## NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers (Ex 3.7) Exercise 3.7

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 written by Vedantu’s academic experts contain the solutions to all the problems of Exercise - 3.7. It is categorically broken down into simple steps with proper explanations for you to grab the concept in an easier way. NCERT Solutions for Class 6 Maths Chapter 3 Playing with numbers is strictly based on the guidelines provided by CBSE. Download Maths Chapter 3 NCERT Solutions and practice all the solutions from the comfort of your home. You can also register Online for NCERT Solutions Class 6 Science tuition on Vedantu.com to score more marks in your examination.

## Download PDF of NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers (Ex 3.7) Exercise 3.7

## Class 6 Maths Chapter 3 – An Overview of Playing with Numbers.

### Exercise 3.7

1. Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times.

Ans: It is given renu purchases two bags of fertilizer of weights 75kg and 69kg.

To find the maximum weight which can measure the weight of the fertilizer an exact number of times, we need to find the Highest Common Factor of 75 and 69.

Now list the factors of 75 and 69.

Factors of \[75 = 5 \times 3 \times 5\]

Factors of \[69 = 23 \times 3\]

We can see that the Highest Common Factor of 75 and 69 is 3.

Therefore the Maximum required weight is 3kg.

2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the maximum distance each should cover so that all can cover the distance in complete steps?

Ans: The step measure of the three boys is given as 63cm, 70cm, and 77cm.

To find the maximum distance each should cover so that all the three boys can cover the distance in complete steps, we need to find the Least Common Multiple of 63, 70 and 77.

Now let us find the LCM(63,70,77).

$ 2\left| \!{\underline {\, {63,70,77} \,}} \right. \\ 3\left| \!{\underline {\, {63,35,77} \,}} \right. \\ 3\left| \!{\underline {\, {21,35,77} \,}} \right. \\ 5\left| \!{\underline {\, {7,35,77} \,}} \right. \\ 7\left| \!{\underline {\, {7,7,77} \,}} \right. \\ 11\left| \!{\underline {\, {1,1,11} \,}} \right. \\ 1\left| \!{\underline {\, {1,1,1} \,}} \right. $

LCM (63,70, 77) = \[7 \times 9 \times 10 \times 11 = 6930cm\]

Therefore, the maximum distance each of the boys should cover is 6930cm.

3. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

Ans: It is given that,

Length of the room = 825cm.

Breadth of the room = 675cm.

Height of the room = 450cm.

To find the measurement of the tape we need to find the Highest Common Factor of 825, 675, 450.

Factors of \[825 = 5 \times 5 \times 3 \times 11\]

Factors of \[675 = 5 \times 5 \times 3 \times 3 \times 3\]

Factors of \[450 = 2 \times 3 \times 3 \times 5 \times 5\]

Therefore the HCF (825, 675, 450) =\[3 \times 5 \times 5 = 75\]

So, the length of the tape that can measure the three dimensions of the room exactly is 75cm.

4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and12.

Ans: To find the smallest 3-digit number that is exactly divisible by 6, 8 and 12 , first we need to find the Least Common Multiple of 6, 8 and 12.

$ 2\left| \!{\underline {\, {6,8,12} \,}} \right. \\ 2\left| \!{\underline {\, {3,4,6} \,}} \right. \\ 2\left| \!{\underline {\, {3,2,3} \,}} \right. \\ 3\left| \!{\underline {\, {3,1,3} \,}} \right. \\ 1\left| \!{\underline {\, {1,1,1} \,}} \right. $

Therefore the LCM (6,8,12) = \[2 \times 2 \times 2 \times 3 = 24\]

The smallest three digit number divisible by 24 is the number that is exactly divisible by 6, 8 and 12.

120 is the smallest 3-digit number divisible by 24.

Hence, the smallest 3-digit number that is exactly divisible by 6,8,12 is 120.

5. Determine the largest 3-digit number which is exactly divisible by 8, 10 and12.

Ans: To find the largest 3-digit number that is exactly divisible by 8, 10 and 12, we need to find the Least Common Multiple of 8, 10 and 12.

$ 2\left| \!{\underline {\, {8,10,12} \,}} \right. \\ 2\left| \!{\underline {\, {4,5,6} \,}} \right. \\ 2\left| \!{\underline {\, {2,5,3} \,}} \right. \\ 3\left| \!{\underline {\, {1,5,3} \,}} \right. \\ 5\left| \!{\underline {\, {1,5,1} \,}} \right. \\ 1\left| \!{\underline {\, {1,1,1} \,}} \right. $

Therefore the LCM (8,10,12) = \[2 \times 2 \times 2 \times 3 \times 5 = 120\]

The largest 3-digit number divisible by 120 is the number that is exactly divisible by 8, 10 and 12.

960 is the largest 3-digit number divisible by 120

So, the largest 3-digit number that is exactly divisible by 8, 10 and 12 is 960.

6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m. at what time will they change simultaneously again?

Ans: It is given that 3 different traffic lights change after every 48 seconds, 72 seconds and 108 seconds.

In order to find after how many seconds the traffic light changes simultaneously we need to find the Least Common Multiple of 48, 72 and 108.

$ 2\left| \!{\underline {\, {48,72,108} \,}} \right. \\ 2\left| \!{\underline {\, {24,36,54} \,}} \right. \\ 2\left| \!{\underline {\, {12,18,27} \,}} \right. \\ 3\left| \!{\underline {\, {6,9,27} \,}} \right. \\ 3\left| \!{\underline {\, {2,3,9} \,}} \right. \\ 2\left| \!{\underline {\, {2,1,3} \,}} \right. \\ 3\left| \!{\underline {\, {1,1,3} \,}} \right. \\ 1\left| \!{\underline {\, {1,1,1} \,}} \right. $

Therefore the LCM (48,72,108) =\[2 \times 2 \times 2 \times 3 \times 3 \times 2 \times 3 = 432\]sec.

After 432sec the 3 different traffic signals will change simultaneously.

i.e. if the change simultaneously at 7.am then,

432 sec = 7 minutes 12 seconds.

Therefore the time they will change simultaneously again = 7.am+7minutes 12 seconds

= 7:07:12am

The time the 3 different traffic signal changes simultaneously is 7:07:12am

7. Three tankers contain 403 litres 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers the exact number of times.

Ans: It is given that 3 tankers can contain a diesel of 403 litres, 434 litres and 465 litres. In order to find the maximum capacity of the container we need to find the Highest Common Factor of 403, 434 and 465 litres.

Factors of \[403 = 31 \times 13\]

Factors of \[434 = 7 \times 2 \times 31\]

Factors of \[465 = 31 \times 3 \times 5\]

Therefore the HCF (403, 434, 465) = 31.

Thus 31 litres is the maximum capacity of a container that can measure the diesel of three containers an exact number of times.

8. Find the least number which when divided by 6, 15 and 18, leave the remainder 5 in each case.

Ans: To find the least number divided by 6, 15 and 18, we need to find the Least Common Multiple of 6, 15 and 18.

$ 2\left| \!{\underline {\, {6,15,18} \,}} \right. \\ 3\left| \!{\underline {\, {3,15,9} \,}} \right. \\ 3\left| \!{\underline {\, {1,5,3} \,}} \right. \\ 5\left| \!{\underline {\, {1,5,1} \,}} \right. \\ 1\left| \!{\underline {\, {1,1,1} \,}} \right. $

The LCM (6, 15, 18) = \[2 \times 3 \times 3 \times 5 = 90\]

90 is the least number that is divided by 6, 15 and 18.

Therefore, the least number which when divided by 6, 15, and 18 leaves remainder 5 is 90+5 = 95.

9. Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Ans: To find the smallest 4-digit number divisible by 18, 24 and 32 we need to find the Least Common Multiple of 18, 24 and 32.

$ 2\left| \!{\underline {\, {18,24,32} \,}} \right. \\ 2\left| \!{\underline {\, {9,12,16} \,}} \right. \\ 2\left| \!{\underline {\, {9,6,8} \,}} \right. \\ 2\left| \!{\underline {\, {9,3,4} \,}} \right. \\ 2\left| \!{\underline {\, {9,3,2} \,}} \right. \\ 3\left| \!{\underline {\, {9,3,1} \,}} \right. \\ 3\left| \!{\underline {\, {3,1,1} \,}} \right. \\ 1\left| \!{\underline {\, {1,1,1} \,}} \right. $

The LCM (18, 24, 32)=\[2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 288\]

The smallest 4-digit number that is divisible by 288 is the number that is exactly divisible by 18, 24, and 32.

1152 is the smallest 4-digit number divisible by 288.

So the smallest 4-digit number that is divisible by 18, 24 and 32 is 1152.

10. Find the L.C.M. of the following numbers:

Observe a common property in the obtained L.C.Ms. Is L.C.M. the product of two numbers in each case?

9 and 4

Ans: LCM of 9 and 4:

$ 2\left| \!{\underline {\, {9,4} \,}} \right. \\ 2\left| \!{\underline {\, {9,2} \,}} \right. \\ 3\left| \!{\underline {\, {9,1} \,}} \right. \\ 3\left| \!{\underline {\, {3,1} \,}} \right. \\ 1\left| \!{\underline {\, {1,1} \,}} \right. $

LCM (9, 4) = \[2 \times 2 \times 3 \times 3 = 36\]

Therefore LCM (9, 4) = 36

12 and 5

Ans: LCM of 12 and 5:

$ 2\left| \!{\underline {\, {12,5} \,}} \right. \\ 2\left| \!{\underline {\, {6,5} \,}} \right. \\ 3\left| \!{\underline {\, {3,5} \,}} \right. \\ 5\left| \!{\underline {\, {1,5} \,}} \right. \\ 1\left| \!{\underline {\, {1,1} \,}} \right. $

LCM (12, 5) = \[2 \times 2 \times 3 \times 5 = 60\]

Therefore the LCM (12, 5) = 60

6 and 5

Ans: LCM of 6 and 5

$ 2\left| \!{\underline {\, {6,5} \,}} \right. \\ 3\left| \!{\underline {\, {3,5} \,}} \right. \\ 5\left| \!{\underline {\, {1,5} \,}} \right. \\ 1\left| \!{\underline {\, {1,1} \,}} \right. $

The LCM (6, 5) = \[2 \times 3 \times 5 = 30\]

Therefore the LCM (6, 5) = 30

15 and 4

Ans: The LCM of 15 and 4:

$ 2\left| \!{\underline {\, {15,4} \,}} \right. \\ 2\left| \!{\underline {\, {15,2} \,}} \right. \\ 3\left| \!{\underline {\, {15,1} \,}} \right. \\ 5\left| \!{\underline {\, {5,1} \,}} \right. \\ 1\left| \!{\underline {\, {1,1} \,}} \right. $

LCM (15, 4) = \[2 \times 2 \times 3 \times 5 = 60\]

LCM(15, 4) = 60

The common property obtained in all the LCMs is that all the LCM is the multiple of 3.

Yes, the LCMs are the product of two numbers in each case.

11. Find the L.C.M of the following numbers in which one number is the factor of other:

What do you observe in the result obtained?

5, 20

Ans: LCM of 5 and 20:

$ 2\left| \!{\underline {\, {5,20} \,}} \right. \\ 2\left| \!{\underline {\, {5,10} \,}} \right. \\ 5\left| \!{\underline {\, {5,5} \,}} \right. \\ 1\left| \!{\underline {\, {1,1} \,}} \right. $

The LCM (5, 20) =\[2 \times 2 \times 5 = 20\]

LCM (5, 20) = 20

6, 18

Ans: LCM of 6 and 18:

$ 2\left| \!{\underline {\, {6,18} \,}} \right. \\ 3\left| \!{\underline {\, {3,9} \,}} \right. \\ 3\left| \!{\underline {\, {1,3} \,}} \right. \\ 1\left| \!{\underline {\, {1,1} \,}} \right. $

The LCM (6, 18) = \[2 \times 3 \times 3 = 18\]

Therefore LCM (6, 18) = 18

12, 48

Ans: LCM of 12 and 48:

$ 2\left| \!{\underline {\, {12,48} \,}} \right. \\ 2\left| \!{\underline {\, {6,24} \,}} \right. \\ 2\left| \!{\underline {\, {3,12} \,}} \right. \\ 2\left| \!{\underline {\, {3,6} \,}} \right. \\ 3\left| \!{\underline {\, {3,3} \,}} \right. \\ 1\left| \!{\underline {\, {1,1} \,}} \right. $

The LCM (12, 48) = \[2 \times 2 \times 2 \times 2 \times 3 = 48\]

Therefore the LCM (12, 48) = 48

9, 45

Ans: The LCM of 9 and 25:

$ 3\left| \!{\underline {\, {9,45} \,}} \right. \\ 3\left| \!{\underline {\, {3,15} \,}} \right. \\ 5\left| \!{\underline {\, {1,5} \,}} \right. \\ 1\left| \!{\underline {\, {1,1} \,}} \right. $

The LCM (9, 45) = \[3 \times 3 \times 5 = 45\]

Therefore LCM (9, 45) = 45

From the answers of the LCM we can clearly observe that the LCM of the two given numbers is equal to that of the larger number given.

### Class 6 Maths Chapter 3 – An Overview of Playing with Numbers.

Playing with numbers deals with the basic and important concepts of Maths like Factors and Multiples, test for the divisibility of numbers, prime numbers, composite numbers, Prime factorisation, LCM and HCF. To master these concepts you can download Maths NCERT Solutions Chapter 3 and go through all the Exercises from the Chapter Playing With Numbers and apply the methods given in the Exercise to successfully solve any word-based or expression based numerical. You need frequent practice and timely revision to gain the speed in solving problems related to this Chapter. Vedantu aims to clear all your doubts and strengthen your knowledge of the different concepts covered in the Chapter by referring to our NCERT Solutions for Class 6 Maths Chapter 3.

Each numerical is explained step by step to make it easy for you to understand them and grasp the logic behind each step given in NCERT solution. Not only this but you will also find many helpful suggestions and alternative techniques to solve similar problems accurately and with more confidence.

NCERT Maths Chapter 3 - Playing with numbers is divided into seven sections namely –

Exercise 3.1 - Finding Factors and Multiples of the given numbers.

Exercise 3.2 - Recognising and finding out Prime and Composite Numbers.

Exercise 3.3 - Checking the divisibility of the numbers.

Exercise 3.4 - Finding common factors and multiples.

Exercise 3.5 - Finding the prime factors of the given numbers.

Exercise 3.6 - Finding the Highest Common Factors of the Given Numbers.

Exercise 3.7 - Applications of HCF and LCM

Each of these Exercises covers all the aforementioned topics thoroughly and is designed in such a way that you can yourself test your ability to solve the sums on the basis of the knowledge given. By practising the problems on a daily basis, you not only strengthen your grasp of the Chapter but it will further help you to approach different types of problems quickly and with more ease. If you require proper guidance and an effective approach towards solving the problems more conveniently, you can refer to our NCERT Solutions for Class 6 Maths Chapter 3. Doing so, you will be able to solve such problems more accurately and will develop effective shortcut techniques for solving the same with each practice. Improve your problem-solving skills by availing a PDF free download version of NCERT Maths Class 6 Chapter 3 and practise it frequently.

### Maths Class 6 Chapter 3 Exercise 3.7

Chapter 3 Exercise 3.7 of NCERT deals with the application of the lowest common multiple (LCM) and highest common factor (HCF). This Exercise contains numerical related to the types of situations in real life where we make use of the concept of HCF and LCM.

In this chapter, you will first learn to find out the common factors of two or three numbers by simple division or prime factorisation method and then you will be taught to conclude by recognising the highest common factor. You will also learn the major difference between LCM and HCF as most people often get confused between the two.

The previous Exercises primarily deal with solving simple problems related to HCF only. It consists of basic questions like finding HCF of two or three numbers and finding HCF of two consecutive or two co-prime numbers. The Exercise- 3.6 acts as a recapitulation of the concept of factorisation and divisibility of numbers which will help you solve application-based problems in Exercise 3.7. As discussed before the solutions of these questions are written by the experts as per the latest CBSE syllabus to maintain the standard.

Class 6 Maths Chapter 3 Exercise 3.7 comprises of eleven questions concerning the real-life scenarios. The solutions to these eleven questions are written in such a way that you will get an overview of the entire Chapter with better clarification. Our experts are experienced teachers who have been in the field of education since years and came across different kinds of conceptual mistakes students tend to make. Thus with an intention to rectify all the usual common errors by students, our teachers have come up with an interesting pattern to solve the numerical. They have broken the solution into different steps such that each step has the potential to clear the unattended doubt clouded in your mind. Teachers claim that after practising the given eleven solutions you can solve all the types of problems based on this topic as the main aim of the solution is to fetch you the concept in the easiest way.

### NCERT Solutions for Class 6 Maths Chapter 3 - Exercise 3.7

Exercise 3.7 is the last Exercise in Chapter 3. This Exercise concludes the Chapter by dealing with numerical based on all the topics taught in this Chapter. So, you need a thorough revision of the entire Chapter before you start solving the problems from this Exercise. Also, the familiarity of HCF and LCM and their usage in a given problem will help you to solve all the questions effectively related to this Chapter. Regardless, you should focus on attempting to invest your time and dedication towards practising Chapter 3 Class 6 Maths Exercises frequently as this is the basic and will help you in future calculations too.

Below is a summary of all the points you need to remember.

### Things to Remember:

Factors and Multiples: There are few numbers that can be written as the product of two or more numbers. Such numbers are called multiples and the numbers which the multiple is a product of are called factors.

Example: 2x3 = 6.

Here, 2 and 3 are factors of 6 and 6 is a multiple of 2 and 3.

Below are the facts about factors and multiples needed to keep in mind.

A multiple is a number of each of its factors

1 is a factor of every number

Every number is a factor of itself.

Every number is a multiple of itself

Every factor of a number is an exact divisor of the number

Multiple of a number can never be smaller than its factor

Number of factors of a multiple is always finite

Number of multiples of a factor is always infinite.

Prime and Composite numbers: All the numbers are divided into two categories:

Numbers having only 2 factors (i.e, 1 and the number itself) are called Prime Numbers.

Numbers having more than two factors are called Composite Numbers.

NOTE: “Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.”

Tests for Divisibility: There are few short-cut ways to find out if a number is divisible by a particular number or not. This saves your time by preventing you from doing long calculations

Divisibility by 10: If the last digit of a number is zero (i.e, at one’s place) then the number is divisible by 10.

Divisibility by 5: If the last digit of a number is zero or five then the number is divisible by 5.

Divisibility by 2: If the last digit of a number is an even number then the number is divisible by 2.

Divisibility by 3: If the sum of all the digits of a number is a multiple of 3 then the number is divisible by 3.

Divisibility by 6: If a number passes the divisibility test of both 2 and 3 then the number is divisible by 6 also.

Divisibility by 4: If the number formed by the last two digits of a particular number is divisible by 4 then the entire number is also divisible by 4.

Divisibility by 8: If the number formed by the last three digits of a particular number is divisible by 8 then the entire number is also divisible by 8.

Divisibility by 9: If the sum of all the digits of a number is a multiple of 9 then the number is divisible by 9.

Divisibility by 11: A number has odd places (i.e, ones, hundreds, etc) and even places (i.e, tens, thousands, etc). If the difference of the sum of the numbers in odd places and the sum of the numbers in even places is either a 0 or a multiple of 11 then the entire number is divisible by 11.

Co-Prime numbers: If two numbers have only one as a common factor then they are called co-prime numbers.

Prime Factorisation: It is the method of expressing a number as a product of its factors.

If a number ‘x’ is divisible by a number ‘y’ then ‘x’ is divisible by each of the factors of ‘y’ too.

If a number ‘x’ is divisible by two co-prime numbers then ‘x’ is divisible by the product of those two prime numbers also.

If two given numbers are divisible by a number ‘x’ then their sum and difference are also divisible by ‘x’.

HCF: To find the HCF of two or more numbers, first find the factors of the given numbers and then find out the highest of the common factors of those numbers.

LCM: To find the LCM of two or more numbers, first find the factors of the given numbers and then find out the lowest of the common multiples of those numbers.

### Exercise 3.7- Solutions to the Questions

The following offers a fair idea of the different types of questions that are covered in this Exercise and how referring to the Playing With Numbers Class 6 Solutions will benefit you.

Question 1 of Exercise 3.7: The solution to this question relates the theoretical concept to practical concept. In previous Exercises, you have learned to find out the HCF of two or three numbers. This question helps you apply the same method in real-life problems. Here, the weight of fertilizer in Renu’s bag is found out by using the factorization method of HCF. It looks like a potent technique to compute the ‘Highest Common Factor’ or HCF of the given numbers. Regardless, you must approach the sums smartly to get them solved within a few steps. Learn how to do the same by following the different approaches shown in our study solutions like NCERT Maths Class 6 Chapter 3 Exercise 2.6.

Also, by using the tricks we shared in our solutions, you can solve them correctly and enhance your scope of securing full marks. Download our latest NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 and others to learn such tricks and techniques immediately.

Question 2 of Exercise 3.7: This is a similar real scenario-based problem which needs to be solved by the method of finding the lowest common multiple (LCM) and not Highest Common Factor (HCF). Thus this question will help you to differentiate between the application of HCF and LCM as it is important for you to know when to use which method. With the help of these solutions, you can ace your upcoming examination quite easily with all the conveniences by incorporating our solutions named NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 into your revision plan. Download our study solutions from our learning portal with just a click and improve your learning experience without much ado. Download our app now to access quality study solutions formulated by experienced faculty across several cities of India that too for free!

Question 3 of Exercise 3.7: Question 3 and Question 1 are very different from each other wherein Question 1 you have to find the exact value of weight to measure the weight of fertilizers and in Question 3 you are asked to find the length of the tape to measure the dimensions of the room. Now, the most interesting part is that though both the questions seem different they still have something common in them so they both are to be solved by the same method. The only difference between both the questions are, the first question has only two numbers thus it needs to be solved by finding the HCF of both the numbers and the third question has three numbers so it is to be solved by finding the HCF of three numbers.

Find out how each of these questions covers the fundamental concepts of the Chapter by taking a quick look at our CBSE NCERT books for Class 6 Maths online. A quick reference followed by a prompt revision of NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 will come in handy for strengthening your grasp of the Chapter and related concepts.

Our solutions will serve as an excellent tool of year-round revision and will help you ace your Class tests and annual exam. Similarly, by adopting smart and shortcut techniques of solving numerical, you will pave the way for understanding fraction and decimal-based concepts better in higher Classes.

Question 4 of Exercise 3.7: This is comparatively a complicated problem and so it is divided into two parts. First part is the normal way of solving LCM with greater ease and the second part is a little complicated yet conceptual. The solution to this part is given in a very simple way to minimise the room for silly errors. To solve these kinds of numerical problems you should equip yourself with adequate knowledge about the rules to check the divisibility of numbers and prime factorization method. Thus to help you solve and understand the problems with greater convenience, we worked together and prepared NCERT Solutions for Class 6 Maths Chapter 3 to cover each problem of this Exercise in-depth.

Question 5 of Exercise 3.7: This question is solved to develop an effective approach to the numerical without much elaborating the method. This will boost your confidence and help you practise more NCERT Solutions with the skill of solving numerical problems faster and more accurately. To further help you improve your understanding of these types of questions of related concepts, numbers of solved examples from other sources are also offered to you by Vedantu. Moreover, a thorough step by step explanation is provided for each solved example. It helps you understand which methods are to be used to approach different types of questions to get a precise answer. These solutions will clear all the doubts and queries clouding your mind related to HCF and LCM before you move on to the unsolved Exercises. Doing so, you will get the opportunity to equip yourself with the requisite knowledge to solve the Exercises effectively so that you can start the other questions with greater confidence.

Other than that, to make the learning experience more fun and interesting, the Exercise also extends several interesting real-life examples and tidbits on all the concepts covered in this Chapter.

Question 6 of Exercise 3.7: This is an interesting question based on a scenario in which we come across almost every day. This tests your thinking ability to connect the solutions to the theoretical solutions. It is suggested that you should try solving the numerical on your own first and refer to the textbook if facing any problem. Once you are ready with your answer you can compare the method and answer given in NCERT Maths Chapter 3 Exercise 3.7 solutions available to help you understand the difference in your approach to the problem and the standard approach to the same problem. Vedantu also offers online Classes for you to solve all your doubts.

Question 7 of Exercise 3.7: This question is the best example for you if you are looking for alternative shortcut techniques to solve similar numerical problems. To know how to give a different approach to the same numerical problems take a quick look at the NCERT Solutions of Chapter 3 Maths Exercise 3.7 online to get them cleared immediately. In case, you harbour any doubts about this Chapter you can ask Vedantu’s expert teachers online. You would not even require any teacher if you practise these kinds of sums twice or thrice because it will provide you with a better understanding of the Chapter. It is familiar with the concepts used earlier to solve the problems. By solving these questions step by step you can master the concept and application of prime numbers, composite numbers and factorization method. This question aims to broaden the application of the method of prime factorisation to find the LCM and HCF of any given numbers. Moreover, it is advisable to refresh your knowledge of prime factorisation method by revising the Chapter once more before you begin your final revision before Board Examination.

Question 8 of Exercise 3.7: This is the first time you will be introduced to the concept of remainder while solving problems based on HCF and LCM. To be able to solve this particular problem accurately, you must know how to calculate dividend from the given divisor and remainder. These kinds of Word problems seem tricky if one is not aware of the techniques that go into solving them. You should know all the essential tips, tricks and techniques to answer word problem-based questions with more practice and thorough revision. Follow the techniques shared by our expert professionals to solve these type of questions and make the most of your learning with regular practice. Study the shortcut techniques from up close by taking a quick look at the NCERT Solutions for Class 10 Maths Chapter 1 offered online in its PDF format. Considering that such solutions can be downloaded to your phone, you are free to carry the solutions anywhere and read at your convenience. With this digital blessing of education, learning becomes not just convenient, but fun with us, at Vedantu.

Question 9 of Exercise 3.7: This type of question needs a lot of steps to be followed to reach the solution and therefore, it involves the risk of making a lot of silly mistakes. You can minimise the scope of making such mistakes by acquiring a better insight into each of these steps. In-depth understanding of these steps would help you clear all your lingering doubts easily.

As said this question is again divided into 3 parts.

First part is a simple LCM of three numbers each of having two digits. Therefore, the resultant LCM will be of three digits only. But as per the question we need to find a four-digit number.

In the second part, the smallest four-digit number ( i.e, 1000) is divided by the resultant of the first part of the solution. Then remainder needs to be subtracted by the divider.

Third part: To get the final answer the smallest four-digit number is again divided by the difference.

This will give you the smallest four-digit number divisible by all the three given numbers. The solution of this question is broken down and explained very clearly so that even if you get any other solution with a different number of digits or more than three numbers, then you will easily be able to do it.

Some of the numerical problems like this one involve lengthy steps and complex approaches, which is why it is important to be well-versed with the fundamentals of the concepts they are based on. Make it a point to start solving the Exercise from the beginning to strengthen your grasp on the fundamental concepts of the Chapter for solving them effectively. Our study guides like NCERT Solutions for Class 6 Maths Chapter 3 have been engineered by the experts keeping in mind the needs and requirements of both the CBSE board of examination and students.

The well-crafted solutions are updated regularly and incorporate the latest inclusions and suggestions. All of this makes the study materials student-friendly like our study material of Chapter 3 for Maths Class 6 is a potent guide towards success. Thus, enhance your chance of scoring high in your Class 6 board examination 2019-2020 by availing NCERT Solutions for Class 6 Maths Chapter 3 from us!

Question 10 and 11 of Exercise 3.7: Last two questions are very easy as it aims to bring you back to the basic method of finding the lowest common multiple.

If you are facing any kind of problem to solve any question from this Chapter then you can check our online Classes for Class 10 Maths or you can also attend doubt sessions taken by our extremely qualified and experienced teachers. These Classes have been made available online. You can also refer to our latest Playing with Numbers Exercises with Solutions and NCERT Class 10 Maths book PDF to refresh your fundamentals of the Chapter and to revise them at leisure. It must be noted that the numerical problems of this Chapter are mostly based on specific rules namely divisibility rules and other associated concepts. Application of such rules are discussed in details in this Chapter, and a necessary explanation of each step has been provided wherever required.

By studying the pattern we can say that each topic is followed by a compact Exercise which aims to test the knowledge and fetch you the depth of understanding of different topics and concepts that are introduced to you in this Chapter.

Chapter 3 of NCERT Maths Book covers all the topics mentioned above.to bring a detailed NCERT Solution for Class 6 with step by step solutions. Vedantu is making it easier for you to download the most preferred book and access them easily in order to assist yourself academically and score higher in exams. The Solved PDF book for Playing with Numbers has right solution approach to the problems. It also tests your learning and retention skills with complicated problems.

### Importance of Exercise 3.7

Maths is a subject of great understanding and learning the application of concepts for various situations in our life. The co-relation between HCF and LCM states that the product of HCF and LCM of two numbers is always equal to the actual product of two numbers. But the product of HCF and LCM of two or three or more numbers can never be equal to the product of the numbers. This Exercise will take you through various real applications of HCF and LCM.

#### HCF Can be Used:

To divide things into smaller sections

To distribute items among large groups of people.

To arrange something into rows and groups.

LCM Can be Used:

To find out when an event will repeat itself again.

To purchase multiple items in order to have enough.

So far, you have been solving questions where a natural number has to be multiplied with a decimal. However, it this particular question you will be multiplying two decimals to avail accurate solution.

Find out how to solve this particular problem by glancing through the solved examples accompanying Class 6 Maths Chapter 3 Exercise 3.7. It will come in handy for solving similar problems and will also offer you a brief insight into the step by step approach of the same.

You can also take a cue from our NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 and practice the solved problems. Also, our solved study materials can be used as a study guide, with the help of which you can check both your answers and steps applied for each sum. It will further help you to identify your mistakes successfully and will help you to rectify them easily.

Over time, this will help you boost your confidence and further allow you to pick up on the correct approach of solving important questions in exams. Nonetheless, you must ensure that you practice the sums and refer to its solutions regularly to maintain consistency. It will further help you to remember the steps correctly and will allow you to take your exam preparation to the next level.

So, download the latest NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 to learn about the latest inclusions and explanations in detail!

### Vedantu – The Proven Path to Success

At Vedantu, we are an ardent believer of smart work and tend to help those who believe in the same. Our faculty comprises experienced teaching professionals who are adept at learning and possess a greater passion for imparting the same.

All our study solutions have been compiled, keeping in mind the latest curriculum of the CBSE board and are also updated on a timely basis. It ensures that all the needs and requirements of both teachers and students are taken into account and met accordingly. You can also access our Chapter-based solutions for other Exercises as well to gain valuable insight into how to solve sums that are based on these topics. So, by including our study solutions into your study regime to get all your lingering doubts about fractions and decimals absolved effectively.

Each sum of such Exercises has been treated with the utmost care and provides a suitable explanation behind every Mathematical operation. Also, to beat the burden of the last-minute exam revisions, you need a study guide like NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7. It makes the entire process seem less stressful and facilitates quality learning by allowing you to retain information for a long time.

So, make your revision process fun, easy and convenient by adopting effective tips and techniques from our solutions and use them for solving essential questions in exams.

Our study materials like NCERT Solutions for Class 10 Maths Chapter 1 are well-crafted and can be easily accessed by from our online learning portal with utmost convenience. Further, our user-friendly interface and easy to navigate design facilitates user experience and makes it a fun, interesting and feasible way of gaining knowledge. Download Vedantu App now and get prepped for your CBSE Board Examination immediately!