McGraw Hill Math

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 11 Lesson 6 Creating Number Patterns are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 11 Lesson 6 Creating Number Patterns

Calculate

Complete the patterns in each set of tables. Then compare each pattern and write about your comparison.

Question 1.

Compare the patterns. What is their relationship?
Answer:
Each number in the second pattern is 3 more than the number in the first pattern.

Question 2.

Compare the patterns. What is their relationship?
Answer:

Explanation:
The pattern in left hand side table is 8, 7, 6, 5, 4.
The pattern in right hand side table is 5, 4, 3, 2, 1.
Each number in the second pattern is 3 less than the number in the first pattern.

Question 3.

Compare the patterns. What is their relationship?
Answer:

Explanation:
The pattern in left hand side table is 0, 4, 8, 12, 16.
The pattern in right hand side table is 0, 2, 4, 6, 8.
Each number in the second pattern is half the number in the first pattern except zero.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 11 Lesson 5 Analyzing Shapes and Lines are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 11 Lesson 5 Analyzing Shapes and Lines

Question 1.
On the grid, draw lines \(\overline{A B}\), \(\overline{B C}\), and \(\overline{C A}\). What shape did you draw?
____________
What are the coordinates of the shape?
Answer:

Explanation:
After drawing the given lines \(\overline{A B}\), \(\overline{B C}\), and \(\overline{C A}\) on the grid the shape formed with these lines is triangle.
The coordinates of the triangle are A (2, 9) , B (2, 2), C (9, 2).

Question 2.
Draw a rectangle on the grid. Be sure each corner is a point Label the points.
What are the coordinates of the points you drew?

Answer:

Explanation:
A rectangle is drawn on the grid and labelled the points as A, B, C, D.
The coordinates of the rectangle are A (1, 3), B (4, 3), C (4, 1) and D (1, 1).

Question 3.
Kim is building a flower box. She draws lines \(\overline{A B}\), \(\overline{C D}\), and \(\overline{A C}\). She wants to draw a fourth side parallel to \(\overline{A C}\). Draw the line.
What shape is the flower box?

Answer:

Explanation:
Kim drawn a lines \(\overline{A B}\), \(\overline{C D}\), and \(\overline{A C}\). She wants to draw a fourth side parallel to \(\overline{A C}\). After drawing the fourth side the shape of the flower box is parallelogram.
The coordinates of the parallelogram is A (4, 6) B (2, 3) C (9, 6) and D (7, 3).

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 11 Lesson 3 Plotting Points on a Coordinate Grid are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 11 Lesson 3 Plotting Points on a Coordinate Grid

Plot and Label

Plot and label each point on the grid. The first point has been plotted and labeled for you.

Question 1.
A (2, 3)
Answer:

The point A (2, 3) is labeled and plotted on the grid.

Question 2.
B (7, 8)
Answer:

The point B (7, 8) is labeled and plotted on the grid.

Question 3.
C (4, 5)
Answer:

The point C (4, 5) is labeled and plotted on the grid.

Question 4.
D (2, 9)
Answer:

The point D (2, 9) is labeled and plotted on the grid.

Question 5.
E (3, 1)
Answer:

The point E (3, 1) is labeled and plotted on the grid.

Question 6.
F (8, 4)
Answer:

The point F (8, 4) is labeled and plotted on the grid.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 11 Lesson 2 Using Line Plots to Solve Problems are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 11 Lesson 2 Using Line Plots to Solve Problems

Solve

Use the line plot above for exercises 1-6. 

Question 1.
What is the difference between the tallest plant and the shortest plant shown on the line plot?
2\(\frac{1}{4}\)in.

Question 2.
What is the sum of the height of the two tallest plants?
Answer:
The height of the first tallest plant is 43\(\frac{1}{2}\)in.
The height of the second tallest plant is 43\(\frac{1}{4}\)in.
43\(\frac{1}{2}\)in + 43\(\frac{1}{4}\)in = 86\(\frac{3}{4}\)in
The sum of the height of the two tallest plants is 86\(\frac{3}{4}\)in.

Question 3.
How many plants are 42\(\frac{1}{2}\) in. tall? ________
What is the total sum of their height?
Answer:
Four plants are 42\(\frac{1}{2}\) in tall.
The four plants height is each 42\(\frac{1}{2}\) in tall.
42\(\frac{1}{2}\) in + 42\(\frac{1}{2}\) in  + 42\(\frac{1}{2}\) in  + 42\(\frac{1}{2}\) in = 168 4/2 in or 170 in.
The total sum of their height is 170 inches.

Question 4.
What is the total height of the three shortest plants?
Answer:
The first shortest plant is 41\(\frac{1}{4}\) in.
The second shortest plant is 42 in.
The third shortest plant is 42\(\frac{1}{4}\) in.
41\(\frac{1}{4}\) in + 42 in + 42\(\frac{1}{4}\) in = 125 \(\frac{2}{4}\) in or 125\(\frac{1}{2}\) in
The total height of the three shortest plants is 125\(\frac{1}{2}\) in.

Question 5.
Which height is an outlier?
Answer:
The height 41\(\frac{1}{4}\) inches is an outlier.

Question 6.
To find the average height, you would find the sum of all the heights shown and then divide by the number of plants. What is the average height shown on the line plot?
Answer:
The sum of all the heights are as below
41\(\frac{1}{4}\) in + 42 + 42\(\frac{1}{4}\) in + 170 + 84\(\frac{3}{4}\) in + 43 + 43\(\frac{1}{4}\) in + 43\(\frac{1}{2}\) in = 510
Average height shown on the line plot = Sum of all the heights/ Number of plants
Number of plants = 12
Average height shown on the line plot = 510/12 = 42.5 in or 42 9/16 in

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 11 Lesson 1 Making a Line Plot are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 11 Lesson 1 Making a Line Plot

Diagram and Interpret

The list of data shows the mass in kilograms of nine cats.

Question 1.
Draw a line plot to show this data set. Use the line plot for exercises 2-5.
Answer:

Explanation:
The above line plot is numbered from 2 to 5 with an interval of 1/4.
The given data is marked on the line plot with X on top of it and named the line plot as nine cats mass in kilograms.

Question 2.
Which mass is an outlier?
Answer:
The mass 2 1/4 kg is an outlier.

Question 3.
Draw a rectangle around the cluster. Draw an oval around the gap.
Answer:

Explanation:
A rectangle is drawn around the cluster which is 3 1/4 kg to 3 3/4 kg.
A oval should be drawn around the gap. In the line plot we can observe two gaps so, two ovals should be drawn. One oval  is drawn around 2 2/4 kg to 3 kg and second oval is drawn around 4 2/4 kg to 5 kg.

Question 4.
How many cats have a mass of 3\(\frac{1}{4}\) kg?
Answer:
Two cats have a mass of 3\(\frac{1}{4}\) kg.

Question 5.
What is the most common mass in the data set?
Answer:
The most common mass in the data set is 3\(\frac{1}{2}\) kg

Practice the questions of McGraw Hill Math Grade 4 Answer Key PDF Chapter 2 Lesson 3 Adding Whole Numbers to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 4 Answer Key Chapter 2 Lesson 3 Adding Whole Numbers

Add

Question 1.

Answer:

Explanation:
Sum of 491 and 564 we get 1,055.

Question 2.

Answer:

Explanation:
Sum of 4096 and 8322 we get 12,418.

Question 3.

Answer:

Explanation:
Sum of 52,471 and 29,788 we get 82,259.

Question 4.

Answer:

Explanation:
Sum of 70,628 and 4,914 we get 75,542.

Question 5.

Answer:

Explanation:
Sum of 504,105 and 81,739 we get 585,844.

Question 6.

Answer:

Explanation:
Sum of 623,957 and 348,517 we get 972,474.

Question 7.
38,231 + 25,877 = ___________
Answer:
Sum of 38,231 and 25,877 we get 64,108.

Explanation:
38,231 + 25,877 = 64,108.

Question 8.
620,873 + 76,455 = ____
Answer:
Sum of 620,873 and 76,455 we get 697,328.

Explanation:
620,873 + 76,455 = 697,328.

Question 9.
4,564 + 98,100 = __________
Answer:
Sum of 4,564 and 98,100 we get 102,664.

Explanation:
4,564 + 98,100 = 102,664.

Question 10.
408,641 + 449,243 = ____
Answer:
Sum of 408,641 and 449,243 we get 857,884.

Explanation:
408,641 + 449,243 = 857,884.

Question 11.
Look at problem 7 above. Is regrouping needed in the final step? Why or why not?
Answer:
Yes, regrouping is required in final step to get the correct solution of adding 38,231 and 25,877.

Explanation:

38,231 + 25,877 = 64,108.

Practice the questions of McGraw Hill Math Grade 4 Answer Key PDF Chapter 2 Lesson 4 More Practice with Adding Whole Numbers to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 4 Answer Key Chapter 2 Lesson 4 More Practice with Adding Whole Numbers

Add

Question 1.

Answer:

Explanation:
Sum of 673, 822 and 742:
673+ 822 +742
= 1,495 + 742
= 2,237.

Question 2.

Answer:

Explanation:
Sum of 5824, 8237 and 6540:
5,824 + 8,237 + 6,540
= 14,061 + 6540
= 20,601.

Question 3.

Answer:

Explanation:
Sum of 413327, 826 and 70628:
413,327 + 826 + 70,628
= 414,513 + 70,628
= 484,781.

Question 4.

Answer:

Explanation:
Sum of 61,537, 35,481, 322,765 and 5,825:
61,537 + 35,481 + 322,765 + 5,825
= 97,018 + 322,765 + 5,825
= 419783 + 5,825
= 425,608.

Question 5.

Answer:

Explanation:
Sum of 34,693, 83,001, 19,548 and 42,965:
34,693 + 83,001 + 19,548 + 42,965
= 117,694 + 19,548 + 42,965
= 137,242 + 42,965
= 180,207.

Question 6.

Answer:

Explanation:
Sum of 512,868, 5,906, 124,850 and 257,623:
512,868 + 5,906 + 124,850 + 257,623
= 518,774 + 124,850 + 257,623
= 643,624 + 257,623
= 901,247.

Question 7.
Clearwater, Florida, had a population of 107,685 in 2010. Coral Springs had a population of 121,096. Orlando had a population of 238,300. What was the total population of the three communities?
Answer:
Total population of the three communities = 467,081.

Explanation:
Population of Clearwater, Florida in 2010 = 107,685.
Population of Coral Springs in 2010 = 121,096.
Population of Orlando in 2010 = 238,300.
Total population of the three communities = Population of Clearwater, Florida in 2010 + Population of Coral Springs in 2010 + Population of Orlando in 2010
= 107,685 + 121,096 + 238,300
= 228,781 + 238,300
= 467,081.

Practice the questions of McGraw Hill Math Grade 4 Answer Key PDF Chapter 2 Lesson 5 Subtracting Whole Numbers to secure good marks & knowledge in the exams.

McGraw-Hill Math Grade 4 Answer Key Chapter 2 Lesson 5 Subtracting Whole Numbers

Subtract

Use addition to check your answer.

Question 1.

Answer:

Explanation:
Difference between 762 and 328:
762 – 328 = 434.

Question 2.

Answer:

Explanation:
Difference between 8,295 and 3,667:
8,295 – 3,667 = 4,628.

Question 3.

Answer:

Explanation:
Difference between 87,471  and 53,743:
87,471 – 53,743 = 33,728.

Question 4.

Answer:

Explanation:
Difference between 630,119 and 228,754:
630,119 – 228,754 = 401,365.

Question 5.

Answer:

Explanation:
Difference between 926,138 and 67,880:
926,138 – 67,880 = 858,258.

Question 6.

Answer:

Explanation:
Difference between 725,914 and 699,311:
725,914 – 699,311 = 26,603.

Question 7.
238,960 – 729 = ___________
Answer:
Difference between 238,960 – 729 we get 238,231.

Explanation:
238,960 – 729 = 238,231.

Question 8.
46,008 — 19,725 = _____
Answer:
Difference between 46,008 and 19,725 we get 26,283.

Explanation:
46,008 – 19,725 = 26,283.

Question 9.
52,937 — 8,658 = ___________
Answer:
Difference between 52,937 and 8,658 we get 44,279.

Explanation:
52,937 – 8,658 = 44,279.

Question 10.
783,209 — 97,120 = ____
Answer:
Difference between 783,209 and 97,120 we get 686,089.

Explanation:
783,209 – 97,120 = 686,089.

Question 11.
Look at problem 6. Why was the 0 in the hundred thousands place not written ¡n the answer?
Answer:
Difference between 725,914 and 699,311 we get 26,603. Its not written 0 in the hundred thousands place not written in the answer because zero in the left side of the number doesn’t have any value.

Explanation:
725,914 – 699,311 = 26,603.

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 4 Subtracting Fractions with Like Denominators are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 4 Subtracting Fractions with Like Denominators

Solve

Subtract. Give the answer in simplest terms.

Question 1.
\(\frac{3}{4}\) – \(\frac{1}{4}\) ______________
Answer:
\(\frac{2}{4}\) = \(\frac{1}{2}\)

Question 2.
\(\frac{11}{16}\) – \(\frac{3}{16}\) ______________
Answer:
\(\frac{11}{16}\) – \(\frac{3}{16}\) = \(\frac{1}{2}\).

Explanation:
The difference between \(\frac{11}{16}\) – \(\frac{3}{16}\) is \(\frac{11-3}{16}\)
= \(\frac{8}{16}\)
= \(\frac{1}{2}\).

Question 3.
\(\frac{6}{7}\) – \(\frac{3}{7}\) ______________
Answer:
\(\frac{6}{7}\) – \(\frac{3}{7}\) = \(\frac{3}{7}\).

Explanation:
The difference between \(\frac{6}{7}\) – \(\frac{3}{7}\) is \(\frac{6-3}{7}\)
= \(\frac{3}{7}\).

Question 4.
\(\frac{5}{8}\) – \(\frac{3}{8}\) ______________
Answer:
\(\frac{5}{8}\) – \(\frac{3}{8}\) = \(\frac{1}{4}\).

Explanation:
The difference between \(\frac{5}{8}\) – \(\frac{3}{8}\) is \(\frac{5-3}{8}\)
= \(\frac{2}{8}\)
= \(\frac{1}{4}\).

Question 5.
\(\frac{13}{16}\) – \(\frac{7}{16}\) ______________
Answer:
\(\frac{13}{16}\) – \(\frac{7}{16}\) = \(\frac{3}{8}\).

Explanation:
The difference between \(\frac{13}{16}\) – \(\frac{7}{16}\) is \(\frac{13-7}{16}\)
= \(\frac{6}{16}\)
= \(\frac{3}{8}\).

Question 6.
\(\frac{21}{37}\) – \(\frac{4}{37}\) ______________
Answer:
\(\frac{21}{37}\) – \(\frac{4}{37}\) = \(\frac{17}{37}\).

Explanation:
The difference between \(\frac{21}{37}\) – \(\frac{4}{37}\) is \(\frac{21-4}{37}\)
= \(\frac{17}{37}\).

Question 7.

Answer:
\(\frac{17}{31}\) – \(\frac{7}{31}\) = \(\frac{10}{31}\).

Explanation:
The difference between \(\frac{17}{31}\) – \(\frac{7}{31}\) is \(\frac{17-7}{31}\)
= \(\frac{10}{31}\).

Question 8.

Answer:
\(\frac{2}{9}\) – \(\frac{1}{9}\) = \(\frac{1}{9}\).

Explanation:
The difference between \(\frac{2}{9}\) – \(\frac{1}{9}\) is \(\frac{2-1}{9}\)
= \(\frac{1}{9}\).

Question 9.

Answer:
\(\frac{27}{53}\) – \(\frac{6}{53}\) = \(\frac{21}{53}\).

Explanation:
The difference between \(\frac{27}{53}\) – \(\frac{6}{53}\) is \(\frac{27-6}{53}\)
= \(\frac{21}{53}\).

Question 10.

Answer:
\(\frac{19}{29}\) – \(\frac{14}{29}\) = \(\frac{5}{29}\).

Explanation:
The difference between \(\frac{19}{29}\) – \(\frac{14}{29}\) is \(\frac{19-14}{29}\)
= \(\frac{5}{29}\).

Question 11.

Answer:
\(\frac{7}{8}\) – \(\frac{5}{8}\) = \(\frac{1}{4}\).

Explanation:
The difference between \(\frac{7}{8}\) – \(\frac{5}{8}\) is \(\frac{7-5}{8}\)
= \(\frac{2}{8}\)
= \(\frac{1}{4}\).

Question 12.

Answer:
\(\frac{8}{15}\) – \(\frac{1}{15}\) = \(\frac{7}{15}\).

Explanation:
The difference between \(\frac{8}{15}\) – \(\frac{1}{15}\) is \(\frac{8-1}{15}\)
= \(\frac{7}{15}\).

Question 13.

Answer:
\(\frac{24}{35}\) – \(\frac{13}{35}\) = \(\frac{11}{35}\).

Explanation:
The difference between \(\frac{24}{35}\) – \(\frac{13}{35}\) is \(\frac{24-13}{35}\)
= \(\frac{11}{35}\).

Question 14.

Answer:
\(\frac{7}{15}\) – \(\frac{2}{15}\) = \(\frac{1}{3}\).

Explanation:
The difference between \(\frac{7}{15}\) – \(\frac{2}{15}\) is \(\frac{7-2}{15}\)
= \(\frac{5}{15}\)
= \(\frac{1}{3}\).

Question 15.

Answer:
\(\frac{38}{77}\) – \(\frac{25}{77}\) = \(\frac{13}{77}\).

Explanation:
The difference between \(\frac{38}{77}\) – \(\frac{25}{77}\) is \(\frac{38-25}{77}\)
= \(\frac{13}{77}\).

Question 16.

Answer:
\(\frac{7}{24}\) – \(\frac{5}{24}\) = \(\frac{1}{12}\).

Explanation:
The difference between \(\frac{7}{24}\) – \(\frac{5}{24}\) is \(\frac{7-5}{24}\)
= \(\frac{2}{24}\)
= \(\frac{1}{12}\).

Question 17.

Answer:
\(\frac{43}{65}\) – \(\frac{38}{65}\) = \(\frac{1}{13}\).

Explanation:
The difference between \(\frac{43}{65}\) – \(\frac{38}{65}\) is \(\frac{43-38}{65}\)
= \(\frac{5}{65}\)
= \(\frac{1}{13}\).

Question 18.

Answer:
\(\frac{15}{16}\) – \(\frac{11}{16}\) = \(\frac{1}{4}\).

Explanation:
The difference between \(\frac{15}{16}\) – \(\frac{11}{16}\) is \(\frac{15-11}{16}\)
= \(\frac{4}{16}\)
= \(\frac{1}{4}\).

All the solutions provided in McGraw Hill Math Grade 5 Answer Key PDF Chapter 6 Lesson 5 Adding Fractions with Unlike Denominators are as per the latest syllabus guidelines.

McGraw-Hill Math Grade 5 Answer Key Chapter 6 Lesson 5 Adding Fractions with Unlike Denominators

Solve

Find equivalent fractions and then add. Change improper fractions to mixed numbers.

Question 1.
\(\frac{1}{4}\) + \(\frac{1}{5}\) ______________
Answer:
\(\frac{5}{20}\) + \(\frac{4}{20}\) = \(\frac{9}{20}\)

Question 2.
\(\frac{2}{9}\) + \(\frac{1}{6}\) _______________
Answer:
\(\frac{2}{9}\) + \(\frac{1}{6}\) = \(\frac{11}{18}\).

Explanation:
To get the equivalent fraction we will multiply both numerator and denominator by 2 which is \(\frac{4}{18}\) and we will multiply other addend by 3 which is \(\frac{3}{18}\). So the addition of \(\frac{4}{18}\) + \(\frac{3}{18}\)
= \(\frac{4+7}{18}\)
= \(\frac{11}{18}\).

Question 3.
\(\frac{3}{8}\) + \(\frac{1}{3}\) ________________
Answer:
\(\frac{3}{8}\) + \(\frac{1}{3}\) = \(\frac{11}{18}\).

Explanation:
To get the equivalent fraction we will multiply \(\frac{3}{8}\) both numerator and denominator by 3 which is \(\frac{9}{24}\) and we will multiply other addend \(\frac{1}{3}\) by 8 which is \(\frac{8}{24}\). So the addition of \(\frac{9}{24}\) + \(\frac{8}{24}\)
= \(\frac{9+8}{24}\)
= \(\frac{17}{24}\).

Add. Change any improper fractions to mixed numbers.

Question 4.

Answer:
\(\frac{1}{2}\) + \(\frac{3}{7}\) = \(\frac{13}{14}\).

Explanation:
By adding \(\frac{1}{2}\) + \(\frac{3}{7}\) the sum will be \(\frac{7+6}{14}\)
= \(\frac{13}{14}\).

Question 5.

Answer:
\(\frac{2}{3}\) + \(\frac{3}{5}\) = \(\frac{16}{15}\).

Explanation:
By adding \(\frac{2}{3}\) + \(\frac{3}{5}\) the sum will be \(\frac{10+6}{15}\)
= \(\frac{16}{15}\).

Question 6.

Answer:
\(\frac{1}{7}\) + \(\frac{1}{8}\) = \(\frac{15}{56}\).

Explanation:
By adding \(\frac{1}{7}\) + \(\frac{1}{8}\) the sum will be \(\frac{8+7}{56}\)
= \(\frac{15}{56}\).

Question 7.

Answer:
\(\frac{1}{7}\) + \(\frac{1}{6}\) = \(\frac{13}{42}\).

Explanation:
By adding \(\frac{1}{7}\) + \(\frac{1}{6}\) the sum will be \(\frac{7+6}{42}\)
= \(\frac{13}{42}\).