eureka math lesson 15 homework 5.1 answer key

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Eureka Math Grade 5 Module 3 Lesson 15 Answer Key

Engage ny eureka math 5th grade module 3 lesson 15 answer key, eureka math grade 5 module 3 lesson 15 sprint answer key.

Eureka Math Grade 5 Module 3 Lesson 15 Problem Set Answer Key

Solve the word problems using the RDW strategy. Show all of your work. Question 1. In a race, the-second place finisher crossed the finish line 1\(\frac{1}{3}\) minutes after the winner. The third-place finisher was 1\(\frac{3}{4}\) minutes behind the second-place finisher. The third-place finisher took 34\(\frac{2}{3}\) minutes. How long did the winner take? Answer: Fraction of time of Second place finisher crossed the line after = 1\(\frac{1}{3}\) minutes = \(\frac{4}{3}\) Fraction of time of Third place finisher is behind the second place  = 1\(\frac{3}{4}\) minutes  = \(\frac{7}{4}\) Fraction of time the third place finisher took = 34\(\frac{2}{3}\) = \(\frac{105}{3}\) Fraction of time the second place runner took = \(\frac{105}{3}\) – \(\frac{7}{4}\) = \(\frac{420}{12}\) – \(\frac{21}{12}\) = \(\frac{399}{12}\) = \(\frac{133}{4}\) Fraction of time the First place runner took = \(\frac{133}{4}\) – \(\frac{4}{3}\) = \(\frac{399}{12}\) – \(\frac{16}{12}\) = \(\frac{383}{12}\) = 31 \(\frac{11}{12}\) . Therefore the First Runner took = 31 \(\frac{11}{12}\) .minutes.

Question 2. John used 1\(\frac{3}{4}\) kg of salt to melt the ice on his sidewalk. He then used another 3\(\frac{4}{5}\) kg on the driveway. If he originally bought 10 kg of salt, how much does he have left? Answer: Fraction of Salt used by John =1\(\frac{3}{4}\) kg = \(\frac{7}{4}\) kg Fraction of Salt used again =3\(\frac{4}{5}\) kg = \(\frac{24}{5}\) kg Fraction of salt used = \(\frac{7}{4}\)  + \(\frac{24}{5}\) = \(\frac{35}{20}\)  + \(\frac{96}{20}\) = \(\frac{131}{20}\)  = 6 \(\frac{11}{20}\) . Total Salt = 10 kg. Fraction of salt left = 10 – \(\frac{131}{20}\)  = \(\frac{200}{20}\)  – \(\frac{131}{20}\)  = \(\frac{69}{20}\)  = 3\(\frac{9}{20}\)  . Therefore Fraction of salt left = 3\(\frac{9}{20}\)  .

Question 3. Sinister Stan stole 3\(\frac{3}{4}\) oz of slime from Messy Molly, but his evil plans require 6\(\frac{3}{8}\) oz of slime. He stole another 2\(\frac{3}{5}\) oz of slime from Rude Ralph. How much more slime does Sinister Stan need for his evil plan? Answer: Fraction of slime stolen from Messy Molly = 3\(\frac{3}{4}\) = \(\frac{15}{4}\) oz Fraction of slime stolen from Messy Molly again = 2\(\frac{3}{5}\) = \(\frac{13}{5}\) oz Total Fraction Stolen = \(\frac{15}{4}\)  + \(\frac{13}{5}\) = \(\frac{75}{20}\) + \(\frac{52}{20}\) = \(\frac{127}{20}\) = 6\(\frac{7}{20}\) . Fraction of more slime required = 6\(\frac{3}{8}\) – \(\frac{127}{20}\) = \(\frac{51}{8}\) – \(\frac{127}{20}\) = \(\frac{255}{40}\) – \(\frac{254}{40}\) = \(\frac{1}{40}\) . Therefore, Fraction of more slime required = \(\frac{1}{40}\) oz.

Question 4. Gavin had 20 minutes to do a three-problem quiz. He spent 9\(\frac{3}{4}\) minutes on Problem 1 and 3\(\frac{4}{5}\) minutes on Problem 2. How much time did he have left for Problem 3? Write the answer in minutes and seconds. Answer: Time given for 3 problems = 20 minutes Fraction of time Spent on Problem 1 = 9\(\frac{3}{4}\) minutes = \(\frac{39}{4}\) . Fraction of Time spent on Problem 2 = 3\(\frac{4}{5}\) = \(\frac{19}{5}\) . Fraction of Time spent on Problem 3 = x 20 = \(\frac{39}{4}\) + \(\frac{19}{5}\) + x x = 20 – \(\frac{39}{4}\) – \(\frac{19}{5}\) x = \(\frac{400}{20}\) – \(\frac{195}{20}\) – \(\frac{76}{20}\) x = \(\frac{129}{20}\) = 6\(\frac{9}{20}\) . Therefore, Fraction of Time spent on Problem 3 = 6\(\frac{9}{20}\) .

Question 5. Matt wants to shave 2\(\frac{1}{2}\) minutes off his 5K race time. After a month of hard training, he managed to lower his overall time from 21\(\frac{1}{5}\) minutes to 19\(\frac{1}{4}\) minutes. By how many more minutes does Matt need to lower his race time? Answer: Fraction of Time lowered = 21\(\frac{1}{5}\) minutes to 19\(\frac{1}{4}\) minutes. = \(\frac{106}{5}\) – \(\frac{77}{4}\) = \(\frac{424}{20}\) – \(\frac{385}{20}\) = \(\frac{39}{20}\) =1\(\frac{19}{20}\) . Fraction of Time shaved = 2\(\frac{1}{2}\) =\(\frac{5}{2}\) . Fraction of More Time Matt need to lower his race time = \(\frac{5}{2}\) – \(\frac{39}{20}\) = \(\frac{50}{20}\) – \(\frac{39}{20}\) = \(\frac{11}{20}\) = \(\frac{33}{60}\) = 33 minutes .

Eureka Math Grade 5 Module 3 Lesson 15 Exit Ticket Answer Key

Solve the word problem using the RDW strategy. Show all of your work. Cheryl bought a sandwich for 5\(\frac{1}{2}\) dollars and a drink for $2.60. If she paid for her meal with a $10 bill, how much money did she have left? Write your answer as a fraction and in dollars and cents. Answer: Fraction of Cost of sandwich = 5\(\frac{1}{2}\) = \(\frac{11}{2}\) dollar = 5.5 dollar Fraction of Cost of Drink = $2.60. Total Cost = 5.5 +2.60 = 8.1 $. Amount paid = 10$. Money left = 10 – 8.1 = 1.9 $ .

Eureka Math Grade 5 Module 3 Lesson 15 Homework Answer Key

Solve the word problems using the RDW strategy. Show all of your work. Question 1. A baker buys a 5 lb bag of sugar. She uses 1\(\frac{2}{3}\) lb to make some muffins and 2\(\frac{3}{4}\) lb to make a cake. How much sugar does she have left? Answer: Total Quantity of Sugar = 5 lb Fraction of Quantity of Suagr used for muffins = 1\(\frac{2}{3}\) lb = \(\frac{5}{3}\) Fraction of Quantity of Suagr used cake = 2\(\frac{3}{4}\) lb = \(\frac{11}{4}\) Fraction of Quantity of Sugar used = \(\frac{5}{3}\) + \(\frac{11}{4}\) = \(\frac{20}{12}\) + \(\frac{33}{12}\) = \(\frac{53}{12}\) Fraction of Quantity of Sugar left = 5 – \(\frac{53}{12}\) = \(\frac{60}{12}\) – \(\frac{53}{12}\) =\(\frac{7}{12}\) . Therefore, Fraction of Quantity of sugar left = \(\frac{7}{12}\) .

Question 2. A boxer needs to lose 3\(\frac{1}{2}\) kg in a month to be able to compete as a flyweight. In three weeks, he lowers his weight from 55.5 kg to 53.8 kg. How many kilograms must the boxer lose in the final week to be able to compete as a flyweight? Answer: Fraction of weight need to lose in month = 3\(\frac{1}{2}\) = \(\frac{7}{2}\) = 3.5 kg Weight lost in 3 weeks = 55.5 –  53.8 = 1.7 kg Weight need to lose in final week = 3.5 – 1.7 = 1.8 kg.

Question 3. A construction company builds a new rail line from Town A to Town B. They complete 1\(\frac{1}{4}\) miles in their first week of work and 1\(\frac{2}{3}\) miles in the second week. If they still have 25\(\frac{3}{4}\) miles left to build, what is the distance from Town A to Town B? Answer: Fraction of work completed in first week = 1\(\frac{1}{4}\) miles = \(\frac{5}{4}\) Fraction of work completed in second week = 1\(\frac{2}{3}\) miles = \(\frac{5}{3}\) Fraction of work left to built = 25\(\frac{3}{4}\) miles = \(\frac{103}{4}\) Fraction of Distance from Town A to Town B = \(\frac{103}{4}\) + \(\frac{5}{4}\)  + \(\frac{5}{3}\) = \(\frac{108}{4}\) + \(\frac{5}{3}\) = \(\frac{324}{12}\) + \(\frac{20}{12}\) = \(\frac{344}{12}\)= 28\(\frac{2}{3}\) . Therefore, Fraction of Distance from Town A to Town B = 28\(\frac{2}{3}\) miles.

Question 4. A catering company needs 8.75 lb of shrimp for a small party. They buy 3\(\frac{2}{3}\) lb of jumbo shrimp, 2\(\frac{5}{8}\) lb of medium-sized shrimp, and some mini-shrimp. How many pounds of mini-shrimp do they buy? Answer: Quantity of shrimp needed = 8.75 lb =8\(\frac{3}{4}\) = \(\frac{27}{4}\) Quantity of jumbo shrimp = 3\(\frac{2}{3}\) lb = \(\frac{11}{3}\) Quantity of  medium – sized shrimp = 2\(\frac{5}{8}\) lb = \(\frac{21}{8}\) Quantity of mini shrimp = x \(\frac{35}{4}\)  = \(\frac{11}{3}\) + \(\frac{21}{8}\) + x x = \(\frac{210}{24}\)  – \(\frac{88}{24}\) – \(\frac{63}{24}\) x =  \(\frac{59}{24}\) = 2 \(\frac{11}{24}\) Therefore, Quantity of mini shrimp = x = 2 \(\frac{11}{24}\) lb .

Question 5. Mark breaks up a 9-hour drive into 3 segments. He drives 2\(\frac{1}{2}\) hours before stopping for lunch. After driving some more, he stops for gas. If the second segment of his drive was 1\(\frac{2}{3}\) hours longer than the first segment, how long did he drive after stopping for gas? Answer: Total time of the drive = 9 hours . Fraction of Time drived for first segment = 2\(\frac{1}{2}\) hours  = \(\frac{5}{2}\) Fraction of Time of second segment = 1\(\frac{2}{3}\) hours longer than the first segment = \(\frac{5}{3}\) + \(\frac{5}{2}\) = \(\frac{10}{6}\) + \(\frac{15}{6}\) = \(\frac{25}{6}\) =4 \(\frac{1}{6}\) Fraction of Time of first and second segment= \(\frac{5}{2}\) + \(\frac{25}{6}\) = \(\frac{15}{6}\) + \(\frac{25}{6}\) = \(\frac{40}{6}\) = 6\(\frac{4}{6}\) Fraction of Time he drive after stopping gas = 9 – \(\frac{40}{6}\) = \(\frac{54}{6}\) –  \(\frac{40}{6}\) = \(\frac{14}{6}\) = 2\(\frac{2}{6}\) . Therefore, Fraction of Time he drive after stopping gas = third segment = 2\(\frac{2}{6}\) hours .

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