To find Speed when Distance and Time are given

To find Speed when Distance and Time are given – Definition, Tricks, Examples | How to Calculate Speed?

Time Speed and Distance is a popular concept asked in most competitive exams. The concept remains the same and the level of difficulty increases with passing years. Students who need any assistance on the concept of Time Sped and Distance will find it completely handy. We will explain all about the Speed Definition, Speed Time Distance Formula, Tricks and Tips to keep in mind while solving the Problems on Calculating Speed, Procedure on How to Calculate Speed when Time and Distance are given, etc.

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What is Speed?

Speed tells us how fast an object is moving or traveling. One can find the average speed easily by knowing the distance traveled in a particular amount of time. Units of Speed are given by m/sec, km/hr It is derived from the units of distance(m, km) and time(sec, hr).

Distance Speed Time Formula

The mathematical formula for time speed and distance is given by Speed = Distance/Time
You can rearrange the above formula to find the distance, time whenever you need them. To find the unknown quantity you need to know about the other two values.

How to find Speed when Time, Distance are given?

Go through the simple steps listed below to calculate speed when both time and distance are given. They are in the below way

  • Firstly, find out the known parameters in the question.
  • Later, check for the value to be determined among speeds such as average speed, relative speed, etc.
  • Then, use the relevant formula of finding the speed by substituting the known input values.
  • Simplify the equation further and find the speed value in no time.

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Speed, Distance Time Questions

Example 1.
The distance between the two stations is 600 km. A train takes 4 hours to cover the distance. Find the speed of the train in m/sec?

Solution:

Given,
The distance between the two stations is= 600 km
Time taken by train=4 hours
Speed of the train=distance/time=600/4=150 km/hr
We have to convert km/hr to m/sec by multiplying 5/18 with 150.
=150 × 5/18=41.66 m/sec.
Hence, the speed of the train is 41.66 m/sec.

Example 2.
A car travels at a distance of 220 km in 4 hours. Find the speed of the car?

Solution:

Given,
Car travel at a distance=220 km
Time taken to travel the distance=4 hours
Speed of the car=distance/time=220 km/4=55 km/hr
Hence, the speed of the car=55 km/hr

Example 3.
The distance traveled by train is 580 km in 5 hours 30 minutes. Find its speed?

Solution:

The distance traveled by train is= 580 km
Time taken to travel the distance =5 hours 30 minutes=5.5 hours
speed of the train=580 km/5.5 hr
=105.45 km/hr
Hence, the speed of the train is 105.45 km/hr.

Example 4.
Ranjith drives the first 132 km in 2 hours and the next 200 km in 4 hours. What is the average speed for the entire journey?

Solution:

Total time taken=2+4=6
Total distance=132+200=332 km
Average speed=Total distance/total time
=332/6=55.33 km/hr
Hence, the average speed is 55.33 km/hr.

Example 5.
Surendra rides a bike from the office to yoga class at a speed of 40 km per hour and on a return journey at a speed of 50 km per hour. What is Surendra’s average speed of travel?

Solution:

Let the distance between office and yoga class be x.
Time is taken for riding a bike from office to yoga class=x/40
Time taken for the return journey=x/50
Time taken for both the journeys=x/40+x/50
=5X+4X/200
=9x/200
=9x/200 hours
Total distance traveled both ways=2x
Average speed=2x/9x/200 hours
=2x .200/9x
=44.44 km per hour
Hence, the average speed of travel is 44.44 km per hour.

Example 6.
Ajay travels 20 km in 80 minutes. Find the speed?

Solution:

Ajay travel the distance=20 km
time=80 minutes=80/60 hours
We have to find the speed
speed=distance/time
s=20/80/60
s=20 ×60/80=15
Therefore, the cyclist would travel at a speed of 15 km.

Example 7.
A man covers a certain distance on a scooter. Had he moved 4 km/hr faster he would have taken 40 minutes less. If he had moved 3 km/hr slower, he would have taken 30 minutes more. Find the original speed?

Solution:

Speed s1=4 km/hr
Speed s2=3 km/hr
s=2 ×(s1×s2)/s1-s2
=2 × (4×3)/4-3
=2 ×(12)/1
=24 km/hr
Hence, the original speed is 24 km/hr.

Example 8.
The car drives at the speed of 65 km/hr locate a bus 50 meters ahead of him. After 20 seconds, the bus is 50 meters behind. Find the speed of the bus?

Solution:

Let the speed of the bus=X km/hr
Now, in 20 seconds, the car covers the relative distance=(60+40)=100 m
The relative speed of the car= 100/20=5 m/sec
=5 × 18/5
=18 km/hr
Therefore 65-x=18
x= 65-18
=47 km/hr
Hence, the Speed of the bus=47 km/hr.

Example 9.
Sudheer can row a boat in still water at 20 km per hour. He decides to go boating in a river. To row upstream, he takes 4 hours, and to row downstream, he takes 1 ½ hours. Find the Speed of the river?

Solution:

Suppose the Speed of the river is ‘y’ km per hour.
While rowing upstream he takes 4 hrs and while rowing downstream he takes 1 ½ hour.
As the Distance covered is constant the ratio of the net Speeds of the boat while going upstream and downstream will be the inverse of the ratio of the Time taken.
Ratio of Time taken (downstream: upstream) = 1.5/4 =3/8
So the ratio of Speed of boat (downstream: upstream) = 8/3
Speed downstream: 10 + y Speed upstream : 10 – y
(10+y)/((10-y) ) = 8/3
30+3y=80-8y. Thus, 11y=50 & Y=50/11
Hence, theSpeed of the river = 50/11 kmph

Example 10.
Using the speed formula calculate the speed of a person in kilometers per hour if the distance he travels is 30 kilometers in 3 hours?

Solution:

The formula for speed is [Speed = Distance ÷ Time]
Distance = 30 kilometers
Time = 3 hours
Speed = (30 ÷ 3) km/hr
= 10 kilometers per hour
Hence, the speed of a person is 10 kilometers per hour.

Example 11.
A car takes 3 hours to cover a distance if it travels at a speed of 30 mph. What should be its speed to cover the same distance in 1.5 hours?

Solution:

Given,
time=3 hours
Speed=30 mph
Distance covered = 3*30 = 90 miles
The speed required to cover the same distance in 1.5 hours = 90/1.5 = 60 mph

Example 12.
A train leaves from a station and moves at a certain speed. After 1 hour, another train leaves from the same station and moves in the same direction at a speed of 90 mph. If it catches up with the first train in 2 hours, what is the speed of the first train?

Solution:

Let the speed of the first train be ‘s’.
Distance covered by the first train in (1+2) hours = Distance covered by the second train in 2 hours
Therefore, 3s = 90*2
3s=180
s=180/3=60
Hence, the speed of the first train is 60 mph.

Example 13.
A boat covers a certain distance in 3 hours, while it comes back in 4 hours. If the speed of the stream is 5 km per hour, what is the boat’s speed in still water?

Solution:

Let the boat’s speed in still water be x mph.
Upstream speed = x- 5 mph
Downstream speed = x+5 mph
Since distance covered is the same in both ways
4 (x-5) = 3 (x+5)
4x – 20 = 3x+15
x= 35 mph
Hence, the boat speed in still water is 35 mph.

Example 14.
Sanjeev takes 1.25 hours to drive from home to school. The distance is 14 km from home to school. What is Sanjeev’s average speed on her drive to school?

Solution:

Given,
Sanjeev takes time to drive from home to school=1.25 hours
The distance from home to school is=14 km
Speed=?
We know that speed=distance/time
speed=14/1.25=11.2 km/hr
Hence, the speed is 11.2 km/hr.

 

 

 

 

Worksheet on Word Problems on H.C.F. and L.C.M

Worksheet on Word Problems on H.C.F. and L.C.M | HCF and LCM Word Problems with Solutions PDF

Word Problems Involving Highest Common Factor and Least Common Multiple has several questions on finding LCM of Two or More Numbers, HCF of Two or More Numbers, Relationship between H.C.F. and L.C.M, and many more. Before attempting the actual exam students are advised to solve these kinds of Highest Common Factor and Least Common Multiple Worksheets to be familiar with various models being asked in exams. Try to solve the HCF and LCM Questions on your own and then verify with our solutions explained clearly.

Download the Worksheet on Word Problems on HCF and LCM without paying a single penny and use them as a quick reference whenever you need. Get an idea of how to solve the Word Problems on H.C.F and L.C.M by looking at the solved examples available here.

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Example 1.
The length, breadth, height of a room are 4 m 50 cm, 6 m 30 cm, and 8 m 20 cm respectively. Find the longest tape which can measure the dimensions of the room exactly?

Solution:

First, we have to convert the dimensions into cm.
We know that 1 m=100 cm.
Given,
The length of the room=4 m 50 cm=4 (100) cm +50 cm=450 cm
The breadth of the room=6 m 30 cm=6(100)cm+30 cm=630 cm
The height of the room=8 m 20 cm=8(100)cm+20 cm=820 cm
HCF(450,630,820)=longest tape that can measure the dimensions of the room.
HCF(450,630,820)=10 cm

Example 2.
Find the least length of a rope which can be cut into the whole number of pieces of lengths 34 cm, 52 cm, and 78 cm?

Solution:

Given,
length of the pieces=34 cm,52 cm,78 cm
LCM(34,52,78)
34=2 × 17
52=2 × 2 × 13
78=2 × 3 × 13
=22 × 17 × 13 × 3
=2652 cm
Therefore, the length of the rope is 2652 cm.

Example 3.
The product of the two numbers is 5600. If their HCF is 10. Find their LCM?

Solution:
Given,

The product of the two numbers is= 5600
HCF=10
We know that product of two numbers=LCM × HCF
5600=LCM × 10
LCM=5600/10
=560
Therefore, LCM is 560.

Example 4.
Find the multiple of 80 which lies between 400 to 600 which has a tens digit is double than one’s digit?

Solution:

Multiples of 80 are 80,160,240,320,400,480,560,640 etc.
Given,
The number is between 400 to 600 and the tens digit is double than the hundreds digit.
In 480, the tens digit is double than the hundreds digit.
Hence, 480 is the number.

Example 5.
If the HCF and LCM of two numbers are 10, 840. If one of their numbers is 70. Find the other number?

Solution:

Given,
HCF=10
LCM of two numbers=840
One of their numbers=70
We know that product of two numbers=LCM × HCF
70. x=840 × 10
x=8400/70=120
Therefore, the other number is 120.

Example 6.
Find the largest number that divides 184, 244 leaving 4 as the remainder?

Solution:

Given,
The number divides 184 and leaves 4 as the remainder
Therefore, the number divides 184 – 4 = 180 exactly
The number also divides 244 leaving 4 as the remainder
Therefore, the number divides 244 – 4 = 240 exactly
Now we have to find the highest common factor H.C.F of 180 and 240
hcf example 4
Hence, the largest number is 60.

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

Example 7.
90 boys, 70 girls, 10 teachers are equally arranged in buses such that no one is left. Find the highest number of buses required for them?

Solution:

Given,
No. of boys going for the trip=90
No. of girls going for the trip=70
No. of teachers going for the trip=10
The highest number of buses required=HCF(90,70,10)
Factors of 90 are 1,2,3,5,6,9,10,15,18,30,45,90
Factors of 70 are 1,2,5,7,10,14,35,70
Factors of 10 are 1,2,5,10
GCF=10
Therefore, 10 buses required for the trip.

Example 8.
The chairs in the auditorium can be arranged in rows of 43,65. Find the least number of chairs in the auditorium?

Solution:

Given,
The chairs can be arranged in rows=43, 65
The least number of chairs in the auditorium=LCM(43,65)
43=43
65=5 × 13
LCM(43,65)= 43 × 5 × 13
=2,795

Example 9.
Janu wants to plant 32 red rose plants and 48 white rose plants. What is the greatest number of rows possible if each row has the same number of red rose plants and white rose plants?

Solution:

Given,
Janu wants to plant red rose plants=32
Janu want to plant white rose plants=48
We have to find the greatest number of rows and each row should have the same number of red rose plants and white rose plants.
=HCF(32,48)
Factors of 32 are 1,2,4,8,16,32.
Factors of 48 are 1,2,3,4,6,8,12,16,24,48
HCF(32,48)=16
Hence, there will be 16 rows.

Example 10.
Find the multiple of 90  which lies between 200 and 500 where the digits at tens place and hundreds place are consecutive numbers?

Solution:

Multiples of 90 are 90,180,270,360,450,540 etc.
Given,
The multiple lies between 200 and 500 i.e. 270,360,450.
Also given the digits at tens place and hundreds place are consecutive numbers.
Hence, 450 is the number that lies between 200 and 500, and the digits at tens place and hundreds place are consecutive numbers.

 

 

Highest Common Factor

Highest Common Factor (H. C. F) – Definition, Formula, Methods, Examples | How to find the Highest Common Factor?

The Students who believe learning the Highest Common Factors concept is difficult and confusing, will no longer feel HCF. The Highest Common Factor for two or more natural numbers is known as HCF. For better understanding, we have provided various questions with a brief explanation. Practice the problems given on Highest Common Factor (HCF) on this page and enhance your math skills.

Know how questions are framed on the topic of HCF by referring to the Word Problems on HCF and get an idea of how to find HCF quickly using various methods. We will show different methods to solve the problems on Highest Common Factor (HCF).

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Highest Common Factor (H.C.F) – Definition

The largest common factor of all the given numbers is called as Highest Common Factor (H. C. F). The number can be divided into exactly two or more numbers without any remainder. The HCF is also known as the Greater Common Factor (GCF). The easiest way to find the HCF of two given numbers is to create a Tree factor. Let us discuss with an example about HCF.

Example: Find the HCF of 2 and 4
Solution:
Factors of 2 are 2 x 1
Factors of 4 are 2 x 2
Thus the Highest Common Factors of 2 and 4 are 2.

H.C.F Formula

H.C.F is simply the Highest Common Factor. Let a and b be two integers. The formula to find the HCF of a and b is given as,
HCF = Product of two numbers / L.C.M of two numbers
HCF = (a x b) / L.C.M (a,b)
Where L.C.M is the Least Common Multiple.

How to Find HCF (Highest Common Factor)?

There are three methods to find the Highest Common Factor of two or more numbers. The three methods to find the HCF of the integers is as follows,
1. Prime Factorization Method (Factor Tree Method)
2. Division Method
3. Factorization Method

These three different methods are to find the H.C.F of given numbers are explained step by step here.
1. HCF by Prime Factorization (Factor Tree Method):
In calculating HCF by Prime Factorization, we factorize the numbers into Prime factors, which is known as Prime Factors. Follow the below steps to find the HCF of numbers using the Prime Factorization Method,
Step 1: Initially, check whether the given number is divisible by 2.
Step 2: Now, divide until you cannot divide the number any further.
Step 3: Finally, write the numbers as the product of the prime numbers. The product of those common factors is that the Highest Factor of the given numbers.
Example: What is the HCF of 6 and 8?
Solution: The values are 6 and 8.
Factors of 6 are,

2 x 3 = 6
Factors of 8 are,

2 x 2 x 2 = 8
The Common Factor is 2.
Thus, the HCF of 6 and 8 is 2.

2. HCF by Division Method:
You have understood by now the strategy of finding the common highest factor using Prime Factorization. Now, learn here to find HCF using Division Method. The division method is nothing but divides the given number, simultaneously, to get the common factors between them. Follow the steps mentioned below of division method,
Step 1: In the division method, first, treat the smaller number as the divisor and the bigger number as the dividend.
Step 2: Divide the given number until you get the remainder as 0.
Step 3: We are going to get the common prime factors because the factors within the left-hand side divide all the numbers exactly. The product of those common prime factors is that the HCF of the given numbers.
Example: Find the HCF of the 10 and 15?
Solution: Given the values 10 and 15,
Using the division method,

Thus, the HCF of given numbers 10 and 15 is 5.

3. HCF by Factorization Method:
In HCF by finding the Factorization Method, we will find the greatest common factor by listing down the factors of the numbers. Follow the below steps for finding the HCF using the Factorization method,
Step 1: First, Write the factors of the given numbers.
Step 2: Next, find the highest factor among the listed factors which are common for both the given numbers. 
Example: What’s the HCF of 12 and 24?
Solution: Given the values are 12 and 24.
The Factors of 12 are 1, 2, 3, 4, 6
The Factors of 24 are 1, 2, 3, 4, 6
So, the common factors of given numbers are 1, 2, 3, 6.
The Highest factor among them is 6.
Thus, the HCF of given numbers 12 and 24 is 6.

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Properties of HCF

The Properties of HCF are listed in the below lines for quick reference. They are as under
Property 1: The HCF of any two or more numbers isn’t greater than any of the given numbers.
Property 2: If one number is the factor of the other number, their HCF will be the smallest number.
Property 3: The HCF of the numbers is the product of the co-prime factors.
Property 4: HCF of two or more Prime numbers is always 1.

Highest Common Factor Examples

Example 1:
Find the H.C.F of three numbers 27, 21, 24 by Prime Factorization Method?
Solution:
Given the values are 27, 21, 32
Now, first, we have to write the factors of the given three numbers.
The factors of 27 are 1, 3, 9, and 27.
The factors of 21 are 1, 3, 7.
Factors of 24 are 1, 2, 3, 4, 6, 8,12.
Now, we have to write the common factors of all the three given numbers.
Common Factors of 24, 36, 60 = 1, 3
So, the HCF of three numbers 24, 36, 60 is 3.

Example 2: 
Find the HCF of three numbers are 25, 30, 35?

Solution:
As given the question, the values are 25, 30, and 35.
We can find the HCF of 25, 30, and 35 using the Factorization method.
The factors of 25 are 5 x 5.
The factors of 30 are 5 x 6.
The factors of 35 are 5 x 7.
The common factors of 25, 30, and 35 are 5, 6, 7.
Therefore, the highest common factor of given numbers is 5.

Example 3: 
What is the HCF of 48 and 56 using the Prime Factorization Method?

Solution:
Given the values 48 and 56.
Now, we find the HCF using the Prime Factorization Method.

So, the factors of 48 is 2 x 2 x 2 x 2 x 3

Now, the Factors of 56 = 2 x 2 x 2 x 7
The Highest Common Factors (H.C.F) of given numbers are 2 x 2 x 2.
Hence, the HCF of 48 and 56 is 2 x 2 x 2 = 8.

Example 4:
What is the HCF 9, 75 using the Division Method?

Solution:
Given the values are 9 and 75.
Now, using the division method find the HCF value.
First, divide 9 and 75. If the remainder will become the new divisor then 9 will become the new dividend.
Next, proceed with this process till the remainder is zero and therefore the last divisor is going to be the HCF of the given number.

Thus, the HCF of 9 and 75 is 3.

Example 5: 
Write the Highest Common Factor (H.C.F) of 16 and 20?

Solution:
Given the values 16 and 20.
Now, find the factors
The Factors of 16 are 1, 2, 4, 8.
The factors of 20 are 1, 2, 4, 5.
Therefore the common factors of 16 and 20 are 1, 2, 4
The Highest Common Factor (H.C.F) among them is 4.
Therefore, 4 is the H.C.F or G.C.D of given numbers. 

FAQ’s on Highest Common Factor (H.C.F)

1. How do find HCF? 

For finding the HCF of two or more numbers, we have three methods. The methods are,
1. Division Method
2. Factorization Method
3. Prime Factorization Method

2. Write the relation between HCF and LCM?

LCM and HCF are related by the formula:
LCM (a, b) = a × b/HCF(a,b)
Where a and b are two different numbers.

3. What’s Highest Common Factor?

It is the Highest of the common factor of two or more numbers.

4. How do we use Highest Common Factor?

The HCF is used to simplify the fractions. HCF or GCD is that the greatest number that divides exactly into two numbers.

5. What are the tricks and formulas to find HCF?

The formulas and tricks to find HCF is,

  • If the two numbers are said to be co-prime their HCF is 1.
  • Product of given two numbers = product of HCF and LCM.
  • HCF = HCF of Numerators/LCM of denominators.
Word Problems on H.C.F. and L.C.M

Word Problems on H.C.F. and L.C.M | HCF and LCM Problems with Solutions

Get an idea on how to Solve Word Problems on Highest Common Factor and Least Common Multiple. Try to answer the LCM and HCF Questions available here on your own and then verify with ours. Read the Question Carefully and understand what is asked so that you won’t have any difficulty in solving the LCM and HCF Word Problems. Learn the tips & tricks to solve Highest Common Factor and Least Common Multiple Problems easily and answer the questions you come across in your exams with utmost confidence.

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Word Problems Involving H.C.F and L.C.M

Example 1.
Find the multiple of 40 which is in between 400 and 500 where the digits at tens place and hundreds place are equal?
Solution:
Multiples of 40 are 40,80,120,160,200,240,280,320,360,400,440,480,520 etc.
Given the number is in between 400 and 500
Here the numbers between 400 and 500 are 400,440,480.
Also given the digits at tens place and hundreds place are equal.
Here 440 is the number which has the same digit at tens place and hundreds place are equal.

Example 2.
Vijay’s friend asked him to bring the same number of chocolates and biscuits. The store sells chocolates in packs of 50, biscuits in packs of 20. Find how many chocolates Vijay will buy from the store?
Solution:
The store sells chocolates in packs of 50, biscuits in packs of 20.
Vijay has to buy the same number of chocolates and biscuits.
Now we have to find the LCM of Chocolates and Biscuits.
i.e. LCM(50,20)
50=2 × 5× 5 =2 ×  52
20=2 × 2 ×  5=22 × 5
LCM(50,20)=22 × 5 2
=4 × 5=20

Example 3.
Find the least number of square tiles by which the floor of a room of dimensions 1223 cm, 634 cm can be covered completely?
Solution:
We require the least number of square tiles, so each tile must be of maximum dimension.
To get the maximum dimension of tile, we have to find the largest number that exactly divides 1873 cm,958 cm.
HCF(1223, 634)=1 cm
Hence, the side of a square tile is 1 cm.
Required no. of tiles= Area of the floor/Area of a square tile
=1223.634/1.1
=775382

Example 4.
Anjali goes to dancing classes every 5th day. Sandhya goes to the same dance class every 4 th day. How many days will they meet in the dance class in the month of July and august if we start counting from 1st July?
Solution:
Given,
Anjali goes to dancing classes every 5th day.
Sandhya goes to the same dance class every 4th day.
Then from starting the first day, the number of days when they meet=LCM(5,6)=30
So every 20th day they will meet.
They will meet only once on July 20.

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

Example 5.
A certain no. of fruits are stored in groups of 2,4,6,8 with no fruit left behind. Find the no. of fruits?
Solution:
Given,
Fruits are stored in groups of 2,4,6,8.
No. of fruits=LCM(2,4,6,8)
=24 fruits
Hence, there are 24 fruits.

Example 6.
Sameera can jump 5 steps at a time and Nilima can jump 6 steps at a time. On which of the steps will both meet if both start jumping together?
Solution:
Given,
Sameera can jump  at a time= 5 steps
Nilima can jump at a time= 6 steps
if both starts jumping together, they will meet=LCM(5,6)
=30

Example 7.
Satvika has music classes on alternate days, dance class once in 4 days, and yoga class once on 4 days. On the 1st of Dec, she had all the 3 classes. When will she have all the 3classes again?
Solution:
Given,
Satvika has music classes on alternate days i.e. every second day.
dance classes =4 days,
yoga classes=3 days
Satvika will have all the 3 classes =LCM(2,4,3)
=12
Hence, Satvika will have all three classes on 13th December.

Example 8.
50 boys and 30 girls went on an educational trip. The boys and girls have to be divided into separate smaller groups such that each group has the same number of boys and girls. What is the maximum number of students in each group such that no student is left out?
Solution:
hcf example
Given,
No. of boys went on a trip=50
No. of girls went on a trip=30
Maximum number of students in each group=h.c.f(50,30)=2 × 5=10
=10
They are 10 groups each group consists of 5 boys and 3 girls.
The maximum number of students in each group is(5 +3=8) 8.

Example 9.
Find the least number of plants in a group such that they are planted in the rows of 5,10,15?
Solution:
To find the least number of plants first we have to find the LCM of(5,10,15).
LCM(5,10,15)=30
So we need 30 plants to be planted in rows of 5,10,15.

Example 10.
Find the largest number that divides 92,42 leaving 2 as the remainder?
Solution:
Given,
The number divides 92 and leaves 2 as the remainder.
S0 the number divides 92-2=90 exactly
The number divides 42 and leaves 2 as the remainder.
So the number divides 42-2=40 exactly
Now we have to find the HCF(90,40)
HCF example 2
Therefore, the largest number is 10.

Example 11.
Find the largest number that divides 57, 27 leaving 7 as the remainder?
Solution:
Given,
The number divides 57 and leaves 7 as the remainder.
S0 the number divides 57-7=50 exactly
The number divides 27 and leaves 7 as the remainder.
So the number divides 27-7=20 exactly
Now we have to find the HCF(50,20)
hcf example 3
Therefore, the largest number is 10.

Worksheet on H.C.F. and L.C.M

Worksheet on H.C.F. and L.C.M | Free Printable HCF and LCM Worksheet with Solutions

Quick Lessons on HCF and LCM provided with step-by-step solutions help students to gain complete knowledge on the concepts. Numerous Practice Worksheets on HCF and LCM will assist you in learning different methods of finding the highest common factor, least common multiple, finding HCF when LCM is given, and vice versa.

Solving the LCM and HCF Questions available in LCM and HCF Worksheet will enrich your subject knowledge, improves math skills in a fun and interactive manner. You can download, print, and share this handy Worksheet on H.C.F and L.C.M for free and help them learn how to find LCM and HCF using various methods like cake/ladder method, division method, listing multiples, factorization, prime factorization, etc.

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LCM and HCF Questions with Answers

Example 1.
Find the highest common factors of the following by factorization?
(i) 28, 36,12
(ii) 120,68
(iii) 124, 78, 116
(iv) 1228, 386

Solution:

(i) 28, 36,12
The factors of 28 are 1,2,4,7,14,28.
The factors of 36 are 1,2,3,4,6,9,12,18,36.
The factors of 12 are 1,2,3,4,6,12.
4 is the highest common factor.
Hence, HCF is 4.
(ii) 120, 68
The factors of 120 are 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120.
The factors of 68 are 1,2,4,17,34,68.
Hence,4 is the highest common factor.
(iii) 124, 78, 116
The factors of 124 are 1,2,4,31,62,124.
The factors of 78 are 1,2,3,6,13,26,39,78.
The factors of 116 are 1,2,4,29,58,116.
Hence, 2 is the highest common factor.
(iv) 1228,386
The factors of 1228 are 1,2,4,307,614,1228
The factors of 386 are 1,2,193,386
Hence, 2 is the greatest common factor.

Example 2.
(i) Find the HCF of 45, 120 by long division method?
(ii) Find the HCF of 150, 180 by long division method?
(iii) Find the HCF of 25,34,18 by long division method?
(iv) Find the HCF of 100, 30 by the long division method?

Solution:

1. First, Take the largest number 120 as a dividend and the smallest number 45 as the divisor.
2. Take divisor as the new dividend and remainder as the new divisor i.e. divide the first divisor(45) by the first remainder(30).
3. Continue the process until the remainder is zero and the last divisor will be the HCF of the given numbers.
hcf by long division example
Hence, 15 is the HCF of 45,120.

(ii)
1. First, Consider the largest number 180 as a dividend and the smallest number 150 as the divisor.
2. Consider the divisor as the new dividend and the remainder as the new divisor i.e. divide the first divisor(150) by the first remainder(30).
3. Continue the procedure until the remainder is zero and the last divisor will be the HCF of the given numbers.
hcf by long division example
Hence, 30 is the HCF of 150,180.

(iii)
1. First, take the largest number 34 as a dividend and the smallest number 18 as the divisor.
2. Take divisor as the new dividend and remainder as the new divisor i.e. divide the first divisor(18) by the first remainder(16).
3. Continue the process until the remainder is zero and the last divisor will be the HCF of the given numbers.
hcf by long division method example 3a
Now take the remaining value 25 as a dividend and 2 HCF of 134,325 as of the divisor.
hcf by long division method example 3b
Hence, HCF of 325,134,208 is 1.

(iv)
First, take the largest number 100 as a dividend and the smallest number 30 as the divisor.
hcf of long division example 4
Therefore, HCF of 100,30 is 10.

Example 3.
Find the LCM of 18, 24, 32 by using the Cake/Ladder method?

Solution:

lcm by cake and ladder method example
LCM=2 × 9 × 11× 16=3168
Hence,LCM of 18,24,32 is 3168.

Example 4.
Find the LCM of 30,48,55 by using the Cake/Ladder method?

Solution:

lcm by cake and ladder method example2
LCM=2 ×3 × 5 ×1 × 8 × 11=2640
Hence, LCM of 30, 48, 55 is 2640.

Example 5.
Find the LCM of 420,505 by using the Cake/Ladder method?

Solution:

lcm by cake and ladder method example3
LCM is 5 × 84 ×101=42,420
Hence, LCM of 420,505 is 42,420.

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

Example 6.
Find the LCM of 24,34,54 by using the Cake/Ladder method?

Solution:

lcm by cake and ladder method example4
LCM is 2 × 3 × 4 × 17 × 9=3672
Hence Lcm of 24,34,54 is 3672.

Example 7.
The LCM of the two numbers is 462. If the product of two numbers is 924. Find their HCF?

Solution:

Given,
Lcm of two numbers=462
Product of two numbers=924
We know that product of two numbers= LCM × HCF
924=462 × HCF
HCF=924/462=2
Therefore, HCF is 2.

Example 8.
The HCF of the two numbers is 4. If the product of two numbers is 3264. Find their LCM?

Solution:

Given,
HCF of two numbers=4
product of two numbers=3264
We know that product of two numbers= LCM × HCF
3264=LCM × 4
LCM=3264/4=816
Therefore, LCM is 816.

Example 9.
If HCF and LCM of two numbers are 10 and 280. If one of the numbers is 40, Find the other number?

Solution:

Given,
HCF=10
LCM=280
One of the number=40
Let the other number=x
We know that product of two numbers=LCM × HCF
x . 40=280 × 10
x.40=2800
x=2800 ÷40=70
Hence, the other number is 70.

Example 10.
Find the HCF of 20,80 by prime factorization method?

Solution:

prime factorization of 20 is 2 × 2 ×5
prime factorization of 80 is 2 × 2 × 2 × 2 × 5
HCF is 2 × 2 ×5=20
Hence, HCF of 20,80 is 20.

Example 11.
Find the HCF of 10,40 by factorization method?

Solution:

factors of 10 are 1,2,5,10
factors of 40 are 1,2,4,5,8,10,20,40
Here the highest common factor is 10
Hence, HCF is 10.

Example 12.
Find the LCM of 27,33 by using the prime factorization LCM method.

Solution:

27= 3×3×3=33
33=3 × 11
LCM=33 × 11
= 27 × 11
=297
Therefore, the LCM of 27,33 is 297.

Example 13.
Find the LCM of 15,25 by using the listing multiples method?

Solution:

Multiples of 15 are 15,30,45,60,75,90,105
Multiples of 25 are 25,50,75,100,125
Here the least common multiple is 75
Hence, LCM of 15,25 is 75.

Example 14.
Find the LCM of 13,15 by division table method?

Solution:

Lcm by division table method
LCM is 3 × 5 × 13 =195
Hence, LCM of 13,15 is 195.

Repeated Prime Factors

Repeated Prime Factors – Definition, Facts, Examples | How to find the Repeated Prime Factors of a Number?

You can improve your knowledge regarding the concept of finding Prime Factors. A factor is one of two or more numbers that divides a given number without a remainder. Here you will learn about repeated prime factors, how to find repeated prime factors and some example problems on repeated prime factors, and so on.

This page makes it easy for you to understand how to find repeated prime factors as we have compiled the step-by-step solutions for all the questions. Assess your strengths and weaknesses by finding the repeated prime factors as a part of your preparation.

Also, Refer:

Repeated Prime Factors – Definition & Meaning

Repeated Prime Factor means a number can have two or more different repeated factors. Let us consider an example of repeated prime factors is 8 = 2 x 2 x 2. Here 2 is said to be a repeated prime factor.

How do you find Repeated Prime Factors of a Number?

Follow the simple steps listed below to understand how to find repeated prime factors. They are as such
1. First, We will find the prime factors of a given number.
2. Then, write the all-finding prime factors in one place.
3. Now, Count the same number of finding prime factors. If the factor numbers are two or more same numbers then it is a Repeated Prime Factors.

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

Repeated Prime Factors Examples

Example 1: 
Find the Repeated Prime Factors of the following Numbers,
(i) 25 (ii) 15 (iii) 49 (iv) 16 (v) 32

Solution: 
Given the values
(i) The value is 25
Now, write the prime factors of 25 i.e. 5 x 5 and 25 x 1.
Therefore, the Repeated Prime Factors of 25 are 5 x 5.
(ii) 15
Now, writing 15 in terms of factorization we get 3 x 5, and 5 x 3
In this factorization, there are no repeated prime factors.
Therefore, 15 has no Repeated Prime Factors.
(iii) 49
We can write 49 in terms of the factors as 7 x 7, and 49 x 1.
So, the Repeated Prime Factors of 49 are 7 x 7.
(iv) 16
Writing 14 in terms of factors is 4 x 4, and 2 x 2 x 2 x 2.
Hence, the Repeated Prime Factors of 16 are 4 and 2.
(v) 32
We can express 32 in terms of prime factors is 8×4, 4×8, and 2 x 2 x 2 x 2 x 2.
Therefore, the Repeated Prime Factors of 32 is 2.

Example 2:
Write the Repeated Prime Factors of 125 and 128.

Solution:
Given the values is 125 and 128.
Now, we can express 125 in terms of prime factors as 5 x 5.
Another value is 128. We can write 128 in terms of the prime factors as 2 x 2 x 2 x 2 x 2 x 2 x 2.
Therefore, the Repeated Prime Factors of 125 is 5, and the repeated prime factors of 128 are 2.

Example 3:
Write the Repeated Prime Factors of 432, 81, 900.

Solution:
As given in the question, the value is 432.
We can express 432 in terms of prime factors as 4 x 4 x 3 x 3 x 3
Now, we can express 81 in terms of prime factors as 3 x 3 x 3 x 3.
Writing 900 in terms of prime factors as 5 x 5 x 6 x 6.
Therefore, the Repeated Prime Factors of 432 are 4 and 3, 81 Repeated Prime Factor is 3, and Repeated Prime Factors of 900 are 5, 6.

Example 4:
Write the product value of the below of given Repeated Prime Factors,
(i) 2 x 2 x 3 x 3 x 3
(ii) 7 x 7 x 5

Solution:
(i) Given the Repeated Prime Factors,
Now, we write product value of 2 x 2 x 3 x 3 x 3 is 108.
(ii) 7 x 7 x 5
We can write the product value of 7 x 7 x 5 is 245.
Hence, the product values are 108, and 245.

Example 5: 
Find the Prime Factor of 72 using the Division Method. Write its Repeated Prime Factors.

Solution:
Given the value is 72. We can find the prime factors using the division method.
In this method, first, we have to check each number by dividing the composite number.
To get the prime factors of 72, we have to start by dividing them by prime numbers.
72 ÷ 36 = 2
36 ÷ 18= 2
18 ÷ 9 = 2
9 ÷ 3 = 3
3 ÷ 1 = 3
The Prime factors of 72  is 2 x 2 x 2 x 3 x 3.
Therefore, the Repeated Prime Factors of 72 is 2 and 3.

Worksheet on Unitary Method

Worksheet on Unitary Method | Download Free Printable Worksheet on Word Problems on Unitary Method

Unitary Method is the process of finding the value of a single unit from multiple units and vice versa. We use this technique to solve various math calculations. Unitary Method Worksheet helps you solve questions on ratio, proportion, direct, indirect variations, geometry, etc. Worksheet on Unitary Method has problems for finding the missing value, cost of many units or cost of one, etc.  Use the 5th Grade Math Unitary Method Questions over here to get a good grip on the concept as well as to have a strong foundation of basics.

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Unitary Method Questions with Solutions

Example 1.
The cost of 10 mangoes is Rs 200. Find the cost of 6 mangoes?

Solution:  The cost of 10 mangoes is = Rs 200
The cost of 1 mangoe=200/10= Rs 20
The cost of 6 mangoes=6 × 20=Rs 120.
Hence, the cost of 6 mangoes is Rs 120.

Example 2.
Rajesh walks 1 km in 40 minutes. How much time is required for Rajesh to walk 5 km?

Solution:

Given,
The time is taken by Rajesh to walk 1 km= 40 min
The time is taken by Rajesh to walk 5 km= 5 × 40 min =200 min
=3 hours 33 minutes
Therefore, the time taken by Rajesh to walk 5 km in 3 hours 33 minutes.

Example 3.
The annual salary of Shekar is Rs 3,50,000. Find the salary of Shekar for 4 months?

Solution:

Given,
The annual salary of Shekar is =Rs 3,50,000
Since there are 12 months, divide by 12 to get 1 month’s salary.
Salary of Shekar for the month=Rs 3,50,000/12=29,166.66
salary of Shekar for 4 months=4 × 29,166.66
=Rs 116,664.64
Hence, The salary of Shekar for 4 months is Rs 116,664.64.

Example 4.
7 men can complete the work in 30 days working 6 hours per day. If 4 men can complete the same work in 30 days, how many hours does each man has to work per day?

Solution:

Given,
7 men can complete the work in 30 days working 6 hours per day.
No. of hours required for one man to complete the same work is
no. of men. no. of days. no. of hours per day
=7.30.6
=12600 hours
If there are 4 men, the no. of hours required to complete the same work is
1260 ÷ 4=315 hours
Given the no. of days work has to be completed is 30
The no. of hours each man has to work per day is 315 ÷ 30=10.5 hours
Hence, 4 men have to work 10.5 hours per day.

Example 5.
In a toy factory, they make 3000 toys in the month of September. If they make the same quantity of toys every day, then how many toys can they make in a leap year?

Solution:

Given,
Toys made by the factory in the month of september=3000
In any year the month of September has 30 days.
No. of toys made per day=3000 ÷ 30=100 toys
In a leap year, the no. of days will be 366.
No. of toys made=366 × 100=36600 toys
Therefore, No. of toys made in the leap year is 36600.

Example 6.
A worker is paid 1550 Rs for 6 days of work. If he works for 20 days how much he will earn?

Solution:

Given,
The worker will earn money for 6 days of work=Rs 1550
The worker will get money for doing work 1 day=Rs 1550 ÷ 6=258.33
The money he will get for doing work 20 days=258.33 × 20=5166.66
Hence, the worker will get 5166.66 for working 20 days.

Example 7.
A car traveling at a speed of 120 km/hr covers 360 km. How much time will it take to cover 200 km?

Solution:

Given,
The car travels at a speed=120 km/hr
distance=360 km
we know that speed=distance/time
120=360/t
t=360/120=18/6=3 hours
360 km=3 hours
1 km=3/360 hour
200 km=3/360 × 200=1.666 hours
Hence, 1.66 hours are required to cover the distance of 200 km.

Example 8.
Anu finishes his work in 20 days whereas Bhaskar completes the same work in 30 days. Find how many days are required to complete the same work?

Solution:

Given,
Anu finishes his work in 20 days
Bhaskar finishes his work in 30 days
Anu’s one day work=1/20
Bhaskar’s one day work=1/30
Total work done by Anu and Bhaskar in one day=1/20+1/30
LCM of 20,30 is 60
=3+2/60=5/60=1/12
Therefore, Anu and Bhaskar complete the work in 12 days.

Example 9.
If 5 buses can carry 850 people. Find how many buses are required for 1600 people?

Solution:

Given,
5 buses carry people=850
1 bus carry=850 ÷ 5=170
Buses required to carry 1600 people=1600 ÷ 170=9.41=9 buses
Therefore, 9 buses are required to carry 1600 people.

Example 10.
Ram goes to the shop to buy some notebooks. The price of 3 notebooks is Rs 120. Ram, has Rs 300 find how many notebooks Ram can buy? Also, find the price of 8 notebooks?

Solution:

Given,
Price of 3 notebooks=Rs 120
By the unitary method,
Price of one-note book=120 ÷ 3= Rs 60
Also given, Ram has the money=Rs 300
No. of books Ram can buy=300 ÷ 60=5 notebooks
Hence, Ram can buy 5 books.
Price of 8 note books=8 × 60= Rs 480
Therefore, the price of 8 notebooks is Rs 480.

Example 11.
Sita went to the shop to buy sugar. The price of 5 kg sugar is Rs 200. How many Kg of sugar can she buy for Rs 560?

Solution:

Given,
The price of 5 kg sugar=Rs 200
By the unitary method, 1 kg sugar=Rs 200 ÷5=Rs 40
Also given, Sita has the money of Rs 200.
No. of kg of sugar Sita can buy= 560 ÷ 40=14 kg
Therefore, Sita can buy 14 kg of sugar with the money of Rs 560.

Example 12.
The money required for buying 4 biscuit packets is Rs 120. Find the money required for buying 16 biscuit packets?

Solution:

Given,
Money required for buying 4 biscuit packets is = Rs 120
Money required for buying 1 biscuit packet is=120 ÷ 4=Rs 30
money required for buying 16 biscuits packets=16 × 30=Rs 480
Therefore, the money required for buying 16 biscuit packets is Rs 480.

Worksheet on Measurement

Worksheet on Measurement | Free & Printable Measurement Worksheets PDF with Answers

Printable Worksheets on Measurement have problems related to length, mass, volume, unit conversions. Practice using the Measurement Worksheet and get to know various kinds of questions asked on the topic. Interactive Exercises on Measurement are free to print and download. Online Measurement Worksheets PDFs will help you learn how to apply measurements to real-life situations. The Problems provided in the 5th Grade Math are more than just pencil-paper activities and enhances your math skills regarding the concept of measurement.

Do check:

Example 1.
1. Convert a distance of 7 km to meters
2. Convert 60 meters to centimeters

Solution:

1. 7 km
We know that 1 km=1000 meters
Multiply 1000 with 7
7 × 1000=7000 meters
Hence, the distance of 7 km is 7000 meters.
2. 60 meter
we know that 1 meter=100 cm
Multiply 100 with 60
60 × 100=6000 cm
Hence, 60 meters is 6000 cm.

Example 2.
Anu bought  2 kg of apples. Convert kg into grams.

Solution:

Anu bought apples=2 kg
we know that 1 kg=1000 m
To convert kg into grams multiply with 1000.
2 × 1000=2000 grams
Hence, Anu bought 2000 grams.

Example 3.
Convert 25 m into cm?

Solution:

we know that 1 m=100 cm
Multiply with 100 to convert meters into centimeters
25 × 100=2500 cm
Hence, 25 m is 2500 cm.

Fill in the blanks with the correct sign
1. 10g _______ 100 mg
2. 1 kg _______ 2000 g
3. 1 km_______ 200m
4. 23 m______ 2200 cm

Solution:

1. we know that 10 g=10000 mg
So 10 g is greater than(>) 100 mg
2. we know that 1 kg=1000 grams
So 1 kg is less than(<) 2000 g.
3. We know that1 km =1000 m
So 1 km is greater than 200 m.
4. We know that1 m =100 cm
23 m=2300 cm
So 2300 cm is greater than 2200 cm

Solve the following
Example 1.
Add 2 kg 280 g and 4 kg 300 g

Solution:

 Kg       g
2        280
+4        300
___________________
6        580
___________________
Hence, By adding 2 kg 280 g and 4 kg 300 g we get 6 kg 580 grams.

Example 2.
Subtract 3 kg 120 g from 5 kg 780 g

Solution:

Kg      g
5      780
-3     120
______________
2       660
_______________
Hence, By subtracting 3 kg 120 g from 5 kg 780 g we get 2 kg 660 g.

Example 3.
Multiply 160 kg by 16

Solution:

measurment example
Hence, By multiplying 160 kg with 16 we get 2560 kg.

Example 4.
Divide 360 kg by 3

Solution:

measurment division example
Hence by dividing 360 kg with 3 we get 120 as Quotient.

Example 5.
Express 95 l in dl

Solution:

we know that 1 liter=10 deciliter.
Multiply 95 with 10 we get 950
Hence, 95 l is 950 l.

Write the most appropriate unit for the following
1. Ranjith walks 1 ____ every day.
2. Abhi weighs 56 ______
3. Sindhu bought 2 _______ of apples.
4. Hima travels 1 ____ every day.
5. Height of Teja is 5 _____

Solution:

1. km
2. kg
3.kg
4. km
5. feet

Convert the following measurements into the given units
1. Convert 5 kg 80 g into grams?
2. Convert10 m 33 cm into cm?
3. Convert 6354 mg into g and mg?
4. Convert 5890 ml into l and ml?

Solution:

1.5 kg 80 g
We know that 1 kg=1000 g
=5 × 1000 g+80 g
=5000 g+80 g=5080 g
Hence, 5 kg 80 g is converted to 5080 g.
2. 10 m 33 cm
we know that 1m=100 cm
10 m=10 × 100=1000 cm
=1000cm + 33 cm=1033 cm
Hence, 10 m 33 cm is converted to 1033 cm.
3. 6354 mg
We know that 1 mg=1000 g
=6 g 354 mg
Hence, 6354 mg is converted to 6 g 354 mg.
4. 5890 ml
we know that 1 lit=1000 ml
=5 l 890 ml
Hence, 5890 is converted to 5 liters 890 ml

Word Problems on Measurement of Length, Mass, Volume
Example 1.
Srinivas bought 5 kg of mangoes. His mother distributed 4 kg of mangoes. Find how many kgs of mangoes were left after the distribution?

Solution:

Given,
Srinivas bought mangoes=5 kg
Mangoes distributed by his mother=4 kg
No. of mangoes left=5 kg-4 kg=1 kg
Hence, No. of mangoes left after the distribution is 1 kg.

Example 2.
A milk vendor has three cans of milk which consists of 4 l, 5l, 10l. Find how many liters of milk the vendor has to transfer into the big milk can?

Solution:

Given,
A milk vendor has three cans of milk=4 l,5 l,10 l
No. of liters of milk the vendor has=4+5+10=19 liters
Hence, The vendor has to transfer 19 liters of milk into the big milk can.

Example 3.
Bhaskar bought 5 kg of sweets and distributed 3 kg’s of sweets to his friends. How many kg of sweets are left after the distribution?

Solution:

Given,
Bhaskar bought sweets=5 kg
Bhaskar distributed sweets to his friends=3 kg
No. of kg of sweets left after distribution=5 kg-3 kg=2 kg
Hence, Sweets left after the distribution is 2 kg.

Example 4.
In a fruit shop, there are 40 liters of banana juice. After selling, 5 liters of juice are left. Find how many liters of juice are sold?

Solution:

Given,
No. of liters of banana juice=40 liters
No. of liters of juice left=5 liters
No. of liters of juice sold=40-5=35 liters
Hence, No. of liters of juice sold is 35 liters.

Example 5.
Ajay walks 1 km daily for good health. Find how many kilometers can he walk for 30 days?

Solution:

Given,
Ajay walks dialy=1 km
No. of kilometers he walks for 30 days=30 × 1=30 km.
Therefore, No. of kilometers Ajay walks daily is 30 km.

Example 6.
Sridevi has 3 liters of milk for a recipe. She used 2 liters of milk. How many liters of milk are left?

Solution:

Given,
No. of liters of milk Sridevi has=3 liters
Sridevi used milk for the recipe=2 liters
No. of liters of milk left=3 liters-2 liters=1 liter
Therefore, 1 liter of milk is left.

 

Worksheet on Word Problems on Measurement

Worksheet on Word Problems on Measurement | Measurement Word Problems Worksheet PDF

Worksheet on Word Problems on Measurement has questions on measuring length in both customary and metric units. These Math Worksheets featuring Measurement will let students explore the concepts of length, mass, volume, etc. Students can use the details from the 5th Grade Math Worksheet on Word Problems Measurement to solve and learn measuring capacities, mass, volume concept-related problems. The Fun and Interactive Measurement Word Problems Worksheet makes it easy for you to grasp the concept.

Also, check:

Measurement Word Problems with Solutions

Example 1.
A parking lot has 28 parking spots side by side. Each parking spot is 10 feet wide. What is the width of the parking lot?

Solution:

Given, No. of parking spots side by side in parking lot=28
Each parking spot=10 feet
Width of the parking lot=28 × 10=280 feet
280 feet=98.33 yards
Therefore, the width of the parking lot is 98.33 yards.

Example 2.
Each roll of the string is 80 yards. In a box, Sunil packed 200 rolls of strings. Find the total length of the string in yards?

Solution:

Given,
Each roll of the string = 80 yards
No. of string rolls in the box=200
The total length of the string rolls=200 × 80=16000 yards.
Therefore, the total length of the string is 16000 yards.

Example 3.
Sita is going to tailoring classes.  Sita’s teacher told her to bring 2 pieces of cloth. They must be 1.5 m long for stitching. Sita found an old cloth at home that is 200 cm long. Find how much extra cloth is needed for Sita?

Solution:

Given,No. of pieces of cloth=2
Length of the pieces=1.5 m
Sita’s teacher told her to bring cloth=2 × 1.5 m=  3 m=300 cm (since 1m=100 cm)
Sita found an old cloth at home=200 cm
extra cloth is needed for Sita=300-200=100 cm
Hence, Sita needs 100 cm extra cloth.

Example 4.
The fan is 8 feet 6 inches from the floor. Lakshmi can reach up to 6 feet 4 inches when she raises the arm. If she stands on a stool that is 8 inches tall, Can she reach the fan?

Solution:

Given,
The distance from fan to the floor= 8 feet 6 inches
She can reach up to the fan=6 feet 4 inches
Lakshmi stands on a stool=8 inches
Lakshmi can reach the fan=6 feet 4 inches+8 inches
=6 feet 12 inches
=7 feet
Therefore, Lakshmi can not reach the fan.

Example 5.
Vijaya wants to make a tower of erasers of a height of 30 cm. If the thickness of each eraser is 2 mm. How many erasers are needed to make the desired height?

Solution:

Given,
The total length of the tower=30 cm
The thickness of each eraser=2 mm
First, convert the length of the tower to mm.
We know that 1 cm=10 mm
30 cm=30 ×10mm=300 mm
No. of erasers needed=Length of the tower/thickness of each book
=300/2=150 erasers.
Hence, the total no. of erasers needed is 150 erasers.

Example 6.
Ajay has to take three spoons of syrup each day for his health problem. If the capacity of the spoon is 15 ml, what quantity of syrup is taken by him in 10 days?

Solution:

Given,
The capacity of the spoon is=15 ml
Quantity of syrup taken by him in ten days=15 ml × 10 days=150 ml
Hence, the quantity of syrup taken by him in ten days is 150 ml.

Example 7.
How many 20 cm pieces can be cut out of a 10 m long cloth?

Solution:

Given,
cloth length=10 m=1000 cm
Length of pieces =20 cm
No. of pieces can be cut=1000/20=50
Therefore, the total no. of pieces is 50.

Example 8.
A worker transferred 20 bags of rice weighing 40 kg into the truck. The other grocery’s weight is 150 kg. The weight of the empty truck is 1590 kg. What will be the weight of the truck with the bags and the groceries?

Solution:

 No. of kg of rice=20 × 40=800 kg
The groceries weight=150 kg
weight of the empty truck=1590 kg
Weight of the truck with bags and the groceries=800 kg+150 kg+1590 kg
=2,540 kg
Hence, the weight of the truck with the bags and the groceries is 2,540 kg.

Example 9.
From a cloth of 30 cm, two pieces of length 5 cm,8 cm are cut off. What is the length of the remaining cloth?

Solution:

Given, Length of the cloth=30 cm
Length of two pieces=5 cm, 8 cm
Length of the remaining cloth=30 cm-(5 cm+8 cm)
=30 cm-13 cm
=17 cm
Therefore, the length of the remaining cloth is 17 cm.

Example 10.
8 inches of ribbon is used for wrapping up each present. How many presents can be wrapped with 7 yards of ribbons?

Solution:

Given, Length of the ribbon used for wrapping the present=8 inches
No. of  presents can be wrapped with the length of ribbons=7 yards
7 yards=252 inches
252 ÷ 7=36
Hence, 36 presents can be wrapped with 7 yards of ribbons.

Example 11.
The diameter of the coin is 20 mm. Suresh lined up 50 coins from the left end of the paper to the right end of the paper. what is the width of the paper in cm?

Solution:

Given, The diameter of the coin =20 mm
No. of coins lined up from left end to right end=50 coins
Width of the paper=no. of coins × diameter of the coin
=50 × 20 mm
=1000 mm=100 cm
Therefore, the width of the paper is 100 cm.

 

 

Prime Factors

Prime Factors – Definition, Methods, Examples | How to find Prime Factors of a Number?

Prime Factors are simply factors of number which are Prime Numbers. The ‘Prime Factors’ are composed of two distinct words one is prime and another word is factors and they are very important in mathematics. The number that may be multiplied together to induce another number could be a factor. Suppose when the division of a number with all the factors results in 0 remainders, that factor is an exact divisor of that number.

In this article, you will learn about Prime Factors, finding prime factors of a number using the division method, some examples explaining how to find the prime factor, how to find the prime factorization of a number using the Factor Tree method, and so on.

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Prime Factors – Definition

Prime Factors means a group of prime numbers that are multiplied together to get the actual number. In other words, Prime factors are factors of a number that are, themselves, prime numbers. We have many methods for finding prime factors, but the most commonly used method is finding prime factors using the factor tree method.

Consider a value is 6, the prime factor of value 6 is 3 x 2. So, the prime factors of 6 are 3 and 2. Now, take another value 8, the prime factors of value 8 are, 4×2 and 2x2x2. Here will take only 2 as a prime factor as 4 is not a prime number.

The Prime Factors Table for 1-100 Numbers is as shown below,

How to find Prime Factors of a Number through Division Method?

Following are the steps for finding the prime factor of any given number. Remember and follow these steps

Step 1: First, by dividing the given number by the first prime number 2.
Step 2: Continue dividing by 2 until you get a decimal or remainder. Then divide the next prime numbers like 3, 5, 7, etc. Until the numbers left are prime numbers.
Step 3: Now, write the number as a product of prime numbers.

The method of finding prime factors of any given number is called Prime Factorization. Prime Factorization means any number written in the form of multiplication of their prime factors and the original number is evenly divisible by these factors. In terms of the factor tree, prime factors can be easily understood.

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How to find Prime Factors of a Number using the Tree Method?

Tree method factorization is a straightforward method. The below provided guidelines in order to write down the prime factors using the tree method. They are,

  • First, take the given number as the root of the tree.
  • Next, write the factors of the given number as branches of the tree. You can break the number as any of the factors.
  • Then factorize the composite numbers and write the factors pairs as the branches.
  • You can do this process until you get the prime factors.

Finding Prime Factors Examples

Example 1: 
Find the Prime Factors of the below-given Numbers,
(i) 8 (ii)4 (iii)13 (iv)14

Solution:
(i)Given Value is 8
Firstly, write the factors of 8 i.e. 4*2
As 4 is not a prime number we will write 4 again in terms of its factors i.e. 2*2. Therefore, 8 can be expressed in terms of prime factors as 2.
Now, pick the prime numbers among the factor list. Therefore, prime factors of 8 are 2
(ii) 4
Writing 4 in terms of factorization is 2×2.
So, 2 is a prime factor of 4.
(iii) 13
We can write 13 in terms of the factors as 1×13 and 13×1. Since 1 is neither prime nor composite number 13 is the Prime factor.
(iv) 14
Given Value is 14
We can express 14 in terms of factors as 2* 7 or 14*1
However, we will consider 2 and 7 as prime factors but not 14 and 1 as 14 is not a prime number.

Example 2:
Find the Prime Factors of 27 using the Factor Tree Method?

Solution:
The given value is 27, using the Factor Tree Method we can find the prime factors.
First, 27 is factored into two factors 3 and 9. 9 is again factored and written as factors of 3 and 3.

So, Writing the Prime Factors of 27 we have 3x3x3.

Example 3:
Find the Prime Factors of 56 using the Factor Tree Method.

Solution:
Given the value is 56,
Now, we will find the prime factors using the Factor Tree Method.
First, split the number given in to any of two factors. 56 is factored into two factors 2 and 28. 28 is again factorized into two factors 2 and 14. Now on splitting 14 further and writing it as multiples of numbers it might be 7 x 2.
The below tree structure formed for number 56 prime factors.


So, therefore the prime factor of 56 is 2x2x2x7.

Example 4:
Find the Prime Factors 216 using the Factor Division method?

Solution:
Given the value is 216, we can find the prime factors using the division method.
The division method consists of very easy and basic algorithms.
In this method, first, we have to check each number by dividing the composite number.
To get the prime factors of 216, we have to start by dividing them by prime numbers.
216 ÷108 = 2
108 ÷ 54 = 2
54 ÷ 27 = 2
27 ÷ 9 = 3
9 ÷ 3 = 3
3 ÷ 1 = 3
Therefore, the Prime Factor of 216 using the division method is 2 x 2 x 2 x 3 x 3 x 3= 23 x 33