eureka math grade 5 lesson 16 homework 5.1

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5th grade (Eureka Math/EngageNY)

Unit 1: module 1: place value and decimal fractions, unit 2: module 2: multi-digit whole number and decimal fraction operations, unit 3: module 3: addition and subtractions of fractions, unit 4: module 4: multiplication and division of fractions and decimal fractions, unit 5: module 5: addition and multiplication with volume and area, unit 6: module 6: problem solving with the coordinate plane.

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Learn, Practice, Succeed

Learn, Practice, and Succeed from   Eureka Math™   offer teachers multiple ways to differentiate instruction, provide extra practice, and assess student learning. These versatile companions to   A Story of Units®   (Grades K–5) guide teachers in response to intervention (RTI), provide extra practice, and inform instruction.

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The Learn book serves as a student’s in-class companion where they show their thinking, share what they know, and watch their knowledge build every day!

Application Problems:  Problem solving in a real-world context is a daily part of   Eureka Math , building student confidence and perseverance as students apply their knowledge in new and varied ways.

Problem Sets :  A carefully sequenced Problem Set provides an in-class opportunity for independent work, with multiple entry points for differentiation.

Exit Tickets:   These exercises check student understanding, providing the teacher with immediate, valuable evidence of the efficacy of that day’s instruction and informing next steps.

Templates:   Learn   includes templates for the pictures, reusable models, and data sets that students need for   Eureka Math   activities.

With   Practice , students build competence in newly acquired skills and reinforce previously learned skills in preparation for tomorrow’s lesson.   Together,   Learn   and   Practice   provide all the print materials a student uses for their core instruction.

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Eureka Math   fluency activities provide differentiated practice through a variety of formats—some are conducted orally, some use manipulatives, others use a personal whiteboard, or a handout and paper-and-pencil format.

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Eureka Math Grade 5 Module 4 Lesson 16 Answer Key

Engage ny eureka math 5th grade module 4 lesson 16 answer key, eureka math grade 5 module 4 lesson 16 problem set answer key.

Solve and show your thinking with a tape diagram.

Question 1. Mrs. Onusko made 60 cookies for a bake sale. She sold \(\frac{2}{3}\) of them and gave \(\frac{3}{4}\) of the remaining cookies to the students working at the sale. How many cookies did she have left?

Answer: The number of cookies left is 5 cookies.

Explanation: Given that Mrs. Onusko made 60 cookies for a bake sale and she sold \(\frac{2}{3}\) of them. So the number of cookies did she sold is \(\frac{2}{3}\) × 60 which is 40 cookies. So the remaining cookies are 60 – 40 which is 20 cookies. And Mrs. Onusko gave \(\frac{3}{4}\) of the remaining cookies to the students working at the sale, so \(\frac{3}{4}\) × 20 which is 15 cookies. So the number of cookies left is 20 – 15 = 5 cookies.

Question 2. Joakim is icing 30 cupcakes. He spreads mint icing on \(\frac{1}{5}\) of the cupcakes and chocolate on \(\frac{1}{2}\) of the remaining cupcakes. The rest will get vanilla icing. How many cupcakes have vanilla icing?

Answer: The remaining vanilla icing is 9 cupcakes.

Explanation: Given that Joakim is icing 30 cupcakes and he spreads mint icing on \(\frac{1}{5}\) of the cupcakes and chocolate, so the number of mint icing is 30 × \(\frac{1}{5}\) which is 6 cupcakes. And \(\frac{1}{2}\) of the remaining cupcakes have chocolate icing is 30 × \(\frac{1}{2}\) which is 15 cupcakes. And the remaining vanilla icing is 30 – (15 + 6) which is 30 – 21 = 9 cupcakes.

Question 3.

The Booster Club sells 240 cheeseburgers. \(\frac{1}{4}\) of the cheeseburgers had pickles, \(\frac{1}{2}\) of the remaining burgers had onions, and the rest had tomato. How many cheeseburgers had tomato?

Answer: 90 cheeseburgers had tomato.

Explanation: Given that the Booster Club sells 240 cheeseburgers and \(\frac{1}{4}\) of the cheeseburgers had pickles which means the number of cheese that had pickle is 240 × \(\frac{1}{4}\) which is 60 burgers. The number of remaining burgers is 240 – 60 which is 180 cheese burgers. And \(\frac{1}{2}\) of the remaining burgers had onions and rest had tomato which means \(\frac{1}{2}\) × 180 which is 90. Therefore 90 cheeseburgers had tomato.

Question 4. DeSean is sorting his rock collection. \(\frac{2}{3}\) of the rocks are metamorphic, and \(\frac{3}{4}\) of the remainder are igneous rocks. If the 3 rocks left over are sedimentary, how many rocks does DeSean have?

Answer: The number of rocks that DeSean has are 36.

Explanation: Given that DeSean is sorting his rock collection and \(\frac{2}{3}\) of the rocks are metamorphic, and \(\frac{3}{4}\) of the remainder are igneous rocks which means \(\frac{1}{3}\) are igneous rocks, and 3 rocks left over are sedimentary, so the number of rocks that DeSean has are x = \(\frac{2}{3}\)x + \(\frac{3}{4}\) × \(\frac{1}{3}\)x + 3 which is = \(\frac{8}{12}\)x + \(\frac{3}{12}\)x + 3 = \(\frac{11}{12}\)x + 3 \(\frac{1}{12}\)x = 3 x = 36. So the number of rocks that DeSean has are 36.

Question 5. Milan puts \(\frac{1}{4}\) of her lawn-mowing money in savings and uses \(\frac{1}{2}\) of the remaining money to pay back her sister. If she has $15 left, how much did she have at first?

Answer: The money Milan had at first is $40.

Explanation: Given that Milan puts \(\frac{1}{4}\) of her lawn-mowing money in savings and uses \(\frac{1}{2}\) of the remaining money to pay back her sister. Let the money that Milan’s has at first be X, the remaining money will be X – \(\frac{1}{4}\) X = \(\frac{3}{4}\) X. Now we are going to calculate the money to pay back her sister, which is \(\frac{1}{2}\) of the remaining money, which is \(\frac{1}{2}\) × \(\frac{3}{4}\) X which is \(\frac{3}{8}\)X. So the total money Milan had at first = money for saving + money for paying back + the amount of money left X = \(\frac{1}{4}\)X + \(\frac{3}{8}\) X + 15 X – \(\frac{1}{4}\)X – \(\frac{3}{8}\) X = 15 \(\frac{8X – 2X – 3X}{8}\)X = 15 On solving we will get the result as 40. So, the money Milan had at first is $40.

Question 6. Parks is wearing several rubber bracelets. \(\frac{1}{3}\) of the bracelets are tie-dye, \(\frac{1}{6}\) are blue, and \(\frac{1}{3}\) of the remainder are camouflage. If Parks wears 2 camouflage bracelets, how many bracelets does he have on?

Answer: Park has 12 bracelets.

Explanation: Given that Parks is wearing several rubber bracelets and \(\frac{1}{3}\) of the bracelets are tie-dye, \(\frac{1}{6}\) are blue, and \(\frac{1}{3}\) of the remainder are camouflage and Park wears 2 camouflage bracelets. Let the sum of all braclets be X, and Park wears 2 camouflage bracelets, that is, \(\frac{1}{3}\) × (1 – \(\frac{1}{3}\) – \(\frac{1}{6}\)) × X = 2 \(\frac{1}{3}\) × (\(\frac{6}{6}\) – \(\frac{2}{6}\) – \(\frac{1}{6}\)) × X = 2 \(\frac{1}{3}\) × \(\frac{3}{6}\) × X = 2 \(\frac{1}{6}\) × X = 2 X = 2 ÷ \(\frac{1}{6}\) X = 2 × 6 = 12.

Question 7. Ahmed spent \(\frac{1}{3}\) of his money on a burrito and a water bottle. The burrito cost 2 times as much as the water. The burrito cost $4. How much money does Ahmed have left?

Answer: Ahmed have left $12.

Explanation: Given that Ahmed spent \(\frac{1}{3}\) of his money on a burrito and a water bottle and the burrito cost 2 times as much as the water, so the water is the burrito cost divided by 2. So water = \(\frac{4}{2}\) which is $2. And water + burrito is $2 + $4 which is $6. And this $6 is \(\frac{1}{3}\) of the money and Ahmed have left \(\frac{2}{3}\) of the money, so $6 ÷ \(\frac{1}{3}\) = X ÷ \(\frac{2}{3}\) X = 6 × (\(\frac{2}{3}\) ÷ \(\frac{1}{3}\), on solving we will get the result as $12. So the Ahmed have left $12.

Eureka Math Grade 5 Module 4 Lesson 16 Exit Ticket Answer Key

Solve and show your thinking with a tape diagram. Three-quarters of the boats in the marina are white, \(\frac{4}{7}\) of the remaining boats are blue, and the rest are red. If there are 9 red boats, how many boats are in the marina?

Answer: The total number of boats in the marine is 84.

Explanation: Let the number of boats in the marina be X, and the number of white boats be \(\frac{3}{4}\)X. Then the remaining boats will be X – \(\frac{3}{4}\)X which is \(\frac{X}{4}\). And now, there are \(\frac{4}{7}\) of the boats are blue, thus the number of blue boats is \(\frac{4}{7}\) × \(\frac{X}{4}\) which is \(\frac{X}{7}\). And the number of red boats is \(\frac{X}{4}\) – \(\frac{X}{7}\) which is \(\frac{3X}{28}\). And if there are 9 red boats, then \(\frac{3X}{28}\) = 9 and 3X = 9 × 28 on solving X = 84. The total number of boats in the marine is 84.

Eureka Math Grade 5 Module 4 Lesson 16 Homework Answer Key

Question 1. Anthony bought an 8-foot board. He cut off \(\frac{3}{4}\) of the board to build a shelf and gave \(\frac{1}{3}\) of the rest to his brother for an art project. How many inches long was the piece Anthony gave to his brother?

Answer: Anthony gave his brother 8 inches board.

Explanation: Given that Anthony bought an 8-foot board, as 1 foot is 12 inches and 8 feet is 8 × 12 which is 96 inches. Anthony cuts \(\frac{3}{4}\) to build a shelf, so \(\frac{3}{4}\) × 96 which is 72 inches. So the left board after cutting is 96 – 72 which is 24 inches. And Anthony gave \(\frac{1}{3}\) of the leftover of his brother which is \(\frac{1}{3}\) × 24 = 8. So Anthony gave his brother 8 inches board.

Question 2. Riverside Elementary School is holding a school-wide election to choose a school color. Five-eighths of the votes were for blue, \(\frac{5}{9}\) of the remaining votes were for green, and the remaining 48 votes were for red. a. How many votes were for blue?

Answer: The number of blue votes is 180 votes.

Explanation: Given that Five-eighths of the votes were for blue and \(\frac{5}{9}\) of the remaining votes were for green which means \(\frac{5}{9}\) × \(\frac{3}{8}\) which is \(\frac{5}{24}\). And the number of red is 1 – (\(\frac{5}{8}\) + \(\frac{5}{24}\) which is \(\frac{1}{6}\). So the total amount of people is 48 × 6 which is 288 people. So the number of blue votes is 288 × \(\frac{5}{8}\) which is 180 votes.

b. How many votes were for green?

Answer: The number of votes were green is 60 votes.

Explanation: As the total amount of people is 48 × 6 which is 288 people. So the number of green votes is 288 × \(\frac{5}{24}\) which is 60 votes.

c. If every student got one vote, but there were 25 students absent on the day of the vote, how many students are there at Riverside Elementary School?

Answer: The total number of students are there at Riverside Elementary School is 313 students.

Explanation: Given that, If every student got one vote, but there were 25 students absent on the day of the vote. So the total number of students are there at Riverside Elementary School is 288 + 25 which is 313 students.

d. Seven-tenths of the votes for blue were made by girls. Did girls who voted for blue make up more than or less than half of all votes? Support your reasoning with a picture.

Explanation: Less than half of all the girls who voted for blue. Because, as Seven-tenths of the votes for blue were made by girls which means \(\frac{7}{10}\) × 180 = 126 which is less than half

e. How many girls voted for blue?

Answer: The number of girls voted for blue is 126.

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