Vinay Pacha

Practice the questions in the Worksheet on Area for different shapes such as Square, Rectangle, Triangle, Circle, Quadrilateral, Polygons, etc. Amount of surface that a plane figure covers is known as Area. Learn the problem-solving approach used in the Area Worksheets over here and try to apply this knowledge to solve the problems you face in your exams. Firstly, try to solve the questions on your own and assess where you stand in your preparation.

Do Check:

• Measuring Area in Square Units
• Area using Square Paper
• Find the Area of the Shaded Region

Example 1.
Find the area of the rectangle of length 12 cm and breadth 15 cm?

Solution:

The Length of the rectangle=12 cm
The breadth of the rectangle=15 cm
Area of the rectangle=12 × 15
=180
Therefore the area of the rectangle=180 cm.

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Example 2.
Find the length of the agriculture field area 200sq.m and breadth is 10 cm?

Solution:

Area of the agricultural field =200 sq.m
The breadth of the agricultural field = 10 cm
The length of the agricultural field=200/10=20 cm
Therefore, the length of the agricultural field is 20 cm.

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Example 3.
Calculate the area of the floor of a square-shaped room which is made up of 70 square tiles of side 18 cm?

Solution:

Area of one tile=18 × 18=324 sq inches.
we know that the area of the floor has 70 tiles.
The area occupied by 70 tiles in the floor area=70 × 324=22,680 square inches.
Therefore, the area of the floor of the room=22,680 sq m.

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Example 4.
 The area of the square-shaped object 2500 cm². What is the length of its side?

Solution:

The area of the square-shaped object =2500 cm²
We know that A= side²
So side=\(\sqrt{ Area} \)
=\(\sqrt{ 2500} \)
=50 cm
Therefore, the length of the side of the square-shaped object is 50 cm.

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Example 5.
Find the area of the square whose diagonal is 6 cm?

Solution:

Given Diagonal is 6.
Area of the square when its diagonal is given by D²/2.
=6²/2
=18
Therefore, the area of the square is 18 sq units.

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Example 6.
Find the area of the painting square board of side 18 cm?

Solution:

The side of the painting board is 18 cm
The area of the painting square board is 18²
=324
Hence, the area of painting square board is 324 sq units.

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Example 7.
How much money required for tiling the room 48 m long and 53m wide at the rate of 20 per sq m?

Solution:

Length of the room=48
Width of the room=53
Area of the room=48×53 =2544.
The money for tiling the room per square meter =20
The money required for tiling the room =2544 × 20
=50880
Hence, the money required for tiling the room is 50880.

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Example 8.
What is the area of the rectangular board of length 62 m and 65 m wide?

Solution:

length of the rectangular board=62 m
Width of the rectangular board=65 m
Area of the rectangular board=62 × 65
=4030
Therefore, the area of the rectangular board is 4030 sq m.

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Example 9.
Find the area of the square of the side 23 m?

Solution:

The side of the square =23 m
Area of the square=23 × 23=529
Hence, the area of the square=529 sq m.

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Example 10.
Find the area of the rectangle when
1) length=2 m 30 cm, width= 40 cm
2) length=5 m 20 cm , width= 20 cm

Solution:

1) Length of the rectangle=2m 30 cm=2 × 100 +30 cm=230 cm
width of the rectangle= 40 cm
Area of the rectangle=230 × 40= 9200 sq cm
Hence, the area of rectangle=9200 sq cm.
2) Length of the rectangle=5 m 20 cm=5 × 100 + 20 cm=500+20 cm=520 cm
width of the rectangle=20 cm
Area of the rectangle=520 × 20= 10400 sq cm.
Hence, the area of rectangle=10400 sq cm.

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Worksheets on Subtraction of Decimals can be great for students to practice various subtracting decimal numbers problems. The Simple funda to keep in mind while subtracting decimals is that you need to convert them into like decimals and subtract as if they are regular whole numbers and at last place a decimal point in result. Assess your knowledge of the concept of decimals subtraction by answering the questions from Subtracting Decimals Worksheet PDF. You can verify your answers with our solutions provided and learn whether you did it right or not.

Do Check:

• Worksheet on Addition of Decimals
• Worksheet on Multiplication of Decimals
• Word Problems on Addition and Subtraction of Decimals

Decimal Subtraction Worksheets

Example 1.
Subtract the given decimal numbers
45.08
-22.04
———–
———–

Solution:

Start subtracting from right to left just like normal subtraction. Remember to put the decimal point in the answer.
45.08
-22.04
———-
23.04
———-
Therefore, by subtracting 22.04 from 45.08 we get 23.04.

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Example 2.
Subtract 15.871 from 22.653

Solution:

Start subtracting from right to left 3-1=2.
As 7 >5, So you must regroup. Take 1 from 6 then 6 becomes 5. Add 10 to 5, So 5 becomes 15. 15-7=8.
As 8 >5, So you must regroup. Take 1 from 2 then 2 becomes 1. Add 10 to 5, So 5 becomes 15. 15-8=7.
As 5>2, So you must regroup. Take 1 from 2 then 2 becomes 1. Add 10 to 1, So 1 becomes 11. 11-5=6.
1-1=0.
Therefore, by subtracting 15.871 from 22.653 we get 6.782.

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Example 3.
Subtract 16.96 from 38.756 and find the difference?

Solution:

Write the larger number 38.756 on the top and the smaller number 16.96 on the below with the decimals lined up.
Add a zero to the number 16.96 so that the numbers have the same length.
Subtract normally just like ordinary numbers, remember to put the decimal point in the answer.

Hence, By subtracting 16.96 from 38.756 we get 21.796 as the difference.

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Example 4.
Find the difference between 28.39 and 75.26?

Solution:

Write the larger number 75.26 on the top and the smaller number 28.39 below with the decimals lined up.
Subtract normally just like ordinary numbers, remember to put the decimal point in the answer.

Therefore, by subtracting 28.39 from 75.26 we get 46.87.

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Example 5.
Solve the statement 38.86 less than 57.238?

Solution:

Write the larger number 57.238 on the top and the smaller number 46.87 on the below with the decimals lined up.
Add a zero to 38.86 so that the numbers have the same length.
Subtract normally just like ordinary numbers, remember to put the decimal point in the answer.

Hence, 38.86 less than 57.238 by 18.378.

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Example 6.
Solve the statement 98.85 less than 186.625?

Solution:

Write the larger number 186.625 on the top and the smaller number 98.85 on the below with the decimals lined up.
Add a zero to 98.85 so that the numbers have the same length.
Subtract normally just like ordinary numbers, remember to put the decimal point in the answer.

Therefore, 98.85 less than 186.625 by 87.775.

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Example 7.
A milk can of capacity 50 liters has 35.50 liters of milk in the can. How much milk can be poured into the milk can?

Solution:

Milk can have the capacity=50
No. of liters of milk=35.50
No. of liters of milk can be poured into the milk can=50-35.50

Therefore, 14.50 liters of milk can be poured into the milk can.

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Example 8.
Srikanth has 225.50 bags of rice in the godown. He sold 195 bags of rice and the remaining he kept in his house. Find how many bags of rice he kept in his house?

Solution:

No. of bags of rice in the godown=225.50
No. of bags of rice Srikanth sold=195
No. of bags of rice Srikanth kept in his house=225.50-195

Therefore, Srikanth kept 30.50 rice bags in his house.

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Example 9.
Rajesh has 150.80 kg of grapes in his shop. He sold 65.87 kg of grapes by the evening. How many kg of grapes has to be sold?

Solution:

Total no. of kg of grapes Rajesh has=150.80
no. of kg of grapes Rajesh sold by evening=65.87
no. of kg of grapes Rajesh has to sell=150.80-65.87

No. of kg of grapes Rajesh has to sell is 84.93.

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Example 10.
Bhaskar has the money of Rs 500. He bought 30 eggs of Rs 5.30 per egg. He gave to the shopkeeper Rs 500note. How much money will he get in return?

Solution:

Bhaskar gave to the shopkeeper=Rs500
No. of eggs bought by Bhaskar=30
Cost of each egg=5.30
Cost of 30 eggs=30 × 5.30=159.
The money he will get in return=500-159
=341.
Hence, the shopkeeper will return the money of Rs 341.

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Decimal Worksheets are perfect for practicing problems on decimal addition. Adding the Decimal Numbers is just like finding the sum of decimal numbers. The Extensive Collection of Decimal Addition Problems will make it easy to add the decimal numbers in columns. Solve the Adding Decimals Worksheet with Answers provided here on a regular basis and score better grades in your exams. Our Worksheet on Adding Decimals includes questions on the Addition of Decimals with and without Regrouping.

Also, Check:

• Worksheet on Word Problems on Addition and Subtraction of Decimals
• Adding Decimals

Addition of Decimals Worksheets with Answers

Example 1.
Add 3.65, 2.8

Solution:

Given Decimal Numbers are 3.65 and 2.80. Both of them are having the same number of decimal places. Aligning them in columns and performing the addition similar to whole numbers we get as such

Therefore, the addition of 3.65, 2.8 is 6.45.

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Example 2.
Add 2.8,4.532,61.8,712.8

Solution:

Given Decimal Values are 2.8,4.532,61.8,712.8
Check if all the numbers have equal decimal places or not. If not pad them with necessary zeros.
Arranging them in columns we get the decimal addition result of the following numbers as follows

Therefore, the addition of 2.8,4.532,61.8,712.8 is 781.932.

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Example 3.
Add 2.98,1.07,5.038

Solution:

Decimal Values are 2.98, 1.07, 5.038
Padding with required zeros and making them equivalent decimals we get the following result. Adding them all we have the addition as such

Therefore, the addition of 2.98,1.07,5.038 is 9.088.

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Example 4.
Add 159.34,28.16,4.1,1.06

Solution:

Given Decimal Places are 159.34, 28.16, 4.1, 1.06
Aligning them in columns we have the addition of decimals as under

Therefore, the addition of 159.34,28.16,4.1,1.06 is 192.66.

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Example 5.
Add 0.5555,5.555,555.5,55.5

Solution:

Given Decimal Places are 0.5555,5.555,555.5,55.5
Aligning them in columns we have the addition as follows

Hence, the sum of 0.5555,5.555,555.5,55.5 is 617.1105.

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Example 6.
Increase 73.056 by 5.683

Solution:

Given Decimals are 73.056, 5.683. As both the decimal numbers have an equal number of decimal places we need not pad with zeros. Adding them we have the result as under

Therefore, by increasing 73.056 by 5.683 we get 78.739.

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Example 7.
Add 1.09,98.37,576.03,2822.9

Solution:

Given Decimal Places are 1.09,98.37,576.03,2822.9
Padding with required zeros and performing the addition process we have the result as follows

Hence, the addition of 1.09,98.37,576.03,2822.9.

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Example 8.
Find the sum of 0.678+6.78+67.80+678.0

Solution:

Given Decimal Places are 0.678, 6.78, 67.80 and 678.0
Arranging them in columns we get the addition result as under

Therefore, the addition of 0.678+6.78+67.80+678.0 is 753.258.

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Example 9.
Increase 585.03 by 3.756

Solution:

Decimal Values are 585.03, 3.756
Arranging them in columns we get the result as such

Therefore, by increasing 585.03 by 3.756 we get 588.786.

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Example 10.
Find the sum of 2.008,46.08,5312.05

Solution:

Decimal Values are 2.008,46.08,5312.05
Arranging them in decimal places we have the result as follows

Therefore, the addition of 2.008,46.08,5312.05 is 5360.138.

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See your kids excel in mathematics taking the help of the free and extensive problems available on decimal addition and subtraction. Use the interactive exercise Word Problems on Addition and Subtraction and develop personalized learning among your kids. This Worksheet on Adding and Subtracting Decimals has an extensive collection of frequently asked problems in your exams. Assess your strengths and weaknesses using the problems over here regarding decimal addition and subtraction and get a good grip on the concept.

Do Read Similar Articles:

• Worksheet on Concept of Decimal
• Simplify Decimals Involving Addition and Subtraction Decimals
• Adding Decimals
• Subtracting Decimals

Word Problems Involving Addition and Subtraction of Decimals

Example 1.
There are 8.50 liters of milk in the pan. Raju added 5 g of sugar and 1.25 liters of water to the pan. Find how many liters of milk and water are there in the pan?
Solution:
No. of liters of milk in the pan = 8.50
No.of liters of water in the pan = 1.25
No. of liters of milk and water in the pan is
8.50 + 1.25 = 9.75
Hence, There are 9.75 liters of milk and water are there in the pan.

Example 2.
Raju has 14.50 acres of agricultural land. He decided to give his first son Sai 6.50 acres of land and the second son Sudheer 5.50 acres of land. Find how much land does Raju has after distributing his sons?
Solution:
Raju has the agricultural land = 14.50
Raju gave the land for the first son = 6.50
Raju gave the land for the second son = 5.50
No. of acres of land Raju distributed for his sons = 6.50 + 5.50 =12.00
No. of acres of land Raju has after distributing his sons =14.50-12.00 = 2.50
Therefore, Raju has 2.50 acres of agricultural land after distribution.

Example 3.
The price of the sugar last month is Rs 42.50. This month the price of sugar is increased by Rs 2.50. Find out what is the price of the sugar this month?
Solution:
The price of the sugar last month = 42.50
The price of the sugar this month is increased by =  2.50
The price of the sugar this month = 42.50 + 2.50 = 45.00
Hence, the price of the sugar this month is Rs 45.

Example 4.
Karthik wants to go to the temple which is 150.50 km. Karthik stops driving the car after driving 65.80 km, because of the traffic jam. How much distance he has to travel for going to the temple?
Solution:
Karthik wants to go to the temple at a distance = 150.50
Karthik traveled by car up to the distance = 65.80
The distance Karthik has to travel for going to the temple = 150.50-65.80
= 84.70
Therefore, Karthik has to travel 84.70 km for going to the temple.

Example 5.
Praveen wants to buy a house in Banglore. He went to choose the houses. In the first house, the rooms were 28.50 square feet longer. The second house was 2.7 square feet shorter. The third house was 5.6 square feet longer than the first house. What is the difference in feet between the second and third house rooms?
Solution: 
The size of the rooms in the first house was = 28.50
The size of the rooms in the second house was shorter by = 2.7
The size of the rooms in the second house was = 28.50 – 2.7 = 25.8
The size of the rooms in the third house was longer than the first house by = 5.6
The size of the rooms in the third house was = 28.50 +5.6 = 34.1
The difference in feet for the second and third house was = 34.1-25.8 = 8.3
Hence, The difference in feet for the second and third houses was 8.3 square feet.

Example 6.
Varsha had money Rs 750.80. She bought a dress for Rs 530.20. How much money left with Varsha?
Solution:
Varsha had money = 750.80
She bought a dress = 530.20
Money left with Varsha = 750.80-530.20
= 220.60
Hence, Money left with Varsha = 220.60

Example 7.
Pavan went to a store. He bought 2.25 kg of cashews and almonds. If pavan bought 1.25 kg of almonds, how many kg of cashews?
Solution:
Total no. of kg of cashews and almonds = 2.25 kg
Pavan bought almonds = 1.25 kg
No. of kg of cashews = 2.25-1.25 = 1 kg
Hence, the total no. of kg of cashews is 1 kg.

Example 8.
Sindhu purchased a book for Rs 50.50, a pen for Rs 25.50. How much amount did Sindhu spend?
Solution:
Sindhu purchased a book = 50.50
Sindhu purchased a pen = 25.50
The amount Sindhu spend = 50.50 +25.50
= 76
Hence, the amount Sindhu spends is Rs 76.

Example 9.
Harish has some money. He bought a gift for his friend in the amount of Rs 500.50. Harish is left with the amount of Rs 300. Find the amount of money Harish has before spending the money?
Solution:
Harish bought a gift for his friend = 500.50
Harish has left with the money = 300
The amount of money Harish has before spending the money = 500.50+300 = 800.50
Therefore, Harish has 800.50 before spending the money.

Example 10.
In a juice shop, there are 10.25 liters of orange juice and 12.50 liters of grape juice. How many liters of juices are needed to fill an order of 30 liters of juice?
Solution:
No. of liters of Orange juice = 10.25
No.of liters of grape juice = 12.50
Total no. of liters of juices in the shop = 10.25+12.50 = 22.75
No. of liters of juices required to fill an order of 30 liters of juice = 30-22.75 = 7.25
Hence, no. of liters of juice required to fill an order is 7.25 liters.

Multiplication of Decimals is similar to the Multiplication of Whole Numbers. It’s just that you might get confused with the placement of decimal points. To help you clear this ambiguity we have curated the Worksheet on Multiplication of Decimals covering various problems for finding the product of decimals. Try practicing the questions from Multiplying Decimals Worksheet PDF over here and improvise your knowledge on the concept. You can verify your solutions in the Decimal Multiplication Worksheet with Answers and understand where you were lagging.

Do Refer:

• Multiplication of Decimals
• Worksheet on Concept of Decimal

Decimal Multiplication Questions and Answers

Example 1.
Find the Multiplication of 5.35 × 5?

Solution:

There are 2 decimal places in the number 5.35.
Ignore the decimal places and continue the multiplication process i.e. 535 × 5.

Rewrite the product with two decimal places i.e. 26.75.
Hence the product of 5.35 × 5 is 26.75

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Example 2.
Find the product of 35.56 × 4?

Solution:

There are 2 decimal places in the number 35.56.
Ignore the decimal places and continue the multiplication process i.e.3556 × 4.

Rewrite the product with two decimal places i.e. 142.24.
Hence the product of 35.56 × 4 is 142.24.

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Example 3.
Find the product of 44.89 × 12?

Solution:

There are 2 decimal places in the number 44.89.
Ignore the decimal places and continue the multiplication process i.e.4489 × 12.

Rewrite the product with two decimal places i.e. 538.68.
Hence the product of 44.89 × 12 is 538.68.

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Example 4.
Find the multiplication of 68.093 × 18?

Solution:

There are 3 decimal places in the number 68.093.
Ignore the decimal places and continue the multiplication process i.e.68.093 × 18.

Rewrite the product with three decimal places i.e. 1225.674.
Therefore, the product of 68.093 × 18 is 1225.674.

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Example 5.
Find the multiplication of 563.32 × 15?

Solution:

There are 2 decimal places in the number 563.32.
Ignore the decimal places and continue the multiplication process i.e.563.32 × 15.

Rewrite the product with two decimal places i.e. 8449.80.
Therefore, the product of 563.32 × 15 is 8449.80.

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Example 6.
Arjun wants to gift a toy to his 15 friends. The cost of each toy car is Rs 128.25. Find the cost of 15 toys cars?

Solution:

The cost of each toy car =Rs 128.25
The cost of 15 toy cars=15 × 128.25

Hence, the cost of 15 toy cars is1923.75.

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Example 7.
Priya went to the shop and purchased 4 dresses at the rate of 580.80 per dress. Find the money spent by Priya?

Solution:

No. of dresses purchased by Priya=4
The amount spent by Priya per dress=580.80
The money spent by priya=580.80 × 4

Therefore, the money spent by Priya is 2323.20.

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Example 8.
Vijay runs 5.8 km in 1 hour. Find how many kilometers can Vijay run in 5 hours?

Solution:

No. of kilometers Vijay can run in 1 hour=5.8 km
No. of kilometers Vijay can run in 5 hours =5.8 × 5
=29 km
Therefore, Vijay can run 29 km in 5 hours.

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Example 9.
The price of a 1-meter cloth is 280.60. Sailaja needs 15 meters of cloth for stitching the dress. Find how much money needed by Sailaja?

Solution:

The price of 1 m cloth = 280.60
The price of 15 m cloth=280.60 × 15

Therefore, Sailaja needs money of Rs 4209.00.

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Example 10.
Srinivas went out with his friends. The price of 1 glass of grape juice is 50.50. Find how much money spent by Srinivas for giving 12 glasses of grape juice to his friends?

Solution:

The price of 1 glass of grape juice=50.50
The money spent by Srinivas for 12 glasses of grape juice=50.50 × 12

Therefore, the Money spent by Srinivas for 12 glasses of grape juice is 606.00.

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Below you will find a variety of questions on the concept of the decimal. Practice various questions from the Free Printable Concept of Decimal Worksheet and enhance your subject knowledge. Solving the Decimal Problems will give you related knowledge on topics such as ascending, descending order of decimals, expressing decimals in short form or expanded form, finding the number of places in decimals, converting fractions to decimals, and vice versa.

Download the Decimal Worksheet with Answers for free of cost and practice offline too after your class or during your homework. Test your understanding using Decimal Worksheets with Solutions and remediate the knowledge gap accordingly and score better grades in exams.

Do Refer:

• Worksheet on Word Problems on Addition and Subtraction of Decimals
• Fraction as Decimal
• Concept of Decimal
• Multiplication of Decimals

I) Express each of the following in Decimals and find the number of decimal places
1) 35/100
2) 6/100
3)79/1000
4) 234/1000
5) 5643/1000

Solution:

1) 35/100
=0.35
The number of decimal places is 2.
2) 6/100
=0.06
The number of decimal places is 2.
3) 79/1000
=0.079
The number of decimal places is 3.
4) 234/1000
=0.234
The number of decimal places is 3.
5) 5643/1000
=5.643
The number of decimal places is 3.

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II) Express each of the following in decimals
1) 195/2
2) 17/5
3) 237/5
4) 1568/3
5) 2457/6

Solution:

1) 195/2
It is the same as 195 ÷ 2.
= 97.5
2)17/5
It is the same as 17 ÷ 5.
=3.4
3)237/5
It is the same as 237 ÷ 5.
=47.4
4)1568/3
It is the same as 1568 ÷ 3.
=522.66
5)2457/6
It is the same as 2457 ÷ 6.
=409.5

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III) Express each of the following as fractions
1) 5.68
2) 0.006
3)0.28
4)16.54
5)207.4

Solution:

1) 5.68
Write the decimal number 5.68 as a fraction with 1 in the denominator i.e. 5.68/1.
Multiply top and bottom by 100 to remove two decimal places.
=5.68/1 ×100/100
=568/100
Therefore, 5.68 can be expressed as a fraction 568/100.
2)0.006
Write the decimal number 0.006 as a fraction with 1 in the denominator i.e. 0.006/1.
Multiply top and bottom by 1000 to remove two decimal places.
=0.006/1 × 1000/1000
=6/1000=3/500
3)0.28
Write the decimal number 0.28 as a fraction with 1 in the denominator i.e. 0.28/1.
Multiply top and bottom by 100 to remove two decimal places.
=0.28/1 ×100/100
=28/100
4)16.54
Write the decimal number 16.54 as a fraction with 1 in the denominator i.e. 16.54/1.
Multiply top and bottom by 100 to remove two decimal places.
=16.54/1 × 100/100
=1654/100
5)207.4
Write the decimal number 207.4 as a fraction with 1 in the denominator i.e. 207.4/1.
Multiply top and bottom by 100 to remove two decimal places.
=207.4/1 × 100/100
=2074/10

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IV) Write the place and place values of underlined digits
1) 62.320
2) 7.0195
3) 834.632
4) 9272.21
5) 101.87

Solution:

1) The place value of 3 is tens,0.3.
2) The place value of 1 is a hundred, 0.01.
3) The place value of 8 is hundreds,800.
4) The place value of 1 is hundreds,0.01.
5) The place value of 1 is hundreds,100.

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V) Write the fractional number and decimal number for the shaded part of each figure

1)

2)

3)

Solution:

1)In this table, the fractional part for the shaded part is 20/64 and the decimal number is 0.3125
2) In this table, the fractional part for the shaded part is 25/64  and the decimal number is 0.39062
3) In this table, the fractional part for the shaded part is 35/64 and the decimal number is 0.54687

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VI) Express the Decimals in Words
1) 5.32
2) 3.89
3) 0.84
4) 7.19
5) 1.86
6) 897.5
7)32.54

Solution:

1) Five point three two
2) Three-point eight nine
3) Zero point eight four
4) seven-point one nine
5) one point eight six
6) eight hundred ninety-seven point five
7) thirty-two point fifty-four

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VII) Express in numerals
1) Fifty-eight point two four
2) twenty-one point zero six
3) sixty-five point seven eight
4) seven hundred eighty point zero five
5) nine hundred twenty point seven eight
6) five hundred point zero five
7) seventeen point eight three

Solution:

1) 58.24
2) 21.06
3) 65.78
4) 780.05
5) 920.78
6) 500.05
7)17.83

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VIII) Express the following in the Expanded form
1) 4. 759
2) 38. 62
3) 975.562
4)0.12
5) 152.18

Solution:

1) 4+ 0.7+0.05 +0.009
2) 30+8+0.6+0.02
3)900+70+5+0.5+0.06+0.002  
4) 0.1+0.02
5) 100+50+2+0.1+0.08

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IX) Express in the short form
1) 50+5+0.8+0.06
2) 200+30+3+0.7+0.02
3) 0.5+0.07
4) 3+0.5+0.04+0.002
5) 70+5+0.8+0.01

Solution:

1) 55.86
2) 233.72
3) 0.57
4) 3.542
5) 75.81

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X) Arrange in ascending order
1) 0.1, 0.32, 0.23, 0.2, 0.45, 0.5
2) 2.15, 2.78, 2.32, 2.02, 2.12, 2.6
3) 3.56, 3.6, 3.89, 3.98, 3.72
4) 6.78,6.07,6.62,6.69,6.63

Solution:

1) 0.1,0.2,0.23,0.32,0.45,0.5
2) 2.02, 2.12, 2.15, 2.32, 2.6, 2.78
3) 3.56, 3.6, 3.72, 3.89, 3.98
4) 6.07, 6.62,6.63,6.69,6.78

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XI) Arrange in descending order
1) 4.53,4.72,4.61,4.02,4.35,4.25,4.06,4.23
2)13.02,13.08,13.53,13.73,13.83,13.33
3)10.50,10.05,10.52,10.51,10.55,10.58
4) 8.01,8.05,8.03,8.09,8.12,8.15,8.74,8.78

Solution:

1) 4.72, 4.61, 4.53, 4.35, 4.25, 4.23, 4.06, 4.02
2) 13.83, 13.73, 13.53, 13.33, 13.08, 13.02
3) 10.58, 10.55, 10.52, 10.51, 10.50, 10.05
4) 8.78, 8.74, 8.15, 8.12, 8.09, 8.05, 8.03, 8.01

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In this article, you will learn how to perform the division of metric measures. Metric measures are divided in the same way as we divide ordinary numbers. First, we will place the digits in columns and then divide as usual. Measurement of the metric system is the way of measuring distance, calculating the height, measuring the weight, and so on. The standard units of length, weight, and capacity are meter, gram, and liter respectively.

On this page, you will learn about the meaning of metric measures, how to divide metric units, solved example problems on the division of metric measures, and so on.

Read More Articles:

• Addition of Metric Measures
• Subtraction of Metric Measures
• Multiplication of Metric Measures
• Metric Measures

What is Metric System? | Metric System – Definition

Based on each object we can measure its length, weight, volume, and time in a different way. The metric system is a system used for measuring distance, length, volume, weight, and temperature. Based on three basic units we are able to measure almost everything in the world.

For measuring length we can use the meter, for measuring mass we can use a kilogram and for the time we can use a second, for measuring capacity we can use liter.

How to Divide Metric Measurements?

Metric Division Measurement is also known as repeated subtraction division. It is a way of understanding division in which you divide an amount into groups of a given size. For example, if you are thinking about division this way, then 16 ÷ 4 means 16 things divided evenly into groups of 4, and we wish to know how many groups we can make.

Worked Out Examples on Dividing Metric Measures

Example 1: 
Divide the value 4 km 8hm 7dam 6 m by 2
Solution:
As given in the question, the value is 4 km 8hm 7dam 6 m by 2
The division is also known as repeated subtraction. First, we place the digits in columns and then divide as usual.
So, we can perform the division perform on the given number 4 km 8hm 7dam 6 m by 2 as follows.
After performing the division we get the remainder as 0, and the quotient as 2433.
Therefore, the final value of the given number is 2 km 4 hm 6dam 6 m.

Example 2:
Ramu has 18 chocolates. He will distribute it among 6 friends equally. How much did each get?
Solution:
As given in the question,
No. of chocolates Ramu has =  18 chocolates.
He will distribute 18 chocolates to 6 friends equally.
Now, we will find how many chocolates each gets.
We know the division law, using that we can find how many chocolates each person gets easily.
So, the given numbers are 18, 6. Now, divide 18 with 6.
18 / 6 = 3.
Therefore, Ramu distributed 18 chocolates among 6 friends equally. So, each friend gets 3 chocolates.

Example 3:
Divide the value  8 kg 4 hg 3dag 6 g by 6
Solution:
As given in the question, the value is 8 kg 4 hg 3dag 6 g by 6
The division is also known as repeated subtraction. First, we place the digits in columns and then divide as usual.
So, we can perform the division perform on the given number i.e. 8 kg 4 hg 3dag 6 g by 6.

The below figures show, how to perform the division of metric measures step-by-step process.

Therefore, the value 8kg 4hg 3dag 6g is divided by 6, to get the final value is 1kg 4hg 6dag

Example 4:
Divide the value 19kl 5hl 9 l by 3
Solution:
As given in the question, the value is 19kl 5hl 9 l by 3.
Using the division rule, we can easily find the value. In division, first, we should place all digits in columns and then divide as usual.
So, now we can perform the division operation of the given number 19kl 5hl 9l by 3.
19kl 5hl 9l / 3 = 653
Therefore, after division, the final value of the given number 19kl 5hl 9l by 3 is 6kl 5 hl 3l

Worksheet on Word Problems on Addition and Subtraction of Decimals has extensive problems on adding, subtracting decimals. Learn how to add and subtract decimals by referring to the Addition and Subtraction of Decimals Worksheet. All the Questions existing cover numerous models asked on the topic and keeps you up to date with the trends. Make sure to attempt the Adding and Subtracting Decimals Word Problems on a daily basis to get a good hold of the concept. You can verify your solutions as we have provided step-by-step solutions for all the problems.

Do Check:

• Multiplication of Decimals
• Adding Decimals
• Subtracting Decimals
• Simplify Decimals Involving Addition and Subtraction Decimals

Word Problems Involving Addition and Subtraction of Decimals

Example 1.
Krishna traveled 8. 37 km by car and 2.20 km by bike to visit his friend. How many kilometers did he travel to visit his friend?

Solution:

Krishna traveled by car=8.37
Krishna traveled by bike=2.20
The total distance he traveled to visit his friend =8.37 +2.20

Therefore, the total distance he traveled to visit his friend is 10.57 km.

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Example 2.
Geetha bought a drawing book for Rs 55.50 and sketch pens for 43.50. Find the total cost of a drawing book and sketch pens?

Solution:

The cost of a drawing book=55.50
The cost of sketch pens=43.50
The total cost of a drawing book and sketch pens =55.50 + 43.50

Hence, the total cost of a drawing book and sketch pens is 99.00.

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Example 3.
Lalitha won a cash prize of Rs 5500.00. Lalitha’s mother gave her Rs 1030.30. Find the total amount Lalitha has?

Solution:

Lalitha won a cash prize=5500.00
Lalitha’s mother gave her=1030.30
The total amount Lalitha has

Therefore, Lalitha has an amount of Rs 6530.30.

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Example 4.
Harsha bought two dresses online for Rs 2525.50. He has a discount coupon worth Rs120. Find how much money he has to pay for the purchase?

Solution:

Harsha bought two dresses online=2525.50
Harsha has a discount=120.00
The total money Harsha has to pay is

Therefore, the total money Harsha has to pay is 2405.50.

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Example 5.
Hema has money of Rs 15540.20. She paid the school fees of Rs 8750.00. She bought the school dress and books for Rs 5290.30. Find the money she has left?

Solution:

Hema has money=15540.20
Hema paid schoolfees=8750.00
Hema bought School books and dress= 5290.30
The total money spent by Hema for school fees, dress and books is

The money she has left=15540.20 -14040.30

Hence, the total money she has left is 1499.90.

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Example 6.
Rajeev has a big plastic pipe of length 9.5 inches long and a small plastic pipe is of length 3.2 inches long. Find how many inches of big plastic pipe is more than a small plastic pipe?

Solution:

Rajeev has a big plastic pipe of length=9.5
Rajeev has a small plastic pipe of length=3.2
No. of inches of big plastic pipe is more than small plastic pipe by

Hence, a big plastic pipe is 6.3 inches more than a small plastic pipe.

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Example 7.
At a juice stall, Srinu sold 10.75 liters of orange juice, 12.50 liters of grape juice on Wednesday. Find how many liters of juices sold by Srinu on Wednesday?

Solution:

No. of liters of orange juice sold on wednesday=10.75
No. of liters of grape juice sold on wednesday=12.50
The total no. of juices sold by Srinu on Wednesday is

Hence, the total no. of juices sold by Srinu on Wednesday is 23.25 liters.

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Example 8.
Janardhan has to attend a marriage in Hyderabad which is 200.6 miles away. He traveled 150.4 km by car. How many more miles does Janardhan have to travel to reach Hyderabad?

Solution:

Janardhan has to travel the distance=200.6
Janardhan traveled the distance by car=150.4
The total no. of miles Janardhan has to travel is

Therefore, Janardhan has to travel 50.2 km to reach Hyderabad.

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Example 9.
Sita and Geetha both are friends. They both decided to collect money for the drought relief fund. Sita collected money Rs 4575.40, Geetha collected money Rs 7984.50. Find how much money does Geetha collected more than Sita?

Solution:

Sita collected money= 4575.40
Geetha collected money=7984.50
The amount of money Geetha collected more than Sita

Therefore, Geetha collected money of Rs 3409.10 more than Sita.

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Example 10.
The highest rainfall recorded for Andhra is 45.7 inches during the Rainy season 1830-35. Its lowest rainfall is 15.2 inches was recorded during the rainy season 1847-49. How much more rain did Andhra receive in the year 1830-35 than in 1847-49?

Solution:

The highest rainfall for Andhra in 1830-35 =45.7
The lowest rainfall for Andhra in 1847-49=15.2
The more rainfall Andhra receive in 1830-35 is

Hence, the more rainfall Andhra received in 1830-35 than in 1847-49 is 30.5 inches.

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Practice the questions given in the Worksheet on Word Problems on Division of Mixed Fractions and learn the concept. Test your understanding by answering various problems in the Dividing Mixed Numbers Worksheet and improvise on the areas needed accordingly. Know the Problem Solving Approach used and solve related problems you come across in your exams with utmost confidence. Our Free Worksheets on Mixed Numbers Word Problems are flexible to use and develop excellent study habits among your kids.

Also, Read:

• Worksheet on Word Problems on Addition of Mixed Fractions
• Worksheet on Word Problems on Subtraction of Mixed Fractions
• Worksheet on Word Problems on Multiplication of Mixed Fractions

Dividing Mixed Fractions Worksheets with Answers PDF

Example 1.
Sindhu has 7 mangoes. She wants to share them with her friends. If Sindhu wants to give 1 \(\frac {2}{5} \) of the mango to each of her friends, Find how many friends will get some mango?

Solution:

No. of mangoes Sindhu has=7
Sindhu gives mango to each of her friend=1 \(\frac {2}{5} \)
No. of friends will get Mangoes= 7 ÷1 \(\frac {2}{ 5} \)
=7 ÷ \(\frac {7}{ 5} \)
=\(\frac {7× 5}{ 7 × 1} \)
=\(\frac {35}{7} \)
=5
Therefore, 5 friends will get some mangoes.

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Example 2.
Anand bought 10 gallons of green paint for painting the rooms in the office. If each room requires 1 \(\frac {1}{9} \) of a gallon of paint. How many rooms will get painted?

Solution:

No. of gallons of paint bought for painting the rooms=10
No. of  gallons of paint required for each room=1 \(\frac {1}{9} \)
No. of rooms will get painted=10 ÷1 \(\frac {1}{9} \)
=10 ÷ \(\frac {10}{9} \)
= \(\frac {10× 9}{ 10 × 1} \)
=\(\frac {90}{ 10} \)
=9
Therefore, 9 rooms will get painted.

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Example 3.
If Janaki runs 1 \(\frac {1}{4} \) of a kilometer per hour. How many hours would it take to run 5 km?

Solution:

No. of kilometers Janaki runs per hour=1 \(\frac {1}{4} \)
No. of hours required for running 5 km =5 ÷1 \(\frac {1}{4} \)=5 ÷ \(\frac {5}{4} \)
=5 ×\(\frac {4}{5} \)
=20/5\(\frac {20}{5} \)
=4 hours
Therefore, Janaki requires 4 hours to run 5 km.

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Example 4.
Suppose Grishma has 20 apples and you give each of your friends 1 \(\frac {1}{4} \) of an apple. Find how many friends Grishma has?

Solution:

No. of apples Grishma has=20
Each friend will get an apple=1 \(\frac {1}{4} \)
No. of Friends Grishma has=20 ÷1 \(\frac {1}{4} \)
= 20 ÷ \(\frac {5}{4} \)
=20 × \(\frac {4}{5} \)
=\(\frac {180}{5} \)
=36
Therefore, Grishma has 36 friends.

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Example 5.
Bhaskar got a plant in the garden. It grows 1 \(\frac {3}{4} \) inches every month, How long it will take time for a plant to grow 9 \(\frac {1}{4} \)  inches?

Solution:

The plant grows every month=1 \(\frac {3}{4} \)
Time is taken for the plant to grow 9 \(\frac {1}{4} \) inches = 1 \(\frac {3}{4} \) ÷  9 \(\frac {1}{4} \)
=\(\frac {7}{4} \)÷\(\frac {37}{4} \)
=\(\frac {7 × 4}{37 × 4} \)
=\(\frac {28}{148} \)
=\(\frac {7}{37} \)
Therefore, the time taken for a plant to grow 9 \(\frac {1}{4} \) inches is \(\frac {7}{37} \).

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Example 6.
Priya’s house is under construction, She is going to use tiles for the floor. The tile is in a square shape which is \(\frac {4}{5} \) feet wide. The size of the room is 12 \(\frac {1}{8} \) feet wide, how many tiles are required to cover the width of the room?

Solution:

The width of the tile = \(\frac {4}{5} \)
Size of the room=12 \(\frac {1}{8} \)
Tiles required to cover the width of the room= \(\frac {4}{5} \) ÷12 \(\frac {1}{8} \)
=\(\frac {4}{5} \)  ÷ \(\frac {97}{8} \)
=\(\frac {4 × 8}{5 × 97} \)
=\(\frac {32}{485} \)
Therefore, \(\frac {32}{485} \) tiles required to cover the width of the room.

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Example 7.
Rithu has a material which is of length 1 \(\frac {2}{8} \) m, she wants to cut pieces that are \(\frac {1}{4} \) m long. How many pieces can she cut?

Solution:

Ritu has a material which is of length= 1 \(\frac {2}{8} \)
The length of pieces=\(\frac {1}{4} \)
Total no. of pieces Ritu can cut= 1 \(\frac {2}{8} \)  ÷\(\frac {1}{4} \)
= \(\frac {10}{8} \) ÷ \(\frac {1}{4} \)
=\(\frac {5}{4} \)  ÷ \(\frac {1}{4} \)
=\(\frac {5 × 4}{4× 1} \)
=\(\frac {20}{4} \)
=5
Therefore, the total no. of pieces Ritu can cut is 5.

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Example 8.
Prasanti has 1 \(\frac {1}{3} \) of an hour to answer the problems in the maths examination. If it takes Prasanti \(\frac {1}{3} \)  of an hour to answer one problem, How many problems she can answer in the exam in the time left?

Solution:

Prasanti has the time to answer the exam= 1 \(\frac {1}{3} \)
Time is taken by Prasanti to answer one problem= \(\frac {1}{3} \)
No. of Problems Prasanti can answer = 1 \(\frac {1}{3} \) ÷ \(\frac {1}{3} \)
=\(\frac {4}{3} \) ÷ \(\frac {1}{3} \)
=\(\frac {4 × 3}{3 × 1} \)
=\(\frac {12}{3} \)
=4
Therefore, Prasanti can answer 4 problems in the time left.

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Example 9.
A wood that is 14 \(\frac {1}{3} \) feet long is cut into 5 pieces. Find how many feet long is each piece?

Solution:

Length of the wood = 14 \(\frac {1}{3} \)
No. of pieces of wood=5
Each piece is of length= 14 \(\frac {1}{3} \) ÷ 5
=\(\frac {44}{3} \) ÷ 5
=\(\frac {44 × 1}{3 × 5} \)
=\(\frac {44}{15} \)
=2  \(\frac {14}{15} \)
Therefore, the length of each piece is 2 \(\frac {14}{15} \) .

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Example 10.
Anusha completes \(\frac {1}{4} \) part of the shopping in 1 \(\frac {1}{6} \) hours. How much time she will take to complete the shopping?

Solution:

Time is taken by Anusha to complete the \(\frac {1}{4} \)  part of shopping = 1 \(\frac {1}{6} \)
Time is taken by Anusha to complete the shopping =1 \(\frac {1}{6} \)  ÷ \(\frac {1}{4} \)
= \(\frac {7}{6} \) ÷ \(\frac {1}{4} \)
=\(\frac {7 × 4}{6 × 1} \) =\(\frac {28}{6} \)=4 \(\frac {2}{3} \).
Therefore, Anusha takes 4 \(\frac {2}{3} \) hours to complete the shopping.

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Students will learn how to solve problems on multiplication of mixed fractions by referring to this page. We have covered various models of Mixed Fractions Multiplication Word Problems that range from easy ones to difficult ones. Try to answer all the questions on mixed fractions multiplication on your own and then verify with our solutions provided to test your skills on the concept. Practice as much as you can and enhance your problem-solving ability and speed of attempting the questions too.

Do Check:

• Word Problems on Division of Mixed Fractions
• Word Problems on Subtraction of Mixed Fractions
• Word Problems on Addition of Mixed Fractions

Mixed Numbers Multiplication Word Problems

Example 1.
Sai spent 8 hours selling vegetables in the market. Murali sells for 1\(\frac {1}{ 4} \)  times as many hours as Sai. How many hours did Murali sell vegetables in the market?
Solution:
Time is taken by Sai for selling the vegetables= 8 hours
Murali sells vegetables =1 \(\frac {1}{ 4} \) × 8
=\(\frac {5}{ 4} \) × 8
=\(\frac {5}{ 4} \)× \(\frac {8}{ 1} \)
=\(\frac {40}{ 4} \) =10
Therefore, Murali spent 10 hours selling vegetables in the market.

Example 2.
On Doctor’s advice, Rajesh decided to walk every day. He walks \(\frac {4}{ 7} \) kilometers every day. How many kilometers did he walk in a week?
Solution:
Rajesh walk every day= \(\frac {4}{ 7} \)
No. of kilometers Rajesh walk in a week=7 × \(\frac {4}{ 7} \)
=4 km
Therefore, Rajesh walks 4 km a week.

Example 3.
The heaviest weight in the family of Sarath is 90 kg. Sarath’s weight is \(\frac {1}{ 3} \) of the heaviest weight. Find the weight of Sarath?
Solution:
The heaviest weight in the family = 90 kg
Sarath weight= \(\frac {1}{ 3} \)× 90
=30 kg
Hence, Sarath’s weight is 30 kg.

Example 4.
Supriya is interested in singing and dancing. Supriya practices 28 hours a week with them. She takes \(\frac {1}{ 3} \) of the time for singing and \(\frac {1}{ 4} \) of the time for dancing. How many hours does she take for dancing?
Solution:
Total no. of hours Supriya practice every week= 28
Supriya takes time for singing=\(\frac {1}{ 3} \)
Supriya takes time for dancing=\(\frac {1}{ 4} \)
No. of hours she spends in dancing=\(\frac {1}{ 4} \)× 28
=7
Therefore, Supriya takes 7 hours for dancing.

Example 5.
In an office, there are 2000 employees. Male employees are \(\frac {1}{ 4} \). Find the Female employees in the office?
Solution:
Total no. of employees in the office=2000
No. of male employees=\(\frac {1}{ 4} \)
No. of female employees=1-\(\frac {1}{ 4} \)=\(\frac {3}{4} \)
\(\frac {3}{ 4} \)× 2000=1500.
Therefore, the total no. of female employees in the office is 1500.

Example 6.
Games are conducted for the students in class X. \(\frac {3}{ 8} \) of the total no of students are in class X. \(\frac {1}{ 5} \) of the students in class X participate in games. What fraction of all the students participate in the games?
Solution:
No. of students in the class X= \(\frac {3}{ 8} \)
No. of students participate in the games=\(\frac {1}{ 5} \)
Fraction of all students participate in the games=\(\frac {3}{ 8} \) × \(\frac {1}{5} \)=\(\frac {3}{40} \)
Therefore, the fraction of all the students participating in the exam is \(\frac {3}{40} \) .

Example 7.
Jagadish planted a mango tree and guava tree in the garden. The guava tree is 3 \(\frac {1}{ 5} \) feet tall. The mango tree is 1 \(\frac {1}{ 3} \) as tall as the guava tree. How tall is the mango tree?
Solution:
The height of the guava tree=3 \(\frac {1}{5} \)
The height of the mango tree =1\(\frac {1}{ 3} \) × 3 \(\frac {1}{ 5} \)
=\(\frac {4}{ 3} \) × \(\frac {16}{ 5} \)
=\(\frac {4 × 16}{ 3× 5} \)
=\(\frac {64}{15} \)
=4 \(\frac {4}{15} \)
Therefore, the height of the mango tree is 4 \(\frac {4}{15} \).

Example 8.
The cake needs 3 \(\frac {1}{2} \) cups of flour. Janci is baking 2 \(\frac {1}{ 2} \) cakes. How many cups of flour will she need?
Solution:
No. of cups of flour required for cake=3 \(\frac {1}{ 2} \)
No. of cakes baking= 2 \(\frac {1}{ 2} \)
No. of cups of flour required=2 \(\frac {1}{ 2} \) × 3 \(\frac {1}{ 2} \)
=\(\frac {5}{ 2} \)× \(\frac {7}{ 2} \)
=\(\frac {35}{ 4} \)
=8 \(\frac {3}{ 4} \)
Therefore, 8 \(\frac {3}{ 4} \) cups of flour required for baking 2 \(\frac {1}{ 2} \) cakes.

Example 9.
Sirisha uses \(\frac {2}{5} \) liters of petrol in her Scooty for 1 km. Find how many liters of petrol Sirisha use to drive 1 \(\frac {3}{8} \) km?
Solution:
No. of liters of petrol for 1 km=\(\frac {2} {5} \)
No. of liters of petrol used for 1 \(\frac {3}{ 8} \) km= 1 \(\frac {3}{ 8} \)  × \(\frac {2} {5} \)
=\(\frac {11} {8} \)× \(\frac {2} {5} \)
=\(\frac {22} {40} \)
=\(\frac {11} {10} \)
Therefore, Sirisha uses \(\frac {11} {10} \) lt of petrol for 1 \(\frac {3} {8} \) km.

Example 10.
A restaurant makes sweet lime juice on Friday and uses 3 \(\frac {1} {8} \) kg of sweet limes. As Saturday is a busy day, the restaurant plans to make 8 \(\frac {1} {9} \) as much as sweet lime juice. How many kg of Sweet lime juice is required?
Solution:
No. of kg of sweet limes used= 3 \(\frac {1} {8} \)
No. of kg of sweet limes used to make 8 \(\frac {1} {9} \) lt=3 \(\frac {1} {8} \) ×  8 \(\frac {1} {9} \)
=\(\frac {25} {8} \) ×  \(\frac {73} {9} \)
=\(\frac {25 ×  73} {8 × 9} \)
=\(\frac {1825} {72} \)
=25 \(\frac {25} {72} \)
Therefore, Restaurant uses 25 \(\frac {25} {72} \) kg of sweet limes.

Example 11.
Rakesh can ride a bike 3 \(\frac {2} {5} \) km in 1 hour. If he plans to ride a bike for 4 \(\frac {1} {7} \) hours, how much distance he can travel?
Solution:
No. of km Rakesh can ride a bike= 3 \(\frac {2} {5} \) km
No. of km Rakesh can ride a bike in 4 \(\frac {1} {7} \)hours=4 \(\frac {1} {7} \)  × 3 \(\frac {2} {5} \)
=\(\frac {29} {7} \) × \(\frac {17} {5} \)
=\(\frac {29 × 17} {7 ×  5} \)
=\(\frac {493} {35} \)
=14 \(\frac {3} {35} \)
Therefore, Rakesh can ride a bike in 14 3/35 km.