Math in Focus – Page 25 – Math in Focus Answer Key

Math in Focus Grade 3 Chapter 4 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands

September 28, 2024September 27, 2024 by Hillary

Go through the Math in Focus Grade 3 Workbook Answer Key Chapter 4 Practice 3 Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands to finish your assignments.

Math in Focus Grade 3 Chapter 4 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands

Subtract. Fill in the blanks.

Question 1.
Example
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 1$
Subtract the ones.
9 ones cannot be subtracted from 0 ones.
So, regroup the tens and ones.
7 tens 0 ones
= __ tens ___ ones
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 2$
Subtract the tens.
7 tens cannot be subtracted from 6 tens.
So, regroup the hundreds and tens.
2 hundreds 6 tens
= __ hundred ___ tens
Subtract the hundreds.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 3$
3 hundreds cannot be subtracted from
___ hundred.
So, regroup the thousands and hundreds.
8 thousands 1 hundred
= ___ thousands ___ hundreds
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 4$
Subtract the thousands.

Answer:
8270 – 1379 = 6891.

Explanation:

$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 1$
Subtract the ones.
9 ones cannot be subtracted from 0 ones.
So, regroup the tens and ones.
7 tens 0 ones
= 6 tens 10 ones.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 2$
Subtract the tens.
7 tens cannot be subtracted from 6 tens.
So, regroup the hundreds and tens.
2 hundreds 6 tens
= 1 hundred 16 tens.
Subtract the hundreds.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 3$
3 hundreds cannot be subtracted from 2 hundred.
So, regroup the thousands and hundreds.
8 thousands 1 hundred
= 7 thousands 11 hundreds
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 4$
Subtract the thousands.
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-1$

Subtract. Fill in the blanks.

Question 2.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 5$
Subtract the ones.
9 ones cannot be subtracted from 7 ones.
So, regroup the tens and ones.
5 tens 7 ones
= ___________ tens ___________ ones
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 6$
Subtract the tens.
8 tens cannot be subtracted from ___________ tens.
So, regroup the hundreds and tens.
3 hundreds 4 tens
= __________ hundreds __________ tens
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 7$
Subtract the hundreds.
7 hundreds cannot be subtracted from
___________ hundreds.
So, regroup the thousands and hundreds.
4 thousands 2 hundreds.
= ____ thousands __ hundreds
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 8$
Subtract the thousands.

Answer:
4357 – 1789 = 2568.

Explanation:
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 5$
Subtract the ones.
9 ones cannot be subtracted from 7 ones.
So, regroup the tens and ones.
5 tens 7 ones
= 4 tens 17 ones.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 6$
Subtract the tens.
8 tens cannot be subtracted from 4 tens.
So, regroup the hundreds and tens.
3 hundreds 4 tens
= 2 hundreds 14 tens.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 7$
Subtract the hundreds.
7 hundreds cannot be subtracted from
2 hundreds.
So, regroup the thousands and hundreds.
4 thousands 2 hundreds.
= 3 thousands 12 hundreds.
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-2$

Subtract. Regroup when needed.

Question 3.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 9$
Answer:
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-3$

Explanation:
Step:1
Subtract the ones.
3 ones cannot be subtracted from 4 ones.
So, regroup the tens and ones.
9 tens 3 ones
= 8 tens 13 ones.

Step:2:
Subtract the tens.

Step:3
Subtract the hundreds.

Question 4.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 10$
Answer:
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-4$

Explanation:
Step:1
Subtract the ones.
7 ones cannot be subtracted from 6 ones.
So, regroup the tens and ones.
5 tens 6 ones
= 4 tens 16 ones.

Step:2:
Subtract the tens.
4 tens cannot be subtracted from 8 tens.
So, regroup the hundreds and tens.
5 hundreds 4 tens
= 4 hundreds 14 tens.

Step:3
Subtract the hundreds.

Question 5.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 11$
Answer:
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-5$

Explanation:
Step:1
Subtract the ones.
8 ones cannot be subtracted from 6 ones.
So, regroup the tens and ones.
3 tens 6 ones
= 2 tens 16 ones.

Step:2:
Subtract the tens.
8 tens cannot be subtracted from 2 tens.
So, regroup the hundreds and tens.
4 hundreds 2 tens
= 3 hundreds 12 tens.

Step:3
Subtract the hundreds.

Question 6.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 12$
Answer:
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-6$

Explanation:
Step:1
Subtract the ones.
7 ones cannot be subtracted from 1 ones.
So, regroup the tens and ones.
1 tens 1 ones
= 10 tens 11 ones.

Step:2:
Subtract the tens.
9 tens cannot be subtracted from 0 tens.
So, regroup the hundreds and tens.
1 hundreds 0 tens
= 0 hundreds 10 tens.

Step:3
Subtract the hundreds.
1 hundreds cannot be subtracted from
0 hundreds.
So, regroup the thousands and hundreds.
2 thousands 0 hundreds.
= 1 thousands 10 hundreds.

Step:4
Subtract the thousands.

Question 7.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 13$
Answer:
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-7$

Explanation:
Step:1
Subtract the ones.
3 ones cannot be subtracted from 1 ones.
So, regroup the tens and ones.
9 tens 1 ones
= 8 tens 11 ones.

Step:2:
Subtract the tens.

Step:3
Subtract the hundreds.
5 hundreds cannot be subtracted from 1 hundreds.
So, regroup the thousands and hundreds.
9 thousands 1 hundreds.
= 8 thousands 11 hundreds.

Step:4
Subtract the thousands.

Question 8.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 14$
Answer:
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-8$

Explanation:
Step:1
Subtract the ones.
9 ones cannot be subtracted from 4 ones.
So, regroup the tens and ones.
0 tens 4 ones
= tens 14 ones.

Step:2:
Subtract the tens.
2 tens cannot be subtracted from 0 tens.
So, regroup the hundreds and tens.
7 hundreds 0 tens
= 6 hundreds 9 tens.

Step:3
Subtract the hundreds.

Question 9.
Color the answers from above to find the path to the present.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 15$
Answer:
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-9$

Explanation:
8270 – 1379 = 6891.
4357 – 1789 = 2568.
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-3$
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-4$
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-5$
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-6$
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-7$
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-Subtract. Fill in the blanks-8$

Subtract. Regroup when needed.
Example
3,852 – 1,621 = 2,231 $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 16$

Question 10.
7,162 – 5,002 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 17$
Answer:
7,162 – 5,002 = 2160. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 17$

Explanation:
7,162 – 5,002 = 2160.

Question 11.
7,156 – 43 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 18$
Answer:
7,156 – 43 = 7113. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 18$

Explanation:
7,156 – 43 = 7113.

Question 12.
3,696 – 2,475 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 19$
Answer:
3,696 – 2,475 = 1221. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 19$

Explanation:
3,696 – 2,475 = 1221.

Question 13.
7,342 – 2,502 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 20$
Answer:
7,342 – 2,502 = 4840. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 20$

Explanation:
7,342 – 2,502 = 4840.

Question 14.
8,513 – 566 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 21$
Answer:
8,513 – 566 = 7947. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 21$

Explanation:
8,513 – 566 = 7947.

Question 15.
6,707 – 1,125 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 22$
Answer:
6,707 – 1,125 = 5582. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 22$

Explanation:
6,707 – 1,125 = 5582.

Question 16.
2,152 – 1,648 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 23$
Answer:
2,152 – 1,648 = 504. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 23$

Explanation:
2,152 – 1,648 = 504.

Question 17.
5,261 – 85 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 24$
Answer:
5,261 – 85 = 5176. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 24$

Explanation:
5,261 – 85 = 5176.

Question 18.
9,133 – 7,269 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 25$
Answer:
9,133 – 7,269 = 1864. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 25$

Explanation:
9,133 – 7,269 = 1864.

Question 19.
3,087 – 1,779 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 26$
Answer:
3,087 – 1,779 = 1308. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 26$

Explanation:
3,087 – 1,779 = 1308.

Question 20.
7,965 – 978 = ___ $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 27$
Answer:
7,965 – 978 = 6987. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 27$

Explanation:
7,965 – 978 = 6987.

Write the corresponding letters from Exercise 11 to 20 to find the name of this national treasure.
Question 21.
$Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 28$
Answer:
$Math-in-Focus-Grade-3-Chapter-4-Practice-3-Answer-Key-Subtraction-with-Regrouping-in-Ones-Tens-Hundreds-and-Thousands-21$

Explanation:
3,852 – 1,621 = 2,231 $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 16$
7,162 – 5,002 = 2160. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 17$
7,156 – 43 = 7113. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 18$
3,696 – 2,475 = 1221. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 19$
7,342 – 2,502 = 4840. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 20$
8,513 – 566 = 7947. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 21$
6,707 – 1,125 = 5582. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 22$
2,152 – 1,648 = 504. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 23$
5,261 – 85 = 5176. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 24$
9,133 – 7,269 = 1864. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 25$
3,087 – 1,779 = 1308. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 26$
7,965 – 978 = 6987. $Math in Focus Grade 3 Chapter 3 Practice 3 Answer Key Subtraction with Regrouping in Ones, Tens, Hundreds, and Thousands 27$

Question 22.
Where is this national treasure located?
____ __________
Answer:
New York, USA is this national treasure located.

Explanation:
This national treasure is located in New York, USA.

Math in Focus Grade 5 Chapter 7 Practice 2 Answer Key Equivalent Ratios

September 28, 2024September 27, 2024 by Hillary

Practice the problems of Math in Focus Grade 5 Workbook Answer Key Chapter 7 Practice 2 Equivalent Ratios to score better marks in the exam.

Math in Focus Grade 5 Chapter 7 Practice 2 Answer Key Equivalent Ratios

Write ratios to compare the two sets of items.
$Math in Focus Grade 5 Chapter 7 Practice 2 Answer Key Equivalent Ratios 1$

Question 1.
The ratio of the number of CDs in Group A to the number of CDs in
Group B is ___ : ____
Answer: 4 : 8
Explanation:
Given,
Number of CD’s in Group A is 4,
Number of CD’s in Group B is 8,
So,the ratio of the number of CD’s in Group A to the number of CD’s in Group B is 4 : 8

Question 2.
The ratio of the number of CD-holders in Group A to the number of CD-holders in Group B is ___ : ____
Answer: 1 : 2
Explanation:
Given,
The number of CD – holders in Group A is 1,
The number of CD – holders in Group B is 2,
So the ratio of the number of CD – holders in Group A to the number of CD – holders in Group B is 1 : 2.

Question 3.
____ : ____ = ____ : ____ in simplest form.
Answer: 18 : 12 = 3 : 2
Explanation:
Given,
Both the numbers 18 and 12 can be divisible by 6 we get 3 : 2 to be in the simplest form.

Write ratios to compare the two sets of items.

$Math in Focus Grade 5 Chapter 7 Practice 2 Answer Key Equivalent Ratios 2$

Question 4.
The ratio of the number of pencils in Group A to the number of pencils in
Group B is ___ : _____
Answer: 18 : 27
Explanation:
In Group A there are 6 bundles of pencils each bundle has 3,
We need to multiply 6 with 3 we get 18,
In Group B there are 9 bundles of pencils each bundle has 3,
We need to multiply 9 with 3 we get 27,
So the ratio of the number of pencils in Group A to the number of pencils in Group B is 18 : 27.

Question 5.
The ratio of the number of bundles in Group A to the number of bundles in
Group B is ___ : ____
Answer: 6 : 9
Explanation:
There are 6 bundles of pencils in Group A,
There are 9 bundles of pencils in Group B,
So the ratio of the number of bundles in Group A to the number of bundles in Group B is 6 : 9.

Question 6.
18 : 27 = 6 : 9 = ___ : ____ in simplest form.
Answer: 2 : 3
Explanation:
Given,
Numbers 18 and 27 are divisible by 3 in the form 6 : 9,
To make it into the simplest form we need to divide by 3, we get 2 : 3.

Find the greatest common factor of each set of numbers.

Example :
4 and 6 2

Question 7.
6 and 9 _______
Answer:
The greatest coomon factor of each set of numbers is 3
Explanation:
Both the numbers 6 and 9 are divisible by 3.

Quantities 8.
6 and 18 _____
Answer: 6
Explanation:
There are four common factors of 6 and 18 that are 1,2, 3 and 6,
So the greatest common factor is 6.

Question 9.
12 and 32 ____
Answer:
4
Explanation:
The greatest common factor for 12 and 32 is 4.

Complete.

Question 10.
$Math in Focus Grade 5 Chapter 7 Practice 2 Answer Key Equivalent Ratios 3$
Answer: 9: 15

Quantities 11.
$Math in Focus Grade 5 Chapter 7 Practice 2 Answer Key Equivalent Ratios 4$
Answer: 28:16

Question 12.
4 : 3 = 24 : ___
Answer:24:18

Question 13.
8 : 3 = 64 : ___
Answer: 64:24

Question 14.
4 : 9 = ___ : 45
Answer: 20:45

Question 15.
6 : 7 = 42 : ___
Answer: 42:49

Question 16.
5 : 8 = 45 : ___
Answer: 45:72

Question 17.
9 : 6 = ___ : 54
Answer: 81:54

Complete to express each ratio in simplest form.

Question 18.
$Math in Focus Grade 5 Chapter 7 Practice 2 Answer Key Equivalent Ratios 5$
Answer: 3:2

Question 19.
$Math in Focus Grade 5 Chapter 7 Practice 2 Answer Key Equivalent Ratios 6$
Answer: 5:7

Question 20.
12 : 30 = ___ : 5
Answer: 2:5
Both the numbers 12 and 30 need to be divisble by 6 to get the simplest form of 2 : 5

Question 21.
14 : 28 = 1 : ___
Answer: 1:2

Question 22.
60 : 45 = ___ : 3
Answer: Both the numbers 60 and 45 need to be divisble by 15 to get the simplest form of 4 : 3.

Question 23.
72 : 104 = __ : 13
Answer: Both the numbers 72 and 104 need to be divisble by 8 to get the simplest form of 9 : 13.

Question 24.
6 : 16 = __ : __
Answer: Both the numbers 6 and 16 need to be divisble by 2 to get the simplest form of 3 : 8.

Question 25.
15 : 35 = __ : ___
Answer: Both the numbers 15 and 35 need to be divisble by 5 to get the simplest form of 3 : 7.

Question 26.
4 : 48 = ___, ___
Answer: Both the numbers 4 and 48 need to be divisble by 4 to get the simplest form of 1 : 12.

Question 27.
56 : 21 = __ : ___
Answer: Both the numbers 56 and 21 need to be divisble by 7 to get the simplest form of 8 : 3.

Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers

September 28, 2024September 27, 2024 by Hillary

Practice the problems of Math in Focus Grade 5 Workbook Answer Key Chapter 8 Practice 3 Rewriting Decimals as Fractions and Mixed Numbers to score better marks in the exam.

Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers

Rewrite each decimal as a fraction or mixed number in simplest form.

$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 1$

Question 1.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 2$
Answer:
$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-2$
1.8 in fraction is
1.8 = 9/5

Question 2.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 3$
2.2 = ______
Answer:
$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-3$
2.2 in mixed number is 2 × 1/5

Question 3.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 4$
Answer:
$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-4$
3.5 mixed number is 3×1/2

Question 4.
0.36
Answer:
0.36 in fraction is 9/25

Question 5.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 6$
Answer:
$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-6$
1.35 in mixed number is 1×7/20

Rewrite each decimal as a fraction or mixed number in simplest form.

Question 6.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 7$
Answer:

$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-7$
1.12 in mixed number is 1×3/25

Question 7.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 8$
Answer:
$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-8$
3.57 in mixed number is 3×57/100

Question 8.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 9$
Answer:
$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-9$
0.058 in fraction number is 29/500.

Question 9.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 10$
Answer:
$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-10$
0.169 in fraction number is 169/1000.

Question 10.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 11$
Answer:
$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-11$
1.092 in mixed number is 1×23/250.

Rewrite the decimal as a mixed number in simplest form.

Question 11.
$Math in Focus Grade 5 Chapter 8 Practice 3 Answer Key Rewriting Decimals as Fractions and Mixed Numbers 12$
Answer:
$Math-in-Focus-Grade-5-Chapter-8-Practice-3-Answer-Key-Rewriting-Decimals-as-Fractions-and-Mixed-Numbers-12$
2.235 in mixed number is 2×47/200.

Rewrite each decimal as a fraction or mixed number in simplest form.

Question 12.
7.3
Answer:
7.3 in fraction number is 73/10.

Question 13.
26.9
Answer:
26.9 in mixed number is 269/10.

Question 14.
0.59
Answer:
0.59 in fraction number is 59/100.

Question 15.
15.82
Answer:
15.82 in mixed number is 15×41/50.

Question 16.
1.28
Answer:
1.28 in mixed number is 1×7/25.

Question 17.
4.109
Answer:
4.109 in mixed number is 4109/1000

Question 18.
0.136
Answer:
0.136 in fraction number is 17/125.

Question 19.
3.602
Answer:
3.602 in mixed number is 3×301/500.

Math in Focus Grade 5 Chapter 11 Answer Key Graphs and Probability

September 28, 2024September 27, 2024 by Hillary

This handy Math in Focus Grade 5 Workbook Answer Key Chapter 11 Graphs and Probability provides detailed solutions for the textbook questions.

Math in Focus Grade 5 Chapter 11 Answer Key Graphs and Probability

Put On Your Thinking Cap!

Challenging Practice

Complete.

Question 1.
The table shows the conversion from gallons to pints. Complete the table.
$Math in Focus Grade 5 Chapter 11 Answer Key Graphs and Probability 1$
Answer:

Explanation:
1 gallon = 8 pints
if number of gallons (x) = 1,
number of pints (y) = 1 x 8 = 8
So, substitute the x value and multiply with 8 to complete the table.

Question 2.
Write the equation relating the number of pints (y) to the number of gallons (x).
______________
Answer:
y = x × 8
Explanation:
x – Number of Gallons
y – Number of Pints
y = x × 8

Question 3.
Draw the graph of the equation. Label the axes and the equation.
$Math in Focus Grade 5 Chapter 11 Answer Key Graphs and Probability 2$
Answer:

Explanation:
Above graph shows the conversion between pints and gallons.
Take number of gallons on x-axis and number of pints on y-axis.
Take the difference between each pint as 10.

Use the graph to answer the questions.

Question 4.
How many pints is 3$\frac{1}{2}$ gallons?
Answer:
28 pints
Explanation:
y = 3$\frac{1}{2}$ × 8 = 28 gallons

Question 5.
How many pints is 4$\frac{1}{2}$ gallons?
Answer:
36 gallons
Explanation:
y = 4$\frac{1}{2}$ × 8 = 36 pints

Question 6.
How many gallons is 20 pints?
Answer:
2 $\frac{1}{2}$ gallons
Explanation:
20 =x × 8 = 28 gallons
x = $\frac{20}{8}$
= 2 $\frac{4}{8}$

Question 7.
How many gallons is 44 pints?
Answer:
5$\frac{1}{2}$ gallons
Explanation:
y = x × 8

Complete.

Question 8.
The table shows the conversion from quarts to cups. Complete the table.
$Math in Focus Grade 5 Chapter 11 Answer Key Graphs and Probability 3$
Answer:

Explanation:
1 Quarts = 4 cups
if number of Quarts (x) = 1,
number of cups (y) = 1 x 4 = 4
So, substitute the x value and multiply with 4 to complete the table.

Question 9.
Write the equation relating the number of cups (y) to the number of quarts (x).
Answer:
y = x × 4
Explanation:
x is Number of Quarts
y is Number of Cups
y = x × 4

Put On Your Thinking Cap!

Problem Solving

Solve.

Question 1.
Jim has a dime, a nickel, and a quarter. How many different amounts of money can he form using one or more of these coins?
Answer:
These are the possible coins combinations

Explanation:
A probability is a number that reflects the chance or likelihood that a particular event will occur.
So, Jim has the above combinations of coins.

Question 2.
There are an equal number of red, blue, and green beads in a bag. One bead is picked, its color is noted and the bead is replaced. Then a second bead is picked.
a. Draw a tree diagram to show the outcomes.
Answer:

Explanation:
You might observed total 9 combinations as per the above tree diagram.
Now, probability of picking different color beads, is as shown in the diagram.
As we know probability is a number that reflects the chance or likelihood that a particular event will occur.

b. What is the probability of picking two red beads?
Answer:
$\frac{1}{9}$
Explanation:
You might observed total 9 combinations as per the above tree diagram.
Now, probability of picking two red beads, is as shown in the diagram below.

c. What is the probability of picking one red and one green bead?
Answer:
$\frac{2}{9}$
Explanation:
You might observed total 9 combinations as per the above tree diagram.
Now, probability of picking one red and one green bead, is as shown in the diagram below.

d. What is the probability of picking no red beads?
Answer:
$\frac{4}{9}$

Explanation:
You might observed total 9 combinations as per the above tree diagram.
Now, probability of picking no red, means first select blue and green.

Math in Focus Grade 6 Chapter 10 Lesson 10.4 Answer Key Area of Composite Figures

September 28, 2024September 26, 2024 by Hillary

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Lesson 10.4 Area of Composite Figures to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Lesson 10.4 Answer Key Area of Composite Figures

Math in Focus Grade 6 Chapter 10 Lesson 10.4 Guided Practice Answer Key

Use graph paper. Copy the hexagon and solve.

Question 1.
Divide the hexagon into two identical triangles and a rectangle.
Answer:
$Math-in-Focus-Grade-6-Chapter-8-Lesson-10.4-Answer-Key-Area-of-Composite-Figures img_1$
Explanation:
As shown in the figure, a regular hexagon is divided into two identical traingles and a rectangle.
The two identical traingle are AFE and BCD.
The rectangle formed is ABDE.

Question 2.
Divide the hexagon in another way. Name the polygons that make up the hexagon.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 1$
Answer:
$Math-in-Focus-Grade-6-Chapter-8-Lesson-10.4-Answer-Key-Area-of-Composite-Figures-1$
Explanation:
Mark the center of the hexagon.
Draw a line from the center point to a vertex.
Draw lines from the center point to the third and fifth vertices.
Each part is a rhombus, which is a four-sided figure with equal side lengths and two sets of parallel sides.
A hexagon can form 3 identical rhombus.

Complete.

Question 3.
Trapezoid ABDE is made up of square A BCE and triangle ECD. triangle ECD is 60 square inches. The length of $\overline{C D}$ is 12 inches.

a) Find the height of triangle ECD.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 2$
Area of triangle ECD = $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ in.²
Area of triangle ECD = $\frac{1}{2}$bh
= $\frac{1}{2}$ ∙ $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ ∙ EC
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ = $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ ∙ EC
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ ÷ $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ = $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ ∙ EC ÷ $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ = EC
The height of triangle ECO is î.. inches,
Answer:
In triangle ECD, the base length is equal to 12 in and the area is 60 sq.in.
Area of triangle ECD = $\frac{1}{2}$bh
60 = $\frac{1}{2}$×12×EC
60 = 6×EC
60÷6 = 6×EC÷6
10 = EC
The height of triangle ECD is 10 in.

b) Find the area of square ABCE.
Area of square ABCE = l²
= $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 4$
The area of square ABCE is $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ square inches.
Answer:
A square has all equal sides, so the length of it will be 10 in.
Area of square ABCE = length×length
= 10×10
= 100 sq.in
The area of square ABCE is 100 sq.in

c) Find the areà of trapezoid ABDE.
Area of the trapezoid ABDE
= area of square ABCE + area of triangle ECD
= $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ + $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$
= $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ in.²
The area of trapezoid ABDE is $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 3$ square inches.
Answer:
Area of the trapezoid ABDE
= $\frac{1}{2}$×(sum of bases)×height
Length of BD = BC+CD = 10+12 = 22 in
= $\frac{1}{2}$×(10+22)×10
= $\frac{1}{2}$×32×10
= 16×10
= 160 sq.in

Complete.

Question 4.
The area of parallelogram ABEF is 84 square meters.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 5$
a) Find the area of triangle BDE.
Area of parallelogram ABEF = bh
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ = $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ ∙ h
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ ÷ $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ = $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ h ÷ $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ = h
The height of parallelogram ABEF is also the height of triangle BDE.
Area of triangle BDE = $\frac{1}{2}$bh
= $\frac{1}{2}$ • BD • h
= $\frac{1}{2}$ • ( $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ + $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ ) • $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$
= $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ m²
The area of triangle BDE is $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ square meters.
Answer:
Given that area of parallelogram ABEF is 84 sq.m and base is 5m.
Area of parallelogram ABEF = bh
84 = 5×h
84÷5 = 5×h÷5
16.8 = h
The height of parallelogram ABEF is also the height of triangle BDE which is 16.8 m high.
Area of triangle BDE = $\frac{1}{2}$bh
= $\frac{1}{2}$ × BD × h
= $\frac{1}{2}$ × (BC+CD) × h
= $\frac{1}{2}$ × (12+8) × h
= $\frac{1}{2}$ × 20 × 16.8
= $\frac{1}{2}$ × 336
= 168 sq.m

b) Find the area of trapezoid CDEF.
Parallelogram ABEF and trapezoid CDEF have the same height.
Area of trapezoid CDEF = $\frac{1}{2}$h(b₁ + b₂)
= $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ • $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ • ( $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ + $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ )
The area of trapezoid CDEF is $Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 6$ square meters.
Answer:
Parallelogram ABEF and trapezoid CDEF have the same height. Therefore, the height will be 16.8 m
Since ABEF is a parallelogram, AB and FE will have the same length.
Height = 16.8 m
Sum of bases = 8+5 = 13
Area of trapezoid CDEF = $\frac{1}{2}$h(b₁ + b₂)
= $\frac{1}{2}$×16.8×13
= $\frac{1}{2}$×218.4
= 109.2 sq.m
The area of trapezoid CDEF is 109.2 sq.m.

Math in Focus Course 1B Practice 10.4 Answer Key

Copy each figure and draw straight lines to divide. Describe two ways to find the area of each figure.

Question 1.
Divide the figure into a rectangle and two right triangles.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 7$
Answer:
$Math-in-Focus-Grade-6-Chapter-8-Lesson-10.4-Answer-Key-Area-of-Composite-Figures-7$
Divide the given figure into two equal parts.
It will form two identical rectangles.
Divide the first identical rectangle diagonally into two equal parts, which will form an equilateral triangle.

Question 2.
Divide the figure into a rectangle and two right triangles.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 8$
Answer:
$Math-in-Focus-Grade-6-Chapter-8-Lesson-10.4-Answer-Key-Area-of-Composite-Figures-8$
Explanation:
Draw a perpendicular segment between the base and the opposite vertex. Similarly, draw a line segment to the other end also.
The figure is divided now into three parts.
Two parts will form an right angle triangle and the third will form a rectangle.

Question 3.
Divide the figure into a rectangle and a right triangle.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 9$
Answer:
$Math-in-Focus-Grade-6-Chapter-8-Lesson-10.4-Answer-Key-Area-of-Composite-Figures-9.png$
Explanation:
Draw a line segment parallel to the base length.
The first part formed is the right angle triangle and the second part is the rectangle.

Copy each figure and draw straight lines to divide. Describe a way to find the area.

Question 4.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 10$
Answer:
$Math-in-Focus-Grade-6-Chapter-8-Lesson-10.4-Answer-Key-Area-of-Composite-Figures-10$
Explanation:
If the given figure is divided using straight line, it will form a triangle and a rectangle.
Area of polygon will be the sum of areas of rectangle and area of triangle.
Area of polygon = Area of triangle+Area of rectangle
= ($\frac{1}{2}$ × b × h) + (length×width)
Here, the base of triangle and the width of rectangle will be of same length.

Question 5.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 11$
Answer:
$Math-in-Focus-Grade-6-Chapter-8-Lesson-10.4-Answer-Key-Area-of-Composite-Figures-11$
Explanation:
If the given figure is divided using straight line, it will form 5 identical triangles.
Area of polygon = n×$\frac{1}{2}$×b×h
Here n equals 5
Area of polygon = 5×$\frac{1}{2}$×b×h
Here, the base of trinagle will be equal to side length of the pentagon.

Find the area of each figure.

Question 6.
Parallelogram ABDE is made up of square ACDF, triangle ABC, and triangle FDE. Triangle ABC and triangle FDE are identical. The area of square ACDF is 36 square meters. Find the area of triangle ABC. Then find the area of parallelogram ABDE.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 12$
Answer:
Given that ACDF is a square and its area is 36 sq.m
Area of ACDF = side×side
36 = side×side
6×6 = side×side
Therefore, side of a sqaure will be 6 m.
Here, the side of square is equal to the height of the trinagle.
The height of triangle is 6m and the base is 4m.
Area of triangle = $\frac{1}{2}$×b×h
= $\frac{1}{2}$×4×6
= 24÷2
= 12 sq.m
Area of triangle ABC and DFE will be same as both have the same dimensions.
Area of parallelogram ABDE = Area of ABC + Area of DFE + Area of ACDF
= 12+12+36
= 60 sq.m

Question 7.
Math Journal Describe how you would divide the figure with straight lines. Which sides of the figure would you measure to find its area? Explain your answer.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 13$
Answer:
$Math-in-Focus-Grade-6-Chapter-8-Lesson-10.4-Answer-Key-Area-of-Composite-Figures-13$

Explanation:
Draw a vertical line segment between the two vertex and divide it in two parts.
Two triangles will be formed. Since both the trinagles lie on same base, the base length of these triangles will be same. The perpendicular segment between the base and the opposite vertex will be the height of the triangle.
Measure the drawn line segment to get the base length and the perpendicular line segment to get the height.
Area of triangle can be found using the formula $\frac{1}{2}$×b×h.
Area of polygon will be the sum of areas of two trinagles formed.
Area of polygon = Area of first triangle + Area of second trinangle

Find the area of triangle EBC.

Question 8.
In the figure below, trapezoid ABDE is made up of three triangles, and figure ABCE is a parallelogram. Find the area of triangle EBC if the area of trapezoid ABDE is 180 square centimeters.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 14$
Answer:
Given that trapezoid ABDE is made up of three triangles, ABE BEC and ECD and its area is 180 sq.cm
Figure ABCE is a parallelogram.
Area of triangle ECD:
In triangle ECD, the base measures 6cm and the height measures 12cm.
Area of ECD = $\frac{1}{2}$×b×h
= $\frac{1}{2}$×6×12
= 72÷2
= 36 sq.cm
To find the area of triangle EBC, we need the length of BC, which is the base of the parallelogram ABCE.
To calculate the base of parallelogram ABCE , first we need to find the area of ABCE.
Area of trapezoid = Area of parallelogram ABCE + Area of triangle ECD
180 = Area of parallelogram ABCE + 36
180-36 = Area of parallelogram ABCE + 36 – 36
Area of parallelogram ABCE = 144 sq.cm
We know the height of the polygon is 12cm.
Area of parallelogram ABCE = base × height
144 = base × 12
144 ÷ 12 = (base × 12) ÷ 12
12 = base
Thus, the length of BC will be 12 cm.
Area of triangle EBC = $\frac{1}{2}$×b×h.
= $\frac{1}{2}$×12×12
= 144÷2
= 72 sq.cm

Solve. Use graph paper.

Question 9.
a) Plot the points P (-2, 2), Q (-2, -2), R (-4, -5), S (1, -5), and T (3, -2) on a coordinate plane. Connect the points in order to form figure PQRST.
Answer:
Plot the points P (-2, 2), Q (-2, -2), R (-4, -5), S (1, -5), and T (3, -2) on a coordinate plane.
When connected the points, it forms a polygon as shown below.
$Math in Focus Grade 6 Course 1 B Chapter 10 Lesson 10.4 Answer Key Area of Composite Figures img_1$

b) Find the area of figure PQRST.
Answer:
When the figure PQRST is divided using a line segment. It forms a triangle and a parallelogram.
Area of the polygon PQRST = Area of QTSR + Area of PQT
In the figure, QTSR forms a parallelogram. Let us assume RS as base and the perpendicular segment between the base and the opposite side will be the height.
Base = 5 units
Height = 3 units
Area of QTSR = base×height
= 5×3
= 15 sq.units
In the figure, PQT forms a parallelogram. Let us assume QT as base and the perpendicular segment between the base and the opposite vertex will be the height.
Base = 5 units
Height = 4 units
Area of PQT = $\frac{1}{2}$×b×h
= $\frac{1}{2}$×5×4
= $\frac{1}{2}$×20
= 10 sq.units
Area of the polygon PQRST = 15 + 10
= 25 sq.units

c) Point V lies on along $\overline{Q T}$. The area of triangle PQV is $\frac{2}{5}$ the area of triangle PQT. Give the coordinates of point V. Plot point V on the coordinate plane.
Answer:
Given that the point V lies on along $\overline{Q T}$. The area of triangle PQV is $\frac{2}{5}$ the area of triangle PQT.
Area of PQV = $\frac{2}{5}$×area of triangle PQT
As calculated above, Area of PQT = 10 sq.units
= $\frac{2}{5}$×10
= 20÷5
= 4 sq.units
Let us assume PQ as base which is 5 units.
Area of PQV = $\frac{1}{2}$×b×h
4 = $\frac{1}{2}$×4×h
4 = 2×h
4÷2 = (2×h)÷2
2 = h
The height will be 2 units long.
Therefore the point V can be placed 2 units high from the base along $\overline{Q T}$, which will be (0,-2)

$Math in Focus Grade 6 Course 1 B Chapter 10 Lesson 10.4 Answer Key Area of Composite Figures img_4$

Find the area of trapezoid MNRS.

Question 10.
In the figure below, trapezoid MNRS is made up of trapezoid MNPT, triangle TPQ, and parallelogram TQRS. The area of triangle TPQ is 84 square feet. The lengths of $\overline{N P}$, $\overline{P Q}$, and $\overline{Q R}$ are in the ratio 2 : 1.5 : 1.
Find the area of trapezoid MNRS.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 15$
Answer:
Given that the trapezoid MNRS made up of trapezoid MNPT, triangle TPQ, and parallelogram TQRS.
Area of triangle TPQ = 84 sq.ft and the length of PQ is 12 ft and MS is 18 ft.
Area of TPQ = $\frac{1}{2}$×b×h, where b is length of PQ and h is unknown.
84 = $\frac{1}{2}$×12×h
84 = 6×h
84÷6 = (6×h)÷h
14 = h
The height of triangle will be 14 ft.
The lengths of $\overline{N P}$, $\overline{P Q}$, and $\overline{Q R}$ are in the ratio 2 : 1.5 : 1
Let p be the proportion.
1.5p=12
1.5p÷1.5 = 12÷1.5
p = 8
Length of NP = 2p = 2×8 = 16 ft
Length of QR = 1p = 1×8 = 8 ft
NR = NP+PQ+QR
= 16+12+8
= 36 ft
In trapezoid MNRS, MS and NR will be the two parallel bases.
Area of trapezoid MNRS = $\frac{1}{2}$×(sum of bases)×height
= $\frac{1}{2}$×(18+36)×14
= $\frac{1}{2}$×54×14
= 756÷2
= 378 sq.ft
Thus, the area of trapezoid MNRS = 378 sq.ft

Find the area of triangle BDE.

Question 11.
The figure below is made up of two trapezoids ABEFand BCDE. The area of triangle FGE is 26 square inches, and the area of trapezoid BCDE is 82.5 square inches. BG is equal to GE. Find the area of triangle BDE.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 16$
Answer:
The figure is made up of two trapezoids ABEFand BCDE, two triangles FGE,BCD and BDE.
Area of triangle FGE is 26 sq.in and height is 8 in.
Area of trapezoid BCDE is 82.5 sq.in
BG is equal to GE and length of CD is 2 in.
First we need to find the base length, GE.
Area of triangle FGE = $\frac{1}{2}$×base×height
26 = $\frac{1}{2}$×base×8
26 = 4×base
26÷4 = (4×base)÷4
6.5 = base
Length of GE = Length of BG = 6.5 in
Length of BE = Length of GE + Length of BG
= 6.5+6.5 =13 in
Area of trapezoid BCDE = $\frac{1}{2}$×(sum of bases)×height
82.5 = $\frac{1}{2}$×(BE+CD)×height
82.5 = $\frac{1}{2}$×15×height
82.5 = 7.5×height
82.5÷7.5 = (7.5×height)÷7.5
height = 11 in
The height of tarpezoid and triangle will be of same length.
Now, we can find the area of triangle BDE.
Area of triangle BDE = $\frac{1}{2}$×base×height
= $\frac{1}{2}$×BE×height
= $\frac{1}{2}$×BE×height
= $\frac{1}{2}$×13×11
= 143÷2
= 71.5 sq.in

Find the area of the shaded region.

Question 12.
$\frac{3}{8}$ of the triangle is shaded.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 17$
Answer:
Given trinagle is 28cm wide and 35cm high.
Area of trinagle = $\frac{1}{2}$ × base × height
= $\frac{1}{2}$ × 28 × 35
= $\frac{1}{2}$ × 980
= 980÷2
= 490 sq.cm
The shaded region is $\frac{3}{8}$ of the triangle.
Therefore, area of shaded region will be $\frac{3}{8}$ × 490
= 1470÷8
= 183.75 sq.cm

Find the height of trapezoid PQRS.

Question 13.
Trapezoid PORS s made up of isosceles triangle PQS and triangle SQR. The area of triangle PQS is 16.5 square inches. The areas of triangle PQS and triangle SQR are in the ratio 2 : 3, Find the height of trapezoid PQRS.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 18$
Answer:
Given that the trapezoid PORS s made up of isosceles triangle PQS and triangle SQR.
Area of triangle PQS is 16.5 square inches.
Since PQS is an isosceles triangle, PS will be equal to SQ.
The areas of triangle PQS and triangle SQR are in the ratio 2 : 3.
$\frac{PQS}{SQR}$ = $\frac{2}{3}$
3×PQS = 2×SQR
3×16.5 = 2×SQR
49.5 = 2×SQR
49.5÷2 = (2×SQR)÷2
24.7 = SQR
Area of triangle SQR = 24.7 sq.in
Area of trapezoid = Area of triangle PQS + Area of triangle SQR
= 16.5+24.7
= 41.2 sq.in
Area of trapezoid = $\frac{1}{2}$×(sum of bases)×height
41.2 = $\frac{1}{2}$×(6+9)×height
41.2 = $\frac{1}{2}$× 15 ×height
41.2 = 7.5×height
41.2 ÷ 7.5 = (7.5×height)÷7.5
5.4= height
Thus, the height of trapezoid PQRS = 5.4 in

Find the area of each figure.

Question 14.
In the figure below, trapezoid ABCD is made up of square BCDE and triangles ABF and APE. The area of square BCDE is 576 square feet. The ratio of BF to FE is 2: 1. Find the area of triangle ABF.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 19$
Answer:
Given that the trapezoid ABCD is made up of square BCDE and triangles ABF and APE.
Area of square BCDE = 576 sq.ft
Area of square BCDE = length×length
576 = length×length
24×24 = length×length
Length of square = 24 ft.
The ratio of BF to FE is 2: 1.
$\frac{BF}{FE}$ = $\frac{2}{1}$
1×BF = 2×FE
BE = 2×FE + FE
24 = 3×FE
24÷3 = 3×FE÷3
FE = 8ft
BF = 2×8 = 16ft
Area of triangle ABF = $\frac{1}{2}$ × base × height
= $\frac{1}{2}$ × 16 × 12
= 192÷2
= 96 sq.ft

Brain @ Work

Question 1.
Figure ABCD is made up of square PQRS and four identical triangles. The area of triangle APD
is 49 square feet. The lengths of $\overline{\mathrm{AP}}$ and $\overline{\mathrm{PD}}$ are in the ratio 1: 2. Find the area of figure ABCD.
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 20$
Answer:
Given that the figure ABCD is made up of square PQRS and four identical triangles. The area of triangle APD
is 49 square feet.
The lengths of $\overline{\mathrm{AP}}$ and $\overline{\mathrm{PD}}$ are in the ratio 1: 2.
Let the proportion be ‘p’
$\frac{AP}{PD}$ = 1: 2
PD = 2×AP
Area of triangle APD = $\frac{1}{2}$ × base × height
49 = $\frac{1}{2}$ × PD × AP
49 = $\frac{1}{2}$ × 2×AP×AP
49 = AP×AP
7×7 = AP×AP
AP = 7ft
PD = 2×7 = 14ft
As all are identical triangles, the longest side or base will be twice of the height.
SC = 2×DS, where DS = 7ft
PD = PS+SD
14 = PS+7
14-7 = PS+7-7
PS = 7ft
PQRS is a square, so all sides will be equal.
Area of square PQRS = length×length
= 7×7
= 49 sq.ft
Area of ABCD = Area of four identical triangles + Area of square PQRS
= 4×49 + 49
= 196+49
= 245 sq.ft

Question 2.
The figure is made up of squares BCDE and AEFH. The length of $\overline{D E}$ is 6 centimeters, and the length of $\overline{E F}$ is 12 centimeters.
a) Write the length of $\overline{F G}$ in terms of x.
Answer:
Length of DE = 6 cm
Length of EF = 12 cm
Since AEFH is a square, all sides will be equal.
Therefore, the length of HF will be 12cm
HF = HG + GF
12 = x+GF
12-x = x+GF-x
GF = 12-x
Therefore, the length of $\overline{F G}$ will be 12-x

b) Find the area of the shaded region in terms of x.
Answer:

c) Give the value of x for which the shaded region has the greatest area. What is the shape of the shaded region for the value of x you have given?
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 21$
Answer:

Question 3.
Figure ABCD is a square. Point S is in the middle of $\overline{A D}$, and point T is in the middle of and $\overline{C D}$. What fraction of the square is shaded?
$Math in Focus Grade 6 Chapter 8 Lesson 10.4 Answer Key Area of Composite Figures 22$
Answer:
Given that the figure ABCD is a square.
Point S is in the middle of $\overline{A D}$, and point T is in the middle of and $\overline{C D}$.
To find the fraction of shaded square we need to find the area of trinagle STB.
Let us assume each side length of the square as 1m.
AD=DC=CB=BA=1m
Since S and T are the midpoints of AD and DC.
Length of DS = Length of SA = $\frac{1}{2}$ m
Length of DT = Length of TC = $\frac{1}{2}$ m
ASB forms a right angle triangle; AB²+AS² = SB²
1²+($\frac{1}{2}$)² = SB²
1+$\frac{1}{4}$ = SB²
$\frac{5}{4}$ = SB²
SB = √$\frac{5}{4}$
SB = $\frac{√5}{2}$
In triangle DST
ST²=($\frac{1}{2}$)² + ($\frac{1}{2}$)²
= $\frac{1}{4}$ + $\frac{1}{4}$
= $\frac{2}{4}$
= $\frac{1}{2}$
ST = √$\frac{1}{2}$
Draw a perpendicular segment from BU on ST
$Math in Focus Grade 6 Chapter 9 Lesson 9.2 Guided Practice Answer Key img_20$
Now BUS will be a right angle triangle,
BS²=BU²+US²
Length of SU will be $\frac{1}{2}$ of ST
Length of SU = $\frac{1}{2}$ × $\frac{1}{√2}$ = $\frac{1}{2√2}$
($\frac{√5}{2}$)² = BU²+ ($\frac{1}{2√2}$ )²
$\frac{5}{4}$ = BU²+$\frac{1}{8}$
$\frac{5}{4}$–$\frac{1}{8}$ = BU²
($\frac{5}{4}$×$\frac{2}{2}$)-$\frac{1}{8}$ = BU²
$\frac{10}{8}$–$\frac{1}{8}$ = BU²
$\frac{9}{8}$= BU²
BU=√$\frac{9}{8}$
BU=$\frac{3}{2√2}$
Area of STB:
Length of ST=$\frac{1}{√2}$
Height = $\frac{3}{2√2}$
Area of triangle STB=$\frac{1}{2}$×b×h
=$\frac{1}{2}$×$\frac{1}{√2}$×$\frac{3}{2√2}$
= 2×√2×2√2
= 2×2×2=8(√2×√2=2)
= $\frac{3}{8}$ sq.m

Math in Focus Grade 6 Chapter 10 Review Test Answer Key

September 28, 2024September 26, 2024 by Hillary

Practice the problems of Math in Focus Grade 6 Workbook Answer Key Chapter 10 Review Test to score better marks in the exam.

Math in Focus Grade 6 Course 1 B Chapter 10 Review Test Answer Key

Concepts and Skills

Identify a base and a height of each triangle.

Question 1.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 1$
Answer:
Assume a side as the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle.
Let us assume the side BC as the base, then the segment AD which is drawn perpendicular between the base and the opposite vertex will be the height.
Thus, base of triangle = BC
Height of the triangle = AD.

Question 2.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 2$
Answer:
Assume a side as the base of triangle and the perpendicular segment between the base and the opposite vertex will be the height of the triangle.
Let us assume the side NP as the base, then the segment MQ which is drawn perpendicular between the base and the opposite vertex will be the height.
Thus, base of triangle = NP
Height of the triangle = MQ

Find the area of each figure.

Question 3.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 3$
Answer:
If we assume the line segment measuring 28cm as the base, then the height will be the perpendicular line segment between the base and the opposite vertex which is 8cm.
Area of triangle = $\frac{1}{2}$×28×8
= $\frac{1}{2}$×224
= 224÷2
= 112 sq.cm
Thus, the area for the given polygon will be 112 sq.cm

Question 4.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 4$
Answer:
The given polygon is parallelogram.
Let us assume the base of the parallelogram as 46.3 cm, then the height will be the perpendicular line segment between the base and the opposite vertex which is 15 cm.
Area of parallelogram = base×height
= 46.3×15
= 694.5 sq.cm
Thus, the area for the given polygon will be 694.5 sq.cm

Question 5.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 5$
Answer:
The given polygon is similar to trapezium.
A trapezium has a pair of parallel sides, measuring 50cm and 18cm.
The height will be the perpendicular line segment between the base and the opposite vertex which is 31.5 cm.
Area of trapezoid = $\frac{1}{2}$×(sum of bases)×height
= $\frac{1}{2}$×(50+18)×31.5
= $\frac{1}{2}$×68×31.5
= 2142÷2
= 1071 sq.cm
Thus, the area for the given polygon will be 1071 sq.cm

Find the area of the shaded region.

Question 6.
The area of the regular octagon below is 560 square inches.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 6$
Answer:
A regular octagon will be made up of her 8 identical triangles.
Thus, the area of octagon will be equal to sum of the 8 identical triangles.
Given that the area of the regular octagon is 560 sq.in
Area of octagon = 8 × Area of triangle
560 = 8 × Area of triangle
560÷8 = (8 × Area of triangle)÷8
Area of triangle = 70sq.in
The shaded region is half of the triangle.
Area of shaded region = $\frac{1}{2}$×Area of triangle
= $\frac{1}{2}$×70
= 35 sq.in
Thus, the area of shaded region is 35 sq.in

Problem Solving

Find the area of the shaded region.

Question 7.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 7$
Answer:
The given figure has two triangles and a square.
Length of square is 15 cm.
Area of square will be length×length
= 15×15
= 225 sq.cm
The area of shaded region can be calculated by subtracting the area of square from the area of parallelogram.
So, calculate the area of parallelogram.
Area of parallelogram = base×height
The base of parallelogram is 15+20=35 cm and
the height will be 15 cm.
= 35×15
= 525 sq.cm
Area of shaded region = Area of parallelogram – Area of square
= 525 – 225
= 300 sq.cm

Solve.

Question 8.
Figure ABCD is a parallelogram. BC is 16 centimeters, CD is 12 centimeters, and AH is 10 centimeters.
a) Find the area of parallelogram ABCD.
Answer:
Given that figure ABCD is a parallelogram.
BC is 16 centimeters and AH is 10 centimeters.
Area of parallelogram ABCD = base × height
= BC × AH
= 16×10
= 160 sq.cm

b) Find the length of $\overline{\mathrm{AK}}$. Round your answer to the nearest tenth of a centimeter.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 8$
Answer:
ABCD is a parallelogram with two triangles ABC and ACD.
Area of parallelogram = Area of triangle ABC + Area of triangle ACD
The area of parallelogram is 160 sq.cm
Area of triangle ABC= $\frac{1}{2}$×base×height
= $\frac{1}{2}$×BC×AH
= $\frac{1}{2}$×16×10
= 160÷2
= 80 sq.cm
Area of parallelogram = 80 + Area of triangle ACD
160 = 80 + Area of triangle ACD
160-80 = 80 + Area of triangle ACD – 80
80 = Area of triangle ACD
Length of AD = 12cm
Area of triangle ACD = $\frac{1}{2}$×base×height
80 = $\frac{1}{2}$×12×AK
80 = 6×AK
80÷6 = (6×AK)÷6
13.3 = AK
13.3 when rounded of to nearest tenth will be 13
The length of AK = 13cm

Question 9.
Figure ABCD is a trapezoid. The length of $\overline{\mathrm{BC}}$ is 36 centimeters. The areas of triangles ABC and ACD are in the ratio 1.5:1. Find the length of $\overline{\mathrm{AD}}$.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 9$
Answer:
Given that figure ABCD is a trapezoid.
The length of $\overline{\mathrm{BC}}$ is 36 centimeters.
Area of triangle ABC = $\frac{1}{2}$×base×height
= $\frac{1}{2}$×36×10
= 360÷2
= 180 sq.cm
The areas of triangles ABC and ACD are in the ratio 1.5:1
$\frac{ABC}{ACD}$ = $\frac{1.5}{1}$
Area of triangle ABC = 1.5×Area of triangle ACD
180 = 1.5×Area of triangle ACD
180÷1.5 = (1.5×Area of triangle ACD)÷1.5
120 = Area of triangle ACD
Area of triangle ACD = $\frac{1}{2}$×base×height
120 = $\frac{1}{2}$×AD×10
120 = AD×5
120 ÷ 5 = (AD×5)÷5
24 = AD
Thus, the length of $\overline{\mathrm{AD}}$ is 24 cm.

Question 10.
Parallelogram PRTVis made up of triangle PQV, triangle QUV, and trapezoid QRTU. The area of parallelogram PRTV is 96 square feet. The lengths of $\overline{\mathrm{TU}}$ and $\overline{\mathrm{UV}}$ are equal. Find the area of triangle QUV.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 10$
Answer:
Parallelogram PRTV is made up of triangle PQV, triangle QUV, and trapezoid QRTU.
The area of parallelogram PRTV is 96 sq.ft. The lengths of TU is equal to UV.
Area of parallelogram PRTV = 96
Area of parallelogram PRTV = base×height
96 = base×8
96÷8 = (base×8)÷8
12 = base
TV is the base of the parallelogram PRTV which is equal to 12 ft.
TV = VU+UT
12 = VU+VU
12 = 2×VU
12÷2 = (2×VU)÷2
6 = VU
Length of VU is 6 ft.
Area of triangle QUV = $\frac{1}{2}$×base×height
= $\frac{1}{2}$×UV×8
= $\frac{1}{2}$×6×8
= 48÷2
= 24 ft

Question 11.
Charles drew a regular hexagon and divided it into two identical trapezoids. The side length of the hexagon is 16 centimeters, and the length of the diagonal shown is 32 centimeters. Charles measured the height of one of the trapezoids and found that the height was 13.9 centimeters. Find the area of the hexagon.
$Math in Focus Grade 6 Chapter 10 Review Test Answer Key 11$
Answer:
A regular hexagon and divided it into two identical trapezoids.
The side length of the hexagon is 16 cm and the height of one of the trapezoids is 13.9 cm. The length of the diagonal shown is 32 cm.
The side length makes with the height of the trapezium.
Pythagoras theorem is applied to get the base of the triangle.
base² = 16²-(13.9)²
base² = 256-193.21
base² = 62.79
base will be the square root of 62.79
Base length will be 7.9 cm
The length of opposite diagonal can be calculated by subtracting the sum of base length of two triangle from the length of diagonal.
Length of the parallel side = 32-(7.9+7.9)
= 16.2 cm
Area of trapezoid = $\frac{1}{2}$×(sum of bases)×height
= $\frac{1}{2}$×(32+16.2)×13.9
= $\frac{1}{2}$×48.2×13.9
= $\frac{1}{2}$×669.9
= 334.9 sq.cm
Area of hexagon = Area of trapezoid + Area of trapezoid
= 334.9+334.9
= 669.8 sq.ft
Thus, the area of hexagon is 669.8 sq.ft

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

September 28, 2024September 26, 2024 by 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 $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 1$ with = or ≠, and each $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 2$ with the correct value.

Question 1.
For what value of x will x + 3 = 7 be true?
If x = 1, x + 3 = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 2$ + 3
= $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 2$ ( $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 3$ 7)
If x = 2, x + 3 = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 2$ + 3
= $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 2$ ( $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 3$ 7)
If x = 4, x + 3 = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 2$ + 3
= $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 2$
x + 3 = 7 is true when x = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 2$ .
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 $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 4$ with + or —, and each $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 5$ with the correct value.

Question 8.
Solve x + 8 = 19.
x + 8 = 19
x + 8 $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 4$ $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 5$ = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 4$ $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 5$
x = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 5$
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 $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 6$ with × or ÷, and $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 7$ with the correct value.

Question 13.
Solve 3x = 27.
3x = 27
3x $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 8$ = 27 $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 9$
x = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 10$
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 $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 11$ with +, —, × or ÷, and $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 12$ with the correct value.

Question 18.
Solve x + $\frac{3}{7}$ = $\frac{5}{7}$
x + $\frac{3}{7}$ = $\frac{5}{7}$
x + $\frac{3}{7}$ $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 13$ = $\frac{5}{7}$ $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 14$
x = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 15$
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 + $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 16$ = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 16$
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 $\frac{2}{5}$.
$Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 16$ x = $Math in Focus Grade 6 Chapter 8 Lesson 8.1 Answer Key Solving Algebraic Equations 16$
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).

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

September 28, 2024September 26, 2024 by Hillary

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.
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 1$
Miguel has $Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 2$ baseball cards.
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 3$
b) If Miguel has k baseball cards, express k in terms of h.
k = $Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 4$ + $Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 4$
c) State the independent and dependent variables.
Independent variable: $Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 2$ , dependent variable: $Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 2$
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.
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 5$
Answer:
Equation: q = p – 2,
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-1$

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:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-2$

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.
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 6$
Answer:
Equation: y = x + 6,
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-3$

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.
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 7$
Answer:
Equation: r = 4b,
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-4$

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.
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 8$
Answer:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-5$

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:

$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 9$
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,

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.
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 10$
Answer:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-6$

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:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-7$

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:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-8$

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.
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 11$
Answer:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-9$

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:

$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations 12$
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:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-10$

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:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-11$

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:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-12$

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:
$Math in Focus Grade 6 Chapter 8 Lesson 8.2 Answer Key Writing Linear Equations-13$

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.

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

September 28, 2024September 26, 2024 by Hillary

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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 1$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 2$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 3$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 4$
Answer:
A cuboid can be formed when the above solid is folded.
Therefore, a cuboid is formed from the given net form.

Question 3.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 5$
Answer:
A cube can be formed when the above solid is folded.
Therefore, a cube is formed from the given net form.

Question 4.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 6$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 7$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 8$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 9$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 10$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 11$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 12$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 13$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 14$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 15$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 16$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 17$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 18$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 19$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 20$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 21$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 22$
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.
$Math in Focus Grade 6 Course 1 B Chapter 12 Lesson 12.1 Answer Key Nets of Solids img_1$

Decide if each net will form a prism. Answer Yes or No.
Question 16.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 23$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 24$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 25$
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.
$Math in Focus Grade 6 Chapter 12 Lesson 12.1 Answer Key Nets of Solids 26$
Answer:
$Math-in-Focus-Grade-6-Chapter-12-Lesson-12.1-Answer-Key-Nets-of-Solids-26$

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

September 28, 2024September 26, 2024 by Hillary

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.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 1$
Answer:
It is a Rectangular prism.
It has 2 bases, 6 faces, 12 edges and 8 vertices.

Question 2.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 2$
Answer:
It is a triangular prism.
It has 2 bases, 5 faces, 9 edges and 6 vertices.

Question 3.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 3$
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.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 4$
Area = $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$
= $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ 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.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 6$
Area = $\frac{1}{2}$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$
= $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ 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.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 7$
Area = $\frac{1}{2}$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • ( $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ + $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ )
= $\frac{1}{2}$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$
= $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ 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.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 8$
Area = $\frac{1}{2}$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • ( $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ + $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ )
= $\frac{1}{2}$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$
= $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ 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.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 9$
Volume = $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$
= $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ 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.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 10$
Volume = $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$
= $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ 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.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 11$
Volume = $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$
= $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ 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.
$Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 12$
Volume = $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ • $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$
= $Math in Focus Grade 6 Chapter 12 Answer Key Surface Area and Volume of Solids 5$ 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.