Vinay Pacha

Inverse Variation is nothing but if one increases the other decreases. Practice the questions in the Worksheet on Inverse Variation using Method of Proportion and learn the concept. All the Problems explained in the Worksheet help you solve inverse variation using the method of proportion. Download the easily accessible Unitary Method Inverse Variation Practice Worksheet PDF and solve the different kinds of questions framed on the topic regularly and attempt the exams with confidence.

Do Check:

• Worksheet on Direct Variation using Unitary Method
• Worksheet on Inverse Variation Using Unitary Method

Unitary Method Inverse Variation Questions

I. If 5 men can repair a road in 8 days. In how many days 10 men will repair the road?

Solution:

Given that,
5 men can repair a road in 8 days.
Here time and men are inversely proportional.
t=k/m ->eq1
Substitute m=5,t=8
8=k/5
k=40
By eq1,t=40/m
We have to find out how many days it will take for 10 men.
t=40/10=4
Hence, it will take 4 days to repair the road.

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II. 5 workers can make 60 toys in 4 days. How many workers will be needed to make 216 toys in 6 days?

Solution:

Given that,
5 workers can make 60 toys in 4 days.
In one day 5 workers would make 60/4=15 toys
No. of toys are made by 1 worker in one day=15/5=3 toys.
1 day-> 3 toys
6 days->6*3=18
No. of workers needed to make 216 toys in 6 days is
216/18=12 workers
Therefore, 12 workers are required to make 108 toys.

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III. If 12 cows can graze a field in 10 days. Then how many cows will graze the same field in 6 days.

Solution:

Given that,
12 cows can graze a field in 10 days.
Here cows and days are inversely proportional.
t=k/m->eq1
10=k/12
k=120
subs k in eq1
t=120/m
We have to find out no. of cows required to graze the field.
6=120/m
6m=120
m=120/6=20
Therefore, 20 cows were required to graze the field in 15 days.

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IV. A Bus takes 4 hours to complete a journey if it travels at a speed of 100 km/hr. How long will it take when the Bus travels at a speed of 80 km/hr?

Solution:

Given that,
A Bus takes 4 hours to complete a journey if it moves at a speed of 100 km/hr.
we know that Distance=speed × time
100 × 4= 80 × T
400=80T
T=400/80=5
Therefore, the bus takes 5 hours to complete the journey.

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V. A train moves at a speed of 100 km/hr and covers a certain distance in 8.5 hours. What should be the speed of the train to cover the same distance in 5 hours?

Solution:

Given that,
Speed of the train = 100km/hr
Time = 8.5 hours
We know that Distance=speed × time
Distance = 100 × 8.5
= 850 km
Distance = 850 km
Time = 5 hours
Speed = Distance/Time
= 850/5
= 170 km/hr
Therefore, the speed of the train is 170km/hr to cover the same distance in 5 hours.

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VI. In a software company, 5 employees are required to fix the bug in 9 days. Then how many employees are required to fix the bug in 5 days?

Solution:

Given that,
5 employees are required to fix the bug in 9 days.
Here employees and days are inversely proportional.
t=k/m->eq1
9=k/5
k=45
subs k in eq1
t=45/m
We have to find out no. of employees required to fix the bug.
5=45/m
5m=45
m=45/5=9
Therefore, 9 employees were required to fix the bug.

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VII. If 2 men can survey the population in a town in 6 days. In how many days 6 men will do it?

Solution:

Given that,
2 men can survey the population in a town in 6 days.
Here time and men are inversely proportional.
t=k/m ->eq1
Substitute m=2,t=6
6=k/2
k=12
By eq1,t=12/m
We have to find out how many days it will take for 6 men.
t=12/6=2
Hence, it will take 2 days to repair the road.

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VIII. 4 workers can stitch 48 shirts in 4 days. How many workers will be needed to make 30 shirts in 2 days?

Solution:

Given that,
4 workers can stitch 48 shirts in 4 days.
In one day 4 workers would make 48/4=12 shirts
No. of shirts are made by 1 worker in one day=12/4=3 shirts.
1 day-> 3 shirts
2 days->2*3=6
No. of workers needed to make 30 shirts in 2 days is
30/6=5 workers
Therefore, 5 workers are required to make 108 toys.

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What is the Tabular Data? Tabular Data is nothing but arranging the data systematically and logically in a table form to understand it easily and accurately. Here, even a large amount of data can be arranged in a coordinated manner and easy to read to make students comfortable.

Tabular presentation of data is one of the most used forms to present the data as tables. It makes children easy to read and prepare the questions regarding the data. A tabular data organizes the information in the form of rows and columns. The most important of tabulation is it coordinates the data for decision making and statistical treatment.

What is Tabular Presentation of Data?

Tabular presentation of data is a method to present the data in a simple and significant way. Anything in a tabular presentation of information is arranged in a table, which organizes the data in rows and columns. It is also known as presenting the data using a statistical table. Tabulation is done to coordinate data for additional mathematical treatment and decision-making. The evaluation used in tabular data is of four types. They are as follows:

• Qualitative Classification
• Quantitative Classification
• Temporal Classification
• Spatial Classification

Objectives of Tabulation

The main objectives of tabulation are as follows:

1. The tabular presentation helps in summarizing and simplifying complex data.
2. Tabulation is done to compare different types of data sets and bring out their primary and necessary aspects.
3. An arithmetical analysis is assumed from tabular data.
4. It helps in saving time and space. Presentation of data further helps in the formation of graphs and diagrams for analyzing the data.

Advantages of Tabular Data

The advantages of tabular presentation of data are listed below:

• A table makes the representation of data easy where a large amount of data can be easily summarized in a data table. It is the simple form of presenting data.
• Data tables are mostly used for arithmetical analysis like calculating numerical data and it helps in analyzing the data easily.
• A tabular data makes it easy to compare the data.
• It saves readers space as well as time.
• It helps in easy visualization of data so students show interest in graphs and diagrams over words.

Disadvantages of Tabular Data

Following are the three disadvantages while presenting the data in a tabular form.

• It lacks the description of data where it involves diagrams. It overlooks the important aspects.
• It fails in providing efficient individual items. It gives aggregate data.
• It requires special knowledge to understand the data in a table.

Difference Between Tabular and Textual Presentation of Data

The following are the differences in the two types of presentation of data (tabular & textual):

Tabular Presentation	Textual Presentation
We can find the data easily in a significant manner.	You have to go through each line of the given data.
It is an efficient way and it summarizes the data in rows and columns.	Makes it tedious and confusing to understand any information.
It helps the viewer to understand and to interpret the information better.	It takes time and space to view the information.
In the tabular presentation, we use graphs and diagrams to represent the data.	In the textual presentation, we use words to explain the data.

Components used in Tabular Data

The components used in the data table are as follows:

• Table Number: Indicated on top of each and every table for easy identification and further reference.
• Title: It describes the contents of the table, and explains the brief of the data, and contains classification data.
• Headnote: It defines the purpose of the title. Usually, headnotes present the units of data in brackets.
• Stubs or Row Headings: Stubs describe the information about the data which holds in a particular row. The title of the horizontal rows is known as ‘stubs’.
• Caption or Column Headings: In each column of a table, it is given to explain the figures of the column. The column head is known as a “caption”.
• Body: The body of the table says the complete information of the data. Each item in a body is defined as a “cell”.
• Footnote: Footnotes are rarely used and it is a specific feature and has not to be explained earlier.
• Source: It is a brief statement indicating the source of data presented in the table. It has to be mentioned below the footnote.

FAQ’s on Advantages of Tabular Presentation

1. What is the main advantage of tabular data?
It makes data representation easy. Also, it is clear to analyze and compare the data which saves readers time as well as space.

2. Describe the disadvantage of tabular presentation?
Only figures are indicated in tabular form. The attributes of those figures can not be mentioned in tables. Qualitative information of figures is not mentioned.

3. What are the three methods of representing data?
Three methods used to represent data are textual, tabular, and graphical presentation.

Worksheet on Inverse Variation using Unitary Method has questions on finding the Inverse Variation using Unitary Method. Solve the problems in the Inverse Variation in Unitary Method Worksheet PDF and learn the concept well. Learn the respective formulas of inverse variation by referring to the problems available. Download the Unitary Method Inverse Variation Worksheet for free and prepare anytime and anywhere to master the concept.

See More:

• Worksheet on Direct Variation using Unitary Method
• Worksheet on Direct variation using Method of Proportion

Unitary Method Inverse Variation Worksheet

I. Raju can complete work in 5 days working 8 hours per day. If he works 10 hours per day, how many days will he take to complete the work?

Solution:
Given that,
No. of days Raju can complete work by working 8 hours per day=5 days
This is an inverse proportion because lesser hours per day-> more days to complete the work.
8 hours per day  -> 5 days to complete the work.
1 hour per day   -> 8× 5=40
10 hours per day  -> 40/10=4
Therefore, Raju can complete the work in 4 days by working 10 hours per day.

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II. Alekya takes 30 days to reduce 2 kilograms of his weight by doing 30 minutes of exercise per day. If he does exercise for 1 hour 40 minutes per day, how many days will he take to reduce the same weight?

Solution:

Given that,
No. of days required to reduce 2 kilograms by doing 30 minutes exercise=30 days
Total minutes in 30 days=30 × 30=900 minutes
So 900 minutes of exercise are required to reduce 2 kilograms weight.
Also given, Alekya did exercise for 1 hour 40 minutes.
1 hour 40 minutes=100 minutes
No. of days required=900/100=9 days
Therefore, 9 days are required to reduce 2 kg by doing 1 hour 40 minutes of exercise.

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III. If 7 men can complete the work in 20 hours, how many men will be able to complete the work in 10 hours?

Solution:

Given that,
In 20 hours, work is completed by 7 men.
No. of hours taken by 1 man to complete the work is
= No. of hours ⋅ No. of men
= 20 ⋅ 7
= 140 hours
No. of men required to complete the work in 10 hours is
= 140 / 10
= 14 men
Therefore, 14 men will be able to work in 10 hours.

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IV. A school has 7 periods in a day such that each period is of 30 minutes. If the number of periods is reduced to 6, then how long would each period be?

Solution:

Given that,
A school has 7 periods in a day such that each period is of 30 minutes.
We know that Total time = Number of period × time of each period.
Total time = 7 × 30
Total time = 210 minutes
Also given,
The number of periods is reduced to 6.
We have to calculate the time of each period,
Total time = Number of period × time of each period.
By Substituting the values, we get
210 = 6 × time of each period
210/6 = time of each period
=35 minutes.
Therefore, the time of each period is 35 minutes.

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V. 5 members in a family had enough food for 30 days after 8 days some members went to Delhi and thus the food lasted for 36 more days how many members left the Family?

Solution:

Given that,
5 members of a family had enough food for 30 days.
Let the number of members who went to Delhi be x.
After 8 days, number of members = (5 – x)
Given that food lasted for 36 more days.
Hence 5 × 22 = (5 – x) × 36
⇒ (5 – x) = 110/36=3
⇒ x = 5 – 3 = 2
Hence, 2 members left the family.

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VI. Before going to sleep a man reads 7 pages of a book every day and completes it in 30 days. How many days will he take to complete reading the book, if he reads 10 pages every day?

Solution:

Given that,
A man reads 7 pages of a book every day and completes it in 30 days.
more pages per day—–> fewer days to complete the book (since it was inverse variation)
7 pages per day ——> 30 days
1 page per day ——–> 7 × 30 = 210 days
10 pages per day ——> 210 / 10 = 21 days
Therefore, the man will complete the book in 21 days if he reads 10 pages per day.

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VII. 8 pipes are required to fill a tank in 1 hour 40 minutes. How long will it take if only 6 pipes of the same type are used?

Solution:

Let the desired time to fill the tank be x minutes.
8 pipes are required to fill a tank in 1 hour 40 minutes.
8 pipes -> 1 hour 40 minutes
We know that 1 hour=60 minute.
1 hour 40 minute=60+40=100 minutes.
In 100 mins a tank is filled with 8 pipes.
Lesser the number of pipes more will be the time required by it to fill the tank.
So, this is a case of inverse proportion.
Hence, 100×8=x×6
x=800/6
x=133
Thus, the time is taken to fill the tank with 6 pipes is 133 minutes or 1 hour 33 minutes.

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VIII. A man has enough money to buy 10 kg of mangoes at Rs 80 per kg. How much can he buy, if the price is increased by Rs 2.50 per kg?

Solution:

This is a situation of inverse proportion.
more price -> less kg of apples
Given that, The cost of 10 kg of apples at Rs 80 per kg is
= 10 ⋅ 80
= Rs 800
So, the person has Rs 800.
If the price is increased by Rs 2.50 per kg, then the new price per kg is
= Rs 82.50
No. of pounds of apples he can buy with Rs 82.50 is
= 800 / 82.50
= 9.69 kg
If the price is increased by Rs 2.50 per kg, the person can buy 9.69 kg of apples.

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IX. A car covers a particular distance in 4 hours with a speed of 80 miles per hour. If the speed is increased by 20 miles per hour, find the time taken by the car to cover the same distance?

Solution:

Speed, time are inversely proportional.
Given that,
Time = 4 hours and Speed = 80 mph
We know that Distance = Time ⋅ Speed
Distance = 4 ⋅ 80 = 320 miles
If the given speed of 80 mph is increased by 20 mph, then the new speed will be 100 mph.
The formula to find the time is,
Time = Distance / Speed
Time = 320 / 100
Time = 3.2 hours
The time taken by the car is 3.2 hours, when the speed is increased by 20 mph.

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X.In a camp, there is food provision for 100 persons for 40 days. If 20 more persons join the camp, for how
many days will the provision last?

Solution :
More the persons, the sooner would be the provision exhausted.
So, this is a case of inverse proportion.
Let the required number of days be x.
Hence, 100 × 40=(100+20) × x
100 × 40=120 × x
x=100 × 40/120
x=33
Therefore, Provisions will be last for 33 days.

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XI. A car is travelling 20 km in one hour. Find the distance traveled by car in 15 minutes?

Solution:

We know that 1 hour = 60 min
In 60 min car travels =20 km
In 1 min car will travel =20/60
In 12 min car will travel = 20/60 ×15=5 km
Therefore, the car travels 5 km in 12 minutes.

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XII. If Kalpana walks 120 steps to cover a distance of 200 meters, find the distance traveled in 380 steps?

Solution:

Given that,
120 steps cover a distance of 200 meters.
1 step covers=200/120=10/6=5/3
380 steps cover a distance=380 × 5/3
=633 meters
Therefore, 380 steps cover a distance of 633 meters.

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Practice the questions on the Worksheet on Direct Variation using Method of Proportion and learn how various questions are framed on the topic.  Direct Variation using Method of Proportion Worksheets are flexible to download and you can learn at your own pace. Our Practice Worksheets on Direct Variation using the Proportion Method will combine fun with studies and help you keep motivated. You can be well versed with the topic if you solve the questions available in the worksheet meticulously.

Do Check: Worksheet on Direct Variation using Unitary Method

Direct Variation using Method of Proportion Worksheet with Answers

I. The cost of 7 kg of sugar is Rs  210. What will the cost of 13 kg of sugar be?

Solution:

Given,
The cost of 7 kg of sugar is = Rs 210
Let the cost of 13 kg of sugar be x.
Since they vary directly,
7/210=13/x
7x=210 × 13
7x=2730
x=2730/7=390
Hence, the cost of 13 kg of sugar is Rs 390.

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II. If the weight of 4 fruits is 500 g, find how many such fruits will weigh 5.5 kg?

Solution:

Given,
The weight of 4 fruits is= 500 g
The weight of 1 fruit=500/4=125
we know that 1 kg=1000 gm.
5.5 kg=5500 gm
No. of fruits will weigh 5.5 kg=5500/125=44
Hence, 44 fruits will weigh 5.5 kg.

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III. Rakesh gets Rs 3500 for 7 days of work. How many days should he work to get Rs 8000?

Solution:

Given,
Rakesh gets money for 7 days of work=Rs 3500
Two quantities vary directly, (money and working days)
i.e. More money more working days.
Let the number of working days to get the money of Rs 8000 be x.
7/3500=x/8000
56000=3500x
x=56000/3500
x=16
Therefore, Rakesh has to work 16 days to get Rs 8000.

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IV. The cost of  8 notebooks is Rs192, what do 30 notebooks cost?

Solution:

Given,
The cost of  8 notebooks is=Rs 192
Let the cost of 30 notebooks be x.
since they vary directly,
8/192=30/x
8x=30 × 192
8x=5760
x=5760/8
=720
Hence, the cost of 30 notebooks is Rs 720.

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V. A man walks 10,000 steps every day to cover a 1 km distance. If he walks 14,000 steps how much distance will be covered?

Solution:

Given,
A man takes 10000 steps every day to cover a distance=1 km
Let the distance covered for 14000 steps be x.
since they vary directly,
10000/1=14000/x
10000x=14000
x=14000/10000
=1.4
Hence, the man covers 1.4 km if he walks 14,000 steps.

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VI. At a birthday party, Ravi buys 8 gallons of juice for 70 members. how many gallons do they need for 180 members?

Solution:

Given,
Ravi buys juice for 70 members=8 gallons
Let the 180 members need gallons of juice be x.
since they are in direct variation,
70/8=180/x
70x=8 × 180
70x=1440
x=1440/70=20.57
Therefore, Ravi needs 20.57 gallons of juice for 180 members.

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VII. A bus travels 120 km on 5 liters of petrol. How far would it travel in 4 liters of petrol?

Solution:

Given,
A bus travels120 km with no. of liters of petrol=5 liters
Let the bus travels the no. of km with 4 liters of petrol be x.
120/5=x/4
480=5x
x=480/5=96
Therefore, the bus travels 96 km with 4 liters of petrol.

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VIII. 9 workers can dig 7 m long trench in one day. How many workers would dig 25 m long trench of the same type in 1 day?

solution:

Given,
No. of workers can dig 7 m long trench in one day=9
Let the no. of workers dig 25 m long trench in one day=x
Since they are in direct proportion,
7/9=25/x
7x=225
x=225/7=32
Therefore, 32 workers were required to dig a trench in one day.

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IX. If two pens cost Rs 50, how many pens can you buy for Rs 250?

Solution:

Given,
Two pens cost= Rs 50
Let the number of pens can buy for Rs 250 is x.
2/50=x/250
2 × 250=50x
500=50x
x=500/50=10
Therefore, 10 pens can be bought for Rs 250.

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X. 3 bags of wheat weigh 150 kg. How many such bags will weigh 900 kg?

Solution:

Given,
Weight of 3 bags of wheat=150 kg
Let the number of bags weighing 900 kg be x.
since it is direct proportion,
3/150=x/900
2700=150x
x=2700/150
=18
Therefore, 18 bags of wheat will weigh 900kg.

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Practice the questions in the Worksheet on Direct Variation using Unitary Method and learn the concept on a deeper level. Solve the different types of problems based on the Unitary Method and know the approach used. Test your knowledge on the concept of Unitary Method using Direct Variation by solving the problems available in the Unitary Method Direct Variation Worksheet. Simply try to answer them on your own and then cross-check with the solutions provided to improve on the areas accordingly.

Worksheet on Unitary Method using Direct Variation

I. In a toy shop, 20 toys cost Rs. 1,330.50. Find the cost of 16 toys?

Solution:

Given,
Cost of 20 toys=Rs 1,330.50
Cost of 1 toy=Rs 1,330.50/20
=66.525
The cost of 16 toys=66.525 × 16
=1064.4
Hence, the cost of 16 toys is Rs 1064.4.

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II. In a business, if Arjun can earn Rs 4,75,000 in 2.5 years, At the same rate, find his earning for 4 years?

Solution:

Given,
Arjun can earn money in 2.5 years=Rs 4,75,000
Earning money for 1 year is=Rs 4,75,000/2.5
=Rs 190000
Earning money for 4 year is=190000 × 4
=Rs 760000
Therefore, Arjun earns money for 4 years is Rs 760000.

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III. Sunil gets Rs 500  for 7 hours of work. How much money does he get for 3 hours?

Solution:

Given,
Sunil gets money for 7 hours of work=Rs 500
Sunil gets money for 1 hour of work=Rs 500/7=Rs 71.42
Sunil gets money for 3 hours of work= Rs 71.42 × 3
=Rs 214.26
Hence, Sunil gets money for 3 hours is Rs 214.26.

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Iv. Sanjana would like to buy 10 dresses. If the cost of each dress is Rs 700, what is the total cost for 10 dresses?

Solution:

Given,
The cost of each dress is =Rs 700
The total cost for 10 dresses=Rs 700 × 10
=Rs 7000
Therefore, the total cost of 10 dresses is Rs 7000.

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V. Gopal bought a dozen apples for Rs 500. Find the cost of 15 such apples. How many apples can be bought for Rs 83.32?

Solution:

Given,
Gopal bought a dozen apples for= Rs 500
Cost of 1 apple=Rs 500/12=Rs 41.66
Cost of 15 such apples=15 × 41.66=624.9
No. of apples can be bought for Rs 83.32=83.32/41.66=2
Hence, Gopal can buy two apples for Rs 83.32.

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VI. The weight of 50 Handbags is 6 kg.
(a) What is the weight of 80 such Handbags?
(b) How many such Handbags weigh 7.5 Kg?

Solution:

Given,
The weight of 50 Handbags is =6 kg
The weight of 1 handbag is=6/50=0.12
The weight of 80 such Handbags=80 × 0.12=9.6
No. of Handbags weigh 7.5 Kg=7.5/0.12=62.5
Hence, No. of handbags weigh 7.5 kg is 62.5 handbags.

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VII. In 5 weeks, Mahesh raised the fund of Rs 25,539.50 for helping the poor people. How much money will he raise in 20 weeks?

Solution:

Given,
Mahesh raised the money  for helping the poor people in 5 weeks=Rs 25,539.50
Mahesh raised the money in one week= Rs 25,539.50/5=5107.9
Mahesh will raise money in 20 weeks=20 × 5107.9=102158
Hence, Mahesh will raise money in 20 weeks is Rs 102158.

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VIII. Janci ordered 5liters of oil for Rs 550. Then she reduced her order to 3 liters. How much money does she have to pay for 3 liters?

Solution:

Given,
Janci ordered 5liters of oil for= Rs 550
Cost of 1-liter oil=Rs 550/5=Rs110
Cost of 3-liter oil=3 × 110=Rs 330
Therefore, Janci has to pay Rs 330.

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Ix.  A car traveling at a speed of 100 kmph covers 320 km. How much time will it take to cover 220 km?

Solution:

Given,
A car traveling at a speed of 100 kmph covers= 320 km
First, we have to find the time required to cover 320 km.
Speed = Distance/Time
100 = 320/T
T = 3.2 hours
Applying the unitary method,
320 km = 3.2 hours
1 km = 3.2/320 hour=0.01
220 km = 0.01 x 220 = 2.2 hours
Therefore, it will take 2.2 hours to cover 220km.

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X. Ajay finishes his work in 20 days while Bhaskar takes 30 days. How many days will the same work be done if they work together?

Solution:

Given,
Ajay finishes his work in =20 days
Bhaskar finishes his work in=30 days
Ajay’s 1 day work=1/20
Bhaskar’s 1 day work=1/30
Now, total work is done by A and B in 1 day = 1/20 + 1/30
Taking LCM(20, 30), we have,
1 day’s work of A and B = (3+2)/60=5/60=1/12
1 day’s work of (A + B) = 1/12
Thus, A and B can finish the work in 12 days if they work together.

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XI. A train is moving at a uniform speed of 58 km per hour. How far will it go in 10 minutes?

Solution:

Given,
A train is moving at a uniform speed of =58 km per hour
We know that speed=distance/time
First, we have to convert 10 minutes to hours.
To convert minutes to hours we have to divide it by 60.
10/60=0.16
58km=distance/0.16
distance=58 x 0.16=9.28
Hence, the train will move 9.28 km in 10 minutes.

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In this article of ours, we will discuss identifying like and unlike terms. Practice the questions in the Like and Unlike Terms Worksheet and check if the variables and exponent powers are the same or not and decide. Get a good hold of the concept by answering the Worksheet on Like and Unlike Terms and feel simplifying algebraic expressions much easier. Download the Identifying Like and Unlike Terms Worksheet PDF accessible for free of cost and prepare anytime and anywhere.

Read Similar Articles:

• Worksheet on Combining Like Terms
• Worksheet on Addition of Unlike Terms
• Worksheet on Subtraction of Unlike Terms

Like and Unlike Terms Worksheet with Answers

I. List out the like terms from each set:
(i) 5x²y³, 2x²y³, -4x²y², 6x²y³
(ii) 15m²n, 13mn², 4nm², 12m²n
(iii) 3a³b³, 5a²b³, 6a²b³, -4a³b³
(iv) -6a³b²z, 2zb³a², 15a³b²z, 18a³b²
(v) 2a²b²c², 3ac²b², -6a²b²c², 8c²a²b², -12a³b²

Solution:

(i) Given 5x²y³, 2x²y³, -4x²y², 6x²y³
Here the like terms are 5x²y³, 2x²y³, 6x²y³(since the variables and their exponent’s powers are the same).

(ii)Given 15m²n, 13mn², 4nm², 12m²n
Here the like terms are 15m²n, 4nm², 12m²n.

(iii)Given 3a³b³, 5a²b³, 6a²b³, -4a²b³
Here the like terms are 5a²b³, 6a²b³, -4a²b³
.
(iv)Given -6a³b²z, 2zb³a², 15a³b²z, 18a³b²
Here the like terms are -6a³b²z, 15a³b²z.

(v)Given 2a²b²c², 3ac²b², -6a²b²c², 8c²a²b², -12a³b²
Here the like terms are 2a²b²c², -6a²b²c², 8c²a²b².

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II. Group the like terms together:
(i) 2a, -5b, -a, b/3, 3a/4, 6a and b
(ii) 1/4xy, -5xy, 5yz, -2/7yz, yz/3 and xy
(iii) –ab2, -b2a2, 2b2a, -13a2b2 and 4ab2
(iv) 14ax, -5by, by/8, 5xa and 1/3ax
(v) 6m2n, mn2, 4m2n, 3mn2 and 13m2n

Solution:

(i) Given 2a, -5b, -a, b/3, 3a/4, 6a and b
2a,-a,3a/4,6a and -5b,b/3,b.
Hence, By grouping the like terms together we get 2a,-a,3a/4,6a and -5b,b/3,b.

(ii) Given 1/4xy, -5xy, 5yz, -2/7yz, yz/3 and xy
1/4xy, -5xy, xy and  5yz, -2/7yz, yz/3.

(iii) Given –ab2, -b2a2, 2b2a, -13a2b2 and 4ab2
–ab2,4ab2,2b2a and -b2a2, -13a2b2

(iv) Given 14ax, -5by, by/8, 5xa and 1/3ax
14ax, 5xa, 1/3ax and -5by, by/8.

(v) Given 6m2n, mn2, 4m2n, 3mn2 and 13m2n
6m2n,4m2n,13m2n and mn2, 3mn2.

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III. State the number of terms in each of the following expressions:
(i) 15x – y
(ii) 3 × x + y ÷ 3
(iii) 5x – x/b
(iv) x ÷ m × n + r
(v) 2x ÷ 4 + z + 3
(vi) (5c – m + 2) ÷ 4
(vii) a × x × y × z ÷ 15
(viii) m+ n ÷ q
(ix) a + b + c + 15 ÷ p
(x) 2 × b + 2 ÷ y + 3

Solution:

(i)two terms
(ii) two terms
(iii) two terms
(iv) two terms
(v) three terms
(vi) three terms
(vii) one term
(viii) two terms
(ix) four terms
(x) three terms

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Iv. State whether the following statements are true or false:
(i) 5z has two terms 5 and z.
(ii) Expression 5 + p has two terms 5 and p.
(iii) ab and –ba are like terms.
(iv) x2y and –y2x are like terms.
(v) a and –a are like terms.
(vi) –yx and 3xy are unlike terms.
(vii) 15 and 15z are like terms.
(viii) 2mn and 4amn are unlike terms.
(ix) 5n2m and -3mn2 are like terms.
(x) x/3, 2x, x are unlike terms.

Solution:

(i) False
(ii) True
(iii) True
(iv) False
(v) True
(vi) False
(vii) False
(viii) True
(ix) True
(x) False

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Combining Like Terms Worksheet can be quite handy for you if you are dealing with the problems of simplifying algebraic expressions involving the addition or subtraction of polynomials. Worksheet on Combining Like Terms will develop accuracy as well as engages you in logical thinking.

Free math Worksheets on Combining Like Terms with Variables on Both Sides can assist you and guide you to see and review your answer sheets. Download the easily accessible PDF Format Combining Like Terms Worksheets and solve questions for free.

See Similar Articles:

• Worksheet on Adding and Subtracting Like Terms
• Worksheet on Addition of Like Terms
• Worksheet on Subtraction of Like Terms

Combining Like Terms with Variables on Both Sides Worksheet

I. Find the sum of the following:
(i) 4a + b – 5c; -a + 3b +3c and 6a – 2b + 8c
(ii) -5x + y + 4z; 7x + 2y + 4z and 3x + 4y – 2z
(iii) 6p + 2q + r; -p + 8q – 9r and 5p – q +3r
(iv) 6a + 4b + 3c; a + 5b – 2c and 4a – b + 6c

Solution:

(i) Given 4a + b – 5c; -a + 3b +3c and 6a – 2b + 8c
First arrange the like terms together,
=4a+(-a)+6a+b+3b+(-2b)+(-5c)+3c+8c
=4a-a+6a+b+3b-2b-5c+11c
=10a-a+4b-2b-5c+11c
=9a+2b+6c
Hence, the sum of like terms 4a + b – 5c, -a + 3b +3c and 6a – 2b + 8c is 9a+2b+6c.

(ii) -5x + y + 4z; 7x + 2y + 4z and 3x + 4y – 2z
First arrange the like terms together,
=-5x+7x+3x+y+2y+4y+4z+4z-2z
=-5x+10x+7y+8z-2z
=5x+7y+6z
Hence, the sum of like terms -5x + y + 4z; 7x + 2y + 4z and 3x + 4y – 2z is 5x+7y+6z.
(iii) 6p + 2q + r; -p + 8q – 9r and 5p – q +3r
By arranging the like terms together we get,
=6p+(-p)+5p+2q+8q-q+r-9r+3r
=11p-p+5p+10q-q+4r-9r
=10p+9q-5r
Therefore, the sum of like terms 6p + 2q + r; -p + 8q – 9r and 5p – q +3r is 10p+9q-5r.
(iv) Given 6a + 4b + 3c; a + 5b – 2c and 4a – b + 6c
By arranging the like terms together we get,
=6a+a+4a+4b+5b+(-b)+3c+(-2c)+6c
=11a+9b-b+9c-2c
=11a+8b+7c
Therefore, the sum of like terms 6a + 4b + 3c; a + 5b – 2c and 4a – b + 6c is 11a+8b+7c.

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II. Subtract the following:
(i) Subtract -3xy + 4yz – 14zx from 7xy – 9yz + 6zx
(ii) Subtract -8a² + 9a – 6 from 2a² – a + 4
(iii)  subtract a²bc – 8b²ca– 5c²ab from 4a²bc + 3b²ca + c²ab.

Solution:

(i) Given 7xy – 9yz + 6zx take -3xy + 4yz – 14zx
=7xy – 9yz + 6zx-(-3xy+4yz-14zx)
=7xy – 9yz + 6zx + 3xy-4yz + 14zx
Arrange the like terms together,
=7xy+3xy-9yz-4yz+6zx+14zx
=10xy-13yz+20zx
Hence, By subtracting -3xy + 4yz – 14zx from 7xy – 9yz + 6zx we get 10xy-13yz+20zx.
(ii) Given 2a² – a + 4 take -8a² + 9a – 6
=2a² – a + 4 -(-8a² + 9a – 6)
=2a² – a + 4+8a² -9a+6
Arrange the like terms together,
=2a²+8a²-a -9a+4+6
=10a²-10a+10
Therefore, By subtracting -8a² + 9a – 6 from 2a² – a + 4 we get 10a2-10a+10.
(iii) Given 4a²bc + 3b²ca + c²ab take a²bc – 8b²ca– 5c²ab
=4a²bc + 3b²ca + c²ab-(a²bc – 8b²ca– 5c²ab)
=4a²bc + 3b²ca + c²ab-a²bc +8b²ca+5c²ab
By arranging the like terms together,
=4a²bc-a²bc+3b²ca+8b²ca +c²ab +5c²ab
=3a²bc+11b²ca+6c²ab
Therefore, By subtracting a²bc – 8b²ca– 5c²ab from 4a²bc + 3b²ca + c²ab we get 3a²bc+11b²ca+6c²ab.

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III. What should be added to 2x³ – 6x²y + 2xy² – 4y³ to get 3x³ + x²y – 2xy² – 6y³?

Solution:

Let M be the polynomial added to 2x³ – 6x²y + 2xy² – 4y³.
2x³ – 6x²y + 2xy² – 4y³ + M=3x³ + x²y – 2xy² – 6y³
M=3x³ + x²y – 2xy² – 6y³-(2x³ – 6x²y + 2xy² – 4y³)
M=3x³ + x²y – 2xy² – 6y³-2x³ +6x²y -2xy²+4y³
M=x³ +7x²y-4xy²-2y³
Therefore, x³ +7x²y-4xy²-2y³ be added to 2x³ – 6x²y + 2xy² – 4y³ to get 3x³ + x²y – 2xy² – 6y³.

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IV. Subtract 2m – m³ + 2m² from the sum of 9 + 5m² – 2m³+3m and 5m³ – 2m² + m -9.

Solution:

First, we have to find the sum of 9 + 5m² – 2m³+3m and 5m³ – 2m² + m – 9.
By arranging the like terms together, we get
=9-9-2m³+5m³+5m²-2m²+3m+m
=3m³+3m²+4m
Now subtract 2m – m³ + 2m²  from 3m³+3m²+4m.
=3m³+3m²+4m-(2m – m³ + 2m²)
=3m³+3m²+4m-2m+m³ -2m²
=4m³+m²+2m
Therefore, By subtracting 2m – m³ + 2m² from the sum of 9 + 5m² – 2m³+3m and 5m³ – 2m² + m -9 we get 4m³+m²+2m.

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V. What is the value of(7x² −2x+8)−(3x² +8x−7)?

Solution:

Given, 7x² −2x+8-(3x² +8x-7)
=7x² −2x+8-3x² -8x+7
=7x² -3x² -2x-8x+8+7
=4x²-10x+15
Hence, the value of (7x² −2x+8)−(3x² +8x−7) is 4x²-10x+15.

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VI.  How much is 2x+y-3z greater than 5x-2y+8z?

Solution:

2x+y-3z=5x-2y+8z+a
a=2x+y-3z-5x+2y-8z
a=-3x+3y-11z
Hence, 2x+y-3z greater than 5x-2y+8z by -3x+3y-11z.

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VII. Determine the length of the third side of the triangle whose perimeter is 3x²+15xy+3units and the length of the other two sides is 7xy-2x² and 3x²+6xy-3.

Solution:

Given,
Two sides of a triangle are 7xy-2x² and 3x²+6xy-3.
The perimeter of the triangle is 3x²+15xy+3.
To find the length of the third side of the triangle we have to add the other two sides and subtract them from the perimeter.
The sum of the other two sides of the triangle is,
=7xy-2x² + 3x²+6xy-3
=x²+13xy-3
Subtract x²+13xy-3 from the perimeter 3x²+15xy+3, we get
=3x²+15xy+3-(x²+13xy-3)
=3x²+15xy+3-x²-13xy+3
=2x²+2xy+6
Therefore, the length of the third side of the triangle is 2x²+2xy+6.

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VIII. A ribbon of length (7xy-3y+2) units is cut into two parts. If the length of the bigger part is (2xy+y) units, find the length of the smaller part of the ribbon?

Solution:

Given,
Total length of ribbon = (7xy-3y+2) units
The length of the bigger part of the ribbon = (2xy+y) units
To find the length of the smaller part of the ribbon, we need to subtract the length of the larger part from the total length of the ribbon.
=7xy-3y+2-(2xy+y)
=7xy-3y+2-2xy-y
=5xy-4y+2
Therefore, the length of the smaller part of the ribbon is 5xy-4y+2.

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IX. From the sum of 3x + 2y – z and 4x – 3y + 2z, subtract the sum of 4x – y + z and x – y – 4z?

Solution:

The sum of 3x + 2y – z and 4x – 3y + 2z is
=3x + 2y – z +4x-3y+2z
=3x+4x+2y-3y-z+2z
=7x-y+z
The sum of 4x – y + z and x – y – 4z is
=4x – y + z + x – y – 4z
=4x+x-y-y+z-4z
=5x-2y-3z
Subtract 5x-2y-3z from 7x-y+z.
=7x-y+z-(5x-2y-3z)
=7x-y+z-5x+2y+3z
=7x-5x-y+2y+z+3z
=2x+y+4z

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X. The two sides of a rectangle are x³ – 2y³ + z³ and 3x³ + 5y³ + 5z³. Find its perimeter.

Solution:

Given,
The two sides of a rectangle are x³ – 2y³ + z³ and 3x³ + 5y³ + 5z³
we know that the perimeter of the rectangle is 2(l+w).
The sum of the two sides of the rectangle is
=x³ – 2y³ + z³+3x³ + 5y³ + 5z³
=x³+3x³-2y³+ 5y³+z³+5z³
=4x³+3y³+6z³
The perimeter of the rectangle is
=2(4x³+3y³+6z³)
=8x³+6y³+12z³
Hence, the perimeter of the rectangle is 8x³+6y³+12z³.

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If you wanna calculate the time taken by an object to travel a certain distance at a given speed then this article is going to be of great help to you. By the end of this page, you will be familiar with many details such as What exactly is time, its units, what is the formula to find time, steps to follow to find the time when distance and speed are given.

Usually, Time and Speed have an Inverse Relationship i.e. if one increases the other decreases. Know how this is applicable by going through our problems on calculating time in the below modules. Try to answer the Time Speed and Distance Questions over here and know what kind of questions you might come across in your exams.

Do refer to similar articles:

• To find the Distance when Speed and Time are given
• To find Speed when Distance and Time are given

What is the Formula to Find Time?

The mathematical equation stating the Relationship between Speed, Distance, and Time is given by the expression Speed = Distance/Time
Rearranging the above basic formula we can get the formula for Time i.e. Time = Distance/Speed

Units of Time

Time uses three units in most common and they are namely Hours, Minutes, Seconds. To know in detail other units you can check out our other articles on Units of Time.

Problems on Calculating Time

Example 1.
Two trains 120 m,100 in length are running in the same direction. The first runs at the rate of 60 km/s and the second rate at the rate of 45 m/sec. How long will they take to cross each other?
Solution:
Given,
Length of first train=120 m
Length of second train=100 m
Speed of first train=60 m/s
Speed of the second train=45 m/sec
Relative speed=60-45=15 m/sec
Since trains are running in the same direction
As per the formula, L1 +L2/x-y
=120+100/15=220/15
=14.66 seconds
Hence, two trains will take 14.66 seconds to cross each other.

Example 2.
Two trains, 90 m, 100 m in length running in the opposite direction. The first runs at the rate of 20 m/sec and the second runs at the rate of 30 m/sec. How long will they take to cross each other?
Solution:
Given that, Length of the first train=90 m
Length of the second train=100 m
Speed of the first train=20 m/sec
speed of the second train=30 m/sec
Relative speed=30+20=50 m/sec
Since trains are running in the opposite directions,
As per the formula, L1+L2/x+y
=90+100/50
=190/50=3.8 seconds
Hence, two trains would take 3.8 seconds to cross each other.

Example 3.
Sandeep walking his 2/5 of his usual speed reaches the temple 15 min late. Find his usual time to reach the temple?
Solution:
Given that,
new speed a/b is=2/5
change in time=15 min
Usual time to reach the temple=?
As we know the formula, usual time=change in time/(b/a-1)
=15/(5/2-1 )
=15 × 2/3
=30/3=10 min
Hence, Sandeep’s usual time to reach the temple is 10 min.

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Example 4.
A train 300 m long has a speed of 20 m/sec. How long will it take to pass a platform of length 60 m?
Solution:
Given,
Length of the train=300 m
Speed=20 m/sec
The distance here will be the same as the length of the train and the length of the platform.
=300+60=360 m
Therefore, time=360/10=36 seconds.

Example 5.
Amrutha and Sudha have to travel from Hyderabad to Vijayawada in their respective cars. Amrutha is driving at 50 Kmph while Sudha is driving at 100 km per hour. Find the time taken by Sudha to reach Vijayawada if Amrutha takes 10 hours?
Solution:
Given,
Amrutha driving speed=50 km per hour
Sudha driving speed=100 km per hour
Here distance is the same. Time taken is inversely proportional to speed.
Speed of Amrutha and Sudha is in the ratio 50:100 or1:2
So that ratio of time taken by Amrutha to that time taken by Sudha will be in the ratio 2:1.
So if Amrutha takes 10 hours, Sudha takes 5 hours.
Hence, Time taken by Sudha to reach Vijayawada is 5 hours.

Example 6.
A train 200 meters long has a speed of 20 m/sec. How long will it take to pass an electric pole?
Solution:
Given,
Length of the train=200 meters
Speed=20 m/sec
We know that time=distance/speed
The distance here will be the same as the length of the train. i.e. 200 meters.
Therefore, Time=200/20=10 seconds.
Hence, it will take 10 seconds to pass an electric pole.

Example 7.
Anil has to attend the dancing class which is 800 meters away. He is walking at the speed of 1.5 ms. How long will it take to get to class?
Solution:
Given,
Anil has to walk a distance=800 meters
walking speed=1.5 ms
time=?
We know that time=distance/speed
=800/1.5=533.33
Hence, Anil will take 533.33 sec to attend the dance class.

Example 8.
Avinash takes 5 hrs 30 min in going by cycle and comes back by scooter. He would have lost 2 hrs 10 min by going on cycle both ways. How long would it take him to go by scooter both ways?
Solution:
Given,
Avinash takes time in going by cycle and comes back by scooter=5 hrs 30 min
Avinash will lose time by going on cycle both ways=2 hrs 10 min
Time taken by Avinash to go by scooter both ways= 5 hr 30 min-2 hrs 10 min
=3 hr 20 min=3 1/3 hrs.
Hence, Avinash will take 3 2/3 hrs to go by scooter both ways.

Do you have any difficulties in solving the problems related to distance, average distance, the distance between two points, etc? Then, look no further and make use of our article on how to find distance when speed and time are given. We have covered everything such as distance definition, units, formula, procedure on how to find distance along with examples. Refer to the problems on calculating distance here and get an idea on how to approach to solve similar kinds of distance problems in your homework or assignments easily.

Check out More Articles:

• To find Speed when Distance and Time are given
• Worksheet on Expressing Speed in Different Units
• Worksheet on Speed, Distance and Time

Distance – Definition

Distance is defined as the complete path traveled by the object. You can better understand this statement with an example. Let suppose a car travels in the direction of north 8th km and then moves towards east 5 km. here the total distance traveled by car is 13 km as it is a complete path.

Distance Formula

The formula used to determine the distance when speed and time are known is distance = speed*time.  Units for distance are m, km, miles, etc.

How to Calculate Distance?

Follow the simple steps listed below to find distance easily and they are along the lines

• Initially, find what’s known from the given information such as speed, time.
• Check whether both the metrics are given are in the same unit of measurement. If not convert accordingly to the same units.
• Later, substitute the known values speed, time in the formula for distance = speed*time
• Simplify further and find the total distance traveled by the object.

Practice Math Online with Unlimited Questions provided in 5th Grade Math Activity Sheets and become a blossoming mathematician in no time.

Problems on Calculating Distance

Example 1.
Cyclists travel at a speed of 15 km/hr. At what distance would they travel in 80 minutes?
Solution:
Cyclists travel with the speed=15 km/hour
time=80 minutes=80/60 hours
We have to find the distance
distance=speed × time
=15 × 80/60
=20
Therefore, the cyclists would travel a distance of 20 km.

Example 2.
A boy walking at a speed of 5 km/hr reaches his school 10 min late. The next day at a speed of 10 km/hr reaches his school 5 minutes late. Find the distance of his school from his house?
Solution:
Given,
A boy walking at a speed of 5 km/hr reaches his school late by=10 min
The next day boy walks at a speed of 10 km/hr reaches his school late by=5 minutes
Difference between time=10-5=5 min=5/60=1/12 hr
The distance of his school from his house=10 × 5/10-5 × 1/12
=50/5 × 1/12
=10/12
=5/6 =0.833 km

Example 3.
Radha and Sudha are standing at two ends of a room with a width of 60 m. They start walking towards each other along the width of the room with a Speed of 4 m/s and 1 m/s, respectively. Find the total distance traveled by Radha when he meets Sudha for the third time?
Solution:
When Radha meets Sudha for the third time, they together would have covered the distance of 5d, i.e. 5 × 60 m=300 m
The ratio of speeds of Radha and Sudha =4:1
The total distance traveled by them will also be in the ratio 4:1  as the time taken is constant.
So the distance traveled by Radha will be 4/5 × 300=240 m

Example 4.
Mani drives his car at a speed of 80 km per hour. How much distance will he cover in 3 hours 30 minutes?
Solution:
Mani drives his car at a speed=80 km
Time taken=3 hours 30 minutes
=3 1/2 hours
Distance covered in 1 hour=80 km
Distance covered in 3 hours 30 minutes=80 × 3 1/2 km
=80 × 7/2 km
=280 km
Hence, Mani covers a distance of 280 km.

Example 5.
A man travels from his home to the office at 3 km/hr and reaches his office 30 min late. If the Speed had been 9 km/hr he would have reached 20 min early. Find the distance from his home to office?
Solution:
Let the distance between home and office=d
Suppose he reaches the office on time, Time taken=x minutes
Case 1:
When he reaches 30 minutes late, Time taken=x+30
Case 2:
When he reaches the office 20 minutes early=x-20
As the distance traveled is the same, the ratio of the speed in case 1 to case 2 will be inverse of the time taken in both cases Ratio of Speed in both cases = 3:9 = 1:3
The ratio of Time in both cases = 3:1
Therefore (x+30)/(x-20)=3/1
3x+90 = x -20
2x= 110
x=55 minutes
Taking case 1,
We know that speed=distance/time
4= d/(85/60)=> d= 240/85 = 2.82 km

Example 6.
A bus travels at a speed of 65 km/hour. How far will it travel in 48 minutes?
Solution:
Given,
Speed = 65 km/hour
Time = 48 minutes
= 48/60 Hour (Since we know, 1 hour = 60 minutes)
= 4/5 hour
Distance = speed × time
= 65 × (4/5) km
= (65 × 4)/5 km
= 52 km.

Example 7.
A person travels from one place to another at 50 km/hr and returns at 150 km/hr. If the total time taken is 5 hours, then find the Distance?
Solution:
Here the Distance is constant, so the Time taken will be inversely proportional to the Speed.
The ratio of Speed is given as 50:150, i.e. 1:3
So the ratio of Time taken will be 3:1.
Total Time taken = 5 hours; Time taken while going is 3 hours and returning is 1 hour.
We know that distance=speed × time
Hence, Distance = 50 x 3 = 150 km

Example 8.
A scooter travels at a speed of 65 km per hour. What is the distance covered by the scooter in 4 minutes?
Solution:
Given,
speed=65 km per hour
=65 x 1000/60
=6500/6= 1083.33
Distance covered in 4 minutes=4 x 1083=4333.2 metre.

 

The Highest Common Factor (H.C.F) of two numbers is the highest number among all the common factors of the given numbers. The HCF is also known as Greater Common Factor(GCF). Division Method, Prime Factorization Method, and Factorization methods are used to find the Highest Common Factor of the given numbers.

For better understanding, we’ve covered various questions with a quick explanation here. In this article, we have covered different methods questions on HCF to enhance their math skills as well as provide you better practice. You just need to solve the HCF Problems available on this page and test your conceptual knowledge.

Also, Read Some More Article:

• Word Problems on HCF and LCM
• Worksheet on Methods of HCF and LCM
• Worksheet on HCF

HCF Example Problems

Example 1:
Find the Highest Common Factor of 45 and 50.

Solution:
As given in the question the values are 45 and 50.
The factors of 45 are 1, 3, 5, 9.
The factors of fifty are 1, 2, 5, 10
So, the Common Factors of 45 and 50 are 1, 5.
Therefore, the Common Highest Factor among them is 5.

Example 2:
What is the HCF of 18 and 36 using the Prime Factorization Method?

Solution: 
Given the values 18 and 36.
Using Prime Factorization Method, we can find the HCF value.
The Prime Factorization of 18 is 2 x 9
The Prime Factorization of 36 is 2 x 2 x 9
So, the common factors of 18 and 36 are 2 x 9
Thus, the HCF of the two given numbers is 2 x 9 is 18.

Example 3: 
What is the Highest Common Factor of 28 and 44 using the Factorization method?

Solution:
As given the values are 28 and 44.
Using the Factorization method, to find the HCF value.
The factor of 28 are 1, 2, 4, 14
The factors of 44 are 1, 2, 4, 11, and 44.
So, the common factors of given numbers are 1, 2, 4.
Thus, the Highest Common Factor of 28 and 44 is 4.

Example 4:
Find the possible numbers. The HCF of the two numbers is 29 and their sum is 174.

Solution:
Given the HCF of the two numbers is 29 and their sum is 174.
Now, we will find the possible numbers.
Let the numbers be
29(a+b) = 174 (or) 27a + 27b = 174
So, a +b = 174/ 29 = 6.
The values of co-primes with sum 6 is 1,5
Therefore, the possible pairs of numbers are 29 x 1, and 29 x 5. It will become the 29 and 145.
Thus, the possible numbers are 29 and 145.

Practice Math Online with Unlimited Questions provided in 5th Grade Math Activity Sheets and become a blossoming mathematician in no time.

Example 5:  
Find the HCF of 65, 45 using the division method and leaving 5 as the remainder?

Solution:
Given,
The number divides 65 and leaves 5 as the remainder.
So, the number divides 65 – 5 = 60 exactly.
The number divides 45 and leaves 5 as the remainder.
So, the number divides 45 – 5 = 40 exactly.
Now, we have to find the HCF(60, 40)

Therefore, the HCF of a given number 65, 45 is 20.

Example 6: 
Find the HCF OF 9, 84 using the Division Method?

Solution:
Given the value is 9, 84
Using the division Method, we will find the value.
First, divide 9 and 84. So, 9 will be the divisor and 84 will be dividend
Now the remainder will become the new divisor and 9 will become the new dividend.
Proceed with this process till the remainder is zero and the last divisor will be the HCF of the given number.

Thus, the HCF of 9, 84 using the Division Method is 3.

Example 7:
Find the Highest Common Factor 6, 72, 26 using the Prime Factorization Method.

Solution: 
As given in the question, the values are 6, 72, 26.
Now, we will find the HCF value using the Prime Factorization Method.

The factors of 6 are 1, 2, and 3.
Next, 72 factors are

So, the factors of 72 are 1, 3, and 4.

The factors 27 are 1, 3
So, the common factors of 6, 72, 27 are 1, 3.
Thus, the Highest Common Factor among them is 3.

Example 8: 
Find the Highest Common Factors of the following using Factorization,
(i) 81, 99
(ii) 47, 63

Solution:
(i) Given the values 81, 99
Using factorization, we will find the values.
The factors of 81 are 1, 3, 9
The factors of 99 are 1, 3, 9, 11.
The common factors are 1, 3, 9.
Therefore, the Highest Common Factor of the given numbers is 9.

(ii) Given the values are 47, 63
Using factorization, we will find the values.
The factors of 47 are 1, 7.
The factors of 63 are 1, 3, 7, 9, and 63.
The common factors are 1, 7.
Therefore, the Highest Common Factor of the given numbers is 7.

Example 9: 
Find the H.C.F. of three numbers 44, 48, 52 by the prime factorization method.
Solution:
Given the values are 44, 48, 52.
Using the prime factorization method, we have to find the factors.
First, we have to write down the factors of the given numbers.
The factors of 44 are 1, 2, 4, 11.
The factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24.
The factors of 52 are 1, 2, 4, 5,12, 13, and 52.
Now, we have to write the common factors of all three numbers.
Common factors of 44, 48, 52 is 1, 2, 4.
Thus, the HCF of three numbers of given numbers is 4.

Example 10:
John has 16 red pens and Ryan has 24 green pens. They want to arrange the pens in such a way that each row contains an equal number of pens and also each row should have only red pens or green pens. What is the greatest number of pens that can be arranged in each row?

Solution:
As given in the question,
Now, find the highest number of pens which will be arranged in equal rows, we have to seek out the HCF of two numbers.
The factors of 16 are 1, 2, 4, 8, and 16.
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.
The common factors are 1, 2, 4, 8. The highest common factor among them is 8.
Therefore, the greatest number of pens that can be arranged in each row is 8.

FAQ’s on HCF Example Questions

1. What are the methods used for finding the HCF?

There are three methods to find H.C.F of two or more numbers
1. Division Method
2. Factorization method
3. Prime Factorization method

2.What’s Highest Common Factor?

It is the Highest of the common factor of two or more given numbers. It is also named Greatest Common Divisor(GCD).

3. How do we use Highest Common Factor?

The HCF is used to simplify the fractions. The Highest Common Factor (HCF) or GCD is that the greatest number that divides exactly into two numbers.