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

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?
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?
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?
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?
The height 41$$\frac{1}{4}$$ inches is an outlier.
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