Eureka Math Grade 4 Module 5 Assessment Review

Engage NY Eureka Math 4th Class Module 5 Stop of Module Assessment Answer Key

Eureka Math Form 4 Module 5 End of Module Cess Chore Respond Central

Question 1.
a. Partitioning the tape diagram to prove five × \(\frac{two}{iii}\). Partition the number line to show x × \(\frac{i}{3}\) .

Answer:
5  × \(\frac{two}{3}\) = 3.3.

Caption:
In the above-given question,
given that,
partition the tape diagram.
v 10 2/3.
5 x 3 = 15.
15 + 2/3 = 17/iii.
17/three = 5.6
10 x ane/3 = 10/3.
10/three = 3.three.

b. Use the models in a higher place to explicate why 5 × \(\frac{ii}{3}\) = ten × \(\frac{ane}{iii}\).

Respond:
5 x 2/3 = 10 x ane/three.

Caption:
In the to a higher place-given question,
given that,
5 x 2/3 = 10/3.
five ten 2 = 10.
10 x ane/three = 10/3.
10 x 1 = 10.
10/3 = 10/three.
5 x two/three = 10 x i/three.

Question 2.
Fill up in the circles beneath with <, =, or > to make true number sentences. Employ decomposition or multiplication to justify your reply.
a. 7 \(\frac{43}{6}\)

Answer:
7 < 43/6.

Explanation:
In the above-given question,
given that,
employ decomposition or multiplication.
43/6 = 7.1.
vii < 7.i.
vii < 43/6.

b. 11\(\frac{ane}{iii}\) \(\frac{34}{iii}\)

Respond:
11(1/three) = 34/3.

Caption:
In the to a higher place-given question,
given that,
11(1/iii) = xi x 3.
eleven x 3 = 33.
33 + 1/3 = 34/3.
34/iii = xi.3.
34/3 = 34/three.

c. \(\frac{13}{half-dozen}\) \(\frac{38}{12}\)

Answer:
(thirteen/6) < 38/12.

Explanation:
In the above-given question,
given that,
13/6 = two.one.
38/12 = 3.one.
two.1 < 3.1.
thirteen/six < 38/12.

Question 3.
Generate a pattern of at to the lowest degree 13 fractions by adding \(\frac{4}{3}\) to \(\frac{1}{3}\) and then continuing to add \(\frac{iv}{3}\) to each fraction. Circle each fraction equal to a whole number. Write what you notice about the pattern of whole numbers. The commencement ii fractions are written for you.
\(\frac{1}{3}\), \(\frac{5}{3}\),

Answer:
i/3, 5/3, nine/3, 13/3, 17/three, 21/iii, 25/3, 29/three, 33/3, 37/3, 41/3, 45/iii, 49/3.

Explanation:
In the higher up-given question,
given that,
Generate a pattern of at least thirteen fractions by adding iv/3 to1/iii.
ane/three + 4/3 = v/3 = ane.vi.
5/3 + four/three = 9/3 = 3.
9/3 + four/iii = xiii/3 = iv.3.
13/iii + 4/3 = 17/3 = 5.6.
17/3 + 4/3 = 21/iii = 7.
21/three + 4/three = 25/3 = 8.3.
25/3 + iv/iii = 29/iii = 9.half-dozen.
29/3 + iv/iii = 33/3 = 11.
33/3 + iv/3 = 37/iii = 12.3.
37/3 + iv/three = 41/3 = xiii.six.
41/iii + 4/3 = 45/3 = 15.
45/3 + 4/3 = 49/three = 16.3.

Question 4.
Observe each sum or difference.
a. 6\(\frac{4}{ten}\) + 7\(\frac{seven}{10}\)

Reply:
half dozen(four/ten)+vii(7/x) = 14.1.

Explanation:
In the to a higher place-given question,
given that,
Detect the sum.
6(4/10).
6 x 10 = 60.
sixty + 4/x.
64/10 = 6.4.
7(7/10).
7 x ten = 70.
lxx + 7/10.
77/ten = 7.7.
half-dozen.iv+7.7 = 14.1.

b. three\(\frac{iii}{8}\) + half-dozen\(\frac{5}{8}\) + 1\(\frac{seven}{eight}\)

Answer:
3(3/8)+half-dozen(v/8)+1(7/8)= eleven.7.

Caption:
In the in a higher place-given question,
given that,
Find the sum.
3(3/8).
3 x eight = 24.
24 + 3/eight.
27/viii = 3.three.
6(5/eight).
6 10 eight = 48.
48 + 5/8.
53/eight = 6.6.
1(7/8).
1 10 8 = viii.
8 + seven/8.
15/8 = one.eight.
6.half dozen+3.3+ane.8 = 11.7.

c. i\(\frac{9}{12}\) – ane\(\frac{4}{12}\)

Answer:
1(9/12)-1(4/12) = fourteen.ane.

Explanation:
In the above-given question,
given that,
Notice the sum.
1(9/12).
ane 10 12 = 12.
12 + ix/12.
21/12 = 6.4.
1(4/12).
1 x 12 = 12.
12 + iv/12.
xvi/12 = seven.7.
vi.4+7.7 = 14.1.

d. 5\(\frac{ii}{v}\) – 1\(\frac{iii}{5}\)

Answer:
five(2/5)-one(three/five) = 3.8.

Explanation:
In the to a higher place-given question,
given that,
Find the sum.
v(ii/5).
5 10 5 = 25.
25 + ii/5.
27/5 = v.four.
1(3/5).
1 ten 5 = 5.
five + 3/5.
8/five = i.6.
5.4 – i.half dozen = 3.eight.
Question 5.
a. Rewrite 3 × \(\frac{6}{8}\) as the production of a unit fraction and a whole number. Solve.

Answer:
3 x six/eight = iii.75.

Explanation:
In the above-given question,
given that,
3 x 6/8.
8 x 3 = 24.
24 + 6/8.
30/eight = 3.75.

b. Rewrite four × 6\(\frac{2}{3}\) every bit the product of a unit fraction and a whole number. Solve.

Answer:
4 x 6 x 2/3 = 24.6.

Explanation:
In the above-given question,
given that,
iv 10 6 x 2/3.
half-dozen 10 4 = 24.
24 x 3 = 72.
72 + 2/3.
74/3 = 24.half-dozen.

Question 6.
Determine if the following are true or false. Explain how yous know using models or words. Make faux problems true by rewriting the right side of the number sentence.
a. 7\(\frac{one}{iii}\) = seven + \(\frac{i}{three}\)

Answer:
No, the sentence is false.

Explanation:
In the above-given question,
given that,
7 x 1/3 = 7/three.
vii + 1/iii = 8/iii.
7/three = 2.3.
8/3 = two.6.

b. \(\frac{5}{three}\) – \(\frac{3}{3}\) + \(\frac{2}{3}\)

Respond:
5/3 – iii/3 + ii/three = 4/3.
Explanation:
In the higher up-given question,
given that,
5/three – 3/iii + 2/3.
5/3 – iii/3 = 2/iii.
5 – three = 2.
2/3 + 2/three = 4/iii.

c. \(\frac{thirteen}{6}\) – \(\frac{v}{vi}\) = \(\frac{xiii-5}{vi}\)

Answer:
xiii/six – five/6 = 8/6.
Explanation:
In the above-given question,
given that,
xiii/half dozen – v/half-dozen.
13 – 5 = eight.
13/6 – 5/6 = 8/6.

d. \(\frac{xi}{3}\) = 11 + \(\frac{1}{3}\)

Reply:
11/3 = 11 + ane/3.

Explanation:
In the above-given question,
given that,
11/3 = 3.half dozen.
11 + 1/3.
11 x iii = 33.
33 + ane/3 = 34/3.
34/3 = 11.iii.
due east. \(\frac{7}{8}\) + \(\frac{7}{8}\) + \(\frac{vii}{8}\) + \(\frac{7}{eight}\) = four × \(\frac{7}{eight}\)

Answer:
True.

Explanation:
In the above-given question,
given that,
7/8 + seven/eight + vii/eight + 7/eight.
7/8 = 0.875.
0.8 + 0.8 + 0.8 + 0.8.
4 x 7/8 = 3.2.

f. five × 3\(\frac{3}{4}\) = 15 + \(\frac{3}{4}\)

Respond:
Truthful.

Explanation:
In the higher up-given question,
given that,
five x three(3/four).
5 x 3 = 15.
15(3/4) = 15 x 4.
60 + 3/iv.
63/four.
5 x 3(3/iv) = 15 + iii/four.

Question seven.
The chart to the right shows data Amashi collected about butterfly wingspans.

Butterfly	Wingspan (inches)
Monarch	3\(\frac{seven}{8}\)
Milbert'due south Tortoiseshell	v\(\frac{5}{8}\)
Zebra Swallowtail	2\(\frac{ane}{ii}\)
Viceroy	2\(\frac{vi}{8}\)
Postman	3\(\frac{3}{eight}\)
Purple Spotted Swallowtail	2\(\frac{2}{8}\)
Julia	three\(\frac{2}{iv}\)
Southern Dogface	two\(\frac{3}{8}\)
Tiger Swallowtail	iii\(\frac{i}{two}\)
Regal Fritillary	iii\(\frac{iv}{viii}\)

Answer:
Monarch:3(vii/eight) = iii.8.

Explanation:
In the to a higher place-given question,
given that,
the wingspan of the butterflies.
3(7/8) = 3 x 8.
24 + seven/viii.
31/8 = iii.8.

Answer:
Milbert's:5(5/8) = 3.8.

Explanation:
In the above-given question,
given that,
the wingspan of the butterflies.
5(5/8) = 5 x 8.
40 + v/8.
45/8 = 5.half-dozen.

Answer:
Zebra consume tail:2(ane/2) = 3.viii.

Caption:
In the above-given question,
given that,
the wingspan of the butterflies.
2(1/two) = 2 ten 2.
iv + ane/ii.
five/2 = two.5.

Answer:
Viceroy:2(6/8) = 2.75.

Explanation:
In the above-given question,
given that,
the wingspan of the collywobbles.
two(6/eight) = 2 x 8.
16 + half-dozen/8.
22/eight = two.75.

Answer:
Postman:3(iii/8) = three.3.

Explanation:
In the to a higher place-given question,
given that,
the wingspan of the butterflies.
3(iii/8) = 3 x 8.
24 + three/8.
27/viii = iii.3.

Respond:
Majestic:2(2/viii) = 2.2.

Explanation:
In the higher up-given question,
given that,
the wingspan of the butterflies.
2(2/8) = 2 x 8.
16 + two/viii.
18/8 = ii.25.

Answer:
Julia:3(ii/iv) = 3.5.

Explanation:
In the to a higher place-given question,
given that,
the wingspan of the butterflies.
three(2/4) = 3 x iv.
12 + two/4.
14/4 = 3.5.

Answer:
Southern Dog:ii(3/8) = 2.3.

Explanation:
In the above-given question,
given that,
the wingspan of the butterflies.
two(3/8) = two x eight.
16 + 3/viii.
19/8 = 2.3.

Answer:
Tiger:3(1/2) = iii.8.

Explanation:
In the above-given question,
given that,
the wingspan of the butterflies.
3(i/2) = iii x 2.
6 + one/2.
vii/2 = 3.5.

Respond:
Regal:three(iv/viii) = 3.viii.

Explanation:
In the to a higher place-given question,
given that,
the wingspan of the butterflies.
3(4/8) = 3 ten viii.
24 + 4/eight.
28/8 = 3.v.

a. At the lesser of this page, create a line plot to display the data in the table.
b. What is the divergence in wingspan betwixt the widest and narrowest collywobbles on the chart?
c. Three butterflies take the aforementioned wingspan. Explain how you know the measurements are equal.

Answer:
The three butterflies have the same wingspan = 3.five in.

Explanation:
In the higher up-given question,
given that,
the butterflies accept the aforementioned wingspan.
Tiger swallowtail = three.5.
Imperial Fritillary = 3.v
Julia = 3.five.

Solve each problem. Draw a model, write an equation, and write a statement for each.
d. Amashi wants to display a Postman and Viceroy side by side in a photograph box with a width of 6 inches. Will these two butterflies fit? Explain how you know.

Answer:
Yeah, the two collywobbles will fit.

Caption:
In the above-given question,
given that,
Amashi wants to display a postman and viceroy side past side in a photo box with a width of 6 inches.
Viceroy = 2.75 in.
Postman = 3.3 in.
two.75 + 3.3 = 6.0.
so, the ii butterflies will fit.

eastward. Compare the wingspan of the Milbert's Tortoiseshell and the Zebra Swallowtail using >, <, or =.

Respond:
Milbert'southward Tortoiseshell > Zebra Swallowtail.

Explanation:
In the in a higher place-given question,
given that,
the wingspan of two butterflies.
Milbert's Tortoiseshell = 5.6 in.
Zebra Swallowtail = 2.5 in.
v.vi > two.v.
and so Milbert'south > Zebra.

f. The Queen Alexandra Birdwing can have a wingspan that is v times equally broad every bit the Southern Dogface's. How many inches can the Birdwing'due south wingspan be?

Respond:
Alexandra Birdwing = xi.five in.

Caption:
In the in a higher place-given question,
given that,
The Queen Alexandra Birdwing can have a wingspan that is 5 times every bit wide every bit the Southern Dogface's.
Southern Dogface's = two.three in.
2.3 x 5 = 11.5.
Alexandra Birdwing = 11.5 in.

chiliad. Amashi discovered a pattern. She started with ii\(\frac{2}{8}\) inches and added \(\frac{1}{8}\) inch to each measurement. List the adjacent four measurements in her design. Name the five collywobbles whose wingspans match the measurements in her pattern.

Respond:
Zebra Swallow tail, Viceroy, Southern Dogface, Tiger Swallowtail.

Explanation:
In the above-given question,
given that,
Amashi discovered a pattern.
she started with ii(2/8)in.
2 x 2/8 = 8 x two.
8 ten 2 = xvi.
sixteen + ii/8.
18/8 = two.25.
18/8 + ane/8 = 19/8.
19/viii + 1/8 = 20/8.
xx/viii + 1/8 = 21/8.
21/8 + one/eight = 22/8.
22/8 + one/8 = 23/viii.
nineteen/8 = 2.iii.
20/8 = 2.five.
21/8 = 2.vi.
22/8 = 2.7.
23/8 = 2.8.

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Source: https://ccssmathanswers.com/eureka-math-grade-4-module-5-end-of-module-assessment/

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