Grade 4 Math Module 2 End of Module Review

Engage NY Eureka Math 4th Course Module ii End of Module Assessment Answer Cardinal

Eureka Math Class 4 Module 2 End of Module Assessment Task Respond Key

Question 1.
Complete the conversion charts.

Length
iii km	3,000 m
9 km	9,000  m
6 km 435 m	half dozen,435  one thousand
12 km 12 thousand	12,012  m

Explanation:
Given 3 km equally ane kilometer = 1,000 meter
so 3 km = 3 Ten i,000 thou = iii,000 m,

9 km = 9 X 1,000 m = 9,000 m,

6 km 435 m = 6 Ten 1,000 m + 435 m =
six,000 yard + 435 m = 6,435m,

12 km 12 m = 12 X i,000 m + 12 m =
12,000 m + 12 thou = 12,012 m.

Mass
3 kg	3000  g
20 kg 300 thou	20,300 thousand
1 kg 74 m	1,074  k
403 kg iv m	403,004 g

Explanation:
Given 3 kg every bit 1 kilogram = 1,000 gram,
and so 3 kg = three 10 1,000 yard = 3,000 one thousand,

20 kg 300 g  = 20 Ten 1,000 g + 300 g =
20,000 thousand + 300 g = 20,300 grand,

1 kg 74 g = i Ten 1,000 g + 74 g =
1,000 k + 74 thou = 1,074 g,

403 kg iv g = 403 X one,000 g + 4 g =
403,000 g + four g = 403,004 g.

Chapters
4 50	iv,000 mL
48 L 808 mL	48,808  mL
two L 20 mL	2,020 mL
639 L 6 mL	639,006  mL

Caption:
Given four L as 1 liter = i,000 milliliters,
So 4 L = 4 X one,000 mL = four,000 mL,

48 L 808 mL = 48 X 1,000 mL =
48,000 mL, + 808 mL = 48,808 mL,

two L 20 mL = 2 X one,000 mL + xx mL =
2,000 mL + twenty mL = 2,020 mL,

639 L 6 mL = 639 10 one,000 mL + 6 mL =
639,000 mL + 6 mL = 639,006 mL.

Question two.
A student completed the problem below. Check his piece of work. Explicate how you know if each solution is correct or wrong.

a. 24 km = 24,000 m, solution is correct,
b. 16 Fifty = 16,000 mL, solution is right,
c. 38 kg = 38,000 kg not three,800 g and so solution is wrong,

Explanation:
Given a student completed the trouble as shown above.
Checking his piece of work as below
a. 24 km = 24,000 k, solution is correct because
24 km = 24 Ten k 1000 = 24,000 m,

b. 16 L = 16,000 mL, solution is right because
16 L = sixteen X 1000 mL = sixteen,000 mL,

c. 38 kg = 38,000 kg non iii,800 g so solution is incorrect
considering 38 kg = 38 10 g g = 38,000 g not 3,800 k.

Question 3.
Discover the sum or deviation.
a. 493 km 43 1000 + 17 km 57 chiliad

b. 25 kg 32 g – 23 kg 83 m

c. 100 L 99 mL + ii,999 mL

a. 493 km 43 chiliad + 17 km 57 g = 510 km 100 m

Caption:
Given 493 km 43 one thousand + 17 km 57 m every bit
493 km 43 m = 493 X one thousand 1000 + 43 thousand =
493000 m + 43 chiliad = 493,043 m and
17 km 57 chiliad = 17 X grand thousand + 57 thousand =
17,000 m + 57 yard = 17,057 m
So 493,043 k + 17,057 m = 510,100 k or
510,100 m ÷ yard = 510 km 100 m,
therefore, 493 km 43 1000 + 17 km 57 chiliad =510,100 one thousand or 510 km 100 g.

b. 25 kg 32 g – 23 kg 83 grand = 1,949 yard or 1 kg 949 chiliad,

Explanation:
Given 25 kg 32 g – 23 kg 83 thou as
25 kg 32 thou = 25 X 1000 thou + 32 g = 25,032 g and
23 kg 83 g = 23 X thousand g + 83 one thousand = 23,083 g,
Now 25,032 g – 23,083 g = 1,949 1000 or 1,949 1000 ÷ 1000 =
one kg 949 g, therefore 25 kg 32 g – 23 kg 83 1000 = 1,949 m or one kg 949 g.

c. 100 Fifty 99 mL + two,999 mL = 103,098 mL or 103 L 098 mL,

Explanation:
Given 100 L 99 mL + 2,999 mL equally
100 L  99 mL = 100 Ten k mL + 99 mL =
100,000 mL + 99 mL = 100,099 mL ,
And so 100,099 mL + two,999 mL = 103,098 mL or
103,098 mL ÷ thou = 103 L 098 mL,
therefore 100 Fifty 99 mL + 2,999 mL = 103,098 mL or 103 L 098 mL.

Question 4.
Billy is training for a half marathon. For the problems below, use tape diagrams, numbers, and words to explain each answer.
a. Each day, Billy runs on the treadmill for v kilometers and runs on the outdoor track for 6,000 meters. In all, how many meters does Billy run each day?
b. Since Billy has started training, he has also been drinking more h2o. On Sat, he drank ii liters 755 milliliters of water. On Sunday, he drank some more. If Billy drank a total of 4 liters 255 milliliters of water on Saturday and Sunday, how many milliliters of water did Billy drink on Lord's day?
c. Since he began exercising so much for his half marathon, Billy has been losing weight. In his commencement week of training, he lost ii kilograms 530 grams. In the post-obit two weeks of training, he lost one kilogram 855 grams each week. Billy now weighs 61 kilograms 760 grams. What was Billy's weight, in grams, earlier he started training?
Explain your thinking.
a. Baton run each day 11,000 meters or 11 kilometrs,

Statement : Billy run each mean solar day xi thousand meters or eleven kilometers,

b. Number of milliliters of h2o did Billy beverage on Sunday is 1,500 milliliters or i L 500 mL,

Argument : Number of milliliters of water did Baton drink on Dominicus is xv hundred milliliters or ane Liter five hunderd milliliters,

c. Billy'south weight, in grams, before he started grooming 68,000 grams,

Argument: Billy'southward weight, in grams, before he started training is sixty eight yard grams,

Explanation:
Given Billy is training for a one-half marathon.
a. Each mean solar day, Billy runs on the treadmill for 5 kilometers and runs on the outdoor track for vi,000 meters.
In all, Billy run each twenty-four hours is 5 km + 6,000 meters equally
v km = five X grand m = 5,000 thou So five,000 m + 6,000 thou =
11,000 m or xi,000 m ÷ 1,000 = xi km,
used a record diagrams to testify problem as shown higher up, therefore Billy run each 24-hour interval 11,000 meters or 11 kilometrs.

b. Since Billy has started grooming, he has likewise been drinking more water. On Saturday, he drank 2 liters 755 milliliters of water. On Sunday, he drank some more than.
If Billy drank a full of 4 liters 255 milliliters of h2o on Saturday and Sunday, number of milliliters of water did Billy beverage on Sun is
4 L 255 mL – 2 L 755 mL as 4 50 255 mL = 4 Ten 1000 mL + 255 mL =
4,255 mL and 2 L 755 mL = 2 X thousand mL + 755 mL = 2,755 mL,
therefore 4,255 mL – ii,755 mL = 1,500 mL or one,500 mL ÷ i,000 =
1 L 500 mL, used a tape diagrams to show problem equally shown to a higher place,
therefore Number of milliliters of water did Billy drink on Sunday is i,500 milliliters or 1 L 500 mL.

c. Since he began exercising so much for his half marathon,
Billy has been losing weight. In his offset week of training, he lost 2 kilograms 530 grams. In the following two weeks of preparation, he lost i kilogram 855 grams each week.
Billy lost in 3 weeks is 2 kg 530 g + i kg 855 g + 1 kg 855 kg
as 2 kg 530 grand = 2 X m thou + 530 one thousand = two,530 g
1 kg 855 g = one X 1000 one thousand + 855 grand = i,855 g
So lost weight is ii,530  g + ane,855 chiliad + 1,855 g = six,240 chiliad,
Billy now weighs 61 kilograms 760 grams.
Billy'southward weight, in grams, before he started training is
61 kg 760 g + 6,240 g as 61 kg 760 k = 61 10 one thousand g + 760 g =
61,760 g then 61,760 1000 + vi,240 chiliad = 68,000 grams,
used a record diagrams to testify problem as shown above,
therefore, Billy's weight, in grams, before he started training 68,000 grams.

Eureka Math Grade four Module 2 Answer Central

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