## Math in Focus Grade 5 Chapter 9 Practice 2 Answer Key Multiplying by Tens, Hundreds, and Thousands

Practice the problems of Math in Focus Grade 5 Workbook Answer Key Chapter 9 Practice 2 Multiplying by Tens, Hundreds, and Thousands to score better marks in the exam.

## Math in Focus Grade 5 Chapter 9 Practice 2 Answer Key Multiplying by Tens, Hundreds, and Thousands

Complete. Draw chips and use arrows to show how the chips move. Then fill in the blanks.

Question 1.

12 x 10 = 120
2 x 10 = 20
0.2 x 10 = 2
0.12 x 10 = 1.2
Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Multiply.
Question 2.

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 3.

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 4.

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 5.

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 6.
10 × 7.9 = ___

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 7.
10 × 4.8 = ___

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 8.
10 × 27.54 = ___

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 9.
10 × 12.009 = ____

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 10.
0.7 × ___ = 7
0.7 x 10 = 7

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 11.
15.72 × ___ = 157.2
15.72 x 10 = 157.2

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 12.
10 × ___ = 534.2

10 x 53.42

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 13.
___ × 10 = 19.07
1.907 x 10 = 190.7

Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Complete

Example
8 × 50 = (8 × 5) × 10
= 40 × 10
= 400
So, 8 × 50 = 400

Question 14.
0.8 × 50 = (0.8 × 5) × ___
= ___ × 10
= ____
So, 0.8 × 50 = ___
0.8 × 50 = (0.8 × 5) × 10
= 4.0 × 10
= 40
So, 0.8 × 50 = 40
Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 15.
0.88 × 50 = (0.88 × ___) × 10
= ___ × 10
= ____
So, 0.88 × 50 = ___
0.88 × 50 = (0.88 × 5) × 10
= 4.4 × 10
= 44
So, 0.88 × 50 = 44
Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Find each product.

Question 16.
0.9 × 40 = ____
0.9 × 40 = (0.9 × 4) × 10
= 3.6 × 10
= 36
So, 0.9 × 40 = 36
Explanation:
Separate 40 as 4 ones and 1 tenths,
then multiply with ones first and later tens.

Question 17.
1.5 × 60 = ___
1.5 × 60 = (1.5 × 6) × 10
= 9.0 × 10
= 90
So, 1.5 × 60 = 90

Explanation:
Separate 60 as 6 ones and 1 tenths,
then multiply with ones first and later tens.

Question 18.
0.05 × 80 = ____
0.05 × 80 = (0.05 × 8) × 10
= 0.4 × 10
= 4
So, 0.05 × 80 = 4
Explanation:
Separate 80 as 8 ones and 1 tenths,
then multiply with ones first and later tens.

Question 19.
9.17 × 70 = ___
9.17 x 70 = (9.17 x 7) x 10
=64.19 x 10
=641.9
So, 9.17 x 70 = 641.9
Explanation:
Separate 70 as 7 ones and 1 tenths,
then multiply with ones first and later tens.

Question 20.
6.358 × 30 = ___
6.358 x 30 = (6.358 x 3) x 10
=19.074 x 10
=190.74
So,6.358x 30 = 190.74
Explanation:
Separate 30 as 3 ones and 1 tenths,
then multiply with ones first and later tens.

Question 21.
34.6 × 50 = ____
34.6 x 50 = (34.6 x 5) x 10
= 173 x 10
= 1730
So, 34.6 x 50 =1730
Explanation:
Separate 50 as 5 ones and 1 tenths,
then multiply with ones first and later tens.

Question 22.
41.32 × 60 = ___
41.32 x 60 = (41.32 x 6) x 10
= 247.92 x 10
= 2479.2
So, 41.32 x 60= 2479.2
Explanation:
Separate 60 as 6 ones and 1 tenths,
then multiply with ones first and later tens.

Question 23.
23.05 × 40 = ___
23.05 x 40 = (23.05 x 4) x 10
=92.2 x 10
=922
So, 23.05 x 40 = 92.2
Explanation:
Separate 40 as 4 ones and 1 tenths,
then multiply with ones first and later tens.

Question 24.

Explanation:
1.3 x 100 = (1.3 x10) x 10
=13 x 10
=130
So, 1.3 x 100 = 130
Question 25.

Explanation:
6.8 x 100 = (6.8 x 10) x 10
= 68 x 10
= 680
So, 6.8 x 100 = 680

Question 26.

Explanation:
4.196 x 100 = (4.196 x 10) x 10
= 41.96 x 10
= 419.6
So, 4.196 x 100 = 419.6

Question 27.

Explanation:
74.3 x 100 = (74.3 x 10) x 10
= 743 x 10
= 7430
So, 74.3 x 100 = 7430

Question 28.
46.8 × 1oo = ______
Explanation:
46.8 x 100 = (46.8 x 10) x 10
=468 x 10
= 4680

Question 29.
4.68 × 100 = ___
Explanation:
4.68 x 100 = (4.68 x 10) x 10
=46.8 x 10
= 468
So, 4.68 x 100 = 468

Question 30.
5.095 × 100 = ______
Explanation:
5.095 x 100 = (5.095 x 10 )x 10
= 50.95 x 10
= 509.5
So, 5.095 x 100 = 509.5

Question 31.
100 × 50.95 = ____
Explanation:
50.95 x 100 = (50.95 x 10) x 10
= 509.5 x 10
= 5095
So, 50.95 x 100 = 5095

Multiply.

Question 32.

Explanation:
1.8 x 1000 = (1.8 x 10) x 10 x 10
= 18 x 10 x 10
=180 x 10
=1800
So, 1.8 x 1000 = 1800
Question 33.

Explanation:
2.1 x 1000 = (2.1 x 10) x 10 x 10
=21 x 10 x 10
= 210 x 10
= 2100
So, 2.1 x 1000 = 2100

Question 34.

Explanation:
9.097 x 1000 = (9.097 x 10) x 10 x 10
=90.97 x 10 x 10
= 909.7 x 10
=9097
So, 9.097 x 1000 = 9097

Question 35.

Explanation:
7.007 x 1000 = (7.007 x 10) x 10 x 10
= 70.07 x 10 x 10
= 700.7 x 10
= 7007
So, 7.007 x 1000 = 7007

Question 36.
2.74 × 1,000 = ______
Explanation:
2.74 x 1000=  (2.74  x 10) x 10 x 10
=27.4 x 10 x 10
=274 x 10
= 2740

Question 37.
27.4 × 1,000 = ____
Explanation:
27.4 x 1000 = (27.4 x 10) x 10 x 10
=274 x 10 x10
=2740 x 10
27400

Question 38.
1,000 × 10.81 = ______
Explanation:
10.81 x 1000 =(10.81 x 10) x 10 x 10
= 108.1 x10 x 10
= 1081 x 10
= 10810
So, 10.81 x 1000 = 10810

Question 39.
108.1 × 1,000 = ____
Explanation:
108.1 x1000 =(108.1 x 10) x 10 x 10
= 1081 x 10 x 10
= 10810 x 10
= 108100
So, 108.1 x 1000 = 108100

Complete.

Example
1.2 = 0.12 × 10
= 0.012 × 100

Question 40.
360 = 36 × ____
= 3.6 × ____
= 0.36 × _____
360 = 36 × 10
= 3.6 × 100
= 0.36 × 1000
Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 41.
438 = __ × 10
= ___ × 100
= ___ × 1,000
438 = 43.8 × 10
= 4.38 × 100
=0.438 × 1,000
Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Question 42.
7,256 = ______ × 10
= ____ × 100
= ____ × 1,000
7,256 = 725.6 × 10
= 72.56 × 100
= 7.256 × 1,000
Explanation:
To multiply decimals, first multiply as if there is no decimal.
Next, count the number of digits after the decimal in each factor.
Finally, put the same number of digits behind the decimal in the product.

Multiply.

Example
0.3 × 700 = (0.3 × 7) × 1oo
= 2.1 × 1oo = 210
So, 0.3 × 700 = 210

Question 43.
0.003 × 700 = (0.003 × __) × 100
= ____ × 1oo = ____
So, 0.003 × 700 = ___.
0.003 × 700 = (0.003 × 7) × 100
= 0.021 × 1oo = 2.1
So, 0.003 × 700 = 2.1.
Explanation:
The place value of a number is the value represented by a digit in a number based on its position in the number.
While a place value is the value a digit holds to be at the place in the number, on the other hand.
First separate the hundredth place value,
and multiply the given numbers with the number .

Question 44.
0.03 × 2,000 = (0.03 × __) × 1,000
= ______ × 1,000 = ______
So, 0.03 × 2,000 = ____.
0.03 × 2,000 = (0.03 × 2) × 1,000
= 0.06 × 1,000 = 60
So, 0.03 × 2,000 = 60.
Explanation:
The place value of a number is the value represented by a digit in a number based on its position in the number.
While a place value is the value a digit holds to be at the place in the number, on the other hand.
First separate the place value,
and multiply the given numbers with the number .

Question 45.
0.003 × 2,000 = (0.003 × __________) × 1,000
= __________× 1,000 = _______.
So, 0.003 × 2000 = __________.
0.003 × 2,000 = (0.003 × 2) × 1,000
= 0.006× 1,000 = 6
So, 0.003 × 2000 = 6.
Explanation:
The place value of a number is the value represented by a digit in a number based on its position in the number.
While a place value is the value a digit holds to be at the place in the number, on the other hand.
First separate the place value,
and multiply the given numbers with the number .

Find each product.

Question 46.
4.5 × 200 = _______
Explanation:
(4.5 x 2) x 10 x 10
= 9 x 10 x 10
= 90 x 10
= 900
So, 4.5 x 200 = 900

Question 47.
0.49 × 300 = ___
Explanation:
0.49 x 300 = (0.49 x 3) x 10 x 10
= 1.47 x 10 x 10
= 14.7 x 10
= 147
So, 0.49 x  300 = 147

Question 48.
3.148 × 500 = ______
Explanation:
3.148 x 500 = (3.148 x 5) x 10 x 10
= 15.74 x 10 x 10
= 157.4 x 10
= 1574
So, 3.147 x 500 = 1574

Question 49.
2.27 × 700 = ___
Explanation:
2.27 x 700 = (2.27 x 7) x 10 x 10
= 15.89 x10 x10
= 158.9 x 10
= 1589
So , 2.27 x 700 = 1589

Question 50.
900 × 3.18 ______
Explanation:
3.18 x 900 = (3.18 x 9) x 10 x 10
= 28.62 x 10 x 10
= 286.2 x 10
= 2862
So, 3.18 x 900 = 2862

Question 51.
1.8 × 2,000 = ___
Explanation:
1.8 x 2000 = (1.8 x 20) x 10 x 10
= 36 x 10 x 10
=360 x 10
=3600
So, 1.8 x 2000 = 3600

Question 52.
4,000 × 2.5 = _______
Explanation:
2.5 x 4000 = (2.5 x 40) x 10 x 10
= 100 x10 x10
= 1000 x 10
= 10000
So, 2.5 x 4000 = 10000

Question 53.
72.5 × 6,000 = ___
Explanation:
72.5 x 6000 = ( 72.5 x 60) x 10 x 10
= 4350 x10 x10
= 43500 x 10
= 435000
So, 72.5 x 6000 = 435000

Question 54.
1.75 × 8,000 = ______
Explanation:
1.75 x 8000 = (1.75 x 80) x 10 x 10
= 140 x10 x10
= 1400 x10
= 14000
So 1.75 x 8000 = 14000

Question 55.
4.19 × 9,000 = ___
Explanation:
4.19 x 9000 = (4.19 x90) x 10 x 10
= 377.1 x10 x10
= 3771 x 10
= 37710
So, 4.19 x 9000 = 37710

శివ అష్టోత్తర శతనామావళిః

This handy Math in Focus Grade 7 Workbook Answer Key Chapter 10 Probability detailed solutions for the textbook questions.

## Math in Focus Grade 7 Course 2 B Chapter 10 Answer Key Probability

### Math in Focus Grade 7 Chapter 10 Quick Check Answer Key

Solve.

Question 1.
Express 10 ounces out of 25 ounces of baking flour as a fraction in the simplest form.
Explanation:
In mathematics, a fraction is a number that represents a part of a whole. It consists of a numerator and a denominator. The numerator represents the number of equal parts of a whole, while the denominator is the total number of parts that make up said whole.

This can be written as 10/25
– In the fraction 10/25, the numerator is 10 and the denominator is 25.
– A more illustrative example could involve a box with 25 ounces of baking flour. 1 of those 10 ounces would constitute the numerator of a fraction, while the total of 25 ounces that comprises the whole box would be the denominator.
– Note that the denominator of a fraction cannot be 0, as it would make the fraction undefined.

Question 2.
12 out of 40 pieces of fruit in a basket are lemons. What fraction of the pieces of fruit are lemons? Write your answer in the simplest form.
Explanation:
The total number of fruits in a basket=40
The number of lemons is there=12
We need to write the fraction of the pieces of fruit are lemons=X

– In the fraction 12/40, the numerator is 12 and the denominator is 40.
– A more illustrative example could involve a box with 40 pieces of fruits. 12 of those are lemons would constitute the numerator of a fraction, while the total of 40 pieces of fruits that comprises the whole box would be the denominator.
– Note that the denominator of a fraction cannot be 0, as it would make the fraction undefined.
X=12/40

Question 3.
If there are 36 boys in a group of 50 students, what percent of the students are girls?
Explanation:

The total number of students=50
Number of boys  Number of girls

Write each fraction as a percent. Round your answer to 2 decimal places when you can.

Question 4.
$$\frac{5}{8}$$
This can be written in a fraction like 5/8.
When it converted into percentage then it becomes 62.5%.
Calculation:
= (5/8) × 100%
= 62.5%
Therefore, 62.5 rounded to 2 decimal places is 62.50.

Question 5.
$$\frac{2}{5}$$
This can be written in a fraction like 2/5.
When it is converted into percentage then it becomes 40%.
Calculation:
= (2/5) × 100%
= 40%
Therefore, 40 rounded to 2 decimal places is 40.00.

Question 6.
$$\frac{24}{9}$$
Explanation:
This can be written in a fraction like 24/9.
When it is converted into a percentage then it becomes 266.66666667%
Calculation:
= (24/9) × 100%
= 266.66666667%
Therefore, 266.6666 rounded to 2 decimal places is 266.67

Write each percent as a fraction or a mixed number in simplest form.

Question 7.
54%
Explanation:
To convert percentages to fractions:
Step 1: Write down the percent divided by 100 like this: percent/100.
Step 2: If the percent is not a whole number, then multiply both top and bottom by 10 for every number after the decimal point. (For example, if there is one number after the decimal, then use 10, if there are two then use 100, etc.)
Therefore, 54/100=0.54
This can be simplified as:

The mixed number can be written as:
since 27/50 is a proper fraction, it cannot be written as a mixed number.
If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction. In that case, you could convert it into a whole number or mixed number fraction.
Therefore, 27/50 = Proper Fraction.

Question 8.
19.5%
To convert percentages to fractions:
Step 1: Write down the percent divided by 100 like this: percent/100.
Step 2: If the percent is not a whole number, then multiply both top and bottom by 10 for every number after the decimal point. (For example, if there is one number after the decimal, then use 10, if there are two then use 100, etc.)
Therefore, 19.5/100=0.195
This can be simplified as:

The mixed number can be written as:
since 39/200 is a proper fraction, it cannot be written as a mixed number.
If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction. In that case, you could convert it into a whole number or mixed number fraction.
Therefore, 39/200 = Proper Fraction.
The fraction is already in mixed form.

Question 9.
1.4%
To convert percentages to fractions:
Step 1: Write down the percent divided by 100 like this: percent/100.
Step 2: If the percent is not a whole number, then multiply both top and bottom by 10 for every number after the decimal point. (For example, if there is one number after the decimal, then use 10, if there are two then use 100, etc.)
Therefore, 1.4/100=0.014
This can be simplified as:

The mixed number can be written as:
since 7/500 is a proper fraction, it cannot be written as a mixed number.
If the numerator is greater than or equal to the denominator of a fraction, then it is called an improper fraction. In that case, you could convert it into a whole number or mixed number fraction.
Therefore, 7/500 = Proper Fraction.
The fraction is already in mixed form.

Question 10.
115%
Explanation:
To convert percentages to fractions:
Step 1: Write down the percent divided by 100 like this: percent/100.
Step 2: If the percent is not a whole number, then multiply both top and bottom by 10 for every number after the decimal point. (For example, if there is one number after the decimal, then use 10, if there are two then use 100, etc.)
Therefore, 115/100=1.15
This can be simplified as:

The mixed number can be written as:
– Convert the fraction to a mixed number by using long division to find the quotient and remainder.
23÷20=1R3
– The quotient will be the whole number in the fraction, and the remainder will be the numerator in the mixed fraction.
23/20=3/20

Write each percent as a decimal.

Question 11.
28%
Explanation:
Percent” means “per 100” or “over 100”. To convert 28% to a decimal rewrite 28 percent in terms of per 100 or over 100.
28% = 28 over 100 or,
28%= 28/100
28 over 100 is the same as 28 divided by 100. Completing the division we get:
28 ÷ 100 = 0.28
Therefore, we have shown that:
28% = 0.28
Simplified conversion:
Remove the percent sign % and divide by 100.
28 ÷ 100 = 0.28
Shortcut conversion:
Move the decimal point 2 places to the left and remove the percent sign %
28% becomes 0.28.

Question 12.
9%
Explanation:
“Percent” means “per 100” or “over 100”. To convert 9% to a decimal rewrite 9 percent in terms of per 100 or over 100.
9% = 9 over 100 or,
9%=9/100.
9 over 100 is the same as 9 divided by 100. Completing the division we get:
9 ÷ 100 = 0.09
Therefore, we have shown that
9% = 0.09
Simplified conversion:
Remove the percent sign % and divide by 100.
9 ÷ 100 = 0.09
Shortcut conversion:
Move the decimal point 2 places to the left and remove the percent sign %
9% becomes 0.09

Question 13.
34.5%
Explanation:
“Percent” means “per 100” or “over 100”. To convert 34.5% to a decimal rewrite 34.5 percent in terms of per 100 or over 100.
34.5% = 34.5 over 100 or,
34.5%=34.5/100
34.5 over 100 is the same as 34.5 divided by 100. Completing the division we get:
34.5 ÷ 100 = 0.345
Therefore, we have shown that
34.5% = 0.345
Simplified conversion:
Remove the percent sign % and divide by 100.
34.5 ÷ 100 = 0.345
Shortcut conversion:
Move the decimal point 2 places to the left and remove the percent sign %
34.5% becomes 0.345

Question 14.
256%
Explanation:
“Percent” means “per 100” or “over 100”. To convert 256% to a decimal rewrite 256 percent in terms of per 100 or over 100.
256% = 256 over 100 or,
256%=256/100
256 over 100 is the same as 256 divided by 100. Completing the division we get:
256 ÷ 100 = 2.56
Therefore, we have shown that
256% = 2.56
Simplified conversion:
Remove the percent sign % and divide by 100.
256 ÷ 100 = 2.56
Shortcut conversion:
Move the decimal point 2 places to the left and remove the percent sign %
256% becomes 2.56

Solve.

Question 15.
A flower bed contains marigolds and zinnias. The ratio of marigolds to zinnias is 7 to 11.
a) What fraction of the flowers in the garden are marigolds? Write your answer a? a fraction in simplest form.
Explanation:
The total number of flower beds= 7+11=18
The number of marigolds=7
The number of zinnias=11
The ratio of marigolds to zinnias=7:11
In the above-given question, we need to write the fraction of the flowers in the garden are marigolds.
Already we know that the marigolds are 7.
– In fraction 7/18, the numerator is 7 and the denominator is 18.

– A more illustrative example could involve a box with 18. 7 of those are marigolds would constitute the numerator of a fraction, while the total of 18 flower beds that comprises the whole box would be the denominator.
– Note that the denominator of a fraction cannot be 0, as it would make the fraction undefined.

b) What percent of the flowers in the garden are marigolds? Round your answer to the nearest whole number percent.
The fraction of flowers in the garden are marigods=7/18   (check in the above question (a))
When it is converted into percentages then it becomes 38.888888889%
Calculation:
(7/18) × 100%
= 38.888888889%
= 0.38888888889
The nearest whole number percentage of 38.888888889% is 39%.
Note:
– When rounding percentages, we rounded up or down to the nearest number of decimals you wanted.
– We rounded up if the next decimal was five or above, and down if it was four or less.

Question 16.
A bookcase holds 20 history books, 23 science fiction books, and 49 mystery books.
a) What fraction of the books are science fiction books?
Explanation:
The total number of books a bookcase holds=20+23+49=92
The number of history books=20
The number of science fiction books=23
The number of mystery books=49
In mathematics, a fraction is a number that represents a part of a whole. It consists of a numerator and a denominator. The numerator represents the number of equal parts of a whole, while the denominator is the total number of parts that make up said whole.
This can be written as 23/92
– In the fraction 23/92, the numerator is 23 and the denominator is 92.
.
– A more illustrative example could involve a box with 92 books. 23 of those are science fiction books would constitute the numerator of a fraction, while the total of 92 books that comprises the whole box would be the denominator.
– Note that the denominator of a fraction cannot be 0, as it would make the fraction undefined.

b) What percent of the books are science fiction books?
Explanation:
The fraction of science fiction books=23/92   (check in the above question (a))
When it is converted into percentages then it becomes 25%
Calculation:
(23/92) × 100%
= 25%

Solve.

Question 17.
The table shows the mass of 100 steel bars rounded to the nearest kilogram.

a) Draw a histogram to display this information.
A histogram is the graphical representation of data where data is grouped into continuous number ranges and each range corresponds to a vertical bar.
– The horizontal axis displays the number range.
– The vertical axis (frequency) represents the amount of data that is present in each range.
– The number ranges depend upon the data that is being used.
Histogram graph:
A histogram graph is a bar graph representation of data. It is a representation of a range of outcomes into columns formation along the x-axis. in the same histogram, the number count or multiple occurrences in the data for each column is represented by the y-axis. It is the easiest manner that can be used to visualize data distributions. Let us understand the histogram graphby plotting one for the given below example.

b) How many steel bars have a mass from 10 to 39 kilograms?
Explanation:
The number of steel bars of mass from 10-19=15
The number of steel bars of mass from 20-29=33
The number of steel bars of mass from 30-39=18
Add all the number of steel bars=15+33+18
therefore, the number of steel bars of the mass 10-39 kgs=66

c) What percent of the steel bars have a mass of at least 20 kilograms, but less than 50 kilograms?
Explanation:

The number of steel bars of the mass 20-29 kgs=33
The number of steel bars of the mass 30-39 kgs=18
The number of steel bars of the mass 40-49 kgs=24
The total number of steel bars have a mass of at least 20 kgs, but less than 50 kgs=75
The total number of steel bars=15+33+18+24+10=100
Now first write the fraction.
Fraction=75/100
When it is converted into percentages then it becomes 75%
Calculation:
(75/100) × 100%
= 75%
= 0.75

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## Math in Focus Grade 3 Chapter 19 Answer Key Area and Perimeter

This handy Math in Focus Grade 3 Workbook Answer Key Chapter 19 Area and Perimeter provides detailed solutions for the textbook questions.

## Math in Focus Grade 3 Chapter 19 Answer Key Area and Perimeter

Math Journal

Look at John’s answers for the perimeter of the squares and rectangles.

John’s mistakes are circled. Explain why his answers are not correct.
John added only two sides to find perimeter,
the perimeter formulas for rectangles and  square.
The perimeter of a rectangle is the total distance of its outer boundary.
It is twice the sum of its length and width and it is calculated with the help of the formula:
Perimeter = 2(length + width).
Explanation:
The perimeter of a square is defined as the total length that its boundary covers
The formula to calculate the perimeter of a square is as, mathematically expressed as;
Perimeter of square, (P) = 4 × Side

Example The unit for the perimeter of Figure B should be meter (m).

Question 1.
Perimeter of Figure A: _____________________
Explanation:
The perimeter of a rectangle is the total distance of its outer boundary.
It is twice the sum of its length and width and it is calculated with the help of the formula.
Perimeter = 2(length + width).
P=2(6 + 4) = 2 x 10 = 20cm

Question 2.
Perimeter of Figure C: _____________________
Explanation:
The perimeter of a square is defined as the total length that its boundary covers
The formula to calculate the perimeter of a square is as, mathematically expressed as;
Perimeter of square, (P) = 4 × Side
P = 4 x s
= 4 x 5 = 20 cm

Question 3.
Perimeter of Figure E: _____________________
Explanation:
The perimeter of a square is defined as the total length that its boundary covers.
The formula to calculate the perimeter of a square is as, mathematically expressed as;
Perimeter of square, (P) = 4 × Side

Challenging Practice

Complete.

Question 1.
Draw different rectangles with an area of 12 square centimeters. Then draw different rectangles with an area of 9 square centimeters. How many rectangles can you draw for each area?

2 rectangles can be drawn for area of  area 12 square centimeters and,
1 rectangle can be drawn for area of 9 square centimeters.

Explanation:
The area of rectangle (A) is the product of its length ‘a’ and width or breadth ‘b’.
So, Area of Rectangle = (a × b) square units.
The square is a shape with four equal sides.
The area of a square is defined as the number of square units that make a complete square.
It is calculated by using the formula Area = s × s = s2 in square units.
So, area = 9 square centimeters.

Solve.

Question 2.
Karl bends a piece of wire into a square as shown.

Explanation:
The perimeter of a square is defined as the total length that its boundary covers,
The formula to calculate the perimeter of a square is as, mathematically expressed as;
Perimeter of square, (P) = 4 × Side
P = 4 x 8 = 32 cm

Which of these rectangles can he make using the same piece of wire?

Rectangle B and C perimeter is 32cm.
Explanation:
Perimeter of rectangle A
Perimeter = 2(length + width).
P=2(8 + 4) = 2 x 12 = 24cm

Perimeter of rectangle B
Perimeter = 2(length + width).
P=2(10 + 6) = 2 x 16 = 32cm

Perimeter of rectangle C
Perimeter = 2(length + width).
P=2(11 + 5) = 2 x 16 = 32cm

Perimeter of rectangle D
Perimeter = 2(length + width).
P=2(9 + 8) = 2 x 17 = 34cm

Question 3.
Ally wants to build an exercise pen for her pet rabbit. She has 36 feet of fencing to build a rectangular enclosure in her yard. She wants to carefully plan the length and width of the pen, measuring in units of whole feet.
Find all the possible ways that Ally could build her pen and have a perimeter of 36 feet. Fill in the table below.

Explanation:
Ally wants to build an exercise pen for her pet rabbit.
So, Perimeter = 2(length + width)
Perimeter = 2(1 + 17) = 36 ft
Perimeter = 2(2 + 16) = 36 ft
Perimeter = 2(3 + 15) = 36 ft
Perimeter = 2(4 + 14) = 36 ft
Perimeter = 2(5 + 13) = 36 ft
Perimeter = 2(6 + 12) = 36 ft
Perimeter = 2(7 + 11) = 36 ft
Perimeter = 2(8 + 10) = 36 ft
Perimeter = 2(9 + 9) = 36 ft

Question 4.
What are some of the concerns that Ally needs to think of in planning for the exercise pen?
Perimeter and area of exercise pen.
length and width of exercise pen.
Explanation:
The above are some of the concerns that Ally needs to think of in planning for the exercise pen.

Problem Solving

Solve. Look at this pattern.

What is the area of each figure?

Explanation:
In the above picture each area of the square is measured as 1 square centimeter.
In figure A there is only 1 square box.
In figure B there are 3 square boxes.
In figure C there are 5 square boxes.
If the pattern continues, what will the area of Figure E be? Draw Figure E below.

the area of Figure D & E

Explanation:
In the above picture each area of the square is measured as 1 square centimeter.
In figure D there are 7 square units.
In figure E there are 9 square units.

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## Math in Focus Grade 2 Chapter 8 Practice 2 Answer Key Comparing Masses in Kilograms

Practice the problems of Math in Focus Grade 2 Workbook Answer Key Chapter 8 Practice 2 Comparing Masses in Kilograms to score better marks in the exam.

## Math in Focus Grade 2 Chapter 8 Practice 2 Answer Key Comparing Masses in Kilograms

Look at the pictures. Then fill in the blanks.

Question 1.
The mass of the bag of oranges is ____ kilograms.
The measuring scale is showing 2 kg.
Therefore, the mass of the bag of oranges is 2 kilograms.
Definition:
A Metric measure of mass (which we feel as weight).
The abbreviation is kg.
1 kg = 1000 grams.
1 kg = 2.205 pounds (approximately).

Question 2.
The mass of the bag of potatoes is ___ kilograms.
The measuring scale is showing 3 kg.
Therefore, the mass of the bag of potatoes is 3 kilograms.
Definition:
A Metric measure of mass (which we feel as weight).
The abbreviation is kg.
1 kg = 1000 grams.
1 kg = 2.205 pounds (approximately).

Question 3.
Which bag is heavier? The bag of _______________
Answer: The bag of potatoes is heavier.
Explanation:
if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object
The mass of potatoes is 3 kg.
The mass of oranges is 2 kg.
Potatoes is having great density so the bag of potatoes is heavier.
Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

Question 4.
How much heavier? ___ kg
The mass of potatoes is 3 kg.
Potatoes is having great density so the bag of potatoes is heavier.

Question 5.
The total mass of the bag of oranges and the bag of potatoes is ____ kilograms.
Explanation:
The mass of potatoes is 3 kg
The mass of oranges is 2 kg
The total mass of potatoes and oranges=X
X=3+2
X=5
Therefore, the total mass of potatoes and oranges are 5 kilograms.

Look at the pictures. Then answer the questions.

Question 6.
Which is the heaviest? The _______________
if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object
The mass of chicken is 3 kg.
The mass of fish is 2 kg.
The mass of vegetables is 1 kg.
chicken is having great density so the mass of chicken is heavier.
Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

Question 7.
Which is the lightest? The _______________
if two objects are the same size but one is lightest, the lightest one has a lesser density than the heaviest object
The mass of chicken is 3 kg.
The mass of fish is 2 kg.
The mass of vegetables is 1 kg.
vegetables is having low density so the mass of vegetables is lightest.
Therefore, when both objects are dropped from the same height and at the same time, the lightest object should go up.

Question 8.
Order the items from lightest to heaviest.

The mass of chicken is 3 kg.
The mass of fish is 2 kg.
The mass of vegetables is 1 kg.
We need to arrange from the lightest to heaviest.

Question 9.

If the items are put on a balance scale, do you think the picture above is correct? ____
Why or why not? ____
Explanation:
1. We will compare the weights of a chicken on one side and on the other side there will be fish and vegetables.
2. The mass of chicken is 3 kg.
3. The mass of fish is 2 kg and vegetables are 1 kg.
4. The total mass of fish and vegetables are 2+1=3 kg.
5. Now both chicken, fish, and vegetables are the same size.
6. The measuring scale will be at the same level. But here the scale is showing one is upwards and the other one is downwards.
7. So what the picture is showing wrong.

Fill in the blanks.

The pictures show Ally’s and Roger’s mass.

Question 10.
Ally has a mass of ____ kilograms.
Ally has a mass of 54 kilograms.
Explanation:
Mass is the amount of matter or substance that makes up an object. The units of measurement used to express mass are called kilograms (abbreviated kg). Whether you’re on the moon or Earth, your mass doesn’t change – there’s still the same amount of you. Your weight, which is different from your mass, depends on gravity – it only changes on the moon because there isn’t the same gravitational downward pull.

Question 11.
Roger has a mass of ___ kilograms.
Roger has a mass of 74 kilograms.
Explanation:
Mass is the amount of matter or substance that makes up an object. The units of measurement used to express mass are called kilograms (abbreviated kg). Whether you’re on the moon or Earth, your mass doesn’t change – there’s still the same amount of you. Your weight, which is different from your mass, depends on gravity – it only changes on the moon because there isn’t the same gravitational downward pull.

Question 12.
Who is heavier, Roger or Ally? _______________
Explanation:
The mass of Ally is 54 kg
The mass of Roger is 74 kg.
By comparing both of their masses Roger is having the heaviest mass.
Therefore, Roger is heavier.

Question 13.
How much heavier? ____ kg
The mass of Roger is 74 kg.
Question 14.
What is the total mass of Roger and Ally? ____ kg
The mass of Ally is 54 kg
The mass of Roger is 74 kg.
The total mass of Roger and Ally=X
By adding we get the total of their masses.
X=54+74
X=128.
Therefore, both of their total mass is 128 kilograms.

Read each sentence. Write True or False.

Question 15.
The mass of Bag A is 2 kilograms ____
Bag A is placed at one end and on the other end, we placed 2 ‘1’ kgs.
Explanation:
1. Choose a scale that measures in grams. Make sure the scale can handle the size of objects you plan on weighing. Since a gram is a metric unit of measurement, your scale needs to use the metric system. Scales are available in digital and mechanical models.
2. Weigh an empty container first before putting an item in it. If you plan to measure something you can’t put directly on the scale, weigh the container before putting the item in it.
3. Press the tare button to zero out the scale. The mysterious button labelled “tare” on digital scales is a reset button. Press the tare button after each item you measure on the scale. If you weighed a container, you can fill it now.
4. Set the object you wish to measure on the scale. If you measured a container first, you can now put the object you wish to measure inside the container. The scale will then calculate the heaviness of your object.
5. Finish weighing the object on the scale. Wait for the scale’s or needle to come to a stop. When it finishes moving, read the number to find out how heavy the object is. Make sure that the weight is in grams. Then, remove your object and hit the tare button again to reset the scale.

Question 16.
Bag B has the same mass as the total mass of both Bag A and Bag C ____
Explanation:
1. When we placed bag B on the one end and bag A and bag C on the other end.
2. Then watch the weighing balance.
3. Finish weighing the object on the scale. Wait for the scale’s or needle to come to a stop. When it finishes moving, read the number to find out how heavy the object is. Make sure that the weight is in grams. Then, remove your object and hit the tare button again to reset the scale.
4. Both are balanced in one place equally. So their mass is equal. Therefore, it becomes true.

Question 17.
The mass of Bag A is different from the mass of Bag B. ____
Explanation:
1. In the second picture, we kept bag B on one side and bag A and bag C on another side.
2. We already know the mass of bag A is 2 kgs.
3. From this we have to say bag B is having more mass than bag A.
4. So definitely both masses are different from one another.

Question 18.
Bag B is heavier than Bag C. _____
Explanation:
1. In the second picture, we kept bag B on one side and bag A and bag C on another side.
2. We already know the mass of bag A is 2 kgs. We don’t know about the mass of bag C. But from the above picture we can say 1 bag is equal to 2 bags.
3. From this we have to say that bag B is having more mass than bag C.
4. So definitely both masses are different from one another.

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## Math in Focus Grade 2 Chapter 8 Practice 5 Answer Key Real-World Problems: Mass

Practice the problems of Math in Focus Grade 2 Workbook Answer Key Chapter 8 Practice 5 Real-World Problems: Mass to score better marks in the exam.

## Math in Focus Grade 2 Chapter 8 Practice 5 Answer Key Real-World Problems: Mass

Solve.

Question 1.
Angelina has two dogs. The masses of the two dogs are 35 kilograms and 67 kilograms. What is the total mass of the two dogs? The total mass of the two dogs is ___ kilograms.
Explanation:
The mass of one dog=35 kg
The mass of another dog=67 kg
The total mass of two dogs=X
X=35+67
X=102.
Therefore, the total mass of the two dogs is 102 kilograms.

Question 2.
Miguel has a mass of 32 kilograms. He is 5 kilograms lighter than Sal. What is Sal’s mass?
Sal’s mass is ___ kilograms.

Explanation:
The mass of Miguel has=32 kg
The number of kilograms of Miguel lighter than sal=5
The sal’s mass=X
Here Miguel’s mass is 5 kilograms less than sal. Now we have to calculate the sal’s mass.
Add 5 to Miguel’s mass then we get the sal’s original mass.
X=32+5
X=37.
Therefore, the mass of sal’s has 37 kg.

Question 3.
Mr. Souza needs 400 grams of clay to make a small statue. He has only 143 grams of clay. How much more clay does he need? He needs ___ grams more clay.
Explanation:
The number of grams Souza needs clay to make a statue=400
The number of grams she has now=143.
The number of more clay she needs=X
subtract total clay she needs and she has clay now
X=400-143
X=257.
Therefore, 257 grams more is needed.

Solve.

Question 4.
Ali has a mass of 25 kilograms. Tyrone is 6 kilograms heavier than Ali. What is their total mass?
Their total mass is ___ kilograms.
Explanation:
The mass of an Ali =25 kg
The mass of Tyrone=25+6=31.   ( 6kgs heavier than Ali so I added 6 to 25).
The total mass of both=25+31=56.
Therefore, Their total mass is 56 kilograms.

Question 5.
Twyla buys a bag of onions with a mass of 750 grams. She uses 100 grams of the onions for lunch. She uses 480 grams of onions for dinner. What is the mass of the onions that are left?
The mass of onions left is ___ grams.
Explanation:
The mass of onions Twyla buys=750
The number of grams of onions she uses for lunch=100
The number of grams of onions she uses for dinner=480
The total number of grams of onions she used overall=100+480=580
The mass of onions she left with her=X
Subtract the mass of onions she buys and the mass of used onions.
X =750-480
X=170
Therefore, 170 grams of onions she had left with her.

Question 6.
Tim sells 45 kilograms of rice on Monday. He sells 18 kilograms less rice on Tuesday than on Monday. How much rice does he sell in all on the two days?
He sells ___ kilograms of rice in all on the two days.
Explanation:
The number of kilograms of rice Tim sells on monday=45
The number of fewer kilograms Tim sells on Tuesday than on monday=18.
The number of kilograms of rice Tim sells on Tuesday is 45-18=27.
The number of kilograms he sells in all on the two days
Add the kilograms of rice he sells on Monday and Tuesday to get the total.
X=45+27
X=72
Therefore, he sells 72 kgs on those two days.

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Practice the problems of Math in Focus Grade 2 Workbook Answer Key Chapter 8 Mass to score better marks in the exam.

Math Journal

Look at the pictures.

Question 1.
Write sentences to compare the mass of the boxes. Use the words lighter, heavier, lightest and heaviest.

1. The weight of an object is a force that pushes downwards. You can feel the weight of lots of different objects when you hold them. It is the reason we find it difficult to lift some objects at all.
2. Below is a set of scales. Scales can be used to compare the weights of two objects.
3. We place a mass at one end of the scale. There is nothing on the other end.
4. The weight of mass pushes down on this end of the scale and so this left side moves downwards.
5. This causes the other side to rise upwards.

Question 2.
Order the boxes from heaviest to lightest.
___, ____, ____
Explanation:
The heaviest object moves to scale downward.
The lightest object moves upward.

Write True or False.

Example
The stapler is as heavy as the pen.
False

Question 3.
The pen has a mass of 60 grams. _____

Question 4.
The stapler is 70 grams lighter than the pen. ____
Explanation:
The mass of a stapler is 130 grams.
The mass of the pen is 60 grams.
The mass of the stapler is heavier than the mass of the pen.

Question 5.
The book is the heaviest. ___
The mass of the book is about 1 kg.

Question 6.
The total mass of the stapler and the pen is 190 grams. ____
Explanation:
The mass of a stapler is 130 grams.
The mass of the pen is 60 grams.
The total mass=130+60=190.
Therefore, the total mass of both stapler and pen is 190 grams.

Challenging Practice

Fill in the blanks.

Question 1.
In Picture A, 3 are needed to make the needle on the scale point as shown. How many are needed to make the needle point as shown in Picture B?

For picture A, 3 1oog are needed.
For picture B, I represented the below add all the 3’s.
3+3+3=9
therefore, 9 100’s are needed.

Question 2.

The slice of honeydew has a mass of ___ grams.
Explanation:
1 slice of watermelon is 55 grams.
I honeydew and watermelon slice both are having a mass of 100 grams.
We need to calculate honeydew slices. assume it as X.
X=100-55
X=45
Therefore, the honeydew slice is 45 grams.

Question 3.

Box B has a mass of ___ kilograms.
Explanation:
In 1st picture, 22 kgs are given.
The mass of two A’s is 15 kgs.
So the remaining kgs are B.
To calculate the mass of B we need to subtract total kgs and two A’s mass. Assume it as X.
X=22-15
X=7
therefore, the mass of bag B is 7 kgs.

Solve.

a. What is the mass of one ball? ___ g
Explanation:
Already given in the first picture, the total balls and box are having 500 grams. Suppose If we take each ball 200 grams then 2 balls will be 200*2=400.
Therefore, the mass of one ball is 200 grams.

b. What is the mass of the box? ___ g
Explanation:
Already given in the first picture, the total balls and box are having 500 grams. Suppose If we take each ball 200 grams then 2 balls will be 200*2=400. Now coming to the mass of the box is 100 grams.
How 100 grams?
The mass of the two balls is 400 and the remaining 100 grams is the mass of the box.

Chapter Review/Test

Vocabulary

Fill in the blanks with the words below.

Question 1.
A table is ___ a watch.
a table is heavier than a watch.

Question 2.
A ___ is a bigger unit of mass and a ___ is a smaller unit of mass.
A kilogram is a bigger unit of mass.
A Metric measure of mass (which we feel as weight).
The abbreviation is kg.
1 kg = 1000 grams.
1 kg = 2.205 pounds (approximately).
Example: The gold bar has a mass of 1 kg.
A gram is a smaller unit of mass.
A Metric measure of mass (which we feel as weight)
A paperclip weighs about 1 gram
Abbreviation is g
1000 grams = 1 kilogram

Question 3.
To measure how heavy an object is, you find its ____
Mass is a measure of the amount of matter inside an object. An object can be quite small yet has a lot of matter inside of it.
A bar of gold for instance may be quite small and yet has a mass of 2 kilograms (kg)
Talking about kilograms, what is 1 kilogram?
One kilogram is the mass of 1 litre of water in its densest state.
Water is in its densest state when the temperature is close to 4 degrees Celsius.
The following cube will be able to hold 1 litre of water.

The volume of the cube = 10 × 10 × 10 = 1000 cm3
In math 1 liter = 1000 cm3

Question 4.
You use a ___ to measure the mass of an object.
Explanation:
Scales of measurement in math are used to classify and/or quantify variables based on certain properties. Each grade of measurement has relevant properties that are crucial to know. The scale of measurement in math is commonly interpreted in the form of graphs. This can be described as the mechanism of marks at fixed intervals, which clearly explain the link between the units being used and their illustration on the graph. Data of Measurement scales are basically classified under the four scales of measurement that have frequent applications in statistical analysis:
The 4 types of scales of measurement include:-
1. Nominal Scale of Measurement
2. Ordinal Scale of Measurement
3. Interval Scale of Measurement
4. Ratio scales Scale of Measurement.

Question 5.

The spoon is the ___
Explanation:
The cup is having a mass of 250 grams.
The plate is a mass of 400 grams.
The spoon is a mass of 70 grams.
So the spoon is the lightest.

Concepts and Skills

Question 6.
Which bag is heavier? Bag ____
Explanation:
Observe the picture carefully, and then write the masses of their respective bags.

The mass of bag A is 3 kgs
The mass of bag B is 7 kgs.
Here asked the bag which one is heavier.
7>3.
Bag B is having 7kgs which is heavier.

Question 7.
How much heavier is it? ___ kg

The bag B is 7 kgs heavier.

Question 8.
What is the total mass of both bags? ___ kg
The mass of bag A is 3 kgs
The mass of bag B is 7 kgs
The total of both the bags=X
X=3+7
X=10
therefore, both bags combine having 10 kgs.

Question 9.
Which bottle is the lightest? Bottle ____
Explanation:
The mass of bottle A is 350 grams.
The mass of bottle B is 900 grams.
The mass of bottle C is 120 grams.
By comparing all the bottles the lightest one is 120 grams.
So, bottle C is the lightest.

Question 10.
What is the difference in mass between the heaviest and lightest bottle? ___ g
Explanation:

The mass of bottle B is 900 grams which are the heaviest.
The mass of bottle C is 120 grams which are lighter.
To find the difference between both the masses we need to subtract them. Assume it as X. Then we get the answer.
X=900-120
X=780
Therefore, the difference is 780 grams.

Question 11.
What is the total mass of Bottle A and Bottle C? ____ g
Explanation:

The mass of bottle A is 350 grams.
The mass of bottle C is 120 grams.
To find the total we need to add both the masses. Assume it as X.
X=350-120
X=470
Therefore, the total mass is 470 grams.

Question 12.
Order the bottles from lightest to heaviest.

Explanation:

It was like ascending order. It means we need to order from lowest to highest.

Problem Solving

Solve.

Question 13.
Mr. Shepherd has 5 kilograms of rice. He buys another 8 kilograms of rice. How many kilograms of rice does he have?
Mr. Shepherd has ____ kilograms of rice.
Explanation:
The number of kilograms of rice Shepherd has=5
The number of kilograms of rice he buys extra=8
The total number of kilograms he has now=X
To get the total kilograms of rice we need to add both 5 and 8.
X=8+5
X=13
Therefore, 13 kgs of rice he has now.

Question 14.
Claudia has two boxes. The mass of Box A is 980 grams. The mass of Box B is 750 grams. What is the difference in masses between the two boxes?
The difference in masses between the two boxes is ___ grams.
Explanation:
The mass of box A=980 grams.
The mass of box B=750 grams.
The difference between two boxes=X
Subtract both the masses to get the answer.
X=980-750
X=230
Therefore, the difference between the two boxes is 230 grams.

Question 15.
Casey has 500 grams of carrots. He buys another 400 grams of carrots. He uses 725 grams of carrots for a recipe. How many grams of carrots does he have left?
Casey has ___ grams of carrots left.
Explanation:
The number of grams of carrots Casey has=500
The number of grams of carrots he buys extra=400
The total number of grams he has now=500+400=900
The number of grams of carrots he used for recipe=725.
We need to calculate the number of grams of carrots he has left with him after using for the recipe. Assume it X.
Subtract the total number of grams of carrots and use carrots.
X=900-725
X=175
Therefore, Casey has 175 grams of carrots.

Question 16.
Lily’s dog weighs 27 kilograms. Her dog is 2 kilograms heavier than Ben’s dog. Joe’s dog is 5 kilograms heavier than Ben’s dog. What is the mass of Joe’s dog?
The mass of Joe’s dog is ___ kilograms.
Explanation:
The mass of Lily’s dog=27 grams.
The number of kilograms of Lily’s dog is heavier than Ben’s dog=2
The mass of Ben’s dog=27-2=25 grams.
The number of kilograms of Joe’s dog is heavier than Ben’s dog=5
The total mass of Joe’s dog=25+5=30 kilograms.

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Practice the problems of Math in Focus Grade 5 Workbook Answer Key Chapter 6 Area to score better marks in the exam.

Math Journal

Question 1.
Four students found the area of the shaded triangle.

These are their findings.
Zach: 4 × 4 = 16 cm2
Preeti: $$\frac{1}{2}$$ × 5 × 4 = 10 cm2
Brian: $$\frac{1}{2}$$ × 7 × 4 = 14 cm2
James: $$\frac{1}{2}$$ × 3 × 4 = 6 cm2

Zach: ________
Preeti: __________
Brian: ______
James: ________
The area of the shaded triangle is: ________
Zach: He didn’t consider the fraction 1/2 to calculate the area of the triangle.
Preeti: Base of the triangle is not 5. Hence the answer provided is wrong.
Brain: Base of the triangle is not 7. Hence the answer provided is wrong.
James: Base of the triangle is not 3. Hence the answer provided is wrong.
Area of the shaded triangle = 1/2 × b × h
= 1/2 × 4 × 4
= 8 cm2

Question 2.

The area of the shaded triangle is 15 square centimeters. Explain why the area of the rectangle is 30 square centimeters.
Area of the triangle = 1/2 × base × height
Area of the rectangle = base × height
= 2 × (1/2 × b × h)
= 2 × Area of triangle
= 2 × 15 cm2
= 30 cm2

Question 3.
ABCD is a rectangle and BE = EC.

Base and height are same for both the triangles. So, area (1/2 × b × h) of the both triangles will be same.

Challenging Practice

Question 1.
ABCD is a square of side 10 centimeters and BE = EC. Find the area of the shaded triangle.

Explanation:
In the above image we can observe ABCD is a square of side 10 centimeters and BE = EC.
Area = 1/2 x b x h
1/2 x 5 x 10
25 square centimeters
Area of the shaded triangle is 25 square centimeters.

Question 2.
ABCD is a rectangle 18 centimeters by 8 centimeters. AE = ED and AF = FB. Find the area of the shaded triangle.

Explanation:
In the above image we can observe ABCD is a rectangle 18 centimeters by 8 centimeters.  Here AE = ED and
AF = FB.
Area = 1/2 x b x h
1/2 x 4 x 9
18 square centimeters
The area of the shaded triangle is 18 square centimeters.

Question 3.
ABCD is a rectangle of area 48 square inches. The length of CD is 3 times the length of DF. BC = 4 inches.

a. Find the length of DF.

Explanation:
ABCD is a rectangle of area 48 square inches. The length of CD is 3 times the length of DF. Here BC = 4 inches.
The length of DF is 4 inches.

b. Find the area of the shaded triangle.

Explanation:
In the above image, we can observe ABCD is a rectangle of area 48 square inches. The length of CD is 3 times the length of DF. Here BC = 4 inches.
Area = 1/2 x b x h
1/2 x 4 x 4
8 square inches
Area of the shaded triangle is 8 square inches.

Question 4.
ABCD is a rectangle 12 centimeters by 5 centimeters. BE = 4 centimeters. Find the area of the shaded region, ABED.

Explanation:
In the above image, we can observe  ABCD is a rectangle 12 centimeters by 5 centimeters. Here BE = 4 centimeters.
Area of unshaded region = 1/2 x b x h
1/2 x 8 x 5
20 square centimeters
Area of shaded region = 12 x 5 – 20
60 – 20
40 square centimeters
The area of the shaded region is 40 square centimeters.

Question 5.
The side of square ABCD is 8 centimeters. AE = AF = 4 centimeters. Find the area of the shaded triangle, CEF.

Explanation:
In the above image we can observe the side of square ABCD is 8 centimeters. Here AE = AF = 4 centimeters.
Area of a square = 8 x 8 = 64 square centimeters
Area of a triangle a = 1/2 x b x h
a = 1/2 x 4 x 4
a = 8 square centimeters
Area of triangle b = 1/2 x b x h
b = 1/2 x 8 x 4
b = 16 square centimeters
Area of triangle c = 1/2 x b x h
c = 1/2 x 8 x 4
c = 16 square centimeters
Area of shaded triangle = 64 – 8 – 16 – 16
Area of shaded triangle =24 square centimeters

Question 6.
The perimeter of rectangle ABCD is 256 inches. Its length is 3 times as long as its width. Find the area of triangle ABC.

Explanation:
In the above image we can observe the perimeter of the rectangle ABCD is 256 inches. Its length is 3 times as long as its width.
We know that 8 units = 256 inches
4 units = 128 inches
2 units = 64 inches
1 unit = 32 inches
Here width = 32 inches
Length = 32 x 3 = 96 inches
Area of the triangle ABC = 32 x 96 = 3,072 inches

Question 7.
ABCD is a rectangle of area 72 square centimeters. The length of AD is 3 times the length of AE. BF = 8 centimeters.

a. Find the width of the rectangle.

Explanation:
In the above image we can observe ABCD is a rectangle of area 72 square centimeters.
The length of AD is 3 times the length of AE. Here BF = 8 centimeters.
Width of the rectangle = 72/12 = 7 cm

b. Find the area of the shaded region, EBFD.

Explanation:
We know that width is equal to 7cm.
Area of the shaded region EBFD = 8 x 7 = 56 square centimeters

Problem Solving

Question 1.
Look at the pattern of these triangles.

What is the area of Triangle 5 in the pattern? _____
Which triangle in the pattern will have an area of 32,768 square centimeters? _____
The Triangle 5 has the base as 32 cm and height as 32 cm.
The area of the Triangle 5 = 1/2 x base x height
1/2 x 32 x 32
512 square centimeters
The area of the Triangle 5 is 512 square centimeters.

The Triangle 6 has the base as 64 cm and height as 64 cm.
The area of the Triangle 6 = 1/2 x base x height
1/2 x 64 x 64
2,048 square centimeters
The area of the Triangle 6 is 2,048 square centimeters.

The Triangle 7 has the base as 128 cm and height as 128 cm.
The area of the Triangle 7 = 1/2 x base x height
1/2 x 128 x 128
8,192 square centimeters
The area of the Triangle 7 is 8,192 square centimeters.
The Triangle 8 has a base as 256 cm and a height of 256 cm.
The area of the Triangle 8 = 1/2 x base x height
1/2 x 256 x 256
32,768 square centimeters
The area of the Triangle 8 is 32,768 square centimeters.
The triangle 8 in the pattern will have an area of 32,768 square centimeters.

Question 2.
ABCD is a square with sides of 20 centimeters. AX = XB, BY = YC, CZ = ZD, AW = WD. WY and XZ are straight lines. Find the total area of the shaded parts.

Explanation:
In the above image, we can observe ABCD is a square with sides of 20 centimeters.
Here AX = XB, BY = YC, CZ = ZD, AW = WD. WY and XZ are straight lines.
The area of triangle WDC = 1/2 x 10 x 20 = 100 square centimeters
The area of triangle WPQ = 1/2 x 5 x 10 = 25 square centimeters
The area of triangle XPB = 25 square centimeters
Area of the shaded region = Area of the whole square – Area of WDC – Area of WPQ – Area of XPB
20 x 20 – 100 – 25 – 25
400 – 100 – 25 – 25
250 square centimeters
Area of the shaded region = 250 square centimeters.

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## Math in Focus Grade 2 Chapter 8 Practice 1 Answer Key Measuring in Kilograms

Practice the problems of Math in Focus Grade 2 Workbook Answer Key Chapter 8 Practice 1 Measuring in Kilograms to score better marks in the exam.

## Math in Focus Grade 2 Chapter 8 Practice 1 Answer Key Measuring in Kilograms

Fill in the blanks.

Question 1.

Explanation:
1. Choose a scale that measures in grams. Make sure the scale can handle the size of objects you plan on weighing. Since a gram is a metric unit of measurement, your scale needs to use the metric system. Scales are available in digital and mechanical models.
2. Weigh an empty container first before putting an item in it. If you plan to measure something you can’t put directly on the scale, weigh the container before putting the item in it.
3. Press the tare button to zero out the scale. The mysterious button labelled “tare” on digital scales is a reset button. Press the tare button after each item you measure on the scale. If you weighed a container, you can fill it now.
4. Set the object you wish to measure on the scale. Place your object in the centre of the scale. If you measured a container first, you can now put the object you wish to measure inside the container. The scale will then calculate the heaviness of your object.
5. Finish weighing the object on the scale. Wait for the scale’s digital display or needle to come to a stop. When it finishes moving, read the number to find out how heavy the object is. Make sure that the weight is in grams. Then, remove your object and hit the tare button again to reset the scale.

Question 2.

The mass of the package is ____ 1 kilogram.
Explanation:
1. We will compare the weights of pear and 1 kg stone by placing them on each end of the scale.
2. The 1 kg stone pushes the scale down more than the pear pushes down.
3. The side with 1 kg moves downwards, causing the side with the pear to move upwards.
4. The 1 kg stone is heavier than the pear.
5.  The pear is lighter than 1 kg stone.
6. It doesn’t matter what size the object is, the heaviest object moves downwards on the scale.

Question 3.

The mass of the package is ___ 1 kilogram.
Explanation:
The measuring scale is showing 1.5 kg which means higher than 1 kg.
1. Choose a scale that measures in grams. Make sure the scale can handle the size of objects you plan on weighing. Since a gram is a metric unit of measurement, your scale needs to use the metric system. Scales are available in digital and mechanical models.
2. Weigh an empty container first before putting an item in it. If you plan to measure something you can’t put directly on the scale, weigh the container before putting the item in it.
3. Press the tare button to zero out the scale. The mysterious button labelled “tare” on digital scales is a reset button. Press the tare button after each item you measure on the scale. If you weighed a container, you can fill it now.
4. Set the object you wish to measure on the scale. Place your object in the centre of the scale. If you measured a container first, you can now put the object you wish to measure inside the container. The scale will then calculate the heaviness of your object.
5. Finish weighing the object on the scale. Wait for the scale’s digital display or needle to come to a stop. When it finishes moving, read the number to find out how heavy the object is. Make sure that the weight is in grams. Then, remove your object and hit the tare button again to reset the scale.

Fill in the blanks.

Question 4.
The ___ is the lightest.
The pear is the lightest.
Explanation:

1. The 1 kg stone moves downwards so it is heaviest.
2. The pear is in upwards so it is lightest.
3. From this we can say pear is less than 1 kg.
4. Therefore, it is the lightest.

Question 5.
The ___ is the heaviest.
Explanation:

1. On the digital display the scale is showing 1.5 kg means which is more than 1 kg.
2. So the mass of the package is 1.5 kilograms.
3. Therefore, the package is the heaviest.

Read each scale. Then write the mass.

Question 6.

The mass of sugar is 2 kg.

Explanation:
1. Choose a scale that measures in grams. Make sure the scale can handle the size of objects you plan on weighing. Since a gram is a metric unit of measurement, your scale needs to use the metric system. Scales are available in digital and mechanical models.
2. Weigh an empty container first before putting an item in it. If you plan to measure something you can’t put directly on the scale, weigh the container before putting the item in it.
3. Press the tare button to zero out the scale. The mysterious button labelled “tare” on digital scales is a reset button. Press the tare button after each item you measure on the scale. If you weighed a container, you can fill it now.
4. Set the object you wish to measure on the scale. Place your object in the centre of the scale. If you measured a container first, you can now put the object you wish to measure inside the container. The scale will then calculate the heaviness of your object.
5. Finish weighing the object on the scale. Wait for the scale’s digital display or needle to come to a stop. When it finishes moving, read the number to find out how heavy the object is. Make sure that the weight is in grams. Then, remove your object and hit the tare button again to reset the scale.

Question 7.

Explanation:
1. Choose a scale that measures in grams. Make sure the scale can handle the size of objects you plan on weighing. Since a gram is a metric unit of measurement, your scale needs to use the metric system. Scales are available in digital and mechanical models.
2. Weigh an empty container first before putting an item in it. If you plan to measure something you can’t put directly on the scale, weigh the container before putting the item in it.
3. Press the tare button to zero out the scale. The mysterious button labelled “tare” on digital scales is a reset button. Press the tare button after each item you measure on the scale. If you weighed a container, you can fill it now.
4. Set the object you wish to measure on the scale. Place your object in the centre of the scale. If you measured a container first, you can now put the object you wish to measure inside the container. The scale will then calculate the heaviness of your object.
5. Finish weighing the object on the scale. Wait for the scale’s digital display or needle to come to a stop. When it finishes moving, read the number to find out how heavy the object is. Make sure that the weight is in grams. Then, remove your object and hit the tare button again to reset the scale.

Question 8.

The mass of watermelon is 4 kg.

Explanation:
1. Choose a scale that measures in grams. Make sure the scale can handle the size of objects you plan on weighing. Since a gram is a metric unit of measurement, your scale needs to use the metric system. Scales are available in digital and mechanical models.
2. Weigh an empty container first before putting an item in it. If you plan to measure something you can’t put directly on the scale, weigh the container before putting the item in it.
3. Press the tare button to zero out the scale. The mysterious button labelled “tare” on digital scales is a reset button. Press the tare button after each item you measure on the scale. If you weighed a container, you can fill it now.
4. Set the object you wish to measure on the scale. Place your object in the centre of the scale. If you measured a container first, you can now put the object you wish to measure inside the container. The scale will then calculate the heaviness of your object.
5. Finish weighing the object on the scale. Wait for the scale’s digital display or needle to come to a stop. When it finishes moving, read the number to find out how heavy the object is. Make sure that the weight is in grams. Then, remove your object and hit the tare button again to reset the scale.

Question 9.

The mass of the rice bag is 8 kg

Explanation:
1. Choose a scale that measures in grams. Make sure the scale can handle the size of objects you plan on weighing. Since a gram is a metric unit of measurement, your scale needs to use the metric system. Scales are available in digital and mechanical models.
2. Weigh an empty container first before putting an item in it. If you plan to measure something you can’t put directly on the scale, weigh the container before putting the item in it.
3. Press the tare button to zero out the scale. The mysterious button labelled “tare” on digital scales is a reset button. Press the tare button after each item you measure on the scale. If you weighed a container, you can fill it now.
4. Set the object you wish to measure on the scale. Place your object in the centre of the scale. If you measured a container first, you can now put the object you wish to measure inside the container. The scale will then calculate the heaviness of your object.
5. Finish weighing the object on the scale. Wait for the scale’s digital display or needle to come to a stop. When it finishes moving, read the number to find out how heavy the object is. Make sure that the weight is in grams. Then, remove your object and hit the tare button again to reset the scale.

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## Math in Focus Grade 3 Chapter 19 Practice 2 Answer Key Square Units (cm² and in²)

This handy Math in Focus Grade 3 Workbook Answer Key Chapter 19 Practice 2 Square Units (cm2 and in2) detailed solutions for the textbook questions.

## Math in Focus Grade 3 Chapter 19 Practice 2 Answer Key Square Units (cm2 and in2)

Find the area of each shaded figure in square centimeters. Then complete the table.

Question 1.

Explanation:
The simplest and most commonly used area calculations are for squares and rectangles.
To find the area of a rectangle, multiply its height by its width.
For a square you only need to find the length of one of the sides as each side is the same length and,
then multiply this by itself to find the area.

Draw two different figures with the same area on the grids.

Explanation:
In general, the area is defined as the region occupied inside the boundary of a flat object or 2d figure.
The measurement is done in square units with the standard unit being square meters (m 2).
As, it is mentioned  each square 1 unit square and each half square is half square unit.
So, by calculating each 1 square unit and half square units in the figure, we find Area.
The area of each figure A is 8 cm2.
The area of each figure B is 8 cm2.

Question 2.
What is the area of the figures?
Explanation:
Drawings may differ from one to one.
In the above grid each square is equal to 1 unit.
The area of drawn pictures in the grid is 8 square units.
In figure A it has 4 full squares and 8 half squares.
In figure B it has 6 full squares and 4 half squares.
So, the area of both figures = 8 cm2.

Question 3.
The figures are made of square and half-square tiles. Find the area of each figure.

Explanation:
The simplest and most commonly used area calculations are for squares and rectangles.
To find the area of a figures is by adding all the areas covered by the squares and triangles,
two triangles is one square.

Question 4.
Which figure has a larger area? Figure _____________
Figure D has largest area
Explanation:
The simplest area calculations are for squares and rectangles.
To find the area of a figures is by adding all the areas covered by the squares and triangles,
two triangles is one square.
Figure D has largest area of  8 cm2

Question 5.
How can you make both figures have the same area?
By adding 3 square centimeters to the figure C ,
we can make both the figures as same area.
Explanation:
Observe the given figure C and D ,
Figure C has 3 units less than figure D.

Find the area of each shaded figure in square inches. Then complete the table.

Question 6.

Explanation:
The simplest and most commonly used area calculations are for squares and rectangles.
To find the area of a figures is by adding all the areas covered by the squares and triangles,
two triangles is one square.

Draw two different figures with the same area on the grid.

Question 7.

Explanation:
The simplest and most commonly used area calculations are for squares and rectangles.
To find the area of a figures is by adding all the areas covered by the squares and triangles,
two triangles is one square.
The area of each figure A is 8 in2.
The area of each figure B is 8 in2.

Question 8.
The area of each figure is ____________ in.2.
Explanation:
The simplest and most commonly used area calculations are for squares and rectangles.
To find the area of a figures is by adding all the areas covered by the squares and triangles, two triangles is one square
The area of each figure A is 8 in.2.
The area of each figure B is 8 in.2.

Find the area of each shaded figure in square inches. Then complete the table.

Question 9.

Explanation:
The simplest and most commonly used area calculations are for squares and rectangles.
To find the area of a figures is by adding all the areas covered by the squares and triangles,
two triangles is one square.

Question 10.
Figure __________ and Figure __________ have the same area.
Figure A and Figure B have the same area of 11in.2
Explanation:
Compare the figures A and B,
the area of both the figures is same as 11in.2
Question 11.
Figure ___________ has the largest area.
Figure D  has the largest area of 13 in.2
Explanation:
Compare all the given figures,
Area of figure D is the greatest with 13 in.2
Question 12.
Figure __________ has the smallest area.
Figure C has the smallest area of 10 in.2
Explanation:
Compare all the given figures,
Area of figure C is the least with 10 in.2

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## Math in Focus Grade 3 Chapter 4 Answer Key Subtraction up to 10,000

Go through the Math in Focus Grade 3 Workbook Answer Key Chapter 4 Subtraction up to 10,000 to finish your assignments.

## Math in Focus Grade 3 Chapter 4 Answer Key Subtraction up to 10,000

Challenging Practice
Fill in the blanks in each number sentence. Use the numbers in the box.

Question 1.
The difference between two numbers is 42.

The difference between two numbers is 42 are 68 and 26.

Explanation:
Difference:
1. 68 – 42 = 26.       68 – 26 = 42.       82 – 68 = 14.
2. 42 – 26 = 26.
3. 26
4. 82 – 68 = 14.        82 – 42 = 40.       82 – 26 = 56.

Question 2.
The difference between two numbers is 280.

The difference between two numbers is 280 are 476 and 196.

Explanation:
Difference:
1. 400 – 196 = 204.         400 – 129 = 271.           400 – 280 = 120.
2. 196 – 129 = 67.
3. 129
4. 476 – 400 = 76.       476 – 196 = 280.       476 – 129 = 347.        476 – 280 = 196.
5. 280 – 196 =          280 – 129 = 151.

Fill in the missing numbers.
Question 3.

Explanation:
Ones place number: 6 – 4 = 2.
Tens place number: 3 – 2 = 1.
Hundreds place number: ?? – 7 = 9
=> ?? = 9 + 7
=> ?? = 16(10 + 6).
Thousand place number: 5 – 1 = 4.
=> 4 – 2 = 2.

Question 4.

Explanation:
Ones place number: 5 – 3 = 2.
Tens place number: 7 – 4 = 3.
Hundreds place number: ?? – 4 = 7
=> ?? = 7 + 4
=> ?? = 11(1 + 10).
Thousand place number: 8 – 1 = 7.
=> 7 – 3 = 4.

Fill in the missing numbers.

Question 5.

Explanation:
Ones place number: 9 – 5 = 4.
Tens place number: 8 – 9 = 9.
=> (10 + 8) – 9 = 9.
Hundreds place number: 6 – ?? = 9.
=> ?? = 9 + 6
=> ?? = 15(5 + 10).
=> 15 – ?? = 9
=> ?? = 15 – 9
=> ?? = 6.
Thousand place number:
3 – 1 = 2.
=> 2 – 2 = 0.

Question 6.

Explanation:
Ones place number: ?? – 9 = 6.
=> ?? = 6 + 9
=> ?? = 15. 15 = (10 + 5)
Tens place number: 2 – 1 = 1.
=> 1 – 7 = 4.(NO)
=> 11 (10 + 1) – 7 = 4.
Hundreds place number: 3 – 1 = 2.
=> 2 – 8 = 4(NO)
=> (2 + 10 )12 – 8 = 4.
Thousand place number: 7 – 1 = 6.
=> 6 – 3 = 3.

Solve. Use the digits to make 4-digit numbers. Show your work. Do not begin any number with ‘0’.

Question 7.
Subtract the least 4-digit number from the greatest 4-digit number.
_____ – ____ = ____
9999 – 1000
= 8999.

Explanation:
Least 4-digit number = 1000.
Greatest 4-digit number = 9999.
Difference:
Greatest 4-digit number – Least 4-digit number
= 9999 – 1000
= 8999.

Solve.
Question 8.
Write a number greater than 5,632 using the digits 0, 1, 4, 7. Then subtract 5,632 from the number.
____ – ____ = _____
7410 – 5632
= 1778.

Explanation:
Number greater than 5,632 using the digits 0, 1, 4, 7.
Number greater than 5,632 = 7410.
Then subtract 5,632 from the number.
=> Difference:
=> Number greater than 5,632 – 5,632
=> 7410 – 5632
=> 1778.

Problem Solving
Question 1.
The difference between two numbers is 100. One number is more than 90 but less than 100. The other number is between 190 and 200. What are the two possible numbers?
____ and ____
The two possible numbers are 199 and 99.

Explanation:
Let the two numbers be X and Y.
The difference between two numbers is 100.
=> X – Y = 100.
One number is more than 90 but less than 100.
=> One Number = 99.
The other number is between 190 and 200.
=> Other number = 199.
Difference:
199 – 99 = 100.

Question 2.
Lilian went shopping with $1,000. She saw five items on display in a shop window. After buying two items, she had$732 left. Which two items did she buy?
___ and ____

Two items she bought are frock and necklace set.

Explanation:
Cost of frock = $68. Cost of calculator =$32.
Cost of mick = $25. Cost of necklace set =$200.
Cost of Television = $500. Amount of money Lilian went shopping =$1000.
After buying two items, she had $732 left. Amount of money spent = Amount of money Lilian went shopping – Amount of money left after buying 2 items =$1000 – $732 =$268. ($68 +$200) (Cost of frock + Cost of necklace set)

Solve.
Question 3.
Nick and Isaac are at a school fair.
They want to collect points to exchange for these prizes.

At the fair games, Nick has 215 points and Isaac has 78 points. They combine their points to exchange for three prizes.
What are the two sets of three prizes they can get?

a. ________________
Three prizes = 2(Pencils) and Pencil holder.

Explanation:
Points of Nick has = 215.
Points of Isaac has = 78.
Total combined points of Nick and Isaac = Points of Nick has  + Points of Isaac has
= 215 + 78
= 293.
Points of pencil = 30.
Points of pencil holder = 200.
Total points of three items  = 2(Points of pencil) + Points of pencil holder
= 2(30) + 200
= 60 + 200
= 260.

b. ________________
Three things are Pencil, Notebook and pencil holder.

Explanation:
Points of Nick has = 215.
Points of Isaac has = 78.
Total combined points of Nick and Isaac = Points of Nick has  + Points of Isaac has
= 215 + 78
= 293.
Points of pencil holder = 200.
Points of Pencil  = 30.
Points of Notebook = 50.
Total points of three items  = Points of pencil holder + Points of Pencil  + Points of Notebook
= 200 + 30 + 50
= 230 + 50
= 280.