Chapter 5: Ratios and Proportions
Chapter 5 homework solutions.
5.1-5.3 Extra Practice Worksheet
5.1 Ratios and Rates
5.1 Lesson Preso
5.1 Textbook Exercises: page 167
Ba: 1-6, 11-27 odd;
Avg: 1-6, 11-21 odd, 24-28 even, 29, 31;
Adv: 1-6, 18-38 even
5.2 Proportions
5.2 Textbook Exercises page 174
Ba: 1-4, 5-13 odd, 21-27 odd;
Avg 1-4, 5-13 odd, 22-30 even;
Adv: 1-4, 6-14 even, 22-32 even
5.3 Writing Proportions
5.3 Lesson Preso
5.3 Textbook Exercises page 182
Ba: 1-3, 9, 11, 12, 13-23 odd
Avg: 1-3, 8-14 even, 19-23
Adv: 1-3, 8-24 even, 25
5.4 Solving Proportions
5.4 Lesson Preso
5.4 Textbook Exercises page 190
Ba: 1-3, 5-9 odd, 15-21 odd, 22, 23-27 odd
Avg: 1-3, 5-9 odd, 15-21 odd, 22, 29, 30, 32-35
Adv: 1-3, 4-8 even, 14-38 even
5.5 Lesson Preso
5.5 Textbook Exercises page 196
Ba: 1-3, 5-11 odd, 12, 13-17 odd
Avg: 1-3, 5-11 odd, 12, 14-17
Adv: 1-3, 4-18 even
5.6 Direct Variation
5.6 Lesson Preso
5.6 Textbook Exercises page 202
Ba: 1-3, 7-17 odd, 18, 19-25 odd
Avg: 1-3, 7-17 odd, 18-28 even
Adv: 1-3, 8-28 even
Chapter 5 Review
5.1-5.3 Review Quizizz:
5.4-5.6 Review Quizizz:
Chapter 5 Review Quizizz:
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enVision MATH Common Core 5, Grade: 5 Publisher: Scott Foresman Addison Wesley
Envision math common core 5, title : envision math common core 5, publisher : scott foresman addison wesley, isbn : 328672637, isbn-13 : 9780328672639, use the table below to find videos, mobile apps, worksheets and lessons that supplement envision math common core 5., textbook resources.
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Texas Go Math Grade 5 Lesson 5.3 Answer Key Estimate Fraction Sums and Differences
Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Lesson 5.3 Answer Key Estimate Fraction Sums and Differences.
Unlock the Problem
Kimberly will be riding her bike to school this year. The distance from her house to the end of the Street is \(\frac{1}{62}\)mile. The distance from the end of the Street to the school is \(\frac{3}{8}\) mile. About how far is Kimberly’s house from school?
You can use benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
STEP 3: Add the rounded fractions.
Another Way
Use mental math. You can compare the numerator and the denominator to round a fraction and find a reasonable estimate.
Estimate. \(\frac{9}{10}\) – \(\frac{5}{8}\) STEP 1: Round \(\frac{9}{10}\). Think: The numerator is about the same as the denominator. Round the fraction \(\frac{9}{10}\) to __________.
Remember A fraction with the same numerator and denominator, such as \(\frac{2}{2}, \frac{5}{5}, \frac{12}{12}\) or \(\frac{96}{96}\), is equal to 1.
STEP 2: Round \(\frac{5}{8}\) Think: The numerator is about half the denominator. Round the fraction \(\frac{5}{8}\) to ___________.
STEP 1: Round \(\frac{9}{10}\). Think: The numerator is about the same as the denominator. Round the fraction \(\frac{9}{10}\) to \(\frac{10}{10}\)
STEP 2: Round \(\frac{5}{8}\) Think: The numerator is about half the denominator. Round the fraction \(\frac{5}{8}\) to \(\frac{4}{8}\)
Math Talk Mathematical Processes
Explain another way you could use benchmarks to estimate \(\frac{9}{10}\) – \(\frac{5}{8}\). Answer: \(\frac{9}{10}\) – \(\frac{5}{8}\) = \(\frac{1}{6}\) \(\frac{1}{6}\) is very near to \(\frac{1}{5}\) Explanation: Used bench marks to find the sum
Share and Show
Estimate the sum or difference.
Question 1. \(\frac{5}{6}\) + \(\frac{3}{8}\) a. Round \(\frac{5}{6}\) to its closest benchmark. Answer: \(\frac{6}{6}\)
b. Round \(\frac{3}{8}\) to its closest benchmark. Answer: \(\frac{4}{8}\)
c. Add to find the estimate. \(\frac{6}{6}\) +\(\frac{4}{8}\) = 1\(\frac{1}{2}\) Answer: 1\(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Go Math Lesson 5.3 5th Grade Answer Key Question 2. \(\frac{5}{9}\) – \(\frac{3}{8}\) Answer: a. Round \(\frac{5}{9}\) to its closest benchmark. Answer: \(\frac{5}{9}\)
c. Add to find the estimate. \(\frac{5}{9}\) – \(\frac{4}{8}\) = 1\(\frac{1}{18}\) Answer: 1\(\frac{1}{18}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 3. \(\frac{5}{6}\) + \(\frac{2}{5}\) Answer: a. Round \(\frac{5}{6}\) to its closest benchmark. Answer: \(\frac{6}{6}\)
b. Round \(\frac{2}{5}\) to its closest benchmark. Answer: \(\frac{2}{5}\)
c. Add to find the estimate. \(\frac{6}{6}\) +\(\frac{2}{5}\) = 1\(\frac{1}{2}\) Answer: 1\(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 4. \(\frac{9}{10}\) – \(\frac{1}{9}\) Answer: a. Round \(\frac{9}{10}\) to its closest benchmark. Answer: \(\frac{10}{10}\)
b. Round \(\frac{1}{9}\) to its closest benchmark. Answer: \(\frac{0}{9}\)
c. Add to find the estimate. \(\frac{10}{10}\) – \(\frac{0}{9}\) = 1 Answer: 1 Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Problem Solving
Lesson 5.3 Answer Key 5th Grade Go Math Question 5. How do you know whether your estimate for \(\frac{9}{10}\) + 3\(\frac{6}{7}\) would be greater than or less than the actual sum? Explain. Answer: Greater than the actual sum \(\frac{9}{10}\) + 3\(\frac{6}{7}\) = close to bench marks \(\frac{10}{10}\) + 3\(\frac{7}{7}\) = 4 Explanation: Is greater than the actual sum used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 6. Write Math Nick estimated that \(\frac{5}{8}\) + \(\frac{4}{7}\) is about 2. Explain how you know his estimate is not reasonable. Answer: \(\frac{5}{8}\) + \(\frac{4}{7}\) close to benchmarks \(\frac{4}{8}\) + \(\frac{4}{7}\) = 1 Explanation: Nick estimated that \(\frac{5}{8}\) + \(\frac{4}{7}\) is about 2. used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1. so, his estimation is wrong
Question 7. Lisa and Valerie are picnicking in Trough Creek State Park in Pennsylvania. Lisa has brought a salad that she made with \(\frac{3}{4}\) cup of strawberries, \(\frac{7}{8}\) cup of peaches, and \(\frac{1}{6}\) cup of blueberries. About how many total cups of fruit are in the salad? Answer: \(\frac{3}{4}\) + \(\frac{7}{8}\) + \(\frac{1}{6}\) very close to bench marks \(\frac{4}{4}\) + \(\frac{8}{8}\) + \(\frac{0}{6}\) =2 \(\frac{1}{2}\) Explanation: Lisa and Valerie are picnicking in Trough Creek State Park in Pennsylvania. Lisa has brought a salad that she made with \(\frac{3}{4}\) cup of strawberries, \(\frac{7}{8}\) cup of peaches, and \(\frac{1}{6}\) cup of blueberries. 2\(\frac{1}{2}\) total cups of fruit are in the salad
Go Math 5th Grade Lesson 5.3 How to Estimate Fractions Question 9. H.O.T Explain how you know that \(\frac{5}{8}\) + \(\frac{6}{10}\) is greater than 1. Answer: No Explanation: Close to the bench marks \(\frac{8}{8}\) + \(\frac{5}{10}\) = 1 actual sum is greater than 1
Daily Assessment Task
Fill in the bubble completely to show your answer.
Question 10. Mia uses \(\frac{1}{5}\) of a bag of gravel in the morning and \(\frac{11}{12}\) of a bag in the afternoon. About how much gravel does she use in one day? (A) 0 bags (B) \(\frac{1}{2}\) bag (C) 1 bag (D) 2\(\frac{1}{2}\) bags Answer: C \(\frac{1}{5}\) + \(\frac{11}{12}\) nearest benchmarks are \(\frac{0}{5}\) + \(\frac{12}{12}\) = 1 Explanation: Mia uses \(\frac{1}{5}\) of a bag of gravel in the morning and \(\frac{11}{12}\) of a bag in the afternoon. she use 1 bag of gravel
Question 11. Evaluate Reasonableness Hector and Veronica are going hiking. They made a trail mix that has \(\frac{2}{3}\) cup of almonds, \(\frac{7}{8}\) cup of peanuts, and \(\frac{4}{5}\) cup of raisins in it. Hector estimates that they made about 3 cups of trail mix. Is the estimate greater than or less than the actual sum? How do you know? (A) The estimate is greater because each fraction is rounded up to a benchmark. (B) The estimate is less because each fraction is rounded down to a benchmark. (C) The estimate is greater because they really made more than 3 cups. (D) The estimate is less because each fraction is rounded up to a benchmark. Answer: A Explanation: \(\frac{2}{3}\) + \(\frac{7}{8}\) + \(\frac{4}{5}\) rounded to the nearest benchmarks \(\frac{3}{3}\) + \(\frac{8}{8}\) + \(\frac{5}{5}\) = 3 Evaluated Reasonableness Hector and Veronica are going hiking. They made a trail mix that has \(\frac{2}{3}\) cup of almonds, ” \(\frac{7}{8}\) cup of peanuts, and \(\frac{4}{5}\) cup of raisins in it. Hector estimates that they made about 3 cups of trail mix.
Lesson 5.3 Go Math 5th Grade Answer Key Question 12. Multi-Step Amanda picked \(\frac{3}{5}\) pound of blueberries at her local farm yesterday. She used \(\frac{3}{8}\) pound of blueberries. Today she picked \(\frac{4}{5}\) pound of blueberries. About how many pounds of blueberries does Amanda have now? (A) \(\frac{1}{5}\)lb (B) 1 lb (C) \(\frac{1}{2}\)lb (D) 1\(\frac{1}{2}\)lbs Answer: B Explanation: what she bought is that she used yesterday in today marked to nearest benchmarks \(\frac{4}{5}\) is \(\frac{5}{5}\) that is 1
Texas Test Prep
Question 13. Jake added \(\frac{1}{8}\) cup of sunflower seeds and \(\frac{4}{5}\) cup of banana chips to his sundae. Which is the best estimate of the total amount of toppings Jake added to his sundae? (A) about 2 cups (B) about 1 cup (C) about 1\(\frac{1}{2}\) cups (D) about \(\frac{1}{2}\) cup Answer: B Explanation: Jake added \(\frac{1}{8}\) cup of sunflower seeds and \(\frac{4}{5}\) cup of banana chips to his sundae. The best estimate of the total amount of toppings Jake added to his sundae is 1 cup
Texas Go Math Grade 5 Lesson 5.3 Homework and Practice Answer Key
Question 1. \(\frac{3}{8}\) + \(\frac{4}{5}\) = ___________ Answer: \(\frac{3}{8}\) + \(\frac{4}{5}\) rounded to the nearest benchmarks \(\frac{4}{8}\) + \(\frac{5}{5}\) = 1 \(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
5th Grade Go Math Lesson 5.3 Answer Key Question 2. \(\frac{9}{10}\) – \(\frac{3}{8}\) = ___________ Answer: \(\frac{9}{10}\) – \(\frac{3}{8}\) rounded to the nearest benchmarks \(\frac{10}{10}\) – \(\frac{4}{8}\) = \(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 3. \(\frac{5}{8}\) + \(\frac{2}{5}\) = ___________ Answer: \(\frac{5}{8}\) + \(\frac{2}{5}\) rounded to the nearest benchmarks \(\frac{4}{8}\) + \(\frac{2}{5}\) = 1 Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 4. \(\frac{6}{7}\) + \(\frac{3}{5}\) = ___________ Answer: \(\frac{6}{7}\) + \(\frac{3}{5}\) rounded to the nearest benchmarks \(\frac{7}{7}\) + \(\frac{2}{5}\) = 1\(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 5. \(\frac{3}{8}\) – \(\frac{1}{6}\) = ___________ Answer: \(\frac{3}{8}\) – \(\frac{1}{6}\) rounded to the nearest benchmarks \(\frac{4}{8}\) – \(\frac{0}{6}\) = \(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 6. \(\frac{7}{12}\) + \(\frac{1}{7}\) = ___________ Answer: \(\frac{7}{12}\) + \(\frac{1}{7}\) rounded to the nearest benchmarks \(\frac{6}{12}\) + \(\frac{0}{7}\) = \(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Go Math Lesson 5.3 5th Grade Homework Answer Key Question 7. \(\frac{4}{9}\) – \(\frac{5}{8}\) = ___________ Answer: \(\frac{4}{9}\) – \(\frac{5}{8}\) rounded to the nearest benchmarks \(\frac{5}{9}\) – \(\frac{4}{8}\) = 0 Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 8. \(\frac{1}{9}\) + \(\frac{5}{6}\) = ___________ Answer: \(\frac{1}{9}\) + \(\frac{5}{6}\) rounded to the nearest benchmark \(\frac{0}{9}\) + \(\frac{6}{6}\) = 1 Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 9. \(\frac{7}{8}\) + \(\frac{4}{7}\) = ___________ Answer: \(\frac{7}{8}\) + \(\frac{4}{7}\) rounded to the nearest bench mark \(\frac{8}{8}\) + \(\frac{4}{7}\) =1\(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 10. \(\frac{1}{5}\) + \(\frac{3}{8}\) = ___________ Answer: \(\frac{1}{5}\) + \(\frac{3}{8}\) rounded to the nearest benchmark \(\frac{0}{5}\) + \(\frac{4}{8}\) = \(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 11. \(\frac{7}{9}\) – \(\frac{2}{6}\) = ___________ Answer: \(\frac{7}{9}\) – \(\frac{2}{6}\) rounded to the nearest benchmark \(\frac{9}{9}\) – \(\frac{3}{6}\) = \(\frac{1}{2}\) Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Go Math Grade 5 Lesson 5.3 Homework Answer Key Question 12. \(\frac{9}{10}\) – \(\frac{7}{8}\) = ___________ Answer: \(\frac{9}{10}\) – \(\frac{7}{8}\) rounded to the benchmarks \(\frac{10}{10}\) – \(\frac{8}{8}\) = 0 Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 13. Explain how you can estimate the sum of \(\frac{4}{5}\) and \(\frac{1}{6}\). Answer: \(\frac{4}{5}\) + \(\frac{1}{6}\) rounded to the nearest bench marks \(\frac{5}{5}\) + \(\frac{0}{6}\) = 1 Explanation: used benchmarks to find reasonable estimates by rounding fractions to 0, \(\frac{1}{2}\), or 1.
Question 14. Jena uses \(\frac{7}{8}\) cup of raisins for muffins and \(\frac{5}{8}\) cup of raisins for a bowl of oatmeal. Does lena need more than or less than 1 cup of raisins to make muffins and oatmeal? Explain. Answer: more than 1 cup of raisins Explanation: Jena uses \(\frac{7}{8}\) cup of raisins for muffins and \(\frac{5}{8}\) cup of raisins for a bowl of oatmeal. \(\frac{7}{8}\) + \(\frac{5}{8}\) rounded the benhmark \(\frac{8}{8}\) + \(\frac{4}{8}\) = 1\(\frac{1}{2}\)
Question 15. A group of students ate \(\frac{5}{12}\) of a cheese pizza, \(\frac{7}{8}\) of a pepperoni pizza, and \(\frac{5}{8}\) of a veggie pizza. About how many pizzas were eaten? Answer: \(\frac{5}{12}\) + \(\frac{7}{8}\) + \(\frac{5}{8}\) rounded to the nearest benchmark \(\frac{6}{12}\) + \(\frac{8}{8}\) + \(\frac{4}{8}\) = 2 Explanation: A group of students ate \(\frac{5}{12}\) of a cheese pizza, \(\frac{7}{8}\) of a pepperoni pizza, and \(\frac{5}{8}\) of a veggie pizza. 2 pizzas were eaten in whole.
Lesson Check
Question 16. On Saturday, the scouts hiked \(\frac{4}{5}\) mile up the mountain. On Sunday, they hiked \(\frac{1}{4}\) mile up the mountain. About how far did the scouts hike up the mountain in all? (A) \(\frac{1}{2}\) mile (B) 1 mile (C) 1\(\frac{1}{2}\) miles (D) 2 miles Answer: \(\frac{4}{5}\) + \(\frac{1}{4}\) rounded to nearest benchmark \(\frac{5}{5}\) + \(\frac{0}{4}\) is 1 mile Explanation: On Saturday, the scouts hiked \(\frac{4}{5}\) mile up the mountain. On Sunday, they hiked \(\frac{1}{4}\) mile up the mountain. 1 mile far the scouts hike up the mountain in all
Question 17. Which of the following best describes the difference for \(\frac{11}{12}\) – \(\frac{7}{10}\) ? (A) less than \(\frac{1}{2}\) (B) greater than \(\frac{1}{2}\) (C) greater than 1 (D) greater than 1\(\frac{1}{2}\) Answer: A Explanation: \(\frac{11}{12}\) – \(\frac{7}{10}\) is 0 that is less than \(\frac{1}{2}\)
Practice and Homework Lesson 5.3 Answer Key 5th Grade Question 18. Which sum is greatest? Use estimation to decide. (A) \(\frac{2}{7}\) + \(\frac{3}{8}\) (B) \(\frac{1}{10}\) + \(\frac{3}{8}\) (C) \(\frac{1}{6}\) + \(\frac{1}{8}\) (D) \(\frac{2}{9}\) + \(\frac{1}{8}\) Answer: A Explanation: \(\frac{2}{7}\) + \(\frac{3}{8}\) = 1
Question 20. Multi-Step Michaela has \(\frac{11}{12}\) yard of orange fabric and \(\frac{7}{8}\) yard of green fabric. She uses \(\frac{1}{2}\) yard of each color for her sewing project. About how much fabric does Michaela have left if she combines the two colors? (A) 1 yard (B) \(\frac{1}{2}\) yard (C) 1 \(\frac{1}{2}\) yards (D) 2 yards Answer: D \(\frac{11}{12}\) + \(\frac{7}{8}\) rounded to nearest bench marks \(\frac{12}{12}\) + \(\frac{8}{8}\) = 2 Explanation: 2 yards fabric uses Michaela have left if she combines the two colors.
Question 21. Multi-Step Dustin buys \(\frac{9}{10}\) yard of striped fabric. He uses \(\frac{3}{8}\) yard. He buys \(\frac{7}{8}\) yard more. About how much fabric does Dustin have now? (A) 1 yard (B) \(\frac{1}{2}\) yard (C) 1\(\frac{1}{2}\) yards (D) 2 yards Answer: C Explanation: Dustin buys \(\frac{9}{10}\) yard of striped fabric. He uses \(\frac{3}{8}\) yard. He buys \(\frac{7}{8}\) yard more. \(\frac{9}{10}\) + \(\frac{3}{8}\) + \(\frac{7}{8}\) rounded to nearest benchmarks \(\frac{10}{10}\) – \(\frac{4}{8}\) + \(\frac{8}{8}\) = 1\(\frac{1}{2}\) yards
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Eureka Math Grade 5 Module 3 Lesson 5 Answer Key
Engage ny eureka math 5th grade module 3 lesson 5 answer key, eureka math grade 5 module 3 lesson 5 sprint answer key.
Question 1. 4 – \(\frac{1}{2}\) = Answer: 4 – \(\frac{1}{2}\) = 3\(\frac{1}{2}\) Explanation : 4 – \(\frac{1}{2}\) = \(\frac{8}{2}\) – \(\frac{1}{2}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)
Question 2. 3 – \(\frac{1}{2}\) = Answer: 3 – \(\frac{1}{2}\) = 2\(\frac{1}{2}\) Explanation : 3 – \(\frac{1}{2}\) = \(\frac{6}{2}\) – \(\frac{1}{2}\) = \(\frac{5}{2}\) = 2\(\frac{1}{2}\)
Question 3. 2 – \(\frac{1}{2}\) = Answer: 2 – \(\frac{1}{2}\) = 1\(\frac{1}{2}\) Explanation : 2 – \(\frac{1}{2}\) = \(\frac{4}{2}\) – \(\frac{1}{2}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)
Question 4. 1 – \(\frac{1}{2}\) = Answer: 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) Explanation : 1 – \(\frac{1}{2}\) = \(\frac{2}{2}\) – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Question 5. 1 – \(\frac{1}{3}\) = Answer: 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\) Explanation : 1 – \(\frac{1}{3}\) = \(\frac{3}{3}\) – \(\frac{1}{3}\) = \(\frac{2}{3}\)
Question 6. 2 – \(\frac{1}{3}\) = Answer: 2 – \(\frac{1}{3}\) = 1\(\frac{2}{3}\) Explanation : 2 – \(\frac{1}{3}\) = \(\frac{6}{3}\) – \(\frac{1}{3}\) = \(\frac{5}{3}\) = 1\(\frac{2}{3}\)
Question 7. 4 – \(\frac{1}{3}\) = Answer: 4 – \(\frac{1}{3}\) = 3\(\frac{2}{3}\) Explanation : 4 – \(\frac{1}{3}\) = \(\frac{12}{3}\) – \(\frac{1}{3}\) = \(\frac{11}{3}\) = 3\(\frac{2}{3}\)
Question 8. 4 – \(\frac{2}{3}\) = Answer: 4 – \(\frac{2}{3}\) = 3\(\frac{1}{3}\) Explanation : 4 – \(\frac{2}{3}\) = \(\frac{12}{3}\) – \(\frac{2}{3}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)
Question 9. 2 – \(\frac{2}{3}\) = Answer: 2 – \(\frac{1}{3}\) = 1\(\frac{2}{3}\) Explanation : 2 – \(\frac{1}{3}\) = \(\frac{6}{3}\) – \(\frac{1}{3}\) = \(\frac{5}{3}\) = 1\(\frac{2}{3}\)
Question 10. 2 – \(\frac{1}{4}\) = Answer: 2 – \(\frac{1}{4}\) = 1\(\frac{3}{4}\) Explanation : 2 – \(\frac{1}{4}\) = \(\frac{8}{4}\) – \(\frac{1}{4}\) = \(\frac{7}{4}\) = 1\(\frac{3}{4}\)
Question 11. 2 – \(\frac{3}{4}\) = Answer: 2 – \(\frac{3}{4}\) = 1\(\frac{1}{4}\) Explanation : 2 – \(\frac{3}{4}\) = \(\frac{8}{4}\) – \(\frac{3}{4}\) = \(\frac{5}{4}\) = 1\(\frac{1}{4}\)
Question 12. 3 – \(\frac{3}{4}\) = Answer: 3 – \(\frac{3}{4}\) = 2\(\frac{1}{4}\) Explanation : 3 – \(\frac{3}{4}\) = \(\frac{12}{4}\) – \(\frac{3}{4}\) = \(\frac{9}{4}\) = 2\(\frac{1}{4}\)
Question 13. 3 – \(\frac{1}{4}\) = Answer: 3 – \(\frac{1}{4}\) = 2\(\frac{3}{4}\) Explanation : 3 – \(\frac{1}{4}\) = \(\frac{12}{4}\) – \(\frac{1}{4}\) = \(\frac{11}{4}\) = 2\(\frac{3}{4}\)
Question 14. 4 – \(\frac{3}{4}\) = Answer: 4 – \(\frac{3}{4}\) = 3\(\frac{1}{4}\) Explanation : 4 – \(\frac{3}{4}\) = \(\frac{16}{4}\) – \(\frac{3}{4}\) = \(\frac{13}{4}\) = 3\(\frac{1}{4}\)
Question 15. 2 – \(\frac{1}{10}\) = Answer: 2 – \(\frac{1}{10}\) = 1\(\frac{9}{10}\) Explanation : 2 – \(\frac{1}{10}\) = \(\frac{20}{10}\) – \(\frac{1}{10}\) = \(\frac{19}{10}\) = 1\(\frac{9}{10}\)
Question 16. 3 – \(\frac{9}{10}\) = Answer: 3 – \(\frac{9}{10}\) = 2\(\frac{1}{10}\) Explanation : 3 – \(\frac{9}{10}\) = \(\frac{30}{10}\) – \(\frac{9}{10}\) = \(\frac{21}{10}\) = 2\(\frac{1}{10}\)
Question 17. 2 – \(\frac{7}{10}\) = Answer: Answer: 2 – \(\frac{7}{10}\) = 1\(\frac{3}{10}\) Explanation : 2 – \(\frac{7}{10}\) = \(\frac{20}{10}\) – \(\frac{7}{10}\) = \(\frac{13}{10}\) = 1\(\frac{3}{10}\)
Question 18. 4 – \(\frac{3}{10}\) = Answer: 4 – \(\frac{1}{10}\) = 3\(\frac{9}{10}\) Explanation : 4 – \(\frac{1}{10}\) = \(\frac{40}{10}\) – \(\frac{1}{10}\) = \(\frac{39}{10}\) = 3\(\frac{9}{10}\)
Question 19. 3 – \(\frac{1}{5}\) = Answer: 3 – \(\frac{1}{5}\) = 2\(\frac{4}{5}\) Explanation : 3 – \(\frac{1}{5}\) = \(\frac{15}{5}\) – \(\frac{1}{5}\) = \(\frac{14}{5}\) = 2\(\frac{4}{5}\)
Question 20. 3 – \(\frac{2}{5}\) = Answer: 3 – \(\frac{2}{5}\) = 2\(\frac{3}{5}\) Explanation : 3 – \(\frac{2}{5}\) = \(\frac{15}{5}\) – \(\frac{2}{5}\) = \(\frac{13}{5}\) = 2\(\frac{3}{5}\)
Question 21. 3 – \(\frac{4}{5}\) = Answer: 3 – \(\frac{4}{5}\) = 2\(\frac{1}{5}\) Explanation : 3 – \(\frac{4}{5}\) = \(\frac{15}{5}\) – \(\frac{4}{5}\) = \(\frac{11}{5}\) = 2\(\frac{1}{5}\)
Question 22. 3 – \(\frac{3}{5}\) = Answer: 3 – \(\frac{3}{5}\) = 2\(\frac{2}{5}\) Explanation : 3 – \(\frac{3}{5}\) = \(\frac{15}{5}\) – \(\frac{3}{5}\) = \(\frac{12}{5}\) = 2\(\frac{2}{5}\)
Question 23. 3 – \(\frac{1}{8}\) = Answer: 3 – \(\frac{1}{8}\) = 2\(\frac{7}{8}\) Explanation : 3 – \(\frac{1}{8}\) = \(\frac{24}{8}\) – \(\frac{1}{8}\) = \(\frac{23}{8}\) = 2\(\frac{7}{8}\)
Question 24. 3 – \(\frac{3}{8}\) = Answer: 3 – \(\frac{3}{8}\) = 2\(\frac{5}{8}\) Explanation : 3 – \(\frac{3}{8}\) = \(\frac{24}{8}\) – \(\frac{3}{8}\) = \(\frac{21}{8}\) = 2\(\frac{5}{8}\)
Question 25. 3 – \(\frac{5}{8}\) = Answer: 3 – \(\frac{5}{8}\) = 2\(\frac{3}{8}\) Explanation : 3 – \(\frac{5}{8}\) = \(\frac{24}{8}\) – \(\frac{5}{8}\) = \(\frac{19}{8}\) = 2\(\frac{3}{8}\)
Question 26. 3 – \(\frac{7}{8}\) = Answer: 3 – \(\frac{7}{8}\) = 2\(\frac{1}{8}\) Explanation : 3 – \(\frac{7}{8}\) = \(\frac{24}{8}\) – \(\frac{7}{8}\) = \(\frac{17}{8}\) = 2\(\frac{1}{8}\)
Question 27. 2 – \(\frac{7}{8}\) = Answer: 2 – \(\frac{7}{8}\) = 1\(\frac{1}{8}\) Explanation : 2 – \(\frac{7}{8}\) = \(\frac{16}{8}\) – \(\frac{7}{8}\) = \(\frac{9}{8}\) = 1\(\frac{1}{8}\)
Question 28. 4 – \(\frac{1}{7}\) = Answer: 4 – \(\frac{1}{7}\) = 3\(\frac{6}{7}\) Explanation : 4 – \(\frac{1}{7}\) = \(\frac{28}{7}\) – \(\frac{1}{7}\) = \(\frac{27}{7}\) = 3\(\frac{6}{7}\)
Question 29. 3 – \(\frac{6}{7}\) = Answer: 3 – \(\frac{6}{7}\) = 2\(\frac{1}{7}\) Explanation : 3 – \(\frac{6}{7}\) = \(\frac{21}{7}\) – \(\frac{6}{7}\) = \(\frac{15}{7}\) = 2\(\frac{1}{7}\)
Question 30. 2 – \(\frac{3}{7}\) = Answer: 2 – \(\frac{3}{7}\) = Answer: 2 – \(\frac{3}{7}\) = 1\(\frac{4}{7}\) Explanation : 2 – \(\frac{3}{7}\) = \(\frac{14}{7}\) – \(\frac{3}{7}\) = \(\frac{11}{7}\) = 1\(\frac{4}{7}\)
Question 31. 4 – \(\frac{4}{7}\) = Answer: 4 – \(\frac{4}{7}\) = Answer: 4 – \(\frac{4}{7}\) = 3\(\frac{3}{7}\) Explanation : 4 – \(\frac{4}{7}\) = \(\frac{28}{7}\) – \(\frac{4}{7}\) = \(\frac{24}{7}\) = 3\(\frac{3}{7}\)
Question 32. 3 – \(\frac{5}{7}\) = Answer: 3 – \(\frac{5}{7}\) = 2\(\frac{2}{7}\) Explanation : 3 – \(\frac{5}{7}\) = \(\frac{21}{7}\) – \(\frac{5}{7}\) = \(\frac{16}{7}\) = 2\(\frac{2}{7}\)
Question 33. 4 – \(\frac{3}{4}\) = Answer: 4 – \(\frac{3}{4}\) = 3\(\frac{1}{4}\) Explanation : 4 – \(\frac{3}{4}\) = \(\frac{16}{4}\) – \(\frac{3}{4}\) = \(\frac{13}{4}\) = 3\(\frac{1}{4}\)
Question 34. 2 – \(\frac{5}{8}\) = Answer: 2 – \(\frac{5}{8}\) = 1\(\frac{3}{8}\) Explanation : 2 – \(\frac{5}{8}\) = \(\frac{16}{8}\) – \(\frac{5}{8}\) = \(\frac{11}{8}\) = 1\(\frac{3}{8}\)
Question 35. 3 – \(\frac{3}{10}\) = Answer: 3 – \(\frac{3}{10}\) = 2\(\frac{7}{10}\) Explanation : 3 – \(\frac{3}{10}\) = \(\frac{30}{10}\) – \(\frac{3}{10}\) = \(\frac{27}{10}\) = 2\(\frac{7}{10}\)
Question 36. 4 – \(\frac{2}{5}\) = Answer: 4 – \(\frac{2}{5}\) = 3\(\frac{3}{5}\) Explanation : 4 – \(\frac{2}{5}\) = \(\frac{20}{5}\) – \(\frac{2}{5}\) = \(\frac{18}{5}\) = 3\(\frac{3}{5}\)
Question 37. 4 – \(\frac{3}{7}\) = Answer: 4 – \(\frac{3}{7}\) = Answer: 4 – \(\frac{3}{7}\) = 3\(\frac{4}{7}\) Explanation : 4 – \(\frac{3}{7}\) = \(\frac{28}{7}\) – \(\frac{3}{7}\) = \(\frac{25}{7}\) = 3\(\frac{4}{7}\)
Question 38. 3 – \(\frac{7}{10}\) = Answer: 3 – \(\frac{7}{10}\) = 2\(\frac{3}{10}\) Explanation : 3 – \(\frac{7}{10}\) = \(\frac{30}{10}\) – \(\frac{7}{10}\) = \(\frac{23}{10}\) = 2\(\frac{3}{10}\)
Question 39. 3 – \(\frac{5}{10}\) = Answer: 3 – \(\frac{5}{10}\) = 2\(\frac{5}{10}\) Explanation : 3 – \(\frac{5}{10}\) = \(\frac{30}{10}\) – \(\frac{5}{10}\) = \(\frac{25}{10}\) = 2\(\frac{5}{10}\)
Question 40. 4 – \(\frac{2}{8}\) = Answer: 4 – \(\frac{2}{8}\) = 3\(\frac{6}{8}\) Explanation : 4 – \(\frac{2}{8}\) = \(\frac{32}{8}\) – \(\frac{2}{8}\) = \(\frac{30}{8}\) = 3\(\frac{6}{8}\)
Question 41. 2 – \(\frac{9}{12}\) = Answer: 2 – \(\frac{9}{12}\) = 2 – \(\frac{3}{4}\) = 1\(\frac{1}{4}\) Explanation : 2 – \(\frac{3}{4}\) = \(\frac{8}{4}\) – \(\frac{3}{4}\) = \(\frac{5}{4}\) = 1\(\frac{1}{4}\)
Question 42. 4 – \(\frac{2}{12}\) = 3\(\frac{5}{6}\) Answer: 4 – \(\frac{2}{12}\) = 4 – \(\frac{1}{6}\) = 3\(\frac{5}{6}\) Explanation : 4 – \(\frac{1}{6}\) = \(\frac{24}{6}\) – \(\frac{1}{6}\) = \(\frac{23}{6}\) = 3\(\frac{5}{6}\)
Question 43. 3 – \(\frac{2}{6}\) = Answer: 3 – \(\frac{2}{6}\) = 3 – \(\frac{1}{3}\) = 2\(\frac{2}{3}\) Explanation : 3 – \(\frac{1}{3}\) = \(\frac{9}{3}\) – \(\frac{1}{3}\) = \(\frac{8}{3}\) = 2\(\frac{2}{3}\)
Question 44. 2 – \(\frac{8}{12}\) = Answer: 2 – \(\frac{8}{12}\) = 2 – \(\frac{2}{3}\) = 1\(\frac{1}{3}\) Explanation : 2 – \(\frac{2}{3}\) = \(\frac{6}{3}\) – \(\frac{2}{3}\) = \(\frac{4}{3}\) = 2\(\frac{1}{3}\)
Question 1. 1 – \(\frac{1}{2}\) = Answer: 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) Explanation : 1 – \(\frac{1}{2}\) = \(\frac{2}{2}\) – \(\frac{1}{2}\) = \(\frac{1}{2}\)
Question 2. 2 – \(\frac{1}{2}\) = Answer: 2 – \(\frac{1}{2}\) = 1\(\frac{1}{2}\) Explanation : 2 – \(\frac{1}{2}\) = \(\frac{4}{2}\) – \(\frac{1}{2}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)
Question 3. 3 – \(\frac{1}{2}\) = Answer: 3 – \(\frac{1}{2}\) = 2\(\frac{1}{2}\) Explanation : 3 – \(\frac{1}{2}\) = \(\frac{6}{2}\) – \(\frac{1}{2}\) = \(\frac{5}{2}\) = 2\(\frac{1}{2}\)
Question 4. 4 – \(\frac{1}{2}\) = Answer: 4 – \(\frac{1}{2}\) = 3\(\frac{1}{2}\) Explanation : 4 – \(\frac{1}{2}\) = \(\frac{8}{2}\) – \(\frac{1}{2}\) = \(\frac{7}{2}\) = 3\(\frac{1}{2}\)
Question 5. 1 – \(\frac{1}{4}\) = Answer: 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\) Explanation : 1 – \(\frac{1}{4}\) = \(\frac{4}{4}\) – \(\frac{1}{4}\) = \(\frac{3}{4}\)
Question 6. 2 – \(\frac{1}{4}\) = Answer: 2 – \(\frac{1}{4}\) = 1\(\frac{3}{4}\) Explanation : 2 – \(\frac{1}{4}\) = \(\frac{8}{4}\) – \(\frac{1}{4}\) = \(\frac{7}{4}\) = 1\(\frac{3}{4}\)
Question 7. 4 – \(\frac{1}{4}\) = Answer: 4 – \(\frac{1}{4}\) = 3\(\frac{3}{4}\) Explanation : 4 – \(\frac{1}{4}\) = \(\frac{16}{4}\) – \(\frac{1}{4}\) = \(\frac{15}{4}\) = 3\(\frac{3}{4}\)
Question 8. 4 – \(\frac{3}{4}\) = Answer: 4 – \(\frac{3}{4}\) = 3\(\frac{1}{4}\) Explanation : 4 – \(\frac{3}{4}\) = \(\frac{16}{4}\) – \(\frac{3}{4}\) = \(\frac{13}{4}\) = 3\(\frac{1}{4}\)
Question 9. 2 – \(\frac{3}{4}\) = Answer: 2 – \(\frac{3}{4}\) = 1\(\frac{1}{4}\) Explanation : 2 – \(\frac{3}{4}\) = \(\frac{8}{4}\) – \(\frac{3}{4}\) = \(\frac{5}{4}\) = 1\(\frac{1}{4}\)
Question 10. 2 – \(\frac{1}{3}\) = Answer: 2 – \(\frac{1}{3}\) = 1\(\frac{2}{3}\) Explanation : 2 – \(\frac{1}{3}\) = \(\frac{6}{3}\) – \(\frac{1}{3}\) = \(\frac{5}{3}\) = 1\(\frac{2}{3}\)
Question 11. 2 – \(\frac{2}{3}\) = Answer: 2 – \(\frac{2}{3}\) = 1\(\frac{1}{3}\) Explanation : 2 – \(\frac{2}{3}\) = \(\frac{6}{3}\) – \(\frac{2}{3}\) = \(\frac{4}{3}\) = 1\(\frac{1}{3}\)
Question 12. 3 – \(\frac{2}{3}\) = Answer: 3 – \(\frac{2}{3}\) = 2\(\frac{1}{3}\) Explanation : 3 – \(\frac{2}{3}\) = \(\frac{9}{3}\) – \(\frac{2}{3}\) = \(\frac{7}{3}\) = 2\(\frac{1}{3}\)
Question 13. 3 – \(\frac{1}{3}\) = Answer: 3 – \(\frac{1}{3}\) = 2\(\frac{2}{3}\) Explanation : 3 – \(\frac{1}{3}\) = \(\frac{9}{3}\) – \(\frac{1}{3}\) = \(\frac{8}{3}\) = 2\(\frac{2}{3}\)
Question 14. 4 – \(\frac{2}{3}\) = Answer: 4 – \(\frac{2}{3}\) = 3\(\frac{1}{3}\) Explanation : 4 – \(\frac{2}{3}\) = \(\frac{12}{3}\) – \(\frac{2}{3}\) = \(\frac{10}{3}\) = 3\(\frac{1}{3}\)
Question 15. 3 – \(\frac{1}{10}\) = Answer: 3 – \(\frac{1}{10}\) = 2\(\frac{9}{10}\) Explanation : 3 – \(\frac{1}{10}\) = \(\frac{30}{10}\) – \(\frac{9}{10}\) = \(\frac{21}{10}\) = 2\(\frac{1}{10}\)
Question 16. 2 – \(\frac{9}{10}\) = Answer: 2 – \(\frac{9}{10}\) = 1\(\frac{1}{10}\) Explanation : 2 – \(\frac{9}{10}\) = \(\frac{20}{10}\) – \(\frac{9}{10}\) = \(\frac{11}{10}\) = 1\(\frac{1}{10}\)
Question 17. 4 – \(\frac{7}{10}\) = Answer: 4 – \(\frac{7}{10}\) = 3\(\frac{3}{10}\) Explanation : 4 – \(\frac{7}{10}\) = \(\frac{40}{10}\) – \(\frac{7}{10}\) = \(\frac{33}{10}\) = 3\(\frac{3}{10}\)
Question 18. 3 – \(\frac{3}{10}\) = Answer: 3 – \(\frac{3}{10}\) = 2\(\frac{7}{10}\) Explanation : 3 – \(\frac{3}{10}\) = \(\frac{30}{10}\) – \(\frac{3}{10}\) = \(\frac{27}{10}\) = 2\(\frac{7}{10}\)
Question 19. 2 – \(\frac{1}{5}\) = Answer: 2 – \(\frac{1}{5}\) = 1\(\frac{4}{5}\) Explanation : 2 – \(\frac{1}{5}\) = \(\frac{10}{5}\) – \(\frac{1}{5}\) = \(\frac{9}{5}\) = 1\(\frac{4}{5}\)
Question 20. 2 – \(\frac{2}{5}\) = Answer: 2 – \(\frac{2}{5}\) = 1\(\frac{3}{5}\) Explanation : 2 – \(\frac{2}{5}\) = \(\frac{10}{5}\) – \(\frac{2}{5}\) = \(\frac{8}{5}\) = 1\(\frac{3}{5}\)
Question 21. 2 – \(\frac{4}{5}\) = Answer: 2 – \(\frac{4}{5}\) = 1\(\frac{1}{5}\) Explanation : 2 – \(\frac{1}{5}\) = \(\frac{10}{5}\) – \(\frac{4}{5}\) = \(\frac{6}{5}\) = 1\(\frac{1}{5}\)
Question 23. 2 – \(\frac{1}{8}\) = Answer: 2 – \(\frac{1}{8}\) = 1\(\frac{7}{8}\) Explanation : 2 – \(\frac{1}{8}\) = \(\frac{16}{8}\) – \(\frac{1}{8}\) = \(\frac{15}{8}\) = 1\(\frac{7}{8}\)
Question 24. 2 – \(\frac{3}{8}\) = Answer: 2 – \(\frac{3}{8}\) = 1\(\frac{5}{8}\) Explanation : 2 – \(\frac{3}{8}\) = \(\frac{16}{8}\) – \(\frac{3}{8}\) = \(\frac{13}{8}\) = 1\(\frac{4}{8}\)
Question 25. 2 – \(\frac{5}{8}\) = Answer: 2 – \(\frac{5}{8}\) = 1\(\frac{3}{8}\) Explanation : 2 – \(\frac{5}{8}\) = \(\frac{16}{8}\) – \(\frac{5}{8}\) = \(\frac{11}{8}\) = 1\(\frac{3}{8}\)
Question 26. 2 – \(\frac{7}{8}\) = Answer: 2 – \(\frac{7}{8}\) = 1\(\frac{1}{8}\) Explanation : 2 – \(\frac{7}{8}\) = \(\frac{16}{8}\) – \(\frac{7}{8}\) = \(\frac{9}{8}\) = 1\(\frac{1}{8}\)
Question 27. 4 – \(\frac{7}{8}\) = Answer: 4 – \(\frac{7}{8}\) = 3\(\frac{1}{8}\) Explanation : 4 – \(\frac{7}{8}\) = \(\frac{32}{8}\) – \(\frac{3}{8}\) = \(\frac{13}{8}\) = 1\(\frac{4}{8}\)
Question 28. 3 – \(\frac{1}{7}\) = Answer: 3 – \(\frac{1}{7}\) = 2\(\frac{6}{7}\) Explanation : 3 – \(\frac{1}{7}\) = \(\frac{21}{7}\) – \(\frac{1}{7}\) = \(\frac{20}{7}\) = 2\(\frac{6}{7}\)
Question 29. 2 – \(\frac{6}{7}\) = Answer: 2 – \(\frac{6}{7}\) = 1\(\frac{1}{7}\) Explanation : 2 – \(\frac{6}{7}\) = \(\frac{14}{7}\) – \(\frac{6}{7}\) = \(\frac{8}{7}\) = 1\(\frac{1}{7}\)
Question 30. 4 – \(\frac{3}{7}\) = Answer: 4 – \(\frac{3}{7}\) = 3\(\frac{4}{7}\) Explanation : 4 – \(\frac{3}{7}\) = \(\frac{28}{7}\) – \(\frac{3}{7}\) = \(\frac{25}{7}\) = 3\(\frac{4}{7}\)
Question 31. 3 – \(\frac{4}{7}\) = Answer: 3 – \(\frac{4}{7}\) = 2\(\frac{3}{7}\) Explanation : 3 – \(\frac{4}{7}\) = \(\frac{21}{7}\) – \(\frac{4}{7}\) = \(\frac{17}{7}\) = 2\(\frac{3}{7}\)
Question 32. 2 – \(\frac{5}{7}\) = Answer: 2 – \(\frac{5}{7}\) = 1\(\frac{2}{7}\) Explanation : 2 – \(\frac{5}{7}\) = \(\frac{14}{7}\) – \(\frac{5}{7}\) = \(\frac{9}{7}\) = 1\(\frac{2}{7}\)
Question 33. 3 – \(\frac{3}{4}\) = Answer: 3 – \(\frac{3}{4}\) = 2\(\frac{1}{4}\) Explanation : 3 – \(\frac{3}{4}\) = \(\frac{12}{4}\) – \(\frac{3}{4}\) = \(\frac{9}{4}\) = 2\(\frac{1}{4}\)
Question 34. 4 – \(\frac{5}{8}\) = Answer: 4 – \(\frac{5}{8}\) = 3\(\frac{3}{8}\) Explanation : 4 – \(\frac{5}{8}\) = \(\frac{32}{8}\) – \(\frac{5}{8}\) = \(\frac{27}{8}\) = 3\(\frac{3}{8}\)
Question 35. 2 – \(\frac{3}{10}\) = Answer: 2 – \(\frac{3}{10}\) = 1\(\frac{7}{10}\) Explanation : 2 – \(\frac{3}{10}\) = \(\frac{20}{10}\) – \(\frac{3}{10}\) = \(\frac{17}{10}\) = 1\(\frac{7}{10}\)
Question 36. 3 – \(\frac{2}{5}\) = Answer: 3 – \(\frac{2}{5}\) = 2\(\frac{3}{5}\) Explanation : 3 – \(\frac{2}{5}\) = \(\frac{15}{5}\) – \(\frac{2}{5}\) = \(\frac{13}{5}\) = 2\(\frac{3}{5}\)
Question 37. 3 – \(\frac{3}{7}\) = Answer: 3 – \(\frac{3}{7}\) = 2\(\frac{4}{7}\) Explanation : 3 – \(\frac{3}{7}\) = \(\frac{21}{7}\) – \(\frac{3}{7}\) = \(\frac{18}{7}\) = 2\(\frac{4}{7}\)
Question 38. 2 – \(\frac{7}{10}\) = Answer: 2 – \(\frac{7}{10}\) = 1\(\frac{3}{10}\) Explanation : 2 – \(\frac{7}{10}\) = \(\frac{20}{10}\) – \(\frac{7}{10}\) = \(\frac{13}{10}\) = 1\(\frac{3}{10}\)
Question 39. 2 – \(\frac{5}{10}\) = Answer: 2 – \(\frac{5}{10}\) = 1\(\frac{1}{2}\) Explanation : 2 – \(\frac{5}{10}\) = \(\frac{20}{10}\) – \(\frac{5}{10}\) = \(\frac{15}{10}\) = \(\frac{3}{2}\) = 1\(\frac{1}{2}\)
Question 40. 3 – \(\frac{6}{8}\) = Answer: 3 – \(\frac{6}{8}\) = 2\(\frac{1}{4}\) Explanation : 3 – \(\frac{6}{8}\) = \(\frac{24}{8}\) – \(\frac{6}{8}\) = \(\frac{18}{8}\) = 2\(\frac{1}{4}\)
Question 41. 4 – \(\frac{3}{12}\) = Answer: 4 – \(\frac{3}{12}\) = 4 – \(\frac{1}{4}\) = 3\(\frac{3}{4}\) Explanation : 4 – \(\frac{1}{4}\) = \(\frac{16}{4}\) – \(\frac{1}{4}\) = \(\frac{15}{4}\) = 3\(\frac{3}{4}\)
Question 42. 3 – \(\frac{10}{12}\) = Answer: 3 – \(\frac{10}{12}\) = 3 – \(\frac{5}{6}\) = 2\(\frac{1}{6}\) Explanation : 3 – \(\frac{5}{6}\) = \(\frac{18}{6}\) – \(\frac{5}{6}\) = \(\frac{13}{6}\) = 2\(\frac{1}{6}\)
Question 43. 2 – \(\frac{4}{6}\) = Answer: 2 – \(\frac{4}{6}\) = 2 – \(\frac{2}{3}\) = 1\(\frac{1}{3}\) Explanation : 2 – \(\frac{2}{3}\) = \(\frac{6}{3}\) – \(\frac{2}{3}\) = \(\frac{4}{3}\) = 1\(\frac{1}{3}\)
Question 44. 4 – \(\frac{4}{12}\) = Answer: 4 – \(\frac{4}{12}\) = 4 – \(\frac{1}{3}\) = 3\(\frac{2}{3}\) Explanation : 4 – \(\frac{1}{3}\) = \(\frac{12}{3}\) – \(\frac{1}{3}\) = \(\frac{11}{3}\) = 3\(\frac{2}{3}\)
Eureka Math Grade 5 Module 3 Lesson 5 Problem Set Answer Key
Question 2. Mr. Penman had \(\frac{2}{3}\) liter of salt water. He used \(\frac{1}{5}\) of a liter for an experiment. How much salt water does Mr. Penman have left? Answer: Quantity of salt water = \(\frac{2}{3}\) Quantity of salt water used = \(\frac{1}{5}\) Quantity of salt water left = \(\frac{2}{3}\) – \(\frac{1}{5}\) = \(\frac{10}{15}\) – \(\frac{3}{15}\) = \(\frac{7}{15}\) .
Eureka Math Grade 5 Module 3 Lesson 5 Exit Ticket Answer Key
Eureka Math Grade 5 Module 3 Lesson 5 Homework Answer Key
Question 3. Robin used \(\frac{1}{4}\) of a pound of butter to make a cake. Before she started, she had \(\frac{7}{8}\) of a pound of butter. How much butter did Robin have when she was done baking? Give your answer as a fraction of a pound. Answer: Quantity of butter used to make cake = \(\frac{1}{4}\) pound Quantity of butter with Robin before baking cake = \(\frac{7}{8}\) pound . Total Quantity of butter with Robin after baking = \(\frac{7}{8}\) – \(\frac{1}{4}\) pound = \(\frac{7}{8}\) – \(\frac{2}{8}\) = \(\frac{5}{8}\) pound Therefore, Robin have \(\frac{5}{8}\) pound when she was done baking .
Question 4. Katrina needs \(\frac{3}{5}\) kilogram of flour for a recipe. Her mother has \(\frac{3}{7}\) kilogram of flour in her pantry. Is this enough flour for the recipe? If not, how much more will she need? Answer: Quantity of Flour Required for Recipe = \(\frac{3}{5}\) Quantity of Flour with her mother = \(\frac{3}{7}\) Quantity of Flour Enough or not = \(\frac{3}{7}\) – \(\frac{3}{5}\) = \(\frac{15}{35}\) – \(\frac{21}{35}\) = – \(\frac{6}{35}\) that means negative indicate doenot enough. She needs more \(\frac{6}{35}\) Quantity of Flour for the Recipe .
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Eureka Math Grade 5 Module 3 Lesson 7 Problem Set Answer Key. Solve the word problems using the RDW strategy. Show all of your work. Question 1. George weeded \ (\frac {1} {5}\) of the garden, and Summer weeded some, too. When they were finished, \ (\frac {2} {3}\) of the garden still needed to be weeded.
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Engage NY // Eureka Math Grade 5 Module 3 Lesson 7 Homework
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EngageNY/Eureka Math Grade 3 Module 5 Lesson 7For more videos, answer keys, and other resources, please visit http://EMBARC.onlinePLEASE leave a message if a...
Go Math! Practice Book (TE), G5. Name Fraction and Whole Number Multiplication Find the product. Write the product in simplest form. 2. Lesson 7.3 COMMON CORE STANDARD CC5.NF.4a Apply and extend previous understandings of multiplication and division to multiply and divide fractions. — or 33ž 27 or 2— —x 9=10' 10 10 3. 6.
Module 3: Addition and Subtraction of Fractions 1 Lesson 1 Answer Key 5• 3 Lesson 1 Sprint Side A 1. 2 12. 2 23. 4 34. 7 2. 5 13. 2 24. 7 35. 4 3. 2 14. 2 25. 7 36.
Chapter 5 Homework Solutions. 5.1-5.3 Extra Practice Worksheet. 5.1 Ratios and Rates . 5.1 Lesson Preso ... 5.2 Proportions . 5.2 Lesson. 5.2 Textbook Exercises page 174 . Ba: 1-4, 5-13 odd, 21-27 odd; Avg 1-4, 5-13 odd, 22-30 even; Adv: 1-4, 6-14 even, 22-32 even. 5.3 Writing Proportions. 5.3 Lesson Preso ... 7-17 odd, 18, 19-25 odd. Avg: 1-3 ...
Lesson Check (CC.5.NBT.7) 1. Tina divides 21.4 ounces of trail mix equally into 5 bags. How many ounces of trail mix are in each bag? @ 0.428 ounce 4.28 ounces 42.8 ounces 428 ounces Spiral Review (CC.5.NBT.2, CC.5.NBT.6, CC.5.NBT.7) 3. Suzy buys 35 pounds of rice. She divides it equally into 100 bags. How many pounds of rice does Suzy put in ...
1 yellow key, 1 green key 5.3 Enrichment and Extension 1. 60 in.; 5 ft 2. ... For use before Lesson 5.4 Sample answer: The bakery will need to increase ... 14. 18 men; Sample answer: Because 2 out of 7 people at the lecture are men, 2 763 = m where m is the number of men; m = 18
View Common_core_lesson_7_homework_5_3 from LDSP 321 at University of Tennessee, Martin. Common core lesson 7 homework 5 3 Get more information 5. . _ ny5 common core mathematics curriculum 5 lesson ... New york state grade 5 math common core module 4 lesson 26 29 answer. key. Engage ny third grade module 3 lesson 7 back.
Answer Key (Contains all answers except #s 26-28 & 35-41) Answers for 26-28, 35-41 ... Problem Set 5 3.3 Homework: Pg 188: Problem Set 1, 2, 5 Pg 201-202: Problem Set 3, 4, 6, 12 ... Extra Ch. 6 Lessons Homework: Lesson #1 Worksheet Answers Lesson #2 Worksheet Answers Extra Ch. 6 Lessons Review: Worksheet Answers
Students' daily homework will be required reading of at least 30 minutes. Students will have nightly math homework which supports our learning in class. There are a lot of new math concepts in 5th grade and it is important for students' growth and understanding. Additionally, study guides and other assignments may be sent home periodically ...
Chapter 16: Coordinate Geometry. enVision MATH Common Core 5 grade 5 workbook & answers help online. Grade: 5, Title: enVision MATH Common Core 5, Publisher: Scott Foresman Addison Wesley, ISBN: 328672637.
McGraw Hill Math Grade 8 Lesson 11.3 Answer Key Distributive and Identity Properties; McGraw Hill Math Grade 8 Lesson 11.4 Answer Key Properties of Equality and Zero; McGraw Hill Math Grade 8 Lesson 12.1 Answer Key Negative Numbers; McGraw Hill Math Grade 8 Lesson 13.2 Answer Key Solving Equations with Addition and Subtraction
Find step-by-step solutions and answers to Algebra 2 - 9780130625687, as well as thousands of textbooks so you can move forward with confidence. ... Section 7.7: Inverse Relations and Functions. Section 7.8: Graphing Radical Functions. Page 415: Chapter Review. Page 418: Chapter Test. Page 419: Standardized Test Prep. Page 828: Extra Practice ...
Eureka Math Grade 5 Module 3 Lesson 5 Problem Set Answer Key. Question 1. For the following problems, draw a picture using the rectangular fraction model and write the answer. Simplify your answer, if possible. a. b. d.
The source for the homework pages is available here for free. I used the full module PDF:https://www.engageny.org/resource/grade-3-mathematics-module-5
Regina has two electronic files. One has a size of 3.15 MB and the other has a size of 4.89 MB. What is the best estimate of the total size of the two electronic files? 2. Madison is training for a marathon. Her goal is to run 26.2 miles a day. She currently can run 18.5 miles in aday.