lesson 7 homework 1.4 answer key

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Home > CC3 > Chapter 4 > Lesson 4.1.7

Lesson 4.1.1, lesson 4.1.2, lesson 4.1.3, lesson 4.1.4, lesson 4.1.5, lesson 4.1.6, lesson 4.1.7.

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How to Use a Math Medic Answer Key

Written by Luke Wilcox published 3 years ago

Answer key might be the wrong term here. Sure, the Math Medic answer keys do provide the correct answers to the questions for a lesson, but they have been carefully designed to do much more than this. They are meant to be the official guide to teaching the lesson, providing specific instructions for what to do and say to make a successful learning experience for your students.

Before we look at the details of the answer key, let's make sure we understand the instructional model first.

Experience First, Formalize Later (EFFL)

A typical Math Medic lesson always has the same four parts: Activity, Debrief Activity, QuickNotes, and Check Your Understanding. Here are the cliff notes:

Activity: Students are in groups of 2 - 4 working collaboratively through the questions in the Activity. The teacher is checking in with groups and using questions, prompts, and cues to get students to refine their communication and understanding. As groups finish the activity, the teacher asks students to go to the whiteboard to write up their answers to the questions.

Debrief Activity: In the whole group setting, the teacher leads a discussion about the student responses to the questions in the activity, often asking students to explain their thinking and reasoning about their answers. The teacher then formalizes the learning by highlighting key concepts and introducing new vocabulary, notation, and formulas.

QuickNotes: The teacher uses direct instruction to summarize the learning from the activity in the QuickNotes box - making direct connections to the learning targets for the lesson.

Check Your Understanding: Students are then asked to apply their learning from the lesson to a new context in the Check Your Understanding (CYU) problem. This can be done individually or in small groups. The CYU is very flexible in it's use, as it can be used as an exit ticket, a homework problem, or a quick review the next day.

How Do I See EFFL in the Answer Key?

You will see EFFL in the answer key like this:

Anything written in blue is something we expect our students to produce. This might not be quite what we expect by the end of the lesson, but provides us with a starting point when we move to formalization.

Anything written in red is an idea added by the teacher - the formalization of the learning that happened during the Activity. Students are expected to add these "notes" to their Activity using a red pen or marker.

What Do Students Write Down For Notes?

By the end of the lesson, students will have written down everything you see on the Math Medic Answer Keys. The most important transition is when students finish the Activity and we move to Debrief Activity. "Students, now is the time for you to put down your pencils and get out your your red Paper Mate flair pens" We give each student a Paper Mate flair pen at the beginning of the school year and tell them they must cherish and protect it with their life. They all think we should be sponsored by Paper Mate (anyone have any leads on this?)

The lessons you see on Math Medic are all of the notes we use with our students. We do not have some secret collection of guided notes.

Do Students Have Access to Answer Keys?

Yes! Any student can create a free Math Medic account to get access to the answer keys. We often send students to the website when they are absent from a lesson or when we don't quite finish the lesson in class. We are comfortable with students having access to these answer keys because we do not think Math Medic lessons should be used as a summative assessment or be used for a grade (unless it's for completion). Our lessons are meant to be the first steps in the formative process of learning new concepts.

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Eureka Math Grade 7 Module 4 Lesson 7 Answer Key

Engage ny eureka math 7th grade module 4 lesson 7 answer key, eureka math grade 7 module 4 lesson 7 example answer key.

Example 1: A Video Game Markup Games Galore Super Store buys the latest video game at a wholesale price of $30.00. The markup rate at Game’s Galore Super Store is 40%. You use your allowance to purchase the game at the store. How much will you pay, not including tax? a. Write an equation to find the price of the game at Games Galore Super Store. Explain your equation. Answer: Let P represent the price of the video game. Quantity = Percent × Whole P = (100% + 40%)(30) The equation shows that the price of the game at the store is equal to the wholesale cost, which is 100% and the 40% increase. This makes the new price 140% of the wholesale price.

b. Solve the equation from part (a). Answer: P = (100% + 40%)(30) P = (1.40)(30) P = 42 I would pay $42.00 if I bought it from Games Galore Super Store.

c. What was the total markup of the video game? Explain. Answer: The markup was $12.00 because $42 – $30 = $12.

d. You and a friend are discussing markup rate. He says that an easier way to find the total markup is by multiplying the wholesale price of $30.00 by 40%. Do you agree with him? Why or why not? Answer: Yes, I agree with him because (0.40)(30) = 12. The markup rate is a percent of the wholesale price. Therefore, it makes sense to multiply them together because Quantity = Percent × Whole.

→ Which quantity is the whole quantity in this problem? The wholesale price is the whole quantity.

→ How do 140% and 1.4 correspond in this situation? The markup price of the video game is 140% times the wholesale price. 140% and 1.4 are equivalent forms of the same number. In order to find the markup price, convert the percent to a decimal or fraction, and multiply it by the whole.

→ What does a markup mean? A markup is the amount of increase in a price.

Example 2: Black Friday A $300 mountain bike is discounted by 30% and then discounted an additional 10% for shoppers who arrive before 5:00 a.m. a. Find the sales price of the bicycle. Answer: Find the price with the 30% discount: Let D represent the discount price of the bicycle with the 30% discount rate. Quantity = Percent × Whole D = (100% – 30%)(300) D = (0.70)(300) D = 210 $210 is the discount price of the bicycle with the 30% discount rate. Find the price with the additional 10% discount: Let A represent the discount price of the bicycle with the additional 10% discount. A = (100% – 10%)(210) D = (1 – 0.10)(210) D = (0.90)(210) D = 189 $189 is the discount price of the bicycle with the additional 10% discount.

b. In all, by how much has the bicycle been discounted in dollars? Explain. Answer: $300 – $189 = $111. The bicycle has been discounted $111 because the original price was $300. With both discounts applied, the new price is $189.

c. After both discounts were taken, what was the total percent discount? Answer: A final discount of 40% means that you would add 30% to 10% and apply it to the same whole. This is not the case because the additional 10% discount is taken after the 30% discount has been applied, so you are only receiving that 10% discount on 70% of the original price. A 40% discount would make the final price $180 because 180 = (0.60)(300). However, the actual final discount as a percent is 37%. Let P be the percent the sales price is of the original price. Let F represent the actual final discount as a percent. Part = Percent × Whole 189 = P × 300 (\(\frac{1}{300}\))189 = P × 300(\(\frac{1}{300}\)) 0.63 = 63% = P F = 100% – 63% = 37%

d. Instead of purchasing the bike for $300, how much would you save if you bought it before 5:00 a.m.? Answer: You would save $111 if you bought the bike before 5:00 a.m. because $300 – $189 is $111.

Example 3: Working Backward A car that normally sells for $20,000 is on sale for $16,000. The sales tax is 7.5%.

→ What is the whole quantity in this problem? a. The whole quantity is the original price of the car, $20,000. Answer: What percent of the original price of the car is the final price? Quantity = Percent × Whole 16,000 = P(20,000) 16,000(\(\frac{1}{20,000}\)) = P(20,000)(\(\frac{1}{20,000}\)) 0.8 = P 0.8 = \(\frac{80}{100}\) = 80% The final price is 80% of the original price.

b. Find the discount rate. Answer: The discount rate is 20% because 100% – 80% = 20%.

c. By law, sales tax has to be applied to the discount price. However, would it be better for the consumer if the 7.5% sales tax was calculated before the 20% discount was applied? Why or why not? Answer: Apply Sales Tax First Apply the sales tax to the whole. (100% + 7.5%)(20,000) (1 + 0.075)(20,000) (1.075)(20,000) $21,500 is the price of the car, including tax, before the discount. Apply the discount to the new whole. (100% – 20%)(21,500) (1 – 0.2)(21,500) $17,200 is the final price, including the discount and tax.

Apply the Discount First (100% + 7.5%)(16,000) (1 + 0.075)(16,000) (1.075)(16,000) $17,200 is the final price, including the discount and tax.

Because both final prices are the same, it does not matter which is applied first. This is because multiplication is commutative. The discount rate and sales tax rate are both being applied to the whole, $20,000.

d. Write an equation applying the commutative property to support your answer to part (c). Answer: 20,000(1.075)(0.8) = 20,000(0.8)(1.075)

Eureka Math Grade 7 Module 4 Lesson 7 Exercise Answer Key

Exercises 1–3

Exercise 2. A golf store purchases a set of clubs at a wholesale price of $250. Mr. Edmond learned that the clubs were marked up 200%. Is it possible to have a percent increase greater than 100%? What is the retail price of the clubs? Answer: Yes, it is possible. Let C represent the retail price of the clubs, in dollars. C = (100% + 200%)(250) C = (1 + 2)(250) C = (3)(250) C = 750 The retail price of the clubs is $750.

Exercise 3. Is a percent increase of a set of golf clubs from $250 to $750 the same as a markup rate of 200%? Explain. Answer: Yes, it is the same. In both cases, the percent increase and markup rate show by how much (in terms of percent) the new price is over the original price. The whole is $250 and corresponds to 100%. \(\frac{750}{250}\) = \(\frac{3}{1}\) × 100% = 300%. $750 is 300% of $250. 300% – 100% = 200%. From Exercise 2, the markup is 200%. So, percent increase is the same as markup.

Exercise 4. a. Write an equation to determine the selling price in dollars, p, on an item that is originally priced s dollars after a markup of 25%. Answer: p = 1.25s or p = (0.25 + 1)s

d. Interpret the points (0, 0) and (1, r). Answer: The point (0, 0) means that a $0 (free) item will cost $0 because the 25% markup is also $0. The point (1, r) is (1, 1.25). It means that a $1.00 item will cost $1.25 after it is marked up by 25%; r is the unit rate.

Eureka Math Grade 7 Module 4 Lesson 7 Problem Set Answer Key

Question 1. You have a coupon for an additional 25% off the price of any sale item at a store. The store has put a robotics kit on sale for 15% off the original price of $40. What is the price of the robotics kit after both discounts? Answer: (0.75)(0.85)(40) = 25.50. The price of the robotics kit after both discounts is $25.50.

Question 2. A sign says that the price marked on all music equipment is 30% off the original price. You buy an electric guitar for the sale price of $315. a. What is the original price? Answer: \(\frac{315}{1 – 0.30}\) = \(\frac{315}{0.70}\) = 450. The original price is $450.

b. How much money did you save off the original price of the guitar? Answer: 450 – 315 = 135. I saved $135 off the original price of the guitar.

c. What percent of the original price is the sale price? Answer: \(\frac{315}{450}\) = \(\frac{70}{100}\) = 70%. The sale price is 70% of the original price.

Question 3. The cost of a New York Yankee baseball cap is $24.00. The local sporting goods store sells it for $30.00. Find the markup rate. Answer: Let P represent the unknown percent. 30 = P(24) P = \(\frac{30}{24}\) = 1.25 = (100% + 25%). The markup rate is 25%.

Question 4. Write an equation to determine the selling price in dollars, p, on an item that is originally priced s dollars after a markdown of 15%. Answer: p = 0.85s or p = (1 – 0.15)s

c. Interpret the points (0,0) and (1,r). Answer: The point (0, 0) means that a $0 (free) item will cost $0 because the 15% markdown is also $0. The point (1, r) is (1,0.85), which represents the unit rate. It means that a $1.00 item will cost $0.85 after it is marked down by 15%.

Question 5. At the amusement park, Laura paid $6.00 for a small cotton candy. Her older brother works at the park, and he told her they mark up the cotton candy by 300%. Laura does not think that is mathematically possible. Is it possible, and if so, what is the price of the cotton candy before the markup? Answer: Yes, it is possible. \(\frac{6.00}{1 + 3}\) = \(\frac{6}{4}\) = 1.50. The price of the cotton candy before the markup is $1.50.

Question 6. A store advertises that customers can take 25% off the original price and then take an extra 10% off. Is this the same as a 35% off discount? Explain. Answer: No, because the 25% is taken first off the original price to get a new whole. Then, the extra 10% off is multiplied to the new whole. For example, (1 – 0.25)(1 – 0.10) = 0.675 or (0.75)(0.90) = 0.675. This is multiplied to the whole, which is the original price of the item. This is not the same as adding 25% and 10% to get 35% and then multiplying by (1 – 0.35), or 0.65.

Question 7. An item that costs $50.00 is marked 20% off. Sales tax for the item is 8%. What is the final price, including tax? a. Solve the problem with the discount applied before the sales tax. Answer: (1.08)(0.80)(50) = 43.20. The final price is $43.20.

b. Solve the problem with the discount applied after the sales tax. Answer: (0.80)(1.08)(50) = 43.20. The final price is $43.20.

c. Compare your answers in parts (a) and (b). Explain. Answer: My answers are the same. The final price is $43.20. This is because multiplication is commutative.

Question 8. The sale price for a bicycle is $315. The original price was first discounted by 50% and then discounted an additional 10%. Find the original price of the bicycle. Answer: (315 ÷ 0.9) ÷ 0.5 = 700. The original price was $700.

Question 9. A ski shop has a markup rate of 50%. Find the selling price of skis that cost the storeowner $300. Answer: Solution 1: Use the original price of $300 as the whole. The markup rate is 50% of $300 or $150. The selling price is $300 + $150 = $450. Solution 2: Multiply $300 by 1 plus the markup rate (i.e., the selling price is (1.5)($300) = $450).

Question 10. A tennis supply store pays a wholesaler $90 for a tennis racquet and sells it for $144. What is the markup rate? Answer: Solution 1: Let the original price of $90 be the whole. Quantity = Percent × Whole. 144 = Percent(90) \(\frac{144}{90}\) = Percent 1.6 = 160%. This is a 60% increase. The markup rate is 60%. Solution 2: Selling Price = (1 + m)(Whole) 144 = (1 + m)90 1 + m = \(\frac{144}{90}\) m = 1.6 – 1 = 0.6 = 60% The markup rate is 60%.

Question 11. A shoe store is selling a pair of shoes for $60 that has been discounted by 25%. What was the original selling price? Answer: Solution 1: $60 → 75% $20 → 25% $80 → 100% The original price was $80.

Solution 2: Let x be the original cost in dollars. (1 – 0.25)x = 60 \(\frac{3}{4}\) x = 60 (\(\frac{4}{3}\))(\(\frac{3}{4}\) x) = \(\frac{4}{3}\) (60) x = 80 The original price was $80.

Question 12. A shoe store has a markup rate of 75% and is selling a pair of shoes for $133. Find the price the store paid for the shoes. Answer: Solution 1: $133 → 175% $19 → 25% $76 → 100% The store paid $76.

Solution 2: Divide the selling price by 1.75. \(\frac{133}{1.75}\) = 76 The store paid $76.

Question 13. Write 5 \(\frac{1}{4}\)% as a simple fraction. Answer: \(\frac{21}{400}\)

Question 14. Write \(\frac{3}{8}\) as a percent. Answer: 37.5%

Question 15. If 20% of the 70 faculty members at John F. Kennedy Middle School are male, what is the number of male faculty members? Answer: (0.20)(70) = 14. Therefore, 14 faculty members are male.

Question 16. If a bag contains 400 coins, and 33 1/2% are nickels, how many nickels are there? What percent of the coins are not nickels? Answer: (400)(0.335) = 134. Therefore, 134 of the coins are nickels. The percent of coins that are not nickels is 66 1/2%.

Question 17. The temperature outside is 60 degrees Fahrenheit. What would be the temperature if it is increased by 20%? Answer: (60)(1.2) = 72. Therefore, the temperature would be 72 degrees Fahrenheit.

Eureka Math Grade 7 Module 4 Lesson 7 Exit Ticket Answer Key

A store that sells skis buys them from a manufacturer at a wholesale price of $57. The store’s markup rate is 50%. a. What price does the store charge its customers for the skis? Answer: 57 × (1 + 0.50) = 85.50. The store charges $85.50 for the skis.

b. What percent of the original price is the final price? Show your work. Answer: Quantity = Percent × Whole. Let P represent the unknown percent. 85.50 = P(57) 85.50(\(\frac{1}{57}\)) = P(57)(\(\frac{1}{57}\)) 1.50 = P 1.50 = \(\frac{150}{100}\) = 150%. The final price is 150% of the original price

c. What is the percent increase from the original price to the final price? Answer: The percent increase is 50% because 150% – 100% = 50%.

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