, Marvin  burned the most calories per hour.

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