What grade does Eureka math go up to?

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An Elementary, Middle, And High School Math Curriculum . Eureka Math ® is a math program designed to advance equity in the math classroom by helping students build enduring math knowledge.

What is the hardest math in 5th grade?

Some of the hardest math problems for fifth graders involve multiplying: multiplying using square models, multiplying fractions and whole numbers using expanded form, and multiplying fractions using number lines .

What is the hardest math grade?

Generally speaking, the most rigorous math courses in high school include Advanced Placement (AP) Calculus AB and BC, AP Statistics, and for some, Multivariable Calculus (which might be offered at your school or at a local college).

Is Eureka Math good or bad?

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What is the hardest math curriculum?

AP Calculus BC and IB Math HL are particularly challenging because they are designed to prepare students for college-level mathematics. These courses cover a wide range of topics, including calculus, trigonometry, and statistics.

What is the highest level of math in 9th grade?

9th grade math usually focuses on Algebra I , but can include other advanced mathematics such as Geometry, Algebra II, Pre-Calculus or Trigonometry.

What grade does prodigy math go up to?

With 1,500+ curriculum-aligned math skills for 1st to 8th grade , Prodigy Math is so much more than a game. Prodigy Math is an engaging game-based learning platform that's dedicated to improving students' confidence and achievements in math.

What is the highest math class there is?

Wrap up with Calculus , the highest level of math offered by many high schools and often considered the gold standard of pre-college math preparation.

What math do advanced 8th graders take?

The primary strands for an 8th-grade math curriculum are number sense and operations, algebra, geometry, and spatial sense, measurement, and data analysis and probability . While these math strands might surprise you, they are all critical lessons for an 8th-grade math curriculum.

Eureka Math Grade 5 Module 3 Lesson 5 Answer Key (2024)

Engage NY Eureka Math 5th Grade Module 3 Lesson 5 Answer Key

Eureka Math Grade 5 Module 3 Lesson 5 Sprint Answer Key

A
Subtracting Fractions from a Whole Number

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\(\frac{5}{8}\)

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. \(\frac{1}{3}\) – \(\frac{1}{4}\) =
b. \(\frac{2}{3}\) – \(\frac{1}{2}\) =
c. \(\frac{5}{6}\) – \(\frac{1}{4}\) =
d. \(\frac{2}{3}\) – \(\frac{1}{7}\) =
e. \(\frac{3}{4}\) – \(\frac{3}{8}\) =
f. \(\frac{3}{4}\) – \(\frac{2}{7}\) =
Answer:
a.
\(\frac{1}{3}\) – \(\frac{1}{4}\)
L.c.m of 3 and 4 is 12
\(\frac{4}{12}\) – \(\frac{3}{12}\) = \(\frac{1}{12}\)

b.
\(\frac{2}{3}\) – \(\frac{1}{2}\)
lcm of 3 and 2 is 6
\(\frac{4}{6}\) – \(\frac{3}{6}\) = \(\frac{1}{6}\)

c.
\(\frac{5}{6}\) – \(\frac{1}{4}\)
lcm of 6 and 4 is 12
\(\frac{10}{12}\) – \(\frac{3}{12}\) = \(\frac{7}{12}\)

d.
\(\frac{2}{3}\) – \(\frac{1}{7}\)
lcm of 3 and 7 is 21 .
\(\frac{14}{21}\) – \(\frac{3}{21}\) = \(\frac{11}{21}\)

e.
\(\frac{3}{4}\) – \(\frac{3}{8}\)
lcm of 4 and 8 is 8
\(\frac{6}{8}\) – \(\frac{3}{8}\) = \(\frac{3}{8}\)

f.
\(\frac{3}{4}\) – \(\frac{2}{7}\)
lcm of 4 and 7 is 28
\(\frac{21}{28}\) – \(\frac{8}{28}\) = \(\frac{13}{28}\)

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 = \(\frac{2}{3}\) – \(\frac{1}{5}\) = \(\frac{10}{15}\) – \(\frac{3}{15}\)
= \(\frac{7}{15}\) .

Question 3.
Sandra says that \(\frac{4}{7}\) – \(\frac{1}{3}\) = \(\frac{3}{4}\) because all you have to do is subtract the numerators and subtract the denominators. Convince Sandra that she is wrong. You may draw a rectangular fraction model to support your thinking.
Answer:
No, \(\frac{4}{7}\) – \(\frac{1}{3}\) = \(\frac{5}{21}\) not \(\frac{3}{4}\)

Explanation :
\(\frac{4}{7}\) – \(\frac{1}{3}\) =\(\frac{12}{21}\) – \(\frac{7}{21}\) = \(\frac{5}{21}\)
no, first find the l.c.m of the denominators that is lcm of 7 and 3 is 21 . then multiply the denominators to make 21 and and also multiply same number with the numerator . then after making denominators equal subtract the numerators .

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

For the following problems, draw a picture using the rectangular fraction model and write the answer. Simplify your answer, if possible.
a. \(\frac{1}{2}\) – \(\frac{1}{7}\) =
b. \(\frac{3}{5}\) – \(\frac{1}{2}\) =
Answer:
a.
\(\frac{1}{2}\) – \(\frac{1}{7}\)
lcm of 2 and 7 is 14
\(\frac{7}{14}\) – \(\frac{2}{14}\) = \(\frac{5}{14}\)

b.
\(\frac{3}{5}\) – \(\frac{1}{2}\)
lcm of 5 and 2 is 10 .
\(\frac{6}{10}\) – \(\frac{5}{10}\) = \(\frac{1}{10}\)

Eureka Math Grade 5 Module 3 Lesson 5 Homework Answer Key

Question 1.
The picture below shows \(\frac{3}{4}\) of the rectangle shaded. Use the picture to show how to create an equivalent fraction for \(\frac{3}{4}\), and then subtract \(\frac{1}{3}\).
\(\frac{3}{4}\) – \(\frac{1}{3}\) =

Answer:

Explanation :
\(\frac{3}{4}\) – \(\frac{1}{3}\)
l.c.m of 4 and 3 is 12
\(\frac{9}{12}\) – \(\frac{4}{12}\) = \(\frac{5}{12}\)

Question 2.
Find the difference. Use a rectangular fraction model to find common denominators. Simplify your answer, if possible.
a. \(\frac{5}{6}\) – \(\frac{1}{3}\) =
b. \(\frac{2}{3}\) – \(\frac{1}{2}\) =
c. \(\frac{5}{6}\) – \(\frac{1}{4}\) =
d. \(\frac{4}{5}\) – \(\frac{1}{2}\) =
e. \(\frac{2}{3}\) – \(\frac{2}{5}\) =
f. \(\frac{5}{7}\) – \(\frac{2}{3}\) =
Answer:
a.

\(\frac{5}{6}\) – \(\frac{1}{3}\)
Lcm of 6 and 3 is 6 .
\(\frac{5}{6}\) – \(\frac{2}{6}\) = \(\frac{3}{6}\) = \(\frac{1}{2}\)

b.
\(\frac{2}{3}\) – \(\frac{1}{2}\)
lcm of 3 and 2 is 6
\(\frac{4}{6}\) – \(\frac{3}{6}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\)

c.
\(\frac{5}{6}\) – \(\frac{1}{4}\)
lcm of 6 and 4
\(\frac{10}{12}\) – \(\frac{3}{12}\) = \(\frac{7}{12}\)

d.
\(\frac{4}{5}\) – \(\frac{1}{2}\)
lcm of 5 and 2 is 10
\(\frac{8}{10}\) – \(\frac{5}{10}\) = \(\frac{3}{10}\)

e.
\(\frac{2}{3}\) – \(\frac{2}{5}\)
lcm of 3 and 5 is 15
\(\frac{10}{15}\) – \(\frac{6}{15}\)= \(\frac{4}{15}\)

f.
\(\frac{5}{7}\) – \(\frac{2}{3}\)
lcm of 7 and 3 is 21.
\(\frac{15}{21}\) – \(\frac{14}{21}\) = \(\frac{1}{21}\)

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 .

Eureka Math Grade 5 Module 3 Lesson 5 Answer Key (2024)

Engage NY Eureka Math 5th Grade Module 3 Lesson 5 Answer Key

Eureka Math Grade 5 Module 3 Lesson 5 Sprint Answer Key

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

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

Eureka Math Grade 5 Module 3 Lesson 5 Homework Answer Key

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