Hillary

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Lesson 8.1 Solving Algebraic Equations to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations

Math in Focus Grade 6 Chapter 8 Lesson 8.1 Guided Practice Answer Key

Complete each with = or ≠, and each with the correct value.

Question 1.
For what value of x will x + 3 = 7 be true?
If x = 1, x + 3 = + 3
= ( 7)
If x = 2, x + 3 = + 3
= ( 7)
If x = 4, x + 3 = + 3
=
x + 3 = 7 is true when x = .
Answer:
x + 3 = 7 is true when x = 4,

Explanation:
Given to find for what value of x will x + 3 = 7 be true, so
if x =1, x + 3 = 1 + 3 = 4 ≠ 7,
If x = 2, x + 3 = 2 + 3 = 5 ≠ 7,
if x = 3, x + 3 = 3 + 3 = 6 ≠ 7,
if x = 4, x + 3 = 4 + 3 = 7 = 7, therefore x + 3 = 7 is true when x = 4.

Solve each equation using the substitution method.

Question 2.
p + 6 = 13
Answer:
p = 7, p + 6 = 13 is true,

Explanation:
Given equation p + 6 = 13, so solving p = 13 – 6 = 7, So when p = 7 means 7 + 6 = 13 is true therefore p = 7, p + 6 = 13 is true.

Question 3.
r + 4 = 12
Answer:
r = 8, r + 4 = 12 is true,

Equation::
Given equation r + 4 = 12 so solving r = 12 – 4 = 8 therefore when r = 8 means 8 + 4 = 12 is true, therefore r = 8, r + 4 = 12 is true.

Question 4.
k – 10 = 7
Answer:
k = 17 means k – 10 = 7 is true,

Explanation:
Given equation k – 10 = 7 so solving k = 7 + 10 = 17 therefore when k = 17 means 17 – 10 = 7 is true so k = 17., k – 10 = 7 is true.

Question 5.
2m = 6
Answer:
m = 3 means 2m = 6 is true,

Explanation:
Given equation 2m = 6 so solving m =6/2 = 3 therefore when m = 3 means 2 x 3 = 6 is true so m = 3.

Question 6.
4n = 20
Answer:
n = 5 means 4n = 20 is true,

Explanation:
Given equation 4n = 20 so solving n = 20/4 = 5 therefore when n = 5 means 4 X 5 = 20 is true so n = 5.

Question 7.
\(\frac{1}{5}\)z = 3
Answer:
z = 15 means \(\frac{1}{5}\)z = 3,

Explanation:
Given equation \(\frac{1}{5}\)z = 3 so solving x = 3 X 5 = 15 therefore when z = 15 means \(\frac{1}{5}\) X 15 = 3 is true so z = 3.
Complete each with + or —, and each with the correct value.

Question 8.
Solve x + 8 = 19.
x + 8 = 19
x + 8 =
x =
Answer:
x = 11,

Explanation:
Given to solve x + 8 = 19,
x + 8 – 19 = 19 – 19,
x – 11 = 0,
x = 11.

Solve each equation.

Question 9.
f + 5 = 14
Answer:
f = 9,

Explanation:
Given to solve f + 5 = 14,
subtracting both sides by 12 as f + 5 – 14 = 14 – 14,
f – 9 = 0,
therefore f = 9.

Question 10.
26 = g + 11
Answer:
g = 15,

Explanation:
Given to solve 26 = g + 11,
subtracting by 11 on the both sides as 26 – 11 = g + 11 – 11,
15 = g or g = 15.

Question 11.
w – 6 = 10
Answer:
w = 16,

Explanation:
Given to solve w – 6 = 10,
Adding 6 both the sides we get w – 6 + 6 = 10 + 6,
therefore w = 16.

Question 12.
z – 9 = 21
Answer:
z = 30,

Explanation:
Given to solve z – 9 = 21,
Adding 9 both the sides we get z – 9 + 9 = 21 + 9,
therefore z = 30.

Complete each with × or ÷, and with the correct value.

Question 13.
Solve 3x = 27.
3x = 27
3x = 27
x =
Answer:
x = 9,

Explanation:
Given to solve 3x = 27,
Dividing both sides by 3 as 3x ÷ 3 = 27 ÷ 3,
therefore x = 9.

Solve each equation.

Question 14.
6a = 42
Answer:
a = 7,

Explanation:
Given to solve 6a = 42,
Dividing both sides by 6 as 6a ÷ 6 = 42 ÷ 6,
we get a = 7.

Question 15.
65 = 13b
Answer:
b = 5,

Explanation:
Given to solve 65 = 13b,
Dividing both sides by 13 as 65 ÷ 13 = 13b ÷ 13,
we get 5 = b or b = 5.

Question 16.
\(\frac{m}{8}\) = 9
Answer:
m = 72,

Explanation:
Given to solve \(\frac{m}{8}\) = 9,
Multiplying both sides by 8,
\(\frac{m}{8}\) X 8 = 9 X 8,
We get m = 72.

Question 17.
12 = \(\frac{n}{7}\)
Answer:
n = 84,

Explanation:
Given to solve 12 = \(\frac{n}{7}\),
Multiplying both sides by 7,
12 X 7 = \(\frac{n}{7}\) X 7,
we get 84 = n or n = 84.

Complete each with +, —, × or ÷, and with the correct value.

Question 18.
Solve x + \(\frac{3}{7}\) = \(\frac{5}{7}\)
x + \(\frac{3}{7}\) = \(\frac{5}{7}\)
x + \(\frac{3}{7}\) = \(\frac{5}{7}\)
x =
Answer:
x = \(\frac{2}{7}\),

Explanation:
Given to solve x + \(\frac{3}{7}\) = \(\frac{5}{7}\) subtracting \(\frac{3}{7}\) both sides as x + \(\frac{3}{7}\) – \(\frac{3}{7}\) = \(\frac{5}{7}\) – \(\frac{3}{7}\) we have common denominator so x = \(\frac{5 – 3}{7}\) = \(\frac{2}{7}\).

Solve each equation. First tell which operation you will perform on each side of the equation. Write your answer in simplest form.

Question 19.
k + \(\frac{1}{8}\) = \(\frac{7}{8}\)
Answer:
k = \(\frac{6}{8}\) or \(\frac{3}{4}\),

Explanation:
Given to solve k + \(\frac{1}{8}\) = \(\frac{3}{4}\) subtracting \(\frac{1}{8}\) both sides as k + \(\frac{1}{8}\) – \(\frac{1}{8}\) = \(\frac{7}{8}\) – \(\frac{1}{8}\) we have common denominators so x = \(\frac{7 – 1}{8}\) = \(\frac{6}{8}\) further can be divided as both numerator and denominator goes by 2 we get \(\frac{3}{4}\).

Question 20.
4p = \(\frac{3}{4}\)
Answer:
p = \(\frac{3}{16}\),

Explanation:
Given to solve 4p = \(\frac{3}{4}\), Dividing both sides by 4 as
4p ÷ 4 = \(\frac{3}{4}\) ÷ 4,
p = \(\frac{3}{4 X 4}\) = \(\frac{3}{16}\) .

Math in Focus Course 1B Practice 8.1 Answer Key

Solve each equation using the substitution method.

Question 1.
b + 7 = 10
Answer:
b = 3 means b + 7 = 10 is true,

Explanation:
Given equation b + 7 = 10 so solving b = 10 – 7 = 3 when b = 3 means 3 + 7 = 10 is true therefore b = 3.

Question 2.
17 = e + 9
Answer:
e = 8,

Explanation:
Given 17 = e + 9 so solving e = 17 – 9 = 8 when e = 8 means 17 = 8 + 9  is true therefore e = 8.

Question 3.
k – 4 = 11
Answer:
k = 15,

Explanation:
Given k – 4 = 11 so solving k = 11 + 4 = 15 when k = 15 means 15 – 4 = 11 is true therefore k = 15.

Question 4.
42 = 3p
Answer:
p = 14,

Explanation:
Given 42 = 3p so solving 42/3 = p when p = 14 means 42 = 3 X 14 is true therefore p = 14.

Question 5.
8t = 56
Answer:
t = 7,

Explanation:
Given 8t = 56 so solving t = 56/8 = 7 when t = 7 means 8 X 7 = 56 is true therefore t = 7.

Question 6.
\(\frac{1}{4}\)v = 5
Answer:
v = 20,

Explanation:
Given \(\frac{1}{4}\)v = 5 solving v = 5 X 4 = 20 when v = 4 means \(\frac{1}{4}\) X 4 = 5 x 4 = 20 is true therefore v = 4.

Solve each equation using the concept of balancing.

Question 7.
k + 12 = 23
Answer:
k = 11,

Explanation:
Given to solve k + 12 = 23,
subtracting by 12 on the both sides as k + 12 – 12 = 23 – 12,
k = 11.

Question 8.
x- 8 = 17
Answer:
x = 25,

Explanation:
Given to solve x – 8 = 17,
adding 8 by both sides as x – 8 + 8 = 17 + 8,
x = 25.

Question 9.
24 = f – 16
Answer:
f = 40,

Explanation:
Given to solve 24 = f – 16,
adding both sides by 16 as 24 + 16 = f – 16 + 16,
40 = f or f =40.

Question 10.
5j = 75
Answer:
j = 15,

Explanation:
Given to solve 5j = 75,
dividing both sides by 5 as 5j/5 = 75/5,
j = 15.

Question 11.
81 = 9m
Answer:
m = 9,

Explanation:
Given to solve 81 = 9m,
dividing both sides by 9 as 81/9 = 9m/9,
9 = m or m = 9.

Question 12.
\(\frac{r}{6}\) = 11
Answer:
r = 66,

Explanation:
Given to solve \(\frac{r}{6}\) = 11,
multiplying both sides by 6 as \(\frac{r}{6}\) X 6 = 11 X 6,
r = 66.

Solve each equation using the concept of balancing. Write all fraction answers in simplest form.

Question 13.
\(\frac{5}{6}\) = c + \(\frac{1}{6}\)
Answer:
c = \(\frac{4}{6}\) or \(\frac{2}{3}\),

Explanation:
Given to solve \(\frac{5}{6}\) = c + \(\frac{1}{6}\) so subtracting both sides by \(\frac{1}{6}\) we get \(\frac{5}{6}\) – \(\frac{1}{6}\) = c + \(\frac{1}{6}\) – \(\frac{1}{6}\) as we have common denominators we get \(\frac{5 – 1}{6}\) = c,
c = \(\frac{4}{6}\) we further divide as both numerator and denominators goes by 2 we get c = \(\frac{2 X 2}{3 X 2}\), c = \(\frac{2}{3}\).

Question 14.
h + \(\frac{5}{14}\) = \(\frac{11}{14}\)
Answer:
h = \(\frac{6}{14}\) or \(\frac{3}{7}\),

Explanation:
Given to solve h + \(\frac{5}{14}\) = \(\frac{11}{14}\) so subtracting both sides by \(\frac{5}{14}\) we get h + \(\frac{5}{14}\) – \(\frac{5}{14}\) = \(\frac{11}{14}\) – \(\frac{5}{14}\) as we have common denominators we get h = \(\frac{11 – 5}{6}\),
h = \(\frac{6}{14}\) we further divide as both numerator and denominators goes by 2 we get h = \(\frac{3 X 2}{7 X 2}\), h = \(\frac{3}{7}\).

Question 15.
q – \(\frac{3}{10}\) = \(\frac{7}{10}\)
Answer:
q = 1,

Explanation:
Given to solve q – \(\frac{3}{10}\) = \(\frac{7}{10}\) so adding both sides by \(\frac{3}{10}\) we get q – \(\frac{3}{10}\) + \(\frac{3}{10}\) = \(\frac{7}{10}\) + \(\frac{3}{10}\) as we have common denominators we get q = \(\frac{7 + 3}{10}\),
q = \(\frac{10}{10}\) we further divide as both numerator and denominators goes by 10 we get q = \(\frac{1 X 10}{1 X 10}\), q = 1.

Question 16.
7k = \(\frac{4}{7}\)
Answer:
k = \(\frac{4}{49}\),

Explanation:
Given to solve 7k = \(\frac{4}{7}\) dividing both sides by \(\frac{1}{7}\) we get 7k X \(\frac{1}{7}\) = \(\frac{4 X 1}{7 X 7}\), k = \(\frac{4}{49}\).

Question 17.
\(\frac{5}{12}\) = 5d
Answer:
d = \(\frac{1}{12}\),

Explanation:
Given to solve \(\frac{5}{12}\) = 5d so dividing both sides by \(\frac{1}{5}\) we get \(\frac{5}{12}\) X \(\frac{1}{5}\) = 5d X \(\frac{1}{5}\), d = \(\frac{1}{12}\).

Question 18.
\(\frac{1}{2}\)x = \(\frac{1}{4}\)
Answer:
x = \(\frac{1}{2}\),

Explanation:
Given to solve \(\frac{1}{2}\)x = \(\frac{1}{4}\) so multiplying both sides by 2 we get x X \(\frac{2}{2}\) = \(\frac{2}{4}\) we get x = \(\frac{2}{4}\) we further divide as both numerator and denominators goes by 2 we get x = \(\frac{1 X 2}{2 X 2}\), x = \(\frac{1}{2}\).

Question 19.
\(\frac{8}{9}\) = \(\frac{1}{3}\)f
Answer:
f = \(\frac{8}{3}\),

Explanation:
Given to solve \(\frac{8}{9}\) = \(\frac{1}{3}\)f so multiplying both sides by 3 we get \(\frac{8}{9}\) = \(\frac{1}{3}\)f X 3, \(\frac{8 X 3}{3 X 3}\)= f, So f = \(\frac{8}{3}\).

Question 20.
r + 2.1 = 4.7
Answer:
r = 2.6,

Explanation:
Given to solve r + 2.1 = 4.7 subtracting both sides by 2.1 we get r + 2.1 – 2.1 = 4.7 – 2.1, so r = 2.6.

Question 21.
9.9 = x + 5.4
Answer:
x = 4.5,

Explanation:
Given to solve 9.9 = x + 5.4 subtracting both sides by 5.4 we get 9.9 – 5.4 = x – 5.4, so 4.5 = x.

Question 22.
11.2 = f – 1.8
Answer:
f = 13,

Explanation:
Given to solve 11.2 = f – 1.8 adding both sides by 1.8 as 11.2 + 1.8 = f – 1.8 + 1.8 we get 13 = f or f = 13.

Question 23.
j – 3.7 = 20.4
Answer:
j = 24.1,

Explanation:
Given to solve j – 3.7 = 20.4 adding both sides by 3.7 as j – 3.7 + 3.7 = 20.4 + 3.7 we get j = 24.1.

Question 24.
4w = 6.8
Answer:
w = 1.7,

Explanation:
Given to solve 4w = 6.8, dividing both sides by 4 as 4w/4 = 6.8/4 we get w = 1.7.

Question 25.
13.9 = 2.5z
Answer:
z = 5.56,

Explanation:
Given to solve 13.9 = 2.5z, dividing both sides by 2.5 as 13.9/2.5 = 2.5z/2.5 we get 5.56 = z or z = 5.56.

Question 26.
3.2d = 40.8
Answer:
d = 12.75,

Explanation:
Given to solve 3.2d = 40.8 dividing both sides by 3.2 as 3.2d/3.2 = 40.8/3.2 we get d = 12.75.

Question 27.
x + \(\frac{1}{2}\) = 1\(\frac{3}{4}\)
Answer:
x = \(\frac{5}{4}\) or 1\(\frac{1}{4}\),

Explanation:
Given to solve x + \(\frac{1}{2}\) = 1\(\frac{1}{4}\) so subtracting both sides by \(\frac{1}{2}\) we get x = \(\frac{1 X 4 + 3}{4}\) – \(\frac{1}{2}\) we get x = \(\frac{7}{4}\) – \(\frac{1}{2}\) , x = \(\frac{7 –  1 X 2}{4}\) , x = \(\frac{5}{4}\) as numerator is more we write in mixed fraction as x = \(\frac{1 X 4 + 1}{4}\), x = 1\(\frac{1}{4}\).

Question 28.
g + \(\frac{5}{3}\) = 3\(\frac{2}{3}\)
Answer:
g = 2,

Explanation:
Given to solve g + \(\frac{5}{3}\) = 3\(\frac{2}{3}\) so subtracting both sides by \(\frac{5}{3}\) we get g = \(\frac{3 X 3 + 2}{3}\) – \(\frac{5}{3}\) we get g = \(\frac{11}{3}\) – \(\frac{5}{3}\) , g = \(\frac{11 –  5}{3}\) , g = \(\frac{6}{3}\) as numerator and denominator both goes by 3 we get g = \(\frac{2 X 3}{3 X 1}\), g = 2.

Question 29.
2\(\frac{5}{7}\) = p – \(\frac{2}{7}\)
Answer:
p = 3,

Explanation:
Given to solve 2\(\frac{5}{7}\) = p – \(\frac{2}{7}\) so adding both sides by \(\frac{2}{7}\) we get \(\frac{2 X 7 + 5}{7}\) + \(\frac{2}{7}\)= p – \(\frac{2}{7}\) + \(\frac{2}{7}\), we have common denominators so \(\frac{19 + 2}{7}\) = p, \(\frac{21}{7}\) = p as numerator and denominator both goes by 7 we get p = \(\frac{7 X 3}{7 X 1}\), p = 3.

Question 30.
e – \(\frac{18}{11}\) = 1\(\frac{6}{11}\)
Answer:
e = \(\frac{35}{11}\) or 3\(\frac{2}{11}\),

Explanation:
Given to solve e – \(\frac{18}{11}\) = 1\(\frac{6}{11}\) so adding both sides by \(\frac{18}{11}\) we get e = \(\frac{11 X 1 + 6}{11}\) + \(\frac{18}{11}\) we get e = \(\frac{17}{11} \) + \(\frac{18}{11}\) both have common denominators we get e = \(\frac{17 + 18}{11}\) , e = \(\frac{35}{11}\) as numerator is greater than denominator we write in mixed fraction as e = \(\frac{3 X 11 + 2}{11}\), so e = 3\(\frac{2}{11}\).

Question 31.
\(\frac{4}{3}\)y = 36
Answer:
y = 27,

Explanation:
Given to solve \(\frac{4}{3}\)y = 36 multiplying both sides by \(\frac{3}{4}\) we get y = 36 X \(\frac{3}{4}\), y = \(\frac{36 X 3}{4}\) = \(\frac{4 X 9 X 3}{4}\) we get y = 9 X 3 = 27.

Question 32.
\(\frac{9}{10}\) = \(\frac{5}{6}\)v
Answer:
v = \(\frac{54}{50}\) or 1\(\frac{4}{50}\),

Explanation:
Given to solve \(\frac{9}{10}\) = \(\frac{5}{6}\)v multiplying both sides by \(\frac{6}{5}\) we get \(\frac{9}{10}\) X \(\frac{6}{5}\) =v, v = \(\frac{9 X 6}{10 X 5}\) = \(\frac{54}{50}\) as numerator is greater we can further write in mixed fraction as \(\frac{1 X 50 + 4}{50}\) = 1\(\frac{4}{50}\).

Question 33.
\(\frac{2}{3}\)k = 28 ∙ \(\frac{4}{9}\)
Answer:
k = \(\frac{56}{3}\) or 18\(\frac{2}{3}\),

Explanation:
Given to solve \(\frac{2}{3}\)k = 28 . \(\frac{4}{9}\) multiplying both sides by \(\frac{3}{2}\) we get k = \(\frac{28 X 4}{9}\) X \(\frac{3}{2}\), k = \(\frac{28 X 2}{3}\), k = \(\frac{56}{3}\) as numerator is greater we can further write in mixed fraction as \(\frac{18 X 3 + 2}{3}\) = 18\(\frac{2}{3}\).

Solve.

Question 34.
Find five pairs of whole numbers, such that when they are inserted into the equation below, the solution of the equation is 3,
x + =
Answer:
(0,3), (1,4), (2,5), (3,6) and (4,7),

Explanation:
Given to find five pairs of whole numbers such that when they are inserted into the equation x + __ =___  the solution of the equation is 3 so
1. When (0,3) substituting x + 0 = 3 yes true as x = 3,
2. When (1,4) substituting x + 1 = 4 solving we get x = 4 – 1 = 3 which is true,
3. When (2,5) substituting x + 2 = 5 solving we get x = 5 – 2 = 3 which is true,
4. When (3,6) substituting x + 3 = 6 solving we get x = 6 – 3 = 3 which is true,
5. When (4,7) substituting x + 4 = 7 solving we get x = 7 – 4 = 3 which is true therefore 5 pairs of whole numbers are (0,3), (1,4), (2,5), (3,6) and (4,7).

Question 35.
Find five pairs of numbers, such that when they are inserted into the equation below, the solution of the equation is \(\frac{2}{5}\).
x =
Answer:
(\(\frac{2}{5}\), 1), (\(\frac{1}{5}\), 2), (\(\frac{2}{15}\), 3), (\(\frac{2}{20}\),4), (\(\frac{2}{25}\), 5),

Explanation:
Given to find five pairs of numbers such that when they are inserted into the equation ___x = ____ , the solution of the equation is \(\frac{2}{5}\) so
1. When (\(\frac{2}{5}\),1) substituting \(\frac{2}{5}\) X 1 = \(\frac{2}{5}\),
2. When (\(\frac{1}{5}\), 2) substituting \(\frac{1}{5}\) X 2 = \(\frac{2}{5}\),
3. When (\(\frac{2}{15}\), 3) substituting \(\frac{2}{15}\) X 3 = \(\frac{2}{5}\),
4. When (\(\frac{2}{20}\),4) substituting \(\frac{2}{20}\) X 4 = \(\frac{2}{5}\),
5. When (\(\frac{2}{5}\),5) substituting \(\frac{2}{25}\) X 5 = \(\frac{2}{5}\) therefore 5 pairs of numbers are (\(\frac{2}{5}\), 1), (\(\frac{1}{5}\), 2), (\(\frac{2}{15}\), 3), (\(\frac{2}{20}\),4), (\(\frac{2}{25}\), 5).

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 8 Lesson 8.2 Writing Linear Equations to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations

Math in Focus Grade 6 Chapter 8 Lesson 8.2 Guided Practice Answer Key

Complete.

Question 1.
Isaiah has h baseball cards. Miguel has 7 more baseball cards than Isaiah.
a) Write an expression for the number of baseball cards that Miguel has in terms of h.

Miguel has baseball cards.

b) If Miguel has k baseball cards, express k in terms of h.
k = +
c) State the independent and dependent variables.
Independent variable: , dependent variable:
Answer:
a) Miguel has h + 7 baseball cards,
b) k = h + 7,
c) Independent variable: h, dependent variable: k,

Explanation:
Given Isaiah has h baseball cards. Miguel has 7 more baseball cards than Isaiah,
a) Wrote an expression for the number of baseball cards that Miguel has in terms of h as Miguel has h + 7 baseball cards.
b) If Miguel has k baseball cards expressing k in terms of h as k = h + 7.
c) The independent and dependent variables are independent variable is h and dependent variable are k which depends on value of h.

Write an equation for each of the following. Then state the independent and dependent variables for each equation.

Question 2.
Hannah took p minutes to jog around a park. Sofia took 12 minutes longer to jog around the park. If Sofia took t minutes to jog around the park, express t in terms of p.
Answer:
t = p + 12,
Independent  variable : p, Dependent variable: t,

Explanation:
Given Hannah took p minutes to jog around a park. Sofia took 12 minutes longer to jog around the park. If Sofia took t minutes to jog around the park,To jog around the park it is p minutes, and Sofia took 12 more minutes if Sofia took t minutes expressing t in terms of p it is t = p + 12. Here p is independent variable and t is dependent variable on the value of p.

Question 3.
A bouquet of roses costs $30. A bouquet of tulips costs m dollars less. If the cost of one bouquet of tulips is n dollars, express n in terms of m.
Answer:
n = $30 – m,
Independent variable : m, Dependent variable: n,

Explanation:
Given a bouquet of roses costs $30. A bouquet of tulips costs m dollars less. If the cost of one bouquet of tulips is n dollars, expressing n in terms of m as n = $30 – m, Here m is independent variable and n is dependent variable on value of m.

Question 4.
Nathan has 7 boxes of marbles. Each box contains b marbles. If he has c marbles altogether, express c in terms of b.
Answer:
c = 7b,
Independent variable : b, Dependent variable: c,

Explanation:
Given Nathan has 7 boxes of marbles. Each box contains b marbles. If he has c marbles altogether, expressing c in terms of b as c = 7b as Nathan has 7 boxes of b marbles each here b is independent variable and c is dependent variable on the value of b.

Question 5.
A motel charges Mr. Kim x dollars for his stay. Mr. Kim stayed at the motel for 12 nights. If the rate per night for a room is y dollars, express y in terms of x.
Answer:
x = 12y,
Independent variable : b, Dependent variable: c,

Explanation:
Given a motel charges Mr. Kim x dollars for his stay. Mr. Kim stayed at the motel for 12 nights. If the rate per night for a room is y dollars expressing y in terms of x as it is x = 12y where y is independent variable and x is dependent variable on the value of y.

Copy and complete the table. Then use the table to answer the questions.

Question 6.
The width of a rectangular tank is 2 meters less than its length.
a) If the length is p meters and the width is q meters, write an equation relating p and q.

Answer:
Equation: q = p – 2,

Explanation:
Given The width of a rectangular tank is 2 meters less than its length.
a) If the length is p meters and the width is q meters so an equation relating p and q is q = p – 2,
If p =3 then q = 3 – 2 = 1,
If p = 4 then q = 4 – 2 = 2,
If p = 5 then q = 5 – 2 = 3,
If p = 7 then q = 7 – 2 = 5,
If p = 8 then q = 8 – 2 = 6,
Completed the table as shown above.

b) Use the data from a) to plot the points on a coordinate plane. Connect the points with a line.
Answer:

Explanation:
Used the data from a) to plot the points on a coordinate plane as length p meters on x – axis and width q meters on y axis. Connected the points with a line as shown above.

c) The point (5.5, 3.5) is on the line you drew in b). Does this point make sense in the situation?
Answer:
Yes,

Explanation:
The point(5.5,3.5) is on the line which i drew yes this point makes sense as p = 5.5 we have q = p – 2 = 5.5 – 2 = 3.5 which is true.

Copy and complete each table. Then express the relationship between the two variables as an equation.

Question 7.
Paul and Lee went to the library to borrow some books. Paul borrowed 6 more books than Lee.

Answer:
Equation: y = x + 6,

Explanation:
Given Paul and Lee went to the library to borrow some books. Paul borrowed 6 more books than Lee. If x is number of books Lee borrowed and y be books Paul borrowed so the equation is y = x + 6, so
if x = 1 then y = 1 + 6 = 7 books,
if x = 2 then y = 2 + 6 = 8 books,
if x = 3 then y = 3 + 6 = 9 books,
if x = 4 then y = 4 + 6 = 10 books and
if x = 5 then y = 5 + 6 = 11 books shown in the above table.

Question 8.
At a crafts store, Zoey bought some boxes of red beads and some boxes of blue beads. The number of boxes of red beads was 4 times the number of boxes of blue beads.

Answer:
Equation: r = 4b,

Explanation:
Given at a crafts store Zoey bought some boxes of red beads and some boxes of blue beads. The number of boxes of red beads was 4 times the number of boxes of blue beads. Let read beads be r and blue beads be b so the equation is r = 4b,
if b = 2 then r = 4 X 2 = 8 beads,
if b = 3 then r = 4 X 3 = 12 beads,
if b = 4 then r = 4 X 4 = 16 beads,
if b = 5 then r = 4 X 5 = 20 beads shown above in the table.

Use the data in the table to plot points on a coordinate plane. Connect the points to form a line. Then write an equation to show the relationship between the variables.

Question 9.

Answer:

Equation:
d = 50t,

Explanation:
Used the data in the table to plot points on a coordinate plane. Connected the points to form a line as shown above  here d is the distance traveled in miles on y – axis and t is the time taken in t hours on x – axis. As distance traveled is increasing by 50 each hour an equation to show the relationship between the variables is d = 50t.

Math in Focus Course 1B Practice 8.2 Answer Key

Solve.

Question 1.
Joshua is w years old. His brother is 3 years older than he is.
a) If his brother is x years old, express x in terms of w.
Answer:
x = w + 3,

Explanation:
Given Joshua is w years old. His brother is 3 years older than he is means brother is w + 3, So if his brother is x years old expressing x in terms of w as x = w + 3.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : w,
Dependent variable : x,

Explanation:
Here Joshua is w years old his brother is 3 years older than he is and  if his brother is x years old then the independent variable is w and dependent variable is x on the value of w.

Question 2.
Rita has b markers. Sandy has 11 fewer markers than she has.

a) If Sandy has h markers, express h in terms of b.
Answer:
h = b – 11,

Explanation:
As Rita has b markers. Sandy has 11 fewer markers than she has means Sandy has b – 11 markers, If Sandy has h markers expressing h in terms of b as h = b – 11.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : b,
Dependent variable : h,

Explanation:
As Rita has b markers. Sandy has 11 fewer markers than she has means Independent variable is b and dependent variable is h on the value of b.

Question 3.
A small box of cereal weighs k grams. A jumbo box of cereal weighs 5 times as much.

a) If the weight of the jumbo box of cereal is m grams, express m in terms of k.
Answer:

m = 5k,

Explanation:
Given a small box of cereal weighs k grams. A jumbo box of cereal weighs 5 times as much means jumbo box is 5k, If the weight of the jumbo box of cereal is m grams, expressing m in terms of k as m = 5k.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : k,
Dependent variable : m,

Explanation:
As a small box of cereal weighs k grams and a jumbo box of cereal weighs 5 times as much here the independent variable is k and dependent variable is m on the value of k.

Question 4.
The area of Hank’s farm is n acres. The area of Hank’s farm is 8 times as large as the area of Stan’s farm.

a) If s represents the area of Stan’s farm, express s in terms of n.
Answer:
n = 8s,

Explanation:
Given the area of Hank’s farm is n acres. The area of Hank’s farm is 8 times as large as the area of Stan’s farm means Hank’s farm is 8 X Stan’s farm if s represents the area of Stan’s farm expressing s in terms of n is n = 8s.

b) State the independent and dependent variables in the equation.
Answer:
Independent variable : s,
Dependent variable : n,

Explanation:
Given the area of Hank’s farm is n acres. The area of Hank’s farm is 8 times as large as the area of Stan’s farm means Hank’s farm is 8 X Stan’s farm if s represents the area of Stan’s farm expressing s in terms of n here Independent variable is s and dependent variable is n which depends on s.

Question 5.
Ethan scored x points in a game. His younger sister scored 8 points when she played the same game. Their combined score was y points.

a) Write an equation relating x and y.
Answer:
y = x + 8,

Explanation:
Given Ethan scored x points in a game. His younger sister scored 8 points when she played the same game. Their combined score was y points. So an equation relating x and y is y = x + 8.

b) Copy and complete the table to show the relationship between x and y.

Answer:

Explanation:
Completed the table to show the relationship between x and y as y = x + 8,
if x = 10 then y = 10 + 8 = 18,
if x = 11 then y = 11 + 8 = 19,
if x = 12 then y = 12 + 8 = 20,
if x = 13 then y = 13 + 8 = 21,
if x = 14 then y = 14 + 8 = 22 and
if x = 15 then y = 15 + 8 = 23 as shown above.

Question 6.
There are x sparrows in a tree. There are 50 sparrows on the ground beneath the tree. Let y represent the total number of sparrows in the tree and on the ground.
a) Express y in terms of x.
Answer:
y = x + 50,

Explanation:
Given there are x sparrows in a tree. There are 50 sparrows on the ground beneath the tree. Let y represent the total number of sparrows in the tree and on the ground.So expressing y in terms of x is y = x + 50.

b) Make a table to show the relationship between y and x. Use values of x = 10, 20, 30, 40, and 50 in your table.
Answer:

Explanation:
Asked to make a table to show the relationship between y and x. Used values of x = 10, 20, 30, 40, and 50 in my table and completed as shown above in the equation y = x + 50,
When x = 10 then y = 10 + 50 = 60,
when x = 20 then y = 20 + 50 = 70,
when x = 30 then y = 30 + 50 = 80,
when x = 40 then y = 40 + 50 = 90,
when x = 50 then y = 50 + 50 = 100.

c) Graph the relationship between y and x in a coordinate plane.
Answer:

Explanation:
Graphed the relationship between y and x in a coordinate plane as shown above as sparrows x on x -axis and total number of sparrows y on y – axis.

Question 7.
A rectangle has a perimeter of P centimeters. Its width is b centimeters. Its length is double its width.
a) Express P in terms of b.
Answer:
P = 6b,

Explanation:
Given a rectangle has a perimeter of P centimeters. Its width is b centimeters. Its length is double its width.
So expressing P in terms of b as P = 2(2b + b). = 2(3b) = 6b.

b) Copy and complete the table to show the relationship between P and b.

Answer:

Explanation:
Given a rectangle has a perimeter of P centimeters. Its width is b centimeters. Its length is double its width.
So expressed P in terms of b as P = 6b completed the table to show the relationship between P and b above as
if b= 1 then P = 6 X 1 = 6,
if b = 2 then P = 6 X 2 = 12,
if b = 3 then P = 6 X 3 = 18,
if b= 4 then P = 6 X 4 = 24,
if b = 5 then P = 6 X 5 = 30 and
if b = 6 then P = 6 X 6 = 36.

Question 8.
Every month, Amaan spends 60% of what he earns and saves the rest. Amaan earns n dollars and saves r dollars each month.
a) Express r in terms of n.
Answer:

r = n – 0.6n,

Explanation:
Given every month, Amaan spends 60% of what he earns and saves the rest. Amaan earns n dollars and saves r dollars each month. Expressing r in terms of n as r = n – n X 60/100 = n – 0.6n.

b) Make a table to show the relationship between r and n. Use values of n = 100, 200, 400, and 500 in your table.
Answer:

Explanation:
Made a table above to show the relationship between r and n. Using values of n = 100, 200, 400, and 500 in my table and substituting in r = n – 0.6n as
if n = 100 then r = 100 – 0.6 X 100 = 100 – 60 = 40,
if n = 200 then r = 200 – 0.6 X 200 = 200 – 120 = 80,
if n = 300 then r = 300 – 0.6 X 300 = 300 – 180 = 120,
if n = 400 then r = 400 – 0.6 X 400 = 400 – 240 = 160 and
if n = 500 then r = 500 – 0.6 X 500 = 500 – 300 = 200.

c) Graph the relationship between n and r in a coordinate plane.
Answer:

Explanation:
Graphed the relationship between n and r in a coordinate plane as  r = n – 0.6n where earnings n dollars on x – axis and savings r dollars on y – axis as shown above.

d) The point (287.5, 115) is on the line you drew in c). Does this point make sense in the situation? Explain.
Answer:
Yes,

Explanation:
Given the point (287.5, 115) is on the line I drew in c) checking in r = n – 0.6n, if n = 287.5 then r =  287.5 – 0.6 X 287.5 = 287.5 – 172.5 = 115 which matches so this point make sense in the situation.

Question 9.
The side length of a square is t inches. The perimeter of the square is z inches.
a) Express z in terms of t.
Answer:
z = 4t,

Explanation:
Given the side length of a square is t inches. The perimeter of the square is z inches. Expressing z in terms of t as perimeter of sqaure is 4 X side length so it is z = 4t.

b) Make a table to show the relationship between z and t. Use whole number values of t from 1 to 10.
Answer:

Explanation:
Made a table to show the relationship between z and t. Used whole number values of t from 1 to 10 in z – 4t as
if t = 1 then z = 4 X 1 = 4,
if t = 2 then z = 4 X 2 = 8,
if t = 3 then z = 4 X 3 = 12,
if t = 4 then z = 4 X 4 = 16,
if t = 5 then z = 4 X 5 = 20,
if t = 6 then z = 4 X 6 = 24,
if t = 7 then z = 4 X 7 = 28,
if t = 8 then z = 4 X 8 = 32,
if t = 9 then z = 4 X 9 = 36 and
if t = 10 then z = 4 X 10 = 40 shown above.

c) Graph the relationship between z and t in a coordinate plane.
Answer:

Explanation:
Graphed the relationship between z and t in a coordinate plane as the side length of a square is t inches on x – axis and the perimeter of the square is z inches on y – axis as shown above.

d) Use your graph to find the perimeter of the square when the length is 3.5 inches and 7.5 inches.
Answer:
1) 14 inches,
2) 30 inches,

Explanation:
Using my graph the perimeter of the square when the length is 3.5 inches is 4 X 3.5 = 14 inches and when the length is 7.5 inches the perimeter of the square is 4 X 7.5 inches = 30 inches respectively.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.1 Nets of Solids detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.1 Answer Key Nets of Solids

Math in Focus Grade 6 Chapter 12 Lesson 12.1 Guided Practice Answer Key

Hands-On Activity

IDENTIFYING A CUBE FROM A NET

Work in pairs.
Step 1: Trace and cut out each figure.

Step 2: Try and fold them into cubes.

Math Journal Can you fold all the above figures into cubes? Discuss with your partner and explain your thinking.
Answer:
The figure (a) can be folded into cube
The figure (b) can also be folded into cube.
The figure (c) cannot be folded into cube.

Hands-On Activity

IDENTIFYING A PRISM FROM A NET

Work in pairs.
Step 1: Trace and cut out each net along the solid lines, Predict what figure can be formed from the net. Then fold the net to make the figure.

Step 2: Name the solid that each net forms
Answer:
If the figure (a) is folded, it will form a square cube.
If the figure (b) is folded, it will form into a rectangular cuboid.
If the figure (c) is folded, it will form into a triangular prism.
If the figure (d) is folded, it will form into a triangular prism.

Match each solid with its net(s). There may be more than one net of each solid.

Question 1.

Answer:
The first solid can have similar of b,f nets. Hence it can be matched with b,f nets.
The second solid have similar of a,e nets. Hence it can be matched with a,e nets.
The third solid have similar of c,d nets. Hence it can be matched with c,d nets.

Name the solid that each net forms.

Question 2.

Answer:
A cuboid can be formed when the above solid is folded.
Therefore, a cuboid is formed from the given net form.

Question 3.

Answer:
A cube can be formed when the above solid is folded.
Therefore, a cube is formed from the given net form.

Question 4.

Answer:
A triangular prism can be formed when the above solid is folded.
Therefore, a triangular prism is formed from the given net form.

Hand-On Activity

CLASSIFYING PYRAMIDS

Work in pairs
Step 1: Trace, cut out, and fold the nets.

Step 2: Name the solid that each net forms.
Answer:
When the nets in figure(a) is folded, it will form a triangular prism.
When the nets in figure(b) is folded, it will form a square prism.

Match each solid with its net(s). There may be more than one net of each solid.

Question 5.

Answer:
The first solid can have similar of a,c nets. Hence it can be matched with a,c nets.
The second solid can have similar of b nets. Hence it can be matched with b nets.

Math in Focus Course 1B Practice 12.1 Answer Key

Name each solid. In each solid, identify a base and a face that is not a base.

Question 1.

Answer:
The above figure is a pyramid since the sides meet at one point.
There is one base for the given figure, which has four equal sides.A base with four equal sides means the base is a square which is BECD.
A face with three sides means the base is a triangle. There are four faces which are not bases. The triangular faces for the given figure are BAE,ACD,AED and BEA.

Question 2.

Answer:
This figure is a prism since it has two ends that are the same shape.
There are two bases for the given figure, both with three sides. A base with three sides means the base is a triangle. The triangular bases for the given figure will be GFH and JKL.
The faces have four sides or two parallel sides. There are three faces which are not bases. The faces are FJKG, FHLJ and GHLK.

Question 3.

Answer:
The above figure has all equal sides and thus form a cube.
There are two bases for the given figure, both with four equal sides. A base with four equal sides means the base is a square which will be PSRQ and TWVU.
The faces have four equal sides. The square faces will be RSWV, SPTW, PQUT and QRVU.

Question 4.

Answer:
This figure is a prism since it has two ends that are the same shape.
There are two bases for the given figure, both with three sides. A base with three sides means the base is a triangle. The triangular bases for the given figure will be ABC and EFD.
The faces have four sides or two parallel sides. There are three faces which are not bases. The faces are AEDC, AEFB and BFCD.

Name the solid that each net forms.

Question 5.

Answer:
The above given figure has four triangles and a square.
Since all the sides meet a point, it can be a pyramid.
When the above given solid is folded it will form a square pyramid.

Question 6.

Answer:
The above given figure has four rectangles and two squares.
When the above given solid is folded it will form a polyhedron.

Question 7.

Answer:
The above given figure has three rectangles and two triangles.
Since it has two ends that are of the same shape.It can be a prism.
When the above given solid is folded it will form a triangular prism.

Question 8.

Answer:
The above given figure has four triangles.
Since all the sides meet a point, it can be a pyramid.
When the above given solid is folded it will form a triangular pyramid.

Decide if each net will form a cube. Answer Yes or No.

Question 9.

Answer:
No, the above net will not form a cube.
If the above net is folded , it will not form the shape of cube.

Question 10.

Answer:
Yes, the above net can be formed into a cube.
If the above net is folded , it will take the shape of cube.

Question 11.

Answer:
No, the above net will not form a cube.
If the above net is folded , it will not form the shape of cube.

Question 12.

Answer:
Yes, the above net can be formed into a cube.
If the above net is folded , it will take the shape of cube.

Question 13.

Answer:
No, the above net will not form a cube.
If the above net is folded , it will not form the shape of cube.

Question 14.

Answer:
Yes, the above net can be formed into a cube.
If the above net is folded , it will take the shape of cube.

Solve. Use graph paper.

Question 15.
In Exercises 9 to 14, you identified some possible nets for a cube. There are other possible nets. Find all of the other possible nets.
Answer:
Each net has 6 squares that when folded properly form the six faces of a cube.

Decide if each net will form a prism. Answer Yes or No.
Question 16.

Answer:
Since it has two ends that are of the same shape.It can be a prism.
Yes, the above given net can form a prism.

Question 17.

Answer:
Since it has two ends that are of the same shape.It can be a prism.
Yes, the above given net can form a prism.

Question 18.

Answer:
No, the above given net can form a prism.

Copy the net of the rectangular prism shown. Then name the vertices that are not already labeled with a letter. Label the vertices.

Question 19.

Answer:

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Surface Area and Volume of Solids detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Answer Key Surface Area and Volume of Solids

Math in Focus Grade 6 Chapter 12 Quick Check Answer Key

Name each, prism. In each prism identify a base, a face, an edge, and a vertex.

Question 1.

Answer:
It is a Rectangular prism.
It has 2 bases, 6 faces, 12 edges and 8 vertices.

Question 2.

Answer:
It is a triangular prism.
It has 2 bases, 5 faces, 9 edges and 6 vertices.

Question 3.

Answer:
It is a square prism.
It has 2 bases, 6 faces, 12 edges and 8 vertices.

Find the area of each figure.

Question 4.

Area = •
= cm²
Answer:
The given rectangle is 4cm wide and 9cm long.
The formula for finding the area of rectangle is length×breadth.
The area for the given rectangle is 4×9=36 square centimetre.

Question 5.

Area = \(\frac{1}{2}\) • •
= m²

Answer:
The given trinagle measures 12m wide and 10m long.
The formula to find the area of the given triangle is \(\frac{1}{2}\)×base×height
Therefore, the area of the given triangle will be \(\frac{1}{2}\)×12×10 = 60 square metre.
Question 6.

Area = \(\frac{1}{2}\) • • ( + )
= \(\frac{1}{2}\) • •
= ft²
Answer:
The given quadrilateral has two sides of 10 ft and 7 ft with 5ft base.
Area of the trapezium = \(\frac{1}{2}\) × height × (sum of parallel sides)
The area will be \(\frac{1}{2}\)×5×(10+7) = \(\frac{1}{2}\)×5×12
Therefore, the area will be 5×6=30 square ft.

Question 7.

Area = \(\frac{1}{2}\) • • ( + )
= \(\frac{1}{2}\) • •
= in.²
Answer:
The formula to find the area of trapezium = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area = \(\frac{1}{2}\) × 6.5 × (10+6)
= \(\frac{1}{2}\) × 6.5 × 16
= 6.5×8
= 52 square in.

Find the volume of each solid.

Question 8.

Volume = • •
= in.³
Answer:
Volume of the cube = length×width×height
The given cube measures 8 inches in length.
The volume of the given cube wil be 8×8×8=64 cubic.inches

Question 9.

Volume = • •
= in.³
Answer:
The given cuboid has length of 11 in, width of 5 in and height of 6 in.
Volume of the cuboid = length×width×height
The volume of the given cuboid is 11×5×6=330 cubic. inches.

Question 10.

Volume = • •
= ft³
Answer:
The given cube measures 12 inches in length.
Volume of the cube = length×width×height
The volume of the given cube wil be 12×12×12=64 cubic.inches

Question 11.

Volume = • •
= ft³
Answer:
The given cuboid measures 22ft, 13ft and 16ft.
Volume of the cuboid = length×width×height
The volume of the given cuboid is 22×13×16=4576 cubic. inches.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Review Test Answer Key

Concepts and Skills

Match each of the solid figures to its net.

Answer:
The net form of Figure 1 is Figure c.
The net form of Figure 2 is Figure a.
The net form of Figure 3 is Figure b.
The net form of Figure 4 is Figure e.
The net form of Figure 5 is Figure d.

Find the surface area of each solid.

Question 6.

Answer:
Given that the length of the cube is 8 in.
The given figure has 6 faces.
Area of 1 face will be 8×8 = 64 sq.in
Surface area of the given solid = 6×base area
The surface area of the given solid will be 6×64 = 384 sq.in

Question 7.

Answer:
Given solid has four triangular faces and one square base.
Area of base will be 8×8 = 64 sq.m
Area of triangle = \(\frac{1}{2}\) × base × width
Area of triangular face will be \(\frac{1}{2}\) × 8 × 10 = 40 sq.m
There are 4 such triangular faces, 4×40 = 160 sq.m
The surface area of the given figure will be 64+160 = 224 sq.m

Find the volume of each prism.

Question 8.

Answer:
Given cube has length of 7 cm.
Area of base = 7×7 = 49 sq.cm
Volume of the prism = base area × height
Volume of the prism will be 49×7 = 343 cu.cm

Question 9.

Answer:
Area of triangular base = \(\frac{1}{2}\) ×base×height
Area of the base will be \(\frac{1}{2}\) ×6×5 = 15 sq.ft
Volume of the given prism will be 3×15 = 45 cu.ft

Solve.

Question 10.
The solid below is made up of cubes, each of which has an edge length of 3 inches.
a) What is the volume of one cube?
Answer:
Given that the length of the cube is 3 inches.
Area of one cube will be = 3×3 = 9 sq.in
Volume of the cube= base area × height
Volume of one cube will be = 9×3 = 27 cu.in

b) What is the volume of the solid figure?

Answer:
The below solid figure has 10 identical cubes.
Therefore, the volume of the solid figure will be 10×27 = 270 cu.in

Problem Solving

Solve.

Question 11.
A fish tank is 50 centimeters long, 30 centimeters wide, and 40 centimeters high. It contains water up to a height of 28 centimeters. How many more cubic centimeters of water are needed to fill the tank to a height of 35 centimeters?
Answer:
Given that a fish tank is 50 centimeters long, 30 centimeters wide, and 40 centimeters high.
Volume of the tank = length × width × height
Volume of water filled up to a height of 28 centimeters is 50×30×28 = 42000 cu.cm
Volume of water to fill up to 35 cm will be 50×30×35 = 52500 cu.cm
More cubic centimeters of water are needed to fill the tank to a height of 35 centimeters will be 52500-42000 = 10500 cu.cm

Question 12.
Find the surface area of a square pyramid given that its base area is 196 square inches and the height of each of its triangular faces is 16 inches.

Answer:
Given that the base area of the given square pyramid is 196 sq.in.
Area of square = 196
196 can also be written as 14 × 14
length × length = 196
Length of one side of the square will be 14 cm.
Area of trinagular face will be \(\frac{1}{2}\) × 14 × 16
= 7 × 16
= 112 cu.in
Surface area of the given solid is 4×112 + 196
= 448 + 196
= 644 sq.in

Question 13.
The volume of a rectangular prism is 441 cubic feet. It has a square base with edges that are 7 feet long.
a) Find the height of the prism.
Answer:
Given that the volume of a rectangular prism is 441 cubic feet and it has a square base with edges that are 7 feet long.
Volume of the prism = Area of the base × height
Area of the base = 7×7 = 49 sq.ft
441 = 49 × height
height = 441÷49
height = 9 ft.
Therefore, the height of the prism is 9 ft.

b) Find the surface area of the prism.

Answer:
Area of the square base = length×length = 49 sq.ft
Area of the rectangular base = length×width= 7×9 = 63 sq.ft
There are 2 square bases and 4 rectangular faces.
Therefore, the surface area of the prism will be 2×49 + 4×63
= 98 + 252
= 350 sq.ft

Question 14.
The volume of a rectangular tank with a square base is 63,908 cubic centimeters. Its height is 64 centimeters. Find the length of an edge of one of the square bases. Round your answer to the nearest tenth of a centimeter.

Answer:
Given that the volume of a rectangular tank with a square base is 63,908 cubic centimeters and it is 64 cm high.
Volume of a rectangular prism = Area of base × height
63,908 = Area of base × 64
Area of base will be 63908 ÷ 64 = 998.56 sq.cm
Area of the base when rounded to nearest ten will be 1000.
Area of base = length × length
1000 = length × length
10 × 10 = length × length
Length of an edge of one of the square bases will be 10 cm.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.4 Real-World Problem: Surface Area and Volume detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.4 Answer Key Real-World Problem: Surface Area and Volume

Math in Focus Grade 6 Chapter 12 Lesson 12.4 Guided Practice Answer Key

Complete.

Question 1.
Find the volume of water needed to fill three fourths of the aquarium.

Height of water needed = \(\frac{3}{4}\) •
= in.
Volume of water needed = • •
= in.³
The aquarium needs to have cubic inches of water added to it to be \(\frac{3}{4}\) full.
Answer:
Volume = length × width × height
Volume of water needed = 25×12×14
= 300×14
= 4200 cu.in
Height of water needed = \(\frac{3}{4}\) × 4200
= 3 × 150
= 3150 cu.in
The aquarium needs to have 3150 cubic inches of water added to it to be \(\frac{3}{4}\) full.

Question 2.
A metal bar has bases that are parallelograms.

a) Find the volume of the metal bar.
Area of parallelogram
= base of parallelogram • height of parallelogram
= •
= cm²
Volume of metal bar = base of prism • height of prism
= •
= cm³
The volume of the metal bar is cubic centimeters.
Answer:
Area of parallelogram
= base of parallelogram × height of parallelogram
= 8 × 4
= 32 cm²
Volume of metal bar = base of prism × height of prism
= 32 × 24
= 768  cm³
The volume of the metal bar is 480 cubic centimeters.

b) Find the surface area of the metal bar.
Surface area of metal bar
= perimeter of base • height + total area of 2 bases

The surface area of the metal bar is square centimeters.
Answer:
Given parallelogram is 5cm high, 24cm long and 8cm wide.
The perpendicular height of the base is 4cm.
Surface area of metal bar
= perimeter of base • height + total area of 2 bases
= (24+8+8+24) × 5 + (20+20)
= (64×5) + 40
= 320 + 40
= 360 sq.cm

Question 3.
A candle is a square prism. The candle is 15 centimeters high, and its volume is 960 cubic centimeters.

a) Find the length of each side of the square base.
V = Bh
= B •
÷ = • B ÷
= B
Length of each side of base
=
= cm
The length of each side of the square base is centimeters.
Answer:
Given that the square prism is 15cm high and its volume is 960 cubic centimeters.
V = Bh
960 = B × 15
960 ÷ 15 = 15 × B ÷ 15
64 = B
Base is side × side = 8 × 8
The length of each side of the square base is 8 centimeters.

b) Find the surface area of the candle.
Surface area of candle
= perimeter of base • height + area of two bases

The surface area of the candle is square centimeters.
Answer:
= perimeter of base × height + area of two bases
= (8 + 8 +8 + 8) × 15 + (2×64)
= 32 × 15 + 128
= 480 + 128
= 608 sq.cm

Question 4.
A storage chest is a prism with bases that are pentagons. The diagram shows some of the dimensions of the storage chest. The volume of the storage chest is 855 cubic inches.

a) Find the height AS of the prism. Round your answer to the nearest hundredth.
Area of pentagonal base = area of trapezoid + area of rectangle
= \(\frac{1}{2}\) • • ( + ) + •
= +
= in.²

Answer:
Area of pentagonal base = area of trapezoid + area of rectangle
= \(\frac{1}{2}\) × 3× (3+7) + 7× 9
= 3×5 + 63
= 15+63
= 78 sq.in

V = Bh
= • h
÷ = • h ÷
= ≈ h
The height of the prism is approximately inches.
Answer:
855 = 78×h
855 ÷ 78 = 78 × h ÷ 78
10.96 = h
Height of the prism will be 10.96 in.

b) Find the surface area of the prism. Round your answer to the nearest hundredth.
Surface area of prism
= perimeter of base • height + area of two bases

The surface area of the prism is approximately square inches.
Answer:
Area of two bases = 78 + 78 = 156 sq.in
= perimeter of base • height + area of two bases
= (5+3+3+9+7)×10.96 + 156
= 27×10.96 + 156
= 295.92 + 156
= 451.92 sq.in
The surface area of the prism is approximately 500 sq.inf

Math in Focus Course 1B Practice 12.4 Answer Key

Solve.

Question 1.
Savannah has a water bottle that is a rectangular prism. The bottle measures 7 centimeters by 5 centimeters by 18 centimeters and she filled it completely with water. Then, she drank \(\frac{1}{4}\) of the volume of water in her water bottle. How many cubic centimeters of water were left in the water bottle?

Answer:
Given that the bottle measures 7 centimeters by 5 centimeters by 18 centimeters and she filled it completely with water.
Therefore, the volume of the rectangular prism = length×width×height
= 7×5×18
= 630 cu.cm
Volume of water in the bottle is 630 cu.cm
Out of which she drank \(\frac{1}{4}\), so the left over water will be \(\frac{1}{4}\) of total volume
= \(\frac{1}{4}\) × 630
= 157.5 cu.cm
Therefore, 157.5 cu.cm of water will be left in the water bottle.

Question 2.
A rectangular prism has a square base with edges measuring 8 inches each. Its volume is 768 cubic inches.
a) Find the height of the prism.
Answer:
Given that a rectangular prism has a square base with edges measuring 8 inches each and its volume is 768 cubic inches.
Volume of rectangular prism = 768
Area of base × height = 768
8×8× height = 768
height = 768÷64
Height of the prism is 12 in

b) Find the surface area of the prism.
Answer:
Surface area of the base = 8×8 = 64sq.in
Total area of 2 bases = 2×64 = 128 sq.in
Surface area of the prism = perimeter of base × height + total area of 2 bases
= 4×8×12 + 128
= 144 +128
= 272 sq.in

Question 3.
A triangular prism has the measurements shown.
a) Find the volume of the prism.
Answer:
Area of triangular base = \(\frac{1}{2}\)×base×height
Area of triangular base = \(\frac{1}{2}\)×19.6×5
= 49 sq.ft
`Volume of triangular prism = 49×16 = 784 cu.ft

b) Find the surface area of the prism.

Answer:
Area of triangular base = \(\frac{1}{2}\)×base×height
Area of triangular base = \(\frac{1}{2}\)×19.6×5
= 49 sq.ft
Area of two triangular faces will be 2×49 = 98 sq.ft
Area of rectangular face = length×width
Area of first rectangular face = 16×10 = 160 sq.ft
Area of second rectangular face = 14×16 = 192 sq.ft
Area of third rectangular face = 16×19.6 = 313.6 sq.ft
Surface area of prism = 98+160+192+313.6
= 763.6 sq.ft

Question 4.
The volume of Box A is \(\frac{2}{5}\) the volume of Box B. What is the height of Box A if it has a base area of 32 square centimeters?

Answer:
Box B:
Box B is 16cm wide, 8cm long and 10cm high.
Volume of Box B will be = 16×8×10 = 1280 cu.cm
Given that volume of Box A is \(\frac{2}{5}\) the volume of Box B.
Therefore, volume of Box A = \(\frac{2}{5}\)×1280
= 2×256
= 512
Volume of Box A = Area of Base × height
512 = 32 × height
Height = 16 cm.
Therefore height of Box A will be 16cm.

Question 5.
The ratio of the length to the width to the height of an open rectangular tank is 10 : 5 : 8. The height of the tank is 18 feet longer than the width.
a) Find the volume of the tank.
Answer:
Let the width of the tank be ‘a’ ft.
Height of the tank is 18 feet longer than the width, therefore, it will be 18+a
If height equals to 8 parts, width equals 5 parts, then 8 – 5 parts = 18 ft
3 parts = 18 ft.
1 part = 6ft
Therefore, now the length will be 10×6=60 ft, width will be 5×6=30ft and height will be 8×6=48ft.
Volume of the tank will be 60×30×48=86400 cu.ft

b) Find the surface area of the open tank.
Answer:
Since the rectangular tank resembles rectangular prism.
Surface area of rectangular prism = 2×(length×width + width×height + height×length)
The surface area of the open tank = 2×(60×30 + 30×48 + 48×60)
= 12240 sq.ft

Question 6.
Janice is making a gift box. The gift box is a prism with bases that are regular hexagons, and has the dimensions shown in the diagram.

a) Find the height PQ of the prism.
Answer:
Given that the volume of prism is 2835 cu.cm
Volume of prism = Area of hexagonal base × height
6 equilateral triangles form a regular hexagon.
Therefore, Area of hexagonal base = 6×(\(\frac{1}{2}\)×7×6)
Area of hexagonal base = 6×(\(\frac{1}{2}\)×7×6)
= 6×(7×3)
= 6×21
= 126 sq.cm
2835 = 126 × height
Height = 2835÷126
height = 22.5cm
The height PQ of the prism will be 22.5cm

b) Find the surface area of the prism.
Answer:
Surface area of prism will be = Perimeter × height + Area of base
= 7×6 ×22.5 + 126 + 126
= 42×22.5 + 252
= 945 + 252
= 1197 sq.cm
The surface area of the prism will be 1197 sq.cm

Question 7.
Container A was filled with water to the brim. Then, some of the water was poured into an empty Container B until the height of the water in both containers became the same. Find tT.

Answer:
Container A:
Container A is a rectangular prism of 40cm high, 25cm long and 30 cm wide.
Volume of the rectangular prism is = length × width × height
Volume of water in the container will be 40×25×30 = 30000 cu.cm
The volumes of two containers will be same.
Container B:
Therefore the volume of Container B will be 30000 cu.cm
Given that the container B is 18cm long and 25cm wide.
Volume of the rectangular prism is = length × width × height
30000 = length × width × height
30000 = 18 × 25× height
Height = 30000÷450
Height = 66.6cm.
The new height of the water in both containers will be 66.6cm

Brain @ Work

Question 1.
The volume of a cube is 1000 cubic inches. If each of the edges is doubled in length, what will be the volume of the cube?
Answer:
Given that the volume of a cube is 1000 cubic inches.
Volume of a cube = Area of base × height
1000 =length × width × height
1000 = 10 × 10 × 10
Therefore, we can say length of the cube will be 10 in.
If the length of edge is doubled, then it will be 2×10 = 20 in.
Volume of the new cube will be 20×20×20 = 8000 cu.in

Question 2.
The volume of a cube is x cubic feet and its surface area is x square feet, where x represents the same number. Find the length of each edge of the cube.
Answer:
Given that the volume of a cube is x cubic feet and its surface area is x square feet, where x represents the same number.
Volume of the cube = Area of base × height
Surface area will be 6×Area of base
Given surface area is x sq.ft
x = 6 × Area of base
Area of base = \(\frac{x}{6}\)
Volume of the cube = \(\frac{x}{6}\) × height
x = \(\frac{x}{6}\) × height
Height = \(\frac{6x}{x}\)
Height will be 6 ft.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.3 Volume of Prisms detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.3 Answer Key Volume of Prisms

Math in Focus Grade 6 Chapter 12 Lesson 12.3 Guided Practice Answer Key

Find the volume of each rectangular prism.

Question 1.

Length = 5\(\frac{1}{4}\) in.
Width = 6 in.
Height = 12 in.
V = lwh
= • •
= in.³
Answer:
V = lwh
Length = 5\(\frac{1}{4}\) in.
(5×\(\frac{4}{4}\)+\(\frac{1}{4}\)) = \(\frac{21}{4}\)
V = \(\frac{21}{4}\) × 6 × 12 = 378 cu.in

Question 2.

Length = 8 cm
Width = 7.2 cm
Height = 3 cm
V = lwh
= • •
= cm³
Answer:
V = 8 × 7.2 × 3
172.8 cu.cm

Question 3.

Length = 4 ft
Width = 3 ft
Height = 8 \(\frac{1}{3}\) ft
Volume =
= • •
= ft³
Answer:
V = lwh
Height = 8 \(\frac{1}{3}\) ft
(8 × \(\frac{3}{3}\))+\(\frac{1}{3}\) = \(\frac{25}{3}\)
Volume = 4×3×\(\frac{25}{3}\) = 100 cu.ft

Tell whether slices parallel to each given slice will form uniform cross sections. If not, explain why not.

Question 4.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is larger than the area’s of the cube’s face.

Question 5.

Answer:
In the above given figure, the slice parallel to each given slice will form uniform cross-sections.

Question 6.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is smaller than the area’s of the rectangular’s face.

Find the volume of each prism.

Question 7.

Length = 6 cm
Width = 5.5 cm
Height = 9 cm
Area of base = •
= cm²
Volume of prism = •
= cm²
The volume of the prism is cubic centimeters.
Answer:
Length = 6 cm
Width = 5.5 cm
Height = 9 cm
Area of base = 6×5.5 = 33 sq.cm
Volume of prism = Base area of prism × height
Volume of prism = 33×9 = 297 cu.cm
The volume of the prism is 297 cubic centimeters.

Question 8.

Base of triangle = 10 in.
Height of triangle = 3\(\frac{1}{2}\) in.
Height of prism = 14 in.
Area of base = \(\frac{1}{2}\) • •
= in.²
Volume of prism = •
= in.³
The volume of the prism is cubic inches.
Answer:
Base of triangle = 10 in.
Height of triangle = 3\(\frac{1}{2}\) in.
(3×\(\frac{2}{2}\))+\(\frac{1}{2}\) = \(\frac{7}{2}\)
Height of prism = 14 in.
Area of triangular base = \(\frac{1}{2}\) × base × height
Area of base = \(\frac{1}{2}\) × 10 × \(\frac{7}{2}\) = \(\frac{35}{2}\) sq.in
Volume of prism = Base area of prism × height
Volume of prism = \(\frac{35}{2}\) × 14 = 245 cu.in

Question 9.

Length of shorter base of trapezoid = 4 ft
Length of longer base of trapezoid = 10 ft
Height of trapezoid = 2 ft
Height of prism = 12 ft
Area of base = \(\frac{1}{2}\) • ( + )
= \(\frac{1}{2}\) • •
= ft²
Volume of prism = •
= ft³
The volume of the prism is cubic feet.
Answer:
The formula to find the base area = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area of base = \(\frac{1}{2}\) × 2 × (4+10)
= \(\frac{1}{2}\) × 2 × 14
= 14 sq.ft
Volume of prism = Base area of prism × height
Volume of prism = 14×12
= 168 cu.ft

Hands-On Activity

DETERMINING THE RELATIONSHIP BETWEEN VOLUME AND SURFACE AREA OF PRISMS

Materials:

• 27 unit cubes.

Work in pairs.
Step 1: Build the cube and the rectangular prism using unit cubes.

Step 2: Find the volume of the cube. Find the volume of the rectangular prism. What can you say about the volumes of the cube and the rectangular prism?
Answer:
Volume of cube = lwh
Volume of the cube = 2×2×2 = 8 cu.cm
Volume of the rectangular prism = 2×4×1 = 8 cu.cm
The volumes of the cube and the rectangular prism are same.

Step 3: Find the surface area of the cube. Draw its net if it helps you. Find the surface area of the rectangular prism. Draw its net if it helps you. What can you say about the surface areas of the cube and the rectangular prism?
Answer:
Area of one square will be 2×2 = 4 sq.cm
Volume of cuboid = 6 × Area of base
There are 6 faces for a cube, therefore the surface area will be 6×4 = 24 sq.cm
The surface areas of the rectangular prism will be 2(2×4 + 2×1 + 1×4) = 28 sq.cm
The surface areas of the cube and the rectangular prism are different.

Step 4: Now build these rectangular prisms using unit cubes.

Step 5: Find the volume of the cube. Find the volume of the rectangular prism. What can you say about their volumes?
Answer:
Volume of the cube = 3×3×3 = 27 cu.cm
Volume of the rectangular prism = 3×9×1 = 27 cu.cm
The volumes of the cube and the rectangular prism are same.

Step 6: Find the surface area of the cube. Find the surface area of the . rectangular prism. Draw their nets if it helps you. What can you say about their surface areas?
Area of one square will be 3×3 = 9 sq.cm
There are 6 faces for a cube, therefore the surface area will be 6×9 = 54 sq.cm
The surface areas of the rectangular prism will be 2(3×9 + 9×1 + 1×3) = 78 sq.cm
The surface areas of the cube and the rectangular prism are different.

Math Journal Based on the activity, what can you conclude about prisms with the same volume? Discuss with your partner and explain your thinking.
Answer:
Two prisms of different measurements might have the same volume, they might not have the same surface area.
For example: A rectangular prism with side lengths of 1 cm, 2 cm, and 2 cm has a volume of 4 cu cm and a surface area of 16 sq cm. A rectangular prism with side lengths of 1 cm, 1 cm, and 4 cm has the same volume but a surface area of 18 sq cm

Math in Focus Course 1B Practice 12.3 Answer Key

Solve.

Question 1.
A cube has edges measuring 9 inches each. Find the volume of the cube.
Answer:
Given that: A cube has edges measuring 9 inches each.
The base area will be equal to 9×9=81 sq.in
Volume of cube = Area of base × height
Volume of the cube will be 81×9=729 cu.in

Question 2.
A cube has edges measuring 6.5 centimeters each. Find the volume of the cube.
Answer:
Given that: A cube has edges measuring 6.5 cm each.
The base area will be equal to 6.5×6.5=42.25 sq.cm
Volume of cube = Area of base × height
Volume of the cube will be 81×9=274.625 cu.cm

Question 3.
A storage container is shaped like a rectangular prism. The container is 20 feet long, 10 feet wide, and 5\(\frac{1}{2}\) feet high. Find the volume of the storage container.
Answer:
Given that: A rectangular prism is 20 feet long, 10 feet wide, and 5\(\frac{1}{2}\) feet high.
Area of base will be 20×10=200 sq.ft
Volume of prism = Area of base × height
Volume of the storage container will be 200×5\(\frac{1}{2}\)
= 200×(5×\(\frac{2}{2}\) + \(\frac{1}{2}\))
= 200×(\(\frac{10}{2}\) + \(\frac{1}{2}\))
= 200×\(\frac{11}{2}\)
= 100×11
= 1100 cu.ft

Question 4.
Find the volume of the peppermint tea box on the right.

Answer:
Given that: The peppermint tea box is 12.6 cm long, 6.7 cm wide, and 7.8 cm high.
Area of  base will be 12.6×6.7 = 84.42 sq.cm
Volume of prism = Area of base × height
Volume of the box will be 7.8×84.42 = 658.476 cu.cm

Question 5.
The solid below is made of idential cubes. Each cube has an edge length of 2 inches. Find the volume of the solid.

Answer:
Given that: Each cube has an edge length of 2 inches.
Let us first find the volume of a single cube
Area of the base of a single cube will be 2×2=4 sq.in
Volume of cube = Area of base × height
Volume of a single cube will be  4×2=8 cu.in
The given solid is made of idential cubes. There are 9 such cubes.
Therefore, the volume of all the 9 identical cubes or the given solid figure will be 9×8=72 cu.in

Find the volume of the triangular prism.

Question 6.

Answer:
Given that: A triangular prism is 15ft long, 10ft wide and 6ft high.
Area of the triangular prism = \(\frac{1}{2}\) × base × height
Area of the triangular prism will be \(\frac{1}{2}\) × 10 × 6
= 10 × 3
= 30 sq.ft
Volume of prism = Area of base × height
Volume of the given triangular prism will be 30×15=450 cu.ft

Question 7.

Answer:
Given that: A triangular prism is 12cm long, 6.7cm wide and 3cm high.
Area of the triangular prism = \(\frac{1}{2}\) × base × height
Area of the triangular prism will be \(\frac{1}{2}\) × 6.7 × 3
= \(\frac{1}{2}\) × 20.1
= 10.05 sq.cm
Volume of prism = Area of base × height
Volume of the given triangular prism will be 10.05×12=120.6 cu.cm

Tell whether slices parallel to each given slice will form uniform cross-sections. If not, explain why not.

Question 8.

Answer:
In the above given figure, the slice parallel to each given slice will form uniform cross-sections.

Question 9.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is smaller than the area’s of the circular face.

Question 10.

Answer:
No, in the above given figure, the slice parallel to each given slice will not form uniform cross-sections.
Because, the area of the cross-section is smaller than the area’s of the rectangular’s face.

Copy the solid. Draw a slice that has the same cross section as the bases In each prism.

Question 11.

Answer:

The bases for the above given figure is square.
Therefore, a slice that has the same cross section is drawn for the given prism.

Question 12.

Answer:

The bases for the above given figure is hexagon.
Therefore, a slice that has the same cross section is drawn for the given prism.

Question 13.

Answer:

The bases for the above given figure is triangle.
Therefore, a slice that has the same cross section is drawn for the given prism.

Solve.

Question 14.
The bases of the prism shown are trapezoids. Find the volume of the prism.

Answer:
Length of shorter base of trapezoid = 2 m
Length of longer base of trapezoid = 6 m
Height of trapezoid = 2 m
Height of prism = 10 m
The area of the given figure = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area of base = \(\frac{1}{2}\) × 2 × (2+6)
= \(\frac{1}{2}\) × 2 × 8
= 8 sq.m
Volume of prism = base area × height
The volume of prism = 10×8
= 80 cu.m

Question 15.
A cube has a volume of 125 cubic inches. Find the length of its edge.
Answer:
Given that: A cube has a volume of 125 cu.in
Let us assume the length of a cube be ‘a’ unit.
As cube has all equal sides, the base area will be a×a = a² sq.units
Now, the volume will be a²×a = a³ cubic units.
Therefore we can say that the volume of a cube is the length of an edge taken to the third power.
a³ = 125 and 125 in cube can be written as 5³
a³ = 5³
a = 5
Therefore, the length of its edge will be 5 inches.

Question 16.
The volume of a triangular prism is 400 cubic centimeters. Two of its dimensions are given in the diagram. Find the height of a triangular base.

Answer:
Given that: The volume of a triangular prism is 400 cubic centimeters.
The three rectangular faces measures 10cm long and 8cm wide.

The area of three rectangular faces will be 3×10×8 = 240 sq.cm.
Inorder to get the areas of the triangular bases, we need to subtract the area of the rectangular faces from the total volume.
Area of two triangular bases will be 400-240 = 160 sq.cm
Area of one triangular base will be 160÷2 = 80 sq.cm
From the given figure, the base length of the triangle is 10 cm and the height is ‘h’ cm.
Area of triangular base = \(\frac{1}{2}\) × base × height
Area of the triangle is \(\frac{1}{2}\)×10×h = 80
5×h = 80
h = 80÷5
h = 16 cm.

Question 17.
A cross-section of the triangular prism shown below is parallel to a base. The area of the cross-section is 24 square feet. The ratio of DM to MA is 3 : 5 and the length of \(\overline{F O}\) is 6 feet. Find the volume of the triangular prism.

Answer:
Given that the area of cross-section is 24 sq.ft and ratio of DM to MA is 3 : 5.
The length of FO is 6 ft.
By observing the FO length and the ratio, we can say that the FO length is twice of the ratio of DM.
Therefore, the length of OC will be twice of the ratio of MA.
Length of OC = 10 ft.
Total length = FO + OC = 6+10 = 16 ft.
Volume of the triangular prism = Area of triangular base × length
Volume of the triangular prism = 24×16
Volume of the triangular prism = 384 cu.ft

Question 18.
The volume of the rectangular prism shown below is 2,880 cubic inches. The cross-section shown is parallel to a base. The area of the cross-section is 180 square inches. The length of \(\overline{A B}\) is x inches, and the length of \(\overline{B C}\) is 4x inches.
a) Find the length of \(\overline{A C}\).
Answer:
Given that: The volume of the rectangular prism is 2,880 cubic inches and the area of the cross-section is 180 square inches.
The length of AB is x inches and BC is 4x inches.
Length of AC will be AB+BC = x+4x = 5x

b) Find the value of x.

Answer:
Volume of the rectangular prism = 2,880 cubic inches
Volume of the rectangular prism = Area of rectangular base × length
2,880 = 180 × 5x
5x = 2880 ÷ 180
5x = 16
x = \(\frac{16}{5}\)
Length of AB will be \(\frac{16}{5}\) in.
Length of BC will be 4×\(\frac{16}{5}\) = \(\frac{64}{5}\) in.
Length of AC will be \(\frac{80}{5}\) = 16 in.

Question 19.
In the diagram of a cube shown below, points A, B, C, and D are vertices. Each of the other points on the cube is a midpoint of one of its sides. Describe a cross-section of the cube that will form each of the following figures.
a) a rectangle
b) an isosceles triangle
c) an equilateral triangle
d) a parallelogram

Answer:
a) a rectangle
By joining the CDLJ points, a rectangular cross section can be formed.
b) an isosceles triangle
By joining the MHL points, an isosceles triangle cross section can be formed.
c) an equilateral triangle
By joining the CDM points,an equilateral triangle cross section can be formed.
d) a parallelogram
By joining the MHGL points, a parallelogram cross section can be formed.

Solve. Use graph paper.

Question 20.
Points A, B, C, and D form a square. The area of the square is 9 square units.
a) Find the side length of square ABCD.
Answer:
Given that the area of the square is 9 square units.
Area of square = length × length
9 = length × length
3 × 3 = length × length
Length of the square will be 3 units.

b) The coordinates of point A are (2, 6). Points B and C are below \(\overline{A D}\). Point B is below point A, and point D is to the right of point A. Plot the points in a coordinate plane. Connect the points in order to draw square ABCD.
Answer:
The coordinates of point A are (2, 6).
Point B and C are below point A, so they will be (0,2) and C will be (0,8).
D will be (8,6)

c) The points E, F, G, and H also form a square that is the same size as square ABCD. Point E is 4 units to the right of point A, and 3 units up. Points F and G are below \(\overline{E H}\). Point F is below point E, and point H is to the right of point E. Plot the points in the coordinate plane. Draw \(\overline{E H}\) and \(\overline{G H}\) with solid lines, and \(\overline{E F}\) and \(\overline{F G}\) with dashed lines.
Answer:
Point E is 4 units to the right of point A, and 3 units up. So, it will be (6,9).
Points F and G are below, so they will be (6,3) and (12,3).
Point H will be (12,9).

d) Draw \(\overline{A E}\), \(\overline{D H}\), and \(\overline{C G}\) with solid lines, and \(\overline{B F}\) with a dashed line, Use the solid and dashed lines to see the figure as a solid. Name the type of prism formed.
Answer:
After joining AE, DH, CG and BF.Below prism will be formed.
Since all the sides are of same length, it will form a cube.

e) If the height of the prism is 7 units, find the volume of the prism.
Answer:
Area of the base 7×7 = 49 sq.units
Volume of the prism = 49×7 = 343 cu.units

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 12 Lesson 12.2 Surface Area of Solids detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.2 Answer Key Surface Area of Solids

Math in Focus Grade 6 Chapter 12 Lesson 12.2 Guided Practice Answer Key

Complete.

Question 1.
A cube has edges measuring 6 centimeters each. Find the surface area of the cube.

Area of one square face = •
= cm²
Surface area of cube = •
= cm²

Answer:
Each side of a square measures 6cm.
Area of a square base = length×length
Area of one square face = 6×6 = 36 sq.cm
Surface area of cube = Area of base × height
Surface area of cube = 36×6 = 216 sq.cm

Question 2.
A rectangular prism measures 7 inches by 5 inches by 10 inches. Find the surface area of the prism.

Total area of two orange and two purple faces = ( + + + ) •
= •
= in²
Area of two green rectangular bases = 2 • •
= in²
Surface area of rectangular prism = +
= in²

Answer:
Total area of two orange and two purple faces = (length + width + length + width)× height
Total area of two orange and two purple faces = (5+7+5+7)×10 = 24×10 = 240 sq.in
Area of rectangular base = length×width
Area of two green rectangular bases = 2×(5×7) = 70 sq.in
Surface area of rectangular prism = 240+70= 310 sq.in

Question 3.
The triangular prism shown has three rectangular faces. Its bases are congruent right triangles. Find the surface area of the triangular prism.

Total area of orange, purple, and yellow rectangles = ( + + ) •
= •
= cm²
Area of two green triangular bases = 2 • (\(\frac{1}{2}\) • • ) = cm²
Surface area of triangular prism = + = cm²
Answer:
Total area of orange, purple, and yellow rectangles = (length of three rectangles) × width
Total area of orange, purple, and yellow rectangles = (5+13+12)×9 = 30×9 = 270 sq.cm
Area of a triangle = \(\frac{1}{2}\) × base × height
Area of two green triangular bases = 2×(\(\frac{1}{2}\)×12×13) = 156 sq.cm
Surface area of triangular prism = 270+156 = 426 sq.cm

Question 4.
Alicia makes a pyramid that has an equilateral triangle as its base. The other three faces are congruent isosceles triangles. She measures the lengths shown on the net of her pyramid. Find the surface area of the pyramid.

Area of yellow triangle = \(\frac{1}{2}\) • • = in.²
Area of three blue triangles = 3 • \(\frac{1}{2}\) • • = in.²
Surface area of triangular pyramid = + = in.²
Answer:
Area of a triangle = \(\frac{1}{2}\) × base × height
Area of yellow triangle = \(\frac{1}{2}\) × 5.2 × 6 = 15.6 sq.in
Area of three blue triangles = 3 × \(\frac{1}{2}\) × 10 × 6 = 90 sq.in
Surface area of triangular pyramid = 15.6 + 90 = 105.6 sq.in

Math in Focus Course 1B Practice 12.2 Answer Key

Solve.

Question 1.
A cube has edges measuring 6 centimeters each. Find the surface area of the cube.
Answer:
Measurement of each side of a cube is 6 cm.
Area of a square base= length×length
Surface area of one sqaure side will be 6×6 = 36 sq.cm
Surface area of a cube = 6×Area of base
A cube has 6 sides. Therefore, the total surface area will be 6×36 = 216 sq.cm
The surface area of cube will be 216 sq.cm.

Question 2.
The edge length of a cube is 3.5 inches. Find the surface area of the cube.
Answer:
Measurement of each side of a cube is 3.5 in.
Area of a square base= length×length
Surface area of one sqaure side will be 3.5×3.5 = 12.25 sq.in
Surface area of a cube = 6×Area of base
A cube has 6 sides. Therefore, the total surface area will be 6×12.25 = 75 sq.in
The surface area of cube will be 75 sq.in.

Question 3.
A closed rectangular tank measures 12 meters by 6 meters by 10 meters. Find the surface area of the tank.
Answer:
Total area of four rectangular faces = (length + width + length + width)× height
Total area of four rectangular faces = (12+6+12+6)×10 = 36×10 = 360 sq.m
Area of rectangle = length × width
Area of two rectangular bases = 2×(6×12) = 144 sq.m
Surface area of rectangular prism = 360+144= 504 sq.m

Question 4.
A closed rectangular tank has a length of 8.5 feet, a width of 3.2 feet, and a height of 4.8 feet. Find the surface area of the tank.
Answer:
Total area of four rectangular faces = (length + width + length + width)× height
Total area of four rectangular faces = (8.5+3.2+8.5+3.2)×4.8 = 23.4×4.8 = 112.32 sq.feet
Area of rectangle = length × width
Area of two rectangular bases = 2×(8.5×3.2) = 54.4 sq.feet
Surface area of rectangular prism = 112.32+54.4= 166.72 sq.feet

Question 5.
A triangular prism with its measurements is shown. Find the surface area of the prism.

Answer:
Area of rectangle = length × width
Two rectangle measures 13 cm wide and 15 cm long. So, there area will be 2×13×15 = 390 sq.cm
Area of the third rectangle will be 15×10 = 150 sq.cm
Total area of three rectangular faces = 390+150 = 540 sq.cm
Area of triangle = \(\frac{1}{2}\)×base×height
Area of two triangular bases = 2×\(\frac{1}{2}\)×10×12 = 120 sq.cm
Surface area of triangular prism = 540+120 = 660 sq.cm

Question 6.
A triangular prism with its measurements is shown. Find the surface area of the prism.

Answer:
Area of rectangle = length × width
Area of three rectangular faces = 3×6×4.5 = 81 sq.in
Area of triangle = \(\frac{1}{2}\)×base×height
Area of two triangular bases = 2×\(\frac{1}{2}\)×6×5.2 = 31.2 sq.in
Surface area of triangular prism = 81+31.2 = 112.2 sq.in

Solve.

Question 7.
A square pyramid has four faces that are congruent isosceles triangles. Find the surface area of the square pyramid if the base area is 169 square centimeters.

Answer:
The base area of the given figure is 169 sq.cm.
Since it is a square pyramid, the area will be side×side = 169 sq.cm
169 can also be written as 13 × 13
Therefore,  side × side = 13 × 13
Each side will be 13 cm.
Area of triangle = \(\frac{1}{2}\)×base×height
Area of four isosceles triangles faces = 4×\(\frac{1}{2}\)×13×15 = 390 sq.cm
Total surface area will be 169+390 = 559 sq.cm

Question 8.
The faces of this solid consist of four identical trapezoids and two squares. The side lengths of the two squares are 4 centimeters and 8 centimeters. The height of each trapezoid is 12 centimeters. Find the surface area of the solid.

Answer:
Area of square = length×length
Area of first square = 4×4 = 16 sq.cm
Area of second square will be = 8×8 = 64 sq.cm
Total area of two squares base = 16+64 = 80 sq.cm
The formula to find the area of trapezium = \(\frac{1}{2}\) × height × (sum of parallel sides)
Area of four identical trapezoids = 4×\(\frac{1}{2}\) × (4+8) × 12
= 2 × 12 × 12
= 288 sq.cm
The surface area of the solid will be 80+288 = 368 sq.cm

Surface Area of a Trapezoidal Prism = h (b + d) + l (a + b + c + d) square units

Question 9.
Ms. Jones wants to paint the walls of a rectangular room. The height of the room is 8 feet. The floor is 10.5 feet wide and 12 feet long. The doors and windows total 24 square feet and are not going to be painted. Find the total area of the walls that need to be painted.
Answer:
The height of the room is 8 feet and the floor is 10.5 feet wide and 12 feet long.
Surface area of rectangular prism = 2×(length×width + width×height + height×length)
Area of the room will be = 2×(8×10.5 + 8×12 + 12×10.5)
168+192+252 = 612 sq.feet
The doors and windows total 24 square feet and are not going to be painted.As the doors and windows need not be painted, we can remove this area from the total area.
Therefore, the total area of the walls that need to be painted will be 612-24 = 588 sq.feet

Question 10.
The base of a prism has n sides. Write an expression for each of the following.
a) the number of vertices
Answer:
The number of vertices will be 2n.

b) the number of edges
Answer:
The number of edges will be 3n.

c) the number of faces
Answer:
The number of faces will be n+2

Question 11.
The base of a pyramid has m sides. Write an expression for each of the following.
a) the number of vertices
Answer:
The number of vertices will be m+1

b) the number of edges
Answer:
The number of edges will be 2m

c) the number of faces
Answer:
The number of faces will be m+1

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 13 Review Test detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 13 Review Test Answer Key

Concepts and Skills

Copy and complete the table. Use the set of data.

Question 1.
The number of bedrooms in the units of a new apartment building ranged from 1 to 5. The number of bedrooms in each unit is as follows:

Tabulate the data.

Answer:

Explanation:
A Dataset is a set or collection of data.
This set is normally presented in a tabular pattern.
The number of bedrooms in the units of a new apartment building are marked in tally format.
The number of bedrooms in each unit is value noted in frequency in the above table.

Draw a dot plot and a histogram for the set of data. Include a title.

Question 2.
The table below shows the number of hours 30 teachers in a school spent correcting students assignments.

Answer:
Dot plot for the given above table data,

The histogram shows the ages of runners in a marathon. Use the histogram to answer questions 3 and 4.

Explanation:
Given the table below which shows the number of hours 30 teachers in a school spent correcting students assignments.

A dot plot is a graphical display of data using dots as shown below,

The histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.
The below histogram shows the ages of runners in a marathon.

Question 3.
How many data values are there?
Answer:
25

Explanation:
Given the table below which shows the number of hours 30 teachers in a school spent correcting students assignments.

0 to 9 = 1
10 to 19 = 7
20 to 29 = 11
30 to 39 = 5
40 to 49 = 0
50 to 59 = 1
1 + 7 + 11 + 5 + 0 + 1 = 25

Question 4.
Briefly describe the distribution including any outliers in the data.
Answer:
Yes, 11 is outlier.

Explanation:
Given the table below which shows the number of hours 30 teachers in a school spent correcting students assignments.

1, 1, 5, 11
average of 1,1 is 1
average of 5, 11 is 8
range of 1, 8 is 7
8 + 1.5(7) = 18.5
1 + 1.5(7) = 11.5
As, 11 is less then 11.5
So, 11 is outlier.

Outliers should be investigated carefully. Often they contain valuable information about the process under investigation or the data gathering and recording process.

Problem Solving

The data show the lengths (in inches) of 50 trout caught in a lake during a fishing competition. Use the data to answer questions 5 and 6.

Question 5.
Group the data into suitable intervals and tabulate them. Explain your choice of interval.
Answer:

Explanation:
The above group of data is tabulated with an intervals of 2 inches difference, total 50 fishes are  tabulated in the table as shown in the above figure.
The range is 18 – 8 = 10
Choosing an interval of 2 gives six intervals for the data set.
A larger interval will give too few intervals, and will not be able to see the distribution accurately.

Question 6.
Draw a histogram using the interval. Briefly describe the data.
Answer:

Explanation:
During a fishing competition different trots are caught in the lake.
The lengths (in inches) of 50 trout are tabulated in the table with different length intervals.
The range is 18 – 8 = 10
Choosing an interval of 2 gives six intervals for the data set.
A larger interval will give too few intervals, and will not be able to see the distribution accurately.

The data show the distances a golfer hits (in yards) in a long drive championship. Use the data to answer questions 7 and 8.

Question 7.
Group the data into suitable intervals and tabulate them. Explain your choice of interval.
Answer:

Explanation:
The above data show the distances a golfer hits (in yards) in a long drive championship are tabulated with an interval of 10 yards interval.
The range is 278 -240 = 33
Choosing an interval of 10 gives four intervals for the data set.
A larger interval will give too few intervals, and will not be able to see the distribution accurately.

Question 8.
Draw a histogram using the interval. Briefly describe the data.
Answer:

Explanation:
The above data show the distances of a golfer hits (in yards) in a long drive championship are tabulated with an interval of 10 yards interval.
In horizontal axis line distance and on vertical axis line number of hits are recorded.

The table shows the number of cars passing a traffic light during peak hours on a Friday morning. Use the data to answer questions 9 to 11.

Question 9.
How many cars were observed altogether?
Answer:
Total 240 cars passing a traffic light during peak hours on a Friday morning.

Explanation:
At different time intervals cars passing a traffic light during peak hours on a Friday morning are tabulated in the table. The sum of all cars are as given below.
22 + 45 + 64 + 57 + 27 + 25 = 240
Total 240 cars passing a traffic light during peak hours on a Friday morning

Question 10.
Draw a histogram to display the data.
Answer:

Explanation:
The above histogram shows time on horizontal line and number of cars on the vertical line.
At different time intervals cars passing a traffic light during peak hours on a Friday morning are tabulated in the table.
Total 240 cars passing a traffic light during peak hours on a Friday morning are shown in the above histogram diagram.

Question 11.
Describe the distribution of the data. Suggest why the histogram has the shape that it does.
Answer:
The distribution of a data set is the shape of the graph when all possible values are plotted on a frequency graph showing how often they occur.
Usually, we are not able to collect all the data.
Therefore we take a random sample.
This sample is used to make conclusions about the whole data set.

Explanation:
The above histogram shows time on horizontal line and number of cars on the vertical line.
At different time intervals cars passing a traffic light during peak hours on a Friday morning are tabulated in the table.
Total 240 cars passing a traffic light during peak hours on a Friday morning are shown in the above histogram diagram.

The quiz scores of 94 students are shown in the table. Use the data to answer questions 12 to 14.

Question 12.
Find the value of x.
Answer:
x = 7

Explanation:
The given table shows the quiz scores of 94 students.
To find x, first find the sum of the number of students who scored in quiz.
17 + 3 + 5 + 12 + 15 + 17 + 10 + 8 + x = 94
87 + x = 94
x = 94 – 87
x = 7

Question 13.
Draw a histogram to represent the data. Describe the distribution of the data.
Answer:

Explanation:
The distribution of a data set is the shape of the graph when all possible values are plotted on a frequency graph showing how often they occur in the above.
Usually, we are not able to collect all the data.
Therefore we take a random sample.
This sample is used to make conclusions about the whole data set.

Question 14.
If the 5 students who scored 23-25 all scored 26 instead, would this change where most of the data occur? Justify your answer.
Answer:
The changes of data table and the histogram is plotted for the changed values are shown in the below explanation.

Explanation:

If the 5 students who scored 23-25 all scored 26 instead,
the value 5 in the 23-25 is shifted to 26-28 scores student, as shown below.

This handy Math in Focus Grade 6 Workbook Answer Key Chapter 13 Lesson 13.3 Histograms detailed solutions for the textbook questions.

Math in Focus Grade 6 Course 1 B Chapter 13 Lesson 13.3 Answer Key Histograms

Math in Focus Grade 6 Chapter 13 Lesson 13.3 Guided Practice Answer Key

Draw a histogram for each set of data.

Question 1.
In a study of the length of several individuals of one species of fish caught, the following observations were recorded. The lengths were measured to the nearest centimeter.

Answer:

Explanation:
A histogram is an approximate representation of the distribution of numerical data.
As shown in the above table of data the fishes are of different lengths are graphically represented.
In statistics, a histogram is a graphical representation of the distribution of data.
The histogram is represented by a set of rectangles, adjacent to each other,
where each bar represent a kind of data.

Question 2.
The scores obtained by 40 students in a mathematics quiz are recorded in the table below.

Answer:

Explanation:
We know that in statistics, a histogram is a graphical representation of the distribution of data.
The histogram is represented by a set of rectangles, adjacent to each other,
where each bar represent a kind of data.
A histogram is an approximate representation of the distribution of numerical data.
As shown in the above table of data the scores obtained by 40 students in a mathematics quiz.

Draw a histogram for each set of data. Solve.

Question 3.
The cholesterol levels (in milligram per deciliter) of 40 men are listed below.

a) Group the data into suitable intervals and tabulate them. Explain your choice of intervals.
Answer:

Explanation:
Given that, The cholesterol levels (in milligram per deciliter) of 40 men are listed below.

The cholesterol levels (in milligram per deciliter) of 40 men are listed above table with an interval of 50 points.
As shown in the below table 100 – 150, 151 – 200 …… 351 – 400.

b) Draw a histogram using the interval.
Answer:

Explanation:
The histogram diagram for the cholesterol levels (in milligram per deciliter) of 40 men are listed above table with an interval of 50 points as 100 – 150, 151 – 200 …… 351 – 400 and,
minimum of 1 man and maximum of 23 men histogram are plotted in the above diagram.

Question 4.
The speeds in kilometers per hour of 40 cars on a highway were recorded as follows.

a) Group the data into suitable intervals and tabulate them. Explain your choice of intervals.
Answer:

Explanation:
As we know that grouping data means the data given in the form of class intervals such as 0-10 to 91-100 as shown above.
Given that the speeds in kilometers per hour of 40 cars on a highway were recorded with an intervals of 10 kmph difference as shown in the above table.

b) Draw a histogram using the interval.

Answer:
Total 40 cars of different speed histogram diagram

Explanation:
A histogram is a graphical representation of discrete or continuous data.
The area of a bar in a histogram is equal to the frequency.
The y -axis is plotted by frequency density and the x -axis is plotted with the range of values divided into intervals.
Given different care on highway with different speed is recorded in the table with different speed intervals as shown in the above histogram is plotted.

Describe the data.

Question 5.
The histogram shows the number of representatives each state sent to the U.S. Congress in 2011. Briefly describe the data.

Answer:

Explanation:
The histogram shows the number of representatives each state sent to the U.S. Congress in 2011 describe the data in table form.
A histogram is a graphical representation of discrete or continuous data.
The y -axis is plotted by frequency density and the x -axis is plotted with the range of values divided into intervals.
The above table consists of Number of representatives and Number of states and the values with reference to the histogram is tabulated.

Question 6.
The histogram shows the highest temperature (in degrees Fahrenheit) recorded during December for one city. The temperature were recorded to the nearest degree. Briefly describe the data.

Answer:

Explanation:
The histogram shows the highest temperature (in degrees Fahrenheit) recorded during December for one city.
A histogram is a graphical representation of discrete or continuous data.
The area of a bar in a histogram is equal to the frequency.
The y -axis is plotted by frequency density and the x -axis is plotted with the range of values divided into intervals.
So, the temperature were recorded to the nearest degree.
The data is tabulated in the table format.

Math in Focus Course 1B Practice 13.3 Answer Key

Draw a histogram for each set of data. Include a title.

Question 1.
The table shows the number of cans recycled by 25 households in a month.

Answer:

Explanation:
The histogram shows the number of cans recycled by 25 households in a month, data given in the table format.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

Question 2.
The table shows the number of points scored by a football team in 20 games of one season.

Answer:

Explanation:
The histogram shows the number of points scored by a football team in 20 games of one season.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

Question 3.
The table shows the keyboarding speed of 100 students in a beginning keyboarding class.

Answer:

Explanation:
The histogram shows the keyboarding speed of 100 students in a beginning keyboarding class.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

Draw a histogram for the set of data. Include a title. Solve.

Question 4.
The number of sunny days in a year for 200 cities are shown in the table.

a) Find the value of x.
Answer:
48

Explanation:
The sum of the cities in the table are,
21 + 25 + 32 + 31 + 24 + 19 = 152
There must be 200 cities,
then 200 – 152 = 48
x = 48

b) Draw a histogram to represent the data. Briefly describe the data.
Answer:

Explanation:
The histogram shows, the number of sunny days in a year for 200 cities.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

c) What percent of the cities had fewer than 149 sunny days?
Answer:
10.5%

Explanation:
Total number of cities 200.
The number of cities had fewer than 149 sunny days = 21
The percent of the cities had fewer than 149 sunny days,
x = (21 / 200) X 100
x = 10.5%

d) What percent of the cities had more than 184 sunny days?
Answer:
21.5%

Explanation:
Total number of cities 200.
the number of cities had more than 184 sunny days,
24 + 19 = 43
The percent of the cities had more than 184 sunny days,
x = (43 / 200) X 100
x = 21.5%

The histogram shows the number of cars observed at one intersection at different times of the day. Use the histogram to answer questions 5 to 9.

Question 5.
How many observations are there?
Answer:
820 observations at different time intervals.

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.
Number of observations are as follows,
180 + 110 + 130 + 100 + 80 + 220 = 820

Question 6.
How many fewer cars passed the intersection from 6 A.M. to 7:59 A.M. than from 4 P.M. to 5:59 P.M.?
Answer:
40 observations.

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
cars passed the intersection from 6 A.M. to 7:59 A.M =180
cars passed the intersection from 4 P.M. to 5:59 P.M = 220
220 – 180 = 40
40 observations shows fewer cars passed the intersection from 6 A.M. to 7:59 A.M. than 4 P.M. to 5:59 P.M.

Question 7.
How many more cars passed the intersection from 10 A.M. to 11:59 A.M. than from 2 P.M. to 3:59 P.M.?
Answer:
50 cars.

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
cars passed the intersection from 10 A.M. to 11:59 A.M =130
cars passed the intersection from 2 P.M. to 3:59 P.M = 80
So, 130 – 80 = 50
50 cars passed the intersection from 10 A.M. to 11:59 A.M. than from 2 P.M. to 3:59 P.M.

Question 8.
What percent of the number of cars that passed the intersection from 4 P.M. to 5:59 P.M. was observed from 8 A.M. to 9:59 A.M.?
Answer:
50 percent of the number of cars that passed the intersection from 4 P.M. to 5:59 P.M. was observed from 8 A.M. to 9:59 A.M.

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
cars passed the intersection from 8 A.M. to 9:59 A.M =110
cars passed the intersection from 4 P.M. to 5:59 P.M = 220
220 – 110 = 110
110 / 220 = 0.5
0.5 x 100 = 50%
50 percent of the number of cars that passed the intersection from 4 P.M. to 5:59 P.M. was observed from 8 A.M. to 9:59 A.M.

Question 9.
What percent of the total number of cars that passed the intersection from 6 A.M. to 5:59 P.M. was observed from 4 P.M. to 5:59 P.M.? Round your answer to the nearest percent.
Answer:
73%

Explanation:
Given that, the histogram shows the number of cars observed at one intersection at different times of the day.
The total number of cars that passed the intersection from 6 A.M. to 5:59 P.M = 820
The total number of cars that passed the intersection from 4 P.M. to 5:59 P.M = 220
820 – 220 = 600
(600/820)X 100 = 73.17%
Round the answer to the nearest percent 73%
73% percent of the total number of cars that passed the intersection from 6 A.M. to 5:59 P.M. was observed from 4 P.M. to 5:59 P.M.

The histogram shows the number of books students in a class read last month. Use the histogram to answer the question.

Question 10.
Briefly describe the data. Explain whether the histogram shows any outlier of the data set.
Answer:
Yes, 16- 20 books is outlier of the data set.

Explanation:
The above histogram shows the number of books students in a class read last month.
The gap of  no books reading between 16 – 20.
An outlier is an observation that lies an abnormal distance from other values in a random sample from a population.
In a sense, this definition leaves it up to the analyst to decide what will be considered abnormal. Before abnormal observations can be singled out, it is necessary to characterize normal observations.

Draw a histogram for the set of data. Include a title. Solve.

Question 11.
The sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the table.

2 pairs of shoes of at least size 9.5 were sold.
a) Find the values of x and y.
Answer:
x = 1
y = 2

Explanation:
Given that the sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the above table.
The sales figures for 60 pairs of one style of shoe of various sizes are,
8 + 22 + 16 + 4 + 3 + 3 + 1 + x + y = 60
57 + x + y = 60
x + y = 60 – 57
x + y = 3
2 pairs of shoes of at least size 9.5 were sold.
y = 2
x + 2 = 3
x = 3 – 2 = 1
x = 1

b) What fraction of the shoes sold are smaller than size 8?
Answer:
0.783

Explanation:
Given that the sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the table.
The shoes sold are smaller than size 8 = 1 + 8 + 22 + 16 = 47
Totals shoes sold are = 60
The fraction of the shoes sold are smaller than size 8.
47/60 = 0.783

c) Draw a histogram using the intervals 6-6.5, 7-7.5, 8-8.5, and so on. Briefly describe the data.
Answer:

Explanation:
The histogram diagram shown above is the number of shoes sold of different sizes.
on vertical number of shoes sold and on horizontal line the sizes of the shoes sold.
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.

d) If shoes were to be categorized as follows:
small – sizes 6 to 7; medium – sizes 7.5 to 8.5; large – sizes 9 to 10, draw a histogram for the above data using the new categories.
Answer:

Explanation:
As we know the histogram is a graphical display of data using bars of different heights.
In a histogram, each bar groups numbers into ranges.
Taller bars show that more data falls in that range.
Given that the sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the table.
So, shoes were to be categorized as small – sizes 6 to 7; medium – sizes 7.5 to 8.5; large – sizes 9 to 10, and the quantities are added to total for small, medium and large.
Number of small shoes are 31
Number of medium shoes are 23
Number of large shoes are 6 pairs.
The total number of shoes are = 31 + 23 + 6 = 60

e) Compare the two histograms. When would each one be more useful?
Answer:
Both are useful,
for exact sale with reference to size first histogram are very useful.
second histogram tells about the small or medium or large size shoes are sold more or less in quantity.

Explanation:
Given that the sales figures for 60 pairs of one style of shoe of various sizes at a department store are given in the table.
for exact sale with reference to size first histogram are very useful.
second histogram tells about the small or medium or large size shoes are sold more or less in quantity as shoes were to be categorized as small – sizes 6 to 7; medium – sizes 7.5 to 8.5; large – sizes 9 to 10, and the quantities are added to total for small, medium and large.

Question 12.
Math Journal A survey was carried out to find the number of players who scored a certain number of goals during soccer matches in a month. A histogram was drawn to display the results.

Is the histogram drawn correctly? Discuss with your partner and explain your thinking.
Answer:
The histogram is not drawn correctly.

Explanation:
As per the histogram diagram, a survey was carried out to find the number of players who scored a certain number of goals during soccer matches in a month.
No values in histogram are recorded on horizontal line.

Brain @ Work

The table below shows the test scores for all the students in a Spanish I course.

The following grades are used to represent the scores.

Question 1.
Draw a histogram to show the distribution of the scores.
Answer:

Explanation:
Based on the table below shows the test scores for all the students in a Spanish I course.

The above histogram on vertical line frequency is recorded and,
on the horizontal line scores are marked.
Bar graphs are plotted corresponding values given in the table.

Question 2.
Draw a bar graph to display the grades.
Answer:

Explanation:
The following grades are used to represent the scores on histogram as shown above.

In the above histogram on vertical line Grades are marked and on the horizontal line scores are recorded.
Bar graphs are plotted corresponding grades given in the table.

Question 3.
Five students increased their scores from 79 to 89.
a) How would it change the histogram?
Answer:

Explanation:
In the above histogram on vertical line new frequency is recorded after shifting the five students from 79 to 89 scored students and on the horizontal line scores are marked.
Bar graphs are plotted corresponding values given in the table.

five students increased their scores from 79 to 89,
it means that subtract fives students from 73 – 79 and,
add it to 80 – 89 scores as shown in the above table.

b) How would it change the bar graph?
Answer:
Corresponding graph also will change accordingly as shown.

Explanation:
In the above histogram on vertical line new frequency is recorded after shifting the five students from 79 to 89 scored students and on the horizontal line scores are marked.
Bar graphs are plotted corresponding values given in the table.
Five students increased their scores from 79 to 89,
it means the subtract fives students from 73 – 79 and,
add it to 80 – 89 scores as shown in the above table.