Engage NY Eureka Math 5th Grade Module 6 Lesson 26 Answer Key
Eureka Math Grade 5 Module 6 Lesson 26 Problem Set Answer Key
Question 1.
For each written phrase, write a numerical expression, and then evaluate your expression.
a. Three fifths of the sum of thirteen and six
Numerical expression:
\(\frac{13}{5}\) (13 + 6 )
Solution:
\(\frac{13}{5}\) (13 + 6 ) = \(\frac{13}{5}\) (19) = \(\frac{247}{5}\)
b. Subtract four thirds from one seventh of sixty-three.
Numerical expression:
\(\frac{1}{7}\)(63) – \(\frac{4}{3}\)
Solution:
\(\frac{1}{7}\)(63) – \(\frac{4}{3}\) = 9 – \(\frac{4}{3}\) = \(\frac{27 – 4}{3}\) = \(\frac{23}{3}\) = 7\(\frac{2}{3}\)
c. Six copies of the sum of nine fifths and three
Numerical expression:
6(\(\frac{1}{5}\) + 3)
Solution:
6(\(\frac{1}{5}\) + 3) = 6(\(\frac{1 + 15 }{5}\))= 6(\(\frac{16}{5}\)) = \(\frac{96}{5}\)
d. Three fourths of the product of four fifths and fifteen
Numerical expression:
\(\frac{3}{4}\)(\(\frac{4}{5}\) (15)
Solution:
\(\frac{3}{4}\)(\(\frac{4}{5}\) (15)
= \(\frac{3}{4}\) 12
= 9
Question 2.
Write at least 2 numerical expressions for each phrase below. Then, solve.
a. Two thirds of eight
b. One sixth of the product of four and nine
Answer:
a. Numerical expression:
\(\frac{2}{3}\) (8 )or \(\frac{2}{3}\) × 8
Solution:
\(\frac{2}{3}\) × 8 = \(\frac{16}{3}\) = 5 \(\frac{1}{3}\) .
b. Numerical expression:
\(\frac{1}{6}\) (4 × 9 ) or \(\frac{1}{6}\) (9 × 4 )
Solution:
\(\frac{1}{6}\) (4 × 9 ) = \(\frac{1}{2}\) (4 × 3 ) = 2 × 3 = 6
Question 3.
Use <, >, or = to make true number sentences without calculating. Explain your thinking.
a. 217 × (42 + \(\frac{48}{5}\)) 
(217 × 42) + \(\frac{48}{5}\)
b. (687 × \(\frac{3}{16}\)) × \(\frac{7}{12}\) (687 × \(\frac{3}{16}\)) × \(\frac{3}{12}\)
c. 5 × 3.76 + 5 × 2.68 5 × 6.99
Answer:
a. 217 × (42 + \(\frac{48}{5}\)) > (217 × 42) + \(\frac{48}{5}\)
Explanation :
The left number is multiplied by greater number .
b. (687 × \(\frac{3}{16}\)) × \(\frac{7}{12}\) > (687 × \(\frac{3}{16}\)) × \(\frac{3}{12}\)
Explanation :
\(\frac{7}{2}\) is bigger than \(\frac{3}{12}\) . So, Multiplying the same factor by \(\frac{7}{2}\) will give you greater answer than multiplying by \(\frac{3}{2}\) or \(\frac{1}{4}\)
c. 5 × 3.76 + 5 × 2.68 < 5 × 6.99
Explanation :
If you add 3.76 and 2.68 its not as much as 6.99 and using the Distributive property , the first equation could be
5 ( 3.76 + 2.68 ) = 5 (6.44 ) .
Eureka Math Grade 5 Module 6 Lesson 26 Reflection Answer Key
How did the games we played today prepare you to practice writing, solving, and comparing expressions this summer? Why do you think these are important skills to work on over the summer? Will you teach someone at home how to play these games with you? What math skills will you need to teach in order for someone at home to be able to play with you?
Answer:
This Expressions help in comparing the costs of goods while doing shopping and critical thinking skills and ability to utilize math in everyday life.
It is Important to work on this skills on over summer helps in increasing the solving of problems in different approach and constantly Learning .
Yes , Teaching some one to play these games helps me to become perfect in that game .
Math skills that i need is to have patience and understanding in order to Explain clearly to all.
Eureka Math Grade 5 Module 6 Lesson 26 Homework Answer Key
Question 1.
For each written phrase, write a numerical expression, and then evaluate your expression.
a. Forty times the sum of forty-three and fifty-seven
Numerical expression:
Solution:
Answer:
Numerical expression:
40 ( 43 + 57 )
Solution:
40 ( 43 + 57 ) = 40 ( 100 ) = 4000 .
b. Divide the difference between one thousand three hundred and nine hundred fifty by four.
Numerical expression:
Solution:
Answer:
Numerical expression:
\(\frac{1300 – 950}{4}\)
Solution:
\(\frac{1300 – 950}{4}\) = \(\frac{350}{4}\) = \(\frac{175}{2}\) = 87\(\frac{1}{2}\) .
c. Seven times the quotient of five and seven
Numerical expression:
Solution:
Answer:
Numerical expression:
7×(5÷7)
Solution:
7×(5÷7) = 7 (\(\frac{5}{7}\) ) = 5
d. One fourth the difference of four sixths and three twelfths
Numerical expression:
Solution:
Answer:
Numerical expression:
\(\frac{1}{4}\)( \(\frac{4}{6}\) – \(\frac{3}{12}\))
Solution:
\(\frac{1}{4}\)( \(\frac{4}{6}\) – \(\frac{3}{12}\))
= \(\frac{1}{4}\)( \(\frac{2}{3}\) – \(\frac{1}{4}\))
= \(\frac{1}{4}\)( \(\frac{8- 3}{12}\))
= \(\frac{1}{4}\)( \(\frac{5}{12}\))
=\(\frac{5}{48}\)
Question 2.
Write at least 2 numerical expressions for each written phrase below. Then, solve.
a. Three fifths of seven
b. One sixth the product of four and eight
Answer a :
Numerical expression:
\(\frac{3}{5}\)( 7) or \(\frac{3}{5}\) × 7
Solution:
\(\frac{3}{5}\)( 7) = \(\frac{21}{5}\)
Question 3.
Use <, >, or = to make true number sentences without calculating. Explain your thinking.
a. 4 tenths + 3 tens + 1 thousandth 
30.41
b. (5 × \(\frac{1}{10}\)) + (7 × \(\frac{1}{100}\)) 0.507
c. 8 × 7.20 8 × 4.36 + 8 × 3.59
Answer:
a. 4 tenths + 3 tens + 1 thousandth < 30.41
Explanation :
We have thousandth in the left that means 0.001 (43) = 0.043
b. (5 × \(\frac{1}{10}\)) + (7 × \(\frac{1}{100}\)) = 0.507
Explanation :
(5 × \(\frac{1}{10}\)) + (7 × \(\frac{1}{100}\))
= 0.5 + 0.07 = 0.57
we have only \(\frac{1}{100}\) that means only 2 points of decimals
Where as the in 0.507 we have 3 decimal points .
c.
8 × 7.20 < 8 × 4.36 + 8 × 3.59
Explanation :
If you add 4.36 and 3.59 is greater than as 7.20 and using the Distributive property , the Second equation could be
8 ( 4.36 + 3.59 ) = 8 (7.95 ) .