eureka math grade 3 lesson 16 homework 3.5 answer key

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3rd grade (Eureka Math/EngageNY)

Unit 1: module 1: properties of multiplication and division and solving problems with units of 2–5 and 10, unit 2: module 2: place value and problem solving with units of measure, unit 3: module 3: multiplication and division with units of 0, 1, 6–9, and multiples of 10, unit 4: module 4: multiplication and area, unit 5: module 5: fractions as numbers on the number line, unit 6: module 6: collecting and displaying data, unit 7: module 7: geometry and measurement word problems.

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Eureka Math Grade 5 Module 3 Lesson 16 Answer Key

Engage ny eureka math 5th grade module 3 lesson 16 answer key, eureka math grade 5 module 3 lesson 16 problem set answer key.

Question 2. Half of Robert’s piece of wire is equal to \(\frac{2}{3}\) of Maria’s wire. The total length of their wires is 10 feet. How much longer is Robert’s wire than Maria’s? Answer: length of Robert’s wire = R Length of Maria’s wire = M Half Length of Robert’s piece of wire = \(\frac{2}{3}\) Maria’s wire . \(\frac{R}{2}\) = \(\frac{2}{3}\) M \(\frac{R}{M}\) = \(\frac{4}{3}\) 3R = 4M R = \(\frac{4M}{3}\) R + M = 10 feet . \(\frac{4M}{3}\) + M = 10 feet . lcm is 3 4M +3M = 30 7M = 30 M = \(\frac{30}{7}\) R = \(\frac{4(M)}{3}\) = \(\frac{4}{3}\) × \(\frac{30}{7}\) = \(\frac{40}{7}\) . Length of Robert wire = \(\frac{40}{7}\) Length of  Maria wire = \(\frac{30}{7}\) Length of Robert wire more than Maria wire = \(\frac{40}{7}\) – \(\frac{30}{7}\) = \(\frac{10}{7}\) . Therefore, Robert’s wire is \(\frac{10}{7}\) than Maria’s wire .

Question 3. Half of Sarah’s wire is equal to \(\frac{2}{5}\) of Daniel’s. Chris has 3 times as much as Sarah. In all, their wire measures 6 ft. How long is Sarah’s wire in feet? Answer: Length of Sarah’s wire = S Length of Daniel’s wire = D Length of Chris wire = C Half of Sarah’s wire is equal to \(\frac{2}{5}\) of Daniel’s \(\frac{S}{2}\) = \(\frac{2D}{5}\) D = \(\frac{S}{2}\) × \(\frac{5}{2}\) = 5S. Chris has 3 times as much as Sarah C = 3S Total length of wire = 6 ft . C + D + S = 6 3S + 5S + S = 6 9S = 6 S = \(\frac{6}{9}\) = \(\frac{2}{3}\) feet . Therefore, Length of Sarah’s wire in feet = \(\frac{2}{3}\) feet .

Eureka Math Grade 5 Module 3 Lesson 16 Exit Ticket Answer Key

Eureka Math Grade 5 Module 3 Lesson 16 Homework Answer Key

d. 3 roads P, Q and R. P is twice longer than R . Q is half of P . Total length of the roads are 12 miles . which is the shortest road and its length .

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