McGraw Hill Math

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 7 Lesson 2 Word Problems with Fractions and Mixed Numbers are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 7 Lesson 2 Word Problems with Fractions and Mixed Numbers

Solve

Write a fraction and then solve the problem.

Question 1.
Four friends plan to share a 5-pound block of cheese. If they share evenly, how much cheese will each friend get? Sketch a model of the problem.
Answer:
DRAWING

Each will receive 1\(\frac{1}{4}\) pounds of cheese: \(\frac{5}{4}\) = 1\(\frac{1}{4}\)

Question 2.
There are 21 sheep on a farm. The farmer would like to have the same number of sheep in each of his 4 fields. Is this possible? Explain why or why not.
Answer:
No, The farmer cannot have same number of sheep if he wants to use 21 sheep.

Explanation:
If farmer wants to use same number of sheep in 4 of his fields, then farmer have to use 5 sheep per field and leave 1 sheep.
But if he wants to use 21 sheep, it is not possible.

Question 3.
Sheryl has 12 pounds of blueberries. She divides the blueberries evenly into 24 containers. What is the weight of the blueberries in each container?
Answer:
1/2 pound

Explanation:

Each container will receive 1/2 pound of 12 pounds in 24 containers.

Question 4.
Five people move 82 bags of chicken feed. How many bags of chicken feed will each person move if they all move the same number of bags?
Answer:
16.4 bags each

Explanation:
Number of bags = 82
Number of people = 5
Each person move number of bags = 82 ÷ 5 = 16.4

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 7 Lesson 10 Problem Solving are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 7 Lesson 10 Problem Solving

Solve

Question 1.
DeWitt surveys his friends to see how many music videos they each downloaded in a month. DeWitt got the following responses: 0, 3, 6, 7, 9, 2, 5, 3, 8, 0, 10, 3, 5, 7, and 9
Do most of his friends download 0-2 videos, 3-5 videos, 6-8 videos, or 9-11 videos a month?
Make a table to find the solution.

Answer:

Question 2.
Use the data from exercise 1.
Which two intervals have the same number of responses? What is that number?
Answer:
0-2 and 9-11 have same number of responses and it is 3.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 7 Lesson 1 Interpreting Fractions are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 7 Lesson 1 Interpreting Fractions

Solve

Complete the fraction or division shown in the drawing.

Question 1.

Question 2.

Answer:
1/8
Explanation:
The name of the shape is a circle.
It is divided into 8 equal parts and 1 part is shaded among them.
So, the fraction of the shaded part is 1/8

Question 3.

Answer:
3 ÷ 2 

Sketch a drawing to show the following divisions.

Question 4.
\(\frac{15}{3}\)
Answer:
5

Explanation:
(15/3) = 15 ÷ 3 = 5

Question 5.
1 ÷ 12
Answer:
(1/12)

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 17.3 Customary and Metric Units of Temperature to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 17.3 Customary and Metric Units of Temperature

Exercises

CONVERT

Question 1.
150° C = ______________ F
Answer:
302° F,

Explanation:
To convert ° C to ° F, C x (\(\frac{9}{5}\)) + 32 = F,
Given C= 150°c, 150 x (\(\frac{9}{5}\)) + 32 = F,
\(\frac{1350}{5}\) + 32 = F, 270 + 32 = F,
F = 302°, So 150° C = 302 ° F.

Question 2.
150° F = ______________ C
Answer:
65.6° C,

Explanation:
To convert ° F to ° C, (F – 32) x (\(\frac{5}{9}\)) = C,
Given F = 150°F, (150 – 32) x (\(\frac{5}{9}\)) = C,
118 x \(\frac{5}{9}\) = C, \(\frac{590}{9}\) = C,
C = 65.6°, So 150° F = 65.6° C.

Question 3.
0° C = _______________ F
Answer:
32° F,

Explanation:
To convert ° C to ° F, C x (\(\frac{9}{5}\)) + 32 = F,
Given C = 0° C, 0 x (\(\frac{9}{5}\)) + 32 = F,
32° = F, So 0° C = 32° F.

Question 4.
0° F = ____________ C
Answer:
-17.8° C,

Explanation:
To convert ° F to ° C, (F – 32) x (\(\frac{5}{9}\)) = C,
Given F = 0° F, (0 – 32) x (\(\frac{5}{9}\)) = C,
-32 x \(\frac{5}{9}\) = C, \(\frac{-160}{9}\) = C,
C = -17.8 ° F, So 0° F = -17.8° C.

Question 5.
216° F = _____________ C
Answer:
102.2° C,

Explanation:
To convert ° F to ° C, (F – 32) x (\(\frac{5}{9}\)) = C,
Given F = 216° F, (216 – 32) x (\(\frac{5}{9}\)) = C,
184 x \(\frac{5}{9}\) = C, \(\frac{920}{9}\) = C,
C = 102.2° F, So 216° F = 102.2° C.

Question 6.
105° C = ______________ F
Answer:
221° F,

Explanation:
To convert ° C to ° F, C x (\(\frac{9}{5}\)) + 32 = F,
Given C= 105° C, 105 x (\(\frac{9}{5}\)) + 32 = F,
\(\frac{945}{5}\) + 32 = F, 189° + 32° = F,
221° = F, So 105° C = 221° F.

Question 7.
72° F = _____________ C
Answer:
22.2° C,

Explanation:
To convert ° F to ° C, (F – 32) x (\(\frac{5}{9}\)) = C,
Given F = 72° F, (72 – 32) x (\(\frac{5}{9}\)) = C,
40 x \(\frac{5}{9}\) = C, \(\frac{200}{9}\) = C,
C = 22.2°, So 72° F = 22.2° C.

Question 8.
30° C = ______________ F
Answer:
86° F,

Explanation:
To convert ° C to ° F, C x (\(\frac{9}{5}\)) + 32 = F,
Given C= 30° C, 30 x (\(\frac{9}{5}\)) + 32 = F,
\(\frac{270}{5}\) + 32 = F, 54° + 32° = F,
86° = F, So 30° C = 86° F.

Question 9.
-25° F = ______________ C
Answer:
-31.7° C,

Explanation:
To convert ° F to ° C, (F – 32) x (\(\frac{5}{9}\)) = C,
Given F = -25° F, (-25 – 32) x (\(\frac{5}{9}\)) = C,
57 x \(\frac{5}{9}\) = C, \(\frac{285}{9}\) = C,
C = -31.7°, So -25° F = -31.7° C.

Question 10.
35° F = _____________ C
Answer:
1.7° C,

Explanation:
To convert ° F to ° C, (F – 32) x (\(\frac{5}{9}\)) = C,
Given F = 35° F, (35 – 32) x (\(\frac{5}{9}\)) = C,
3 x \(\frac{5}{9}\) = C, \(\frac{15}{9}\) = C,
C = 1.666666°, So 35° F = 1.7° C.

Question 11.
37° C = _____________ F
Answer:
98.6° F,

Explanation:
To convert ° C to ° F, C x (\(\frac{9}{5}\)) + 32 = F,
Given C= 37°C, 37 x (\(\frac{9}{5}\)) + 32 = F,
\(\frac{333}{5}\) + 32 = F, 66.6° + 32° = F,
98.6° = F, So 30° C = 89.6° F.

Question 12.
950° C = _____________ F
Answer:
1742° F,

Explanation:
To convert ° C to ° F, C x (\(\frac{9}{5}\)) + 32 = F,
Given C= 30° C, 950 x (\(\frac{9}{5}\)) + 32 = F,
\(\frac{8550}{5}\) + 32 = F, 1710° + 32° = F,
1742° = F, 30° C = 1742° F.

Question 13.
The melting point of aluminum is 1,219° F and the melting point of nickel is 2,646° F. What is the temperature difference between the two melting points? Express your answer in degrees Celsius.
Answer:
The temperature difference between the two melting points is 775° C,

Explanation:
The melting point of aluminum is 1,219° F and
the melting point of nickel is 2,646° F.
The temperature difference between the
two melting points is 2646 – 1219 = 1427° F.
To convert ° F to ° C, (F – 32) x (\(\frac{5}{9}\)) = C,
Given F = 1,427° F, (1427 – 32) x (\(\frac{5}{9}\)) = C,
1395 x \(\frac{5}{9}\) = C, \(\frac{6975}{9}\) = C,
C = 775°, 1427° F = 775° C.

Question 14.
Jamie took his temperature with a Celsius thermometer, and it read 38° C. What would his temperature be in degrees Fahrenheit?
Answer:
The temperature is 100.4° F,

Explanation:
To convert ° C to ° F, C x (\(\frac{9}{5}\)) + 32 = F,
Given C= 38° C, 38 X (\(\frac{9}{5}\)) + 32 = F,
\(\frac{342}{5}\) + 32 = F, 68.4° + 32° = F,
100.4° = F, 38° C = 100.4° F.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 17.2 Changing from Metric Units to Customary Units to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 17.2 Changing from Metric Units to Customary Units

Exercises

CONVERT

Question 1.
24 mm = _____________ in.
Answer:
0.936 in,

Explanation:
1 mm = 0.039 in, 24 mm = 24 X 0.039 in = 0.936 in.

Question 2.
143 cm = _____________ in.
Answer:
56.342 in,

Explanation:
1 cm = 0.394 in, 143 cm = 143 X 0.394 in = 56.342 in.

Question 3.
3 m = ____________ in.
Answer:
118.11 in,

Explanation:
1 meter = 39.37 in, 3 m = 3 X 39.37 in = 118.11 in.

Question 4.
4 km = _____________ mi
Answer:
2.484 miles,

Explanation:
1 kilometer = 0.621 miles, 4 km = 4 X 0.621 miles = 2.484 miles.

Question 5.
20 g = ____________ oz
Answer:
0.7 oz,

Explanation:
1 gram = 0.035 ounces, 20 g = 20 X 0.035 ounces = 0.7 oz.

Question 6.
6 L = ____________ qt
Answer:
6.342 qt,

Explanation:
1 liter = 1.057 quarts, 6 L = 6 X 1.057 qt = 6.342 qt.

Question 7.
3 kL = ___________ gal
Answer:
792.6 gal,

Explanation:
1 kL = 264.2 gallons, 3 kL = 3 X 264.2 gal = 792.6 gal.

Question 8.
4.5 kg = _____________ lb
Answer:
9.9223 lb,

Explanation:
1 kg = 2.205 lb, 4.5 kg = 4.5 X 2.205 lb = 9.9223 lb.

Question 9.
242 mm = _____________ in.
Answer:
9.44 in,

Explanation:
1 millimeter = 0.039 in, 242 mm = 242 X 0.039 = 9.438 or 9.44 in.

Question 10.
13 kg is about how many oz? _________
Answer:
458.64 oz,

Explanation:
1 kg = 35.28 oz, 13 kg = 13 X 35.28 oz = 458.64 oz.

Question 11.
1100 mm = ___________ in.
Answer:
42.9 in,

Explanation:
1 millimeter = 0.039 in, 1100 mm =1100 X 0.039 in = 42.9 in.

Question 12.
2 kg + 32 g = ______________ oz
Answer:
71.68 oz,

Explanation:
1 kg = 35.28 oz, 2 kg = 2 X 35.28 oz = 70.56 oz,
1 g = 0.035 oz, 32 g = 32 X 0.035 oz = 1.12 oz,
70.56 oz + 1.12 oz = 71.68 oz.

Question 13.
The average player on the school basketball team is 190 centimeters tall. The average volleyball player is 6 ft 4 inches tall. Which team has a taller average height?
Answer:
6 ft 4 in volleyball players are taller on average than
school basketball team,

Explanation:
The average player on the school basketball team is
190 centimeters tall.
The average volleyball player is 6 ft 4 inches tall.
1 cm = 0.0328084 ft, 190 cm = 190 X 0.0328084 ft = 6.233596 ft,
As 6.4 ft > 6.233596 ft, So, 6 ft 4 in volleyball players are taller on
average than school basketball team.

Question 14.
The Booster Club sponsored a 10-km walk to raise money for charity. About how many miles is the length of the walk?
Answer:
6.21 miles is the length of the walk,

Explanation:
The Booster Club sponsored a 10-km walk to raise money for charity.
1 km = 0.621 miles, 10 km = 10 X 0.621 miles = 6.21 miles.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 17.1 Changing from Customary Units to Metric Units to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 17.1 Changing from Customary Units to Metric Units

Exercises

CONVERT

Question 1.
14 in. = ___________cm
Answer:
35.56 cm,

Explanation:
As 1 in = 2.54 cm, 14 in = 14 X 2.54 cm = 35.56 cm.

Question 2.
15 ft = ____________ m
Answer:
4.575 m,

Explanation:
As 1 ft = 0.305 m, 15 ft = 15 X 0.305 m = 4.575 m.

Question 3.
12 yd = ____________ m
Answer:
10.968 m,

Explanation:
1yd = 0.914 m, 12 yds = 12 X 0.914 m = 10.968 m.

Question 4.
4 yd = _____________ m
Answer:
3.656 m,

Explanation:
1 yd = 0.914 m, 4 yds = 4 X 0.914 m = 3.656 m.

Question 5.
3 mi = ____________ km
Answer:
4.827 km,

Explanation:
1 mile = 1.609 km, 3 miles = 3 X 1.609 km = 4.827 km.

Question 6.
4 c = _____________ L
Answer:
0.948 L,

Explanation:
1cup = 0.237 L, 4 cups = 4 X 0.237 L = 0.948 L.

Question 7.
3 pt = _____________ L
Answer:
1.419 L,

Explanation:
1 pt = 0.473 L, 3 pts = 3 X 0.473 L = 1.419 L.

Question 8.
9 qt = ______________ L
Answer:
8.514 L,

Explanation:
1 qt = 0.946 L, 9 qts = 9 X 0.946 L = 8.514 L.

Question 9.
3.5 gal = _____________ L
Answer:
13.2475 L,

Explanation:
1 gal = 3.785 L, 3.5 gals = 3.5 X 3.785 L = 13.2475 L.

Question 10.
11 fl oz = ______________ mL
Answer:
325.314mL,

Explanation:
1 fluid ounce = 29.574mL, 11 fl oz = 11 X 29.574 mL = 325.314 mL.

Question 11.
14 oz = ____________ g
Answer:
396.9 g,

Explanation:
1 ounce = 28.35 g, 14 oz = 14 X 28.35 g = 396.9 g.

Question 12.
3 lbs = _____________ kg
Answer:
1.362 kg,

Explanation:
1 lb = 0.454 kg, 3 lbs = 3 X 0.454 kg = 1.362 kg.

Question 13.
6 oz = ____________ g
Answer:
170.1 g,

Explanation:
1 ounce = 28.35 g, 6 oz = 6 X 28.35 g = 170.1 g.

Question 14.
2 mi = _____________ m
Answer:
3,218.68 m,

Explanation:
1 mile = 1609.34 m, 2 miles = 2 X 1609.34 m = 3,218.68 m.

Question 15.
11 c = ___________ L
Answer:
2.607 L,

Explanation:
1 cup = 0.237 L, 11 cups = 11 X 0.237 L = 2.607 L.

Question 16.
Which would be heavier: 4.8 lb or 2000 g of silver? __________
Answer:
4.8 lbs

Explanation:
Given to find the heavier from 4.8 lb or 2000 g of silver as
1 lb = 453.592 g, So 4.8 X 453.592 g = 2,177.2416 now
Comparing both as 2,177.2416 > 2,000 g, therefore 4.8 lbs.

Question 17.
About how many meters tall is a 120-ft building?
Answer:
36.576 m,

Explanation:
Given to find meters tall of 120 -ft as
1 ft = 0.3048 m so 120 X 0.3048 m = 36.576 m.

Question 18.
Farah’s cooler holds 8 qt of liquid. About how many mL will it hold?
Answer:
7,570.824 mL,

Explanation:
Given Farah’s cooler holds 8 qt of liquid,
As 1 qt = 946.353 mL so it can hold 8 X 946.353 mL = 7570.824 mL.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 16.4 Calculating Perimeter and Area with Metric Units to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 16.4 Calculating Perimeter and Area with Metric Units

Exercises

CALCULATE

Question 1.
What is the perimeter of a square with sides of 10 cm?

What is the area?
Answer:
Perimeter =40 cm; Area = 100 sq cm,

Explanation:
the perimeter of of a square = 4 x side = 4 x 10 cm = 40 cm,
The area of a square = side x side = 10 cm X 10 cm = 100 sq cm.

Question 2.
What is the perimeter of the figure?

Answer:
38 cm,

Explanation:
The perimeter of a a given figure,
P = 6 cm + 4 cm + 3 cm + 2 cm + 9 cm + 14 cm = 38 cm.

Question 3.
A rectangle has sides of 4.4 cm and 6 cm. What is the perimeter of the rectangle? What is the area?
Answer:
Perimeter of a rectangle = 20.8 cm,
Area of a rectangle = 26.4 sq cm,

Explanation:
Perimeter of a rectangle = 2[length + breadth],
= 2[6 cm + 4.4 cm] = 20.8 cm,
Area of a rectangle = length x breadth = 6 cm X 4.4 cm,
= 26.4 sq cm.

Question 4.
What is the perimeter of a rectangle with 2 sides of 12 meters and 2 sides of 6 meters? What is the area?
Answer:
Perimeter of a rectangle = 36 m,
Area of a rectangle = 72 sq m,

Explanation:
Perimeter of a rectangle = 2[length + breadth],
= 2[12 m + 6 m] = 36 m,
Area of a rectangle = length x breadth = 12 m X 6 m = 72 sq m.

Question 5.
What is the area of the triangle?

Answer:
90 sq m,

Explanation:
The are of a triangle = 1/2[base X height] =1/2 [15 m x 12 m],
= 1/2 x 180 sq m = 90 sq m.

Question 6.
What is the area of the triangle?

Answer:
24 sq m,

Explanation:
The area of a triangle = 1/2[base X height] =1/2 [12 m X 4 m],
= 1/2 X 48 sq m = 24 sq m.

Question 7.
Ariel is cleaning a rug that is 4.2 meters long and 3 meters wide. If the dry cleaning store charges $18.00 per square meter to clean it, how much will she pay to clean the rug? What is the perimeter of the rug?
Answer:
She will pay $226.80,
Perimeter of rug is14.4 m,

Explanation:
Length L = 4.2 m, Width W = 3 m, Area of rug = L X W,
A = 4.2 m X  3 m, A = 12.6 sq m,
the dry cleaning store charges $18.00 per square meter,
she pay to clean the rug is 12.6 x $18 = $ 226.80,
The perimeter of the rug is perimeter of a rectangle =
2[Length + Breadth] = 2[4.2 m + 3 m] = 2  X 7.2 m = 14.4 m.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 16.3 Metric Units of Mass to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 16.3 Metric Units of Mass

Exercises

CONVERT

Question 1.
400 g = ____________ kg
Answer:
0.4 kg,

Explanation:
1,000 g = 1 kg, 400 g = 400 ÷ 1,000 = 0.4 kg.

Question 2.
225 kg = ______________ mg
Answer:
225,000,000 mg,

Explanation:
1 kg = 1,000,000 mg, 225 kg = 225 X 1,000,000 mg = 225,000,000 mg.

Question 3.
6600 g = ______________ kg
Answer:
6.6 kg,

Explanation:
1000 g = 1 kg, 6,600 g = 6,600 ÷ 1,000 kg = 6.6 kg.

Question 4.
4505 mg = ______________ g
Answer:
4.505 g,

Explanation:
0.001 mg = 1 g, 4,505 g = 4,505 X 0.001 g = 4.505 g.

Question 5.
21.35 kg = ______________ mg
Answer:
21,350,000 mg,

Explanation:
1 kg =  1,000,000 mg, 21.35 kg = 21.35 X 1,000,000 mg
= 21,350,000 mg.

Question 6.
530 mg = _______________ g
Answer:
0.53 g,

Explanation:
0.001 mg = 1 g, 530 mg = 530 x 0.001 g = 0.53 g.

Question 7.
650 mg = ______________ g
Answer:
0.65 g,

Explanation:
0.001 mg = 1 g, 650 mg = 650 x 0.001 g = 0.65 g.

Question 8.
71 kg = _______________ cg
Answer:
7,100,000 cg,

Explanation:
1 kg = 100,000 cg, 71 kg = 71 X 100,000 cg = 7,100,000 cg.

Question 9.
3.721 kg = ______________ g
Answer:
3,721 g,

Explanation:
1 kg = 1,000 g, 3.721 kg = 3.721 X 1,000 g = 3,721 g.

Question 10.
2313 g = _______________ kg
Answer:
2.313 kg,

Explanation:
1000 g = 1 kg, 2,313 g = 2,313 ÷ 1000 = 2.313 kg.

Question 11.
546 mg = ______________ g
Answer:
0.546 g,

Explanation:
0.001 mg = 1 g, 546 mg = 546 X 0.001 g = 0.546 g.

Question 12.
12,305 g = ______________ kg
Answer:
12.305 kg,

Explanation:
1,000 g = 1 kg, 12,305 g = 12,305 ÷ 1,000 = 12.305 kg.

Question 13.
4430 mg = ______________ g
Answer:
4.43 g,

Explanation:
0.001 mg = 1 g, 4,430 mg = 4,430 X 0.001 = 4.43 g.

Question 14.
12.34 kg = ______________ g
Answer:
12,340 g,

Explanation:
1 kg = 1,000 g, 12.34 kg = 12.34 X 1,000 g = 12,340 g.

Question 15.
2 mg = ______________ kg
Answer:
0.000002 kg,

Explanation:
0.000001 mg = 1 kg, 2 mg = 2 X 0.000001 kg = 0.000002 kg.

Question 16.
300 kg = ______________ mg
Answer:
300,000,000 mg,

Explanation:
1 kg = 1,000,000 mg, 300 kg = 30 X 1,000,000 mg = 300,000,000 mg.

Question 17.
258 mg = ______________ cg
Answer:
25.8 cg,

Explanation:
10 mg = 1 cg, 258 mg = 258 ÷ 10 = 25.8 cg.

Question 18.
65 g = ______________ kg
Answer:
0.65 kg,

Explanation:
1,000 g = 1 kg, 65 g = 65 ÷ 1000 = 0.65 kg.

Question 19.
Larry has 5,599.4 milligrams of saffron seasoning that he brought back from India. He sells 1.06 grams to a local health food restaurant, and he gives his mother .0034 kg. How much saffron does Larry have left?
Answer:
1.1394 g,

Explanation:
Larry has 5,599.4 milligrams of saffron seasoning.
He sells 1.06 grams to a local health food restaurant,
he gives his mother .0034 kg.
Total saffron used by Larry = 1.06 g + 0.0034 kg,
First converting kg to grams 1 kg = 1,000 g,
0.0034 kg = 0.0034 x 1,000 = 3.4 g, 1.06 g + 3.4 g = 4.46 g,
Total saffron left with Larry = Total saffron – used saffron,
= 5,599.4 mg – 4.46 g, first converting mg to g,
1 mg = 0.001 g, 5,599.4 mg = 5,599.4 x 0.001 = 5.5994 g,
Saffron left = 5.5994 g – 4.46 g = 1.1394 g.

Question 20.
Richard keeps three bee hives in his backyard. He collected 1,678.616 grams of honey from the first hive, 1.6 kilograms from the second hive, and 1,660,301 milligrams from the third hive. Which hive produced the most honey?
How much more honey did it produce than the hive that produced the second most amount of honey?
Answer:
First hive produced the most honey than other hives.
18.315 g more than second most hive,

Explanation:
Richard collected 1,678.616 grams of honey from the first hive,
1.6 kilograms from the second hive,
converting kg to grams, 1 kg = 1,000 g,
1.6 kg = 1.6 X 1,000 g = 1,600 g,
1,660,301 milligrams from the third hive.
0.001 mg = 1 g, 1,660,301 mg = 1,660,301 x  0.001 g = 1,660.301 g,
Comparing which hive produced more as
1,678.616 g > 1,660.301 g > 1,600 g,
so first hive produced the most honey than other hives.
Total amount of honey produced than the hive from
the second most amount of honey,
1,678.616 g – 1,660.301 g = 18.315 g.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 16.2 Metric Units of Liquid Volume to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 16.2 Metric Units of Liquid Volume

Exercises

CONVERT

Question 1.
873 mL = ______________ L
Answer:
0.873 L,

Explanation:
1,000 mL = 1L, 873 mL = 873 ÷ 1000 = 0.873 L.

Question 2.
1455 mL = _______________ kL
Answer:
0.001455 kL,

Explanation:
1,000,000 mL = 1 kL, 1455 mL = 1455 ÷ 1,000,000 = 0.001455 kL.

Question 3.
4.75 L = ______________ mL
Answer:
4,750 mL,

Explanation:
1 L = 1,000 mL, 4.75 L = 4.75 x 1000 = 4,750 mL.

Question 4.
7945 mL = ________________ L
Answer:
7.945 L,

Explanation:
1,000 mL = 1 L, 7,945 mL = 7,945 ÷ 1,000 = 7.945 L.

Question 5.
100 mL = _______________ kL
Answer:
0.0001 kL,

Explanation:
1,000,000 mL = 1 kL, 100 mL = 100 ÷ 1,000,000 = 0.0001 kL.

Question 6.
2554 mL = ________________ L
Answer:
2.554 L,

Explanation:
1000 mL = 1L, 2554 mL = 2554 ÷ 1000 = 2.554 L.

Question 7.
.0025 L = _____________ mL
Answer:
2.5 mL,

Explanation:
1 L = 1,000 mL, 0.0025 L = 1,000 x 0.0025 = 2.5 mL.

Question 8.
.007 kL = _______________ mL
Answer:
7,000 mL,
.
Explanation:
1 kL = 1,000,000 mL, 0.007 kL = 1,000,000 x 0.007 = 7000 mL.

Question 9.
Pam’s recipe calls for 3.4 L of olive oil. Unfortunately, all she has is a measuring cup that holds 10 mL. How many times will she have to fill her measuring cup with olive oil to complete the recipe?
Answer:
340 times,

Explanation:
Pam’s recipe calls for 3.4 L of olive oil.
She has a measuring cup that holds 10 mL.
First converting liters to mL 1 L = 1000 mL, 3.4 L = 3,400 mL,
Number of times will she have to fill her measuring cup with
olive oil to complete the recipe 3400 ÷ 10 = 340 times.

Question 10.
Ron mixed 4,500 milliliters of his favorite paint color. How many 1-liter containers can he fill with the mixture?
Answer:
4.5 containers,

Explanation:
Ron mixed 4,500 milliliters of his favorite paint color.
1 L = 1000 mL, 4,500 mL = 4,500 ÷ 1000 = 4.5 mL.

Question 11.
Harriet had 5.768 liters of special-liquid cleanser before she gave 554 mL of it to her mother, and 1.4 mL to her younger sister. How much liquid cleanser does she have left?
Answer:
5.2126 L,

Explanation:
Harriet had 5.768 liters of special-liquid cleanser.
She gave 554 mL of it to her mother, and 1.4 mL
to her younger sister. 554 mL + 1.4 mL = 555.4 mL,
Total liquid cleanser left with her after giving =
5.768 L – 555.4 mL, converting mL into L, 1 L = 1000 mL,
555.4 mL = 555.4 ÷ 1000 = 0.5554 L,
Total liquid 5.768 L – used liquid 0.5554 L = 5.2126 L.

Question 12.
The science teacher instructs the students to add 256 mL of Liquid A to 1.5 L of Liquid B. How much liquid will there be altogether, assuming nothing evaporates or turns to a solid?
Answer:
1,756 mL,

Explanation:
Liquid in A = 256 mL, Liquid in B = 1.5 L, converting 1.5 L into mL,
1 L = 1000 mL, 1.5 L = 1000 x 1.5 = 1500 L,
Total liquid in A & B = 256 + 1500 = 1,756 mL.

Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 16.1 Metric Units of Length to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 8 Answer Key Lesson 16.1 Metric Units of Length

Exercises

CONVERT

Question 1.
210 m = ______________ km
Answer:
0.21 km,

Explanation:
1km = 1000m, 210m = 210 ÷ 1000 = 0.21 km.

Question 2.
145 mm = _____________ cm
Answer:
14.5 cm,

Explanation:
1cm = 10 mm, 145 mm = 145 ÷ 10 = 14.5 cm.

Question 3.
573 m = _______________ cm
Answer:
57,300 cm,

Explanation:
1m = 100 cm, 573m = 100 x 573 = 57300 cm.

Question 4.
26 m = _______________ mm
Answer:
26,000 mm,

Explanation:
1m = 1000 mm, 26m = 1000 x 26 = 26000 mm.

Question 5.
400 cm = _____________ km
Answer:
0.004 km,

Explanation:
1km = 100,000 cm, 400 cm = 400 ÷ 100000 = 0.004 km.

Question 6.
400 m = ______________ cm
Answer:
40,000 cm,

Explanation:
1m = 100 cm, 400m = 100 x 400 = 40,000 cm.

Question 7.
1.6 m = _____________ mm
Answer:
1600 mm,

Explanation:
1m = 1000 mm, 1.6 m = 1000 x 1.6 = 1600 mm.

Question 8.
2.5 km = ______________ m
Answer:
2,500 m,

Explanation:
1 km = 1000m, 2.5 km = 1000 x 2.5 = 2500 m.

Question 9.
116 mm = _____________ m
Answer:
0.116 m,

Explanation:
1 m = 1000 mm, 116 mm = 116 ÷ 1000 = 0.116 m.

Question 10.
1.557 km = ______________ mm
Answer:
1,557,000 mm,

Explanation:
1 kilometer = 1,000,000 mm, 1.557 km = 1,000,000 x 1.557 = 1,557,000.

Question 11.
4355 mm = ______________ cm
Answer:
435.5 cm,

Explanation:
1 mm = 0.1 cm, 4,355 mm = 4355 x 0.1 = 435.5 cm.

Question 12.
.4667 km = _______________ cm
Answer:
46,670 cm,

Explanation:
1 km = 100,000 cm, 0.46670 km = 100,000 x 0.4667 = 46,670 cm.

Question 13.
3.0556 m = ______________ mm
Answer:
3055.6 mm,

Explanation:
1 meter = 1000 mm, 3.0556 m = 1,000 x 3.0556 = 3055.6 mm.

Question 14.
Terry covered the outside edge of his cousin’s monster truck tires with a border of fluorescent yellow paint. If each tire measured 6.25 meters around the outside edge, then how many centimeters of edging did he paint on one tire?
Answer:
625 cm,

Explanation:
Each tire is of 6.25 m, 1 m = 100 cm,
Total cm of paint on one tire = 6.25 x 100 = 625 cm.

Question 15.
The distance from Mike’s house to the school is .775 kilometers. How many meters is that?
Answer:
775 m,

Explanation:
The distance from Mike’s house to the school is .775 kilometers,
1 km = 1,000 m, 0.775 km = 1,000 x 0.775 = 775 km.