Is Every Rational Number a Natural Number?

Every Natural Number is a Rational Number but it’s not the same in the case of Rational Numbers. There is no such thing when it comes to a Rational Number it may or may not be a Natural Number.

We know that 1 = 1/1, 2 = 2/1, ….

In fact, we can say a natural number n can be expressed as n = n/1 which is nothing but the quotient of two integers. Therefore, every natural number is a Rational Number.

On the other hand, 5/2, 3/5, 2/7, 4/20, etc. are all Rational Numbers but aren’t natural numbers.

Therefore, every natural number is a Rational Number but a Rational Number need not be a Natural Number.

Determine Whether the Following Rational Numbers are Natural Numbers or Not

(i) 5/2

5/ 2 is not a natural number.

(ii) 8/2

8/2 is a natural number as on simplifying it we get the result as 4 which is a natural number.

(iii) -20/5

-20/5 isn’t a natural number as on further simplifying we get the result -4 which is an integer but not a natural number.

(iv) -6/-3

-6/-3 is a natural number as we will get the result 2 on simplification which is a natural number.

(v) 1/5

1/5 is not a natural number.

(vi) 0/3

0/3 is not a natural number since 0/3 =0 and 0 is not a natural number.

(vii) 5/5

5/5 is a natural number on simplifying to its lowest term we get 1/1 = 1 and 1 is a natural number.

(viii) 27/45

27/45 is not a natural number as we get 3/5 on reducing to its lowest term which is a rational number but not a natural number.

Thus, by looking after the instances above we can state that Not every Rational Number is a Natural Number.

What is Rational Numbers

What are Rational Numbers?

If you are willing to know What are Rational Numbers and general representation of it have a look at the further modules. The number that can be expressed in the form of a fraction a/b where a, b are integers and the denominator b is non zero is called a rational number.

You can also say that a Rational Number is a number that can be expressed as the quotient of two integers having the condition where the divisor is non zero.

Numerator and Denominator: If a/b is a rational number then integer a is the numerator and b is the denominator.

Examples: 3/2, 8/5, -14/3, -11/5 are all Rational Numbers as they have integers in numerators and denominators and denominators are non zero.

Total Surface Area of Cylinder, Volume of a Cylinder Formulas, Examples

Total Surface Area of a Cylinder: Let us carry out a small activity. Take a ring. We know that ring is circular in shape. Now place another ring on the first ring and so on. You will see that when you place such rings, the structure that you get is a cylinder, which is three dimensional. Let us learn about the volume of a Cylinder and Total Surface area of a Cylinder in this chapter.

Cylinder

Any solid which has two circular ends along with a curved surface area is said to be a cylinder. It has a curved surface in the middle. The base and the top surface are identical. That means the bases are always parallel and congruent to each other. Cylinder has no vertices.

Cylinder
Cylinder

Total Surface Area of Cylinder

The total surface area of cylinder is the area of the circles plus the area of the round section in the midle. The Surface Area (SA) is found with the formula: SA = 2Ï€r2 + 2Ï€rh

Total Surface Area of Cylinder 1
Total Surface Area of Cylinder 2

Find the Total Surface Area of each figure. Remember, d = 2r. Type your answers in the comment box

(i)
formula of total surface area of cylinder
(ii)
Surface Area of Cylinder 1
(iii)
Surface Area of Hollow Cylinder 9
(iv)
total surface area of a cylinder calculator
(v)
total surface area of a cylinder formula
(vi)
total surface area of a cylinder
(vii)
total surface area of cylinder formula
(viii)
total surface area of hollow cylinder
(ix)
total surface area of right circular cylinder

Calculate the total surface area of a cylinder for the given problems and type your answers in the comment box.

Volume of Cylinder

The Volume of Cylinder is the product of the area of the base (B) times the height. The formula for the Volume of the Cylinder is V = B x H. Volume is expressed in cubic units , or units3
Volume of a Cylinder

Find the Volume of each figure. Remember, d = 2r. Type your answers in the comment box

(i)
Volume of Cylinder 9
(ii)
Volume of Cylinder 8
(iii)

(iv)
volume of a cylinder in litres
(v)
volume of a cylinder formula
(vi)
volume of a cylinder calculator
(vii)
volume of a circle cylinder
(viii)
how to calculate volume of a cylinder
(ix)
formula for volume of a cylinder

Calculate the volume of cylinder for the given problems and type your answers in the comment box.

Maths

Vedic Maths

Vedic Maths | 2 Second Maths Tricks

Vedic Maths: Vedic Maths book was written by Swami Bharati Krishna Tirtha, who was an Indian monk. It was first published in 1965. Veda is basically a Sanskrit word, which means Knowledge. It has a list of mental calculation techniques which are said to be present in the Vedas. This book is actually mentioned as Vedic Mathematics. This has been criticized by academics, who have also opposed its formation in the Indian school curriculum.

Vedic Maths Tricks

This Mathematics is a collection of Techniques/Sutras to solve arithmetical problems in an easy and faster way. It consists of 16 Sutras and 13 sub-sutras which can be used for questions involved in arithmetic, algebra, geometry, calculus, conic. The sutras are basically the formulas which we use in the mathematical calculation.

Vedic Maths book was previously included in the syllabus of Madhya Pradesh and Uttar Pradesh Board affiliated Schools. Some of the schools and organizations run by Hindu nationalist groups, also have included methods in their syllabus, even those groups which are outside India. The Hindu nationalists also tried to include these curricula in the NCERT books.

In earlier classes, we studied that during the period of dwelling in Sringeri Matha, Mysore Swami Bharati Krishna Tirtha did an arduous ascetic fervor for eight years. In the highest position of accomplishment, he got the perceptive vision of mathematics formulae mentioned in Vedas, the holiest scriptures and expressed his spiritual feelings in the form of Mantras (formulae).
These mantras were named Vedic Mathematical Formulae, which is exactly true. According to the scholars of Vedas, the knowledge of Vedas is beyond human power. The knowledge of Vedas cannot be obtained only by thinking or learning. It is an accomplish-feeling and expression it in the form of Mantras. In this perspective, the formulae formed by Swamiji are Vedic Mathematics Mantras.

Importance of Vedic Mathematics
To get a proper solution of any mathematical problems when a regular and inure practice of Vedic mathematical formulae is made, the concentration and memory is developed also a fierceness is seemed in his thinking and meditation. Due to freshness, simplicity and agreeability of Vedic mathematics create a feeling of inquisitiveness in the human heart.
The inquisitiveness makes him alert and aware and steps by step his inner soul begin to be vigilant. In fact, Vedic mathematics is a science of making someones inner should awake. This awareness of inner soul becomes the basis of the development of the human mind and his personality.

Important Definitions Related to Vedic Mathematics

  • Preceding Number (Purvena): The previous number is called the preceding number e.g., in 785, the Previous number of 5 is 8.
  • Param Mitra Number or Complementary Number: Two numbers having the sum of 10 are called Param Mitra number or complementary numbers.
    e.g.,Param Mitra number of 2 = 8
    Param Mitra number of 5 = 5
    Param Mitra number of 9 = 1
  • Disjunctor Number: The number which is subtracted from other numbers.
  • Detachable Number: The number from which a number is subtracted.
  • Formula Ekadhikena Purvene: To obtain the next number to a given number.
    A dot (.) is put on the given digit or number of ekadhik to show this sutra e.g., ekadhik of 15 = 15 + 1 = 16
    Ekadhik of digit 5 in 2536 = 2636.
  • Nikhilum Digit: Except unit digit, all the digits are called Nikhilam digit e.g., in 963, 9, 6 are Nikhilam digit.
  • Charam Digit: Unit digit of a number is called charam digit e,g., 3 is charam digit in 963.
  • Vinculum number (negative digits): The digit having bar above on it is called vinculum number. e.g., -5 = \(\bar { 5 }\).
  • Base: 10, 100, 1000, etc. are called base. These are multiple of 10.
  • Sub-base: Sub-base is the multiplier of base mostly it is a digit ends with zero, e.g., In 10 × 8, 10 is based and 8 is sub-base.
  • Deviation: When base or sub-base is subtracted from the given number the difference is called deviation.
    Deviation = Number – Base (or sub-base)
    If number > base/sub-base, then deviation will be +ve
    If number < base/sub-base, then deviation will be -ve.
    A deviation has as many digits as the number of zeros in the base.
  • Param Digit: Digit 9 is called Param digit or Brahm digit.
  • Whole Number: Ten is called the whole number.

Main Operations: Practice and Extension
Addition: In our previous class, we learned and practiced the addition of real numbers with the help of Sutra Ekadhikena Poorvena. In the practice, problems related to perfect numbers (i.e., Measurement units km-m) were solved. In fact, using the Sutra Ekadhikena Purvena all the problems of addition can be solved, measurement unit, money (rupees and paisa), weight (kilogram-gram) capacitance (liter-milliliter), time (hours, minutes, seconds), decimal fractions real numbers, distance (km, m, cm), etc.
Step 1: Informing column, according to the unit of measurement there is certain numbers in unit column. For example one rupee = 100 paise, so there will be two unit columns. 5 paisa will be written as paisa 05. Similarly, 1 kilometer = 1000 meters, so there will be three unit columns in problem and 84 meters will be written as 084 meters, so there will be three unit columns in problem and 84 meters will be written as 084 meters.
Step 2: Having completed forming the column, operation of addition is finished with the help of formula.
Step 3: When the sums are related to time whose unit is an hour, minute and second, the base in the first column (10) and in second 6 should be taken. In the column of an hour the base is always 10.
Example 1.
Vedic Maths Addition and Subtraction Tricks 1
Hints:
(i) 65 gram and 85 grams will be written as 065 grams ans 085 grams.
(ii) Start adding from the top of unit column.
(iii) 5 + 5 = 10, so mark the more than, sign above 8, it comes before 5, remainder = 10 – 10 = 0
(iv) Write 0 + 2 + 5 = 7 at the place of answer.
(v) Keep it up.

Addition Orally (Sutra Ekadikena Poorvena + Soonyant Sankhya)
After some regular practice of above sutra, sums with large numbers can be solved rapidly or orally. Use of a round figure is a special method of ancient Indian Mathematics which is easier and very affected in the operation of addition. In the method, two digits at one and tens place are added in a special manner. Three digits at one’s tens and hundreds can also be added accordingly.
Method: Make one number of two in a round figure maintain its deficiency in the second given number. Add obtained new numbers. If die obtained sum result is more than 100, put an ekadhikena mark above it. Add the remainder into next number. At last, write final remainder as an answer. Revise this operation for the next columns. Look at the example given below:
Example 1.
Add 35 and 58.
Solution:
Convert these 58 into a round figure we need 2.
This deficiency will maintain in 35.
So 35 + 58 = 33 + (2 + 58) = 33 + 60 = 93
Example 2.
Add 19 and 65.
Solution:
19 + 65 = (19 + 1) + 64 = 20 + 64 = 84
Note: Similarly, many numbers can be added.
Example 3.
Add.
Vedic Maths Addition and Subtraction Tricks 1
Hints:
(i) 98 + 89 = 98 + 2 + 87 = 100 + 87 = 187
On the digit 7 before 89 more than one mark.
(ii) Remainder 87 + 15 = 87 + 3 + 12 = 90 + 12 = 102
On the digit 7 before 15, more than mark.
(iii) Remainder 02 + 37 + 39 and 39 + 76 = 35 + 4 + 76 = 35 + 80 = 15 + 20 + 80 = 115
above the digit 9 before 76 more than one mark and 15 is put at the place of answer.
(iv) Now complete the operation as given above.

Example 4.
Add
Vedic Maths Addition and Subtraction Tricks 1
Hints
(i) 34 + 59 = 33 + 1 + 59 = 33 + 60 = 93
(ii) Remainder 93 + 32 = 93 + 7 + 25 = 100 + 25 = 125.
So more than one mark on digit 9 in third column.
(iii) Remainder 25 + 47 = 22 + 3 + 47 = 22 + 50 = 72
(iv) Remaining operation given above.

Subtraction:
In the previous class, we used two Vedic methods to solve the subtractions.
1. The method based on the Sutra Ekadhikena Poorvena Mitra Ank.
2. The method based on the Sutra Ekanyunen Poorvena Param Mitram,
Every problem of subtraction of measure unit of the real number can be solved with the help of the first method.
So, we consider again on this method. We know that two digits of a number are Param Mitra (close friend) to each other if their sum is equal to or more than ten.
The number from which a number is subtracted is called detachable number and the number which is subtracted is called disjunctor. The method can be cleared in an easy way with the help of examples given below.
Example 1.
Subtract by Vedic Method.
Vedic Maths Addition and Subtraction Tricks 1
Hint
(i) 3 cannot be subtracted from 0, so add 7 the Param Mitra Anka of 3, to 0. Write the sum below at the answer place and mark more than one sign on previous digit 6.
(ii) 6 = 7 cannot be subtracted from 0, so add the digit 3, the Param Mitra and of 7 into 0. Write it at the answer place and mark the sign of more than one on previous digit 2.
(iii) Write 8 – \(\dot { 2 }\) = 5, at the answer place, So the remainder = 537.
Example 2.
Subtract the following by Vedic Method:
Vedic Maths Addition and Subtraction Tricks 1
Hints:
(i) Measurement unit in time column-wise base is different.
(ii) In the column of minute and second there will be two bases.
(a) Base in the unit column of both = 10
(b) Base in tens column of both = 6
(iii) Base in the column of hour = 10
(iv) The base of getting complement digits in the tens column of minute and second = 6 and in remaining base 10.
Hence, the answer is 25H 57min. 45sec.
Example 3.
Subtract by Vedic Method:
Vedic Maths Addition and Subtraction Tricks 1
Hints:
(i) Arrange the column numbers in meter and centimeter.
(ii) Centimeter column: 6 cannot be subtracted from 5, thus add 4 the Param Mitra Ank of 6 to 5.
(iii) Write the sum = 9 at the answer place and mark more than one sign on 4 the pre dissociator digit.
(iv) \(\dot { 4 }\) = 5 cannot be subtracted from 3, so add 5 the param Mitra and to 3, the pre-dissociator digit.
(v) Write sum = 8 below and mark the sign of more than one on the pre-dissociator 5.
(vi) Write 7 – \(\dot { 5 }\) = 1 below
(vii) 7 cannot be subtracted from 6, so adding 6 + 3 = 9 below and mark the sign of more than one the pre-disjunctor 3.
(viii) Write 4 – \(\dot { 3 }\) = 0 below.
(ix) All the next operations will be done in the same way.
Remainder = 9 km 91 m 89 cm.

Multiplication:
In the previous class, we studied four methods of multiplication operations based on the Vedic Sutras. We should have the perfect practice of these sutras so that we might be able to choose a proper sutra for a proper solution of any type of problem-related to multiplication at once. Look at the example given below.
Example 1.
Multiply the following :
686 × 614 (Sutra Ekadhikena purvena)
Solution:
686 × 614
sum = 100 = 6 × 7 / 86 × 14 = 42 / 1204 = 421204
Hints:
(i) Sum of lost’digit = 86 + 14 = 100
(ii) Remaining nikhilam digits are mutually equal = 5
(iii) Four digits in right side = 1204.
Hence, 686 × 614 = 421204.

Example 2.
Choose the best Sutra to get an easy and rapid solution of588 × 512.
Solution (i):
May the Sutra Ekadhikena Poorvena be the best formula for solving it?
The sum of the digits at the place of ones and tens = 88 + 12 = 100 and two Nikhilam digits are 5 each. So the Sutra of effective here.
According to the Sutra
588 × 512 = 5 × 6 / 88 × 12 (Four digits in RHS)
= 301056
Solution (ii): Justification of the Sutra Nikhilam Upadhara
Vedic Maths Addition and Subtraction Tricks 1
Hints:
(i) Base = 100
(ii) Sub-base = 100 × 5
(iii) Sub-base digit = 5
(iv) Differences = +88 and +12
(v) Two digits in RHS and the Sutra is effective
Solution (iii): The Sutra Ekanayunena Purvena cannot be applied for finding the solution of 588 × 512, as there is not a digit of 9 in both numbers.
Solution (iv): 588 × 512 can be solved by the Sutra Urdhava-tiryagbhyam
There are three columns in the problem. So five groups will be formed i.e., having obtained five products, they will be written in a special manner and then added.
Vedic Maths Addition and Subtraction Tricks 1
Result 1. Observing the first, second and fourth solutions it is certain that 588 × 512 = 301056.
2. In the first solution, we get the result easier, therefore Sutra Ekadhikena Poorvena is the best.

Example 3.
Which of the Sutra is the best to solve 842 × 858?
Solution:
(i) In this problem the Sutra Ekadhikena Poorvena is not effective as the product of RHS i.e., 42 × 58 cannot be obtained easily.
(ii) The Sutra Nikhilam Adhara cannot be effective here as if the base 1000 is considered, then the difference respectively will be -158 and -142. The Sutra Nikhilam-Upadhara is not effective here as Upadhar (Sub-base) = 800, we get the difference 42 and 58 respectively.
(iii) The Sutra Ekanyunena Poorvena is not effective here.
(iv) The Sutra Urdhva-tiryagbhyam is effective and the best here. If the number is large, the calculation is difficult. So we can think over this new formula.
(v) The New option: To get the solution of 842 × 858, two Sutras are used respectively, first the Sutra Ekadhikena Poorvena and then the Sutra Urdhva-tiryagbhyam.
Vedic Maths Addition and Subtraction Tricks 1
(v) Extension of the Operation of Multiplication (Sutra Urdhvatiryagbhyam + Viloknam)
Using the formulae Urdhva-tiryagbhyam and vinculum, the product of two larger number can be obtained id easily. Using the vinculum siitra first of all. The digits more than 5 are converted into smaller digits i.e., (0, 1, 2, 3, 4, 5) after that the product is calculated with the Sutra Urdhva-tiryagbhyam and at last obtained product, including negative sign is again turned into dying normal digits. this method can be cleared by observing the examples given below.
Example 1.
Solve 842 × 858
Vedic Maths Addition and Subtraction Tricks 1
Hints
(i) Convert greater digits into smaller ones.
(ii) Multiple by Urdhva-tiryagbhyam Sutra.
(iii) Turn the obtained negative signs of the product in the normal digits by the method of Nikhilam.

Example 2.
Find the value of 966 × 973.
Vedic Maths Addition and Subtraction Tricks 1
Note: Using the Sutra Urdhva-tiryagbhyam, we should try to get the product of large number orally and express it into one line.

Division:
In the previous class, the following three methods were used for division in detail:
(a) Sutra Nikhilam
The Nikhilam Sutra based method is effective only when the digits of divisor are greater than 5 and the complementary number of division to the respect of the base 10 or the power of 10 is known. In this method, the main operation is performed by the complementary number.

(b) Sutra Paravartya Yojayet
Sutra Paravartya Yojayet will be effective only when the digits of divisor are smaller than 5 or can be put also the digit from the left-hand side is 1 or can be brought and the base = 10, or the difference of the divisor comparably the power of 10 (not sub-base). Only this method out of three can be used for the division in algebra.

(c) Sutra Urdhva-Tiryabhyam
By the method of dhvanjank based on the Urdhva-tiryagbhyam, every type of problem-related to division can be solved. In this method, it is most important to choose the Mukhyank and dhvajank properly. The dhvanjank may have so many digits as you wish. There may be more than one digits in Mukhyank. As many digits must be put in the dividend at the place of ones in die third section as there are in the dhvajank and remaining in the middle section. We shall clear the method with the help of the following examples:
Example 1.
Solve by the dhvajank method
989765 ÷ 87
Vedic Maths Addition and Subtraction Tricks 1
Hints:
(i) Divisior = 87, Mukhyank = 8 and dhvajank = 7
(ii) In third column one digit of dividend = 5.
(iii) 9 ÷ 8, first digit of quotient = 1 and remainder = 1.
(iv) New dividend = 18, the corrected dividend.
(v) 11 ÷ 8, second digit of quotient = 1, remainder = 3.
(vi) New dividend = 37, the corrected dividend.
(vii) 30 ÷ 8, third digit of quotient = 3, remainder = 6.
(viii) New dividend = 66, the corrected dividend.
(ix) 45 ÷ 8, forth digit of quotient = 5, remainder = 5.
(x) New dividend = 55, the corrected dividend or the final remainder
Quotient = 1135, remainder = 20

Example 2.
Solve: 13579 ÷ 975 (dhvajank method)
Solution: 13579 ÷ 975
Vedic Maths Addition and Subtraction Tricks 1
Hint:
(i) 13 ÷ 9, first digit of quotient = 1, remainder = 4.
(ii) New dividend = 45,corrected dividend = 45 – 1 × 7 = 38
(iii) 38 ÷ 9, second digit of quotient = 4, remainder = 2.
(iv) New dividend = 27
Corrected dividend = 27 – (4 × 7 + 1 × 5) = 27 – 33 = -6
Since, we get the corrected dividend is negative so the second digit of quotient must be 3 instead of 4.
This is why the terms (iii) and (iv) are rejectable.
(v) Again 38 ÷ 9, the second digit of quotient = 3, remainder = 11
(vi) New dividend = 1179. So the corrected dividend or the final dividend or final remainder.
= 1179 – (3 × 7 + 1 × 5) × 10 – 3 × 5 = 1179 – 260 – 15 = 904
Hence, the quotient = 13 and remainder = 904

Example 3.
21015 ÷ 879 (dhvanjank method)
Solution:
Divisor = 897, Mukhyank = 8 and dhvanjank = 79.
Since there are larger digits in the dhvanjank, so the divisor 879 will be converted into smaller digits comparatively by Nikhilam (Vinculum) Method
Vedic Maths Addition and Subtraction Tricks 1
Hints:
(i) 21 ÷ 9, first digit of quotient = 2, remainder = 3.
(ii) New dividend = 309, corrected dividend = 30 – 2 × \(\bar { 2 }\) = 34
(iii) 34 ÷ 9, second digit of quotient = 3, remainder = 7.
(iv) New dividend = 715, corrected dividend or final remainder
= 715 – (3 × \(\bar { 2 }\) + 2 × \(\bar { 1 }\))10 – 3 × \(\bar { 1 }\)
= 715 + 80 + 3 = 798
Hence, quotient = 23, remainder = 798.

Example 4.
7453 ÷ 79 (dhwanjank method)
Hints:
(i) Divisor 79 = 8 \(\bar { 1 }\), mukhyank = 8, dhvanjank = \(\bar { 1 }\)
(ii) 74 ÷ 8, first digit of quotient = 9, remainder = 2
(iii) New dividend = 25, corrected dividend = 25 + 9 = 34
(iv) 34 ÷ 8, second digit of quotient = 4, remainder = 2.
(v) New dividend or final remainder = 23 + 4 = 27.
Hence quotient = 94, remainder = 27
Note:
1. See the construct (iii)
New dividend = 25, corrected dividend = 25 – \(\bar { 9 }\) × 1 = 25 + 9 = 34.
New dividend + pre digit of the quotient.
2. If 9 is at the place of ones in the divisor, then the corrected dividend = New dividend = pre quotient digit may be taken.
3. In the problem in the which the one’s place unit, of divisor, is 1, the corrected dividend = new dividend + the digit of pre quotient is taken.
4. There is no need to write the hints for the two types of above problems.

Example 5.
Solve 43758972 ÷ 81 (dhvajank method)
Solution:
43758972 ÷ 81
Vedic Maths Addition and Subtraction Tricks 1

Addition and Subtraction Vedic Maths Tricks

Vedic Maths Addition and Subtraction Tricks 1

Vedic Maths Addition and Subtraction Tricks 2 Vedic Maths Addition and Subtraction Tricks 3 Vedic Maths Addition and Subtraction Tricks 4 Vedic Maths Addition and Subtraction Tricks 5 Vedic Maths Addition and Subtraction Tricks 6

Vedic Maths Advantages and Uses

Once the student understands the system of mental mathematics, they will become more creative and start thinking logically. This maths is very flexible for them. The students can easily play with numbers with the help of this system.

Regular mathematical methods are sometimes complex and time-consuming. But if use Vedic Mathematic’s Procedure and Techniques, some of the calculations such as, sets of given data,  can be done very fast. Some of the more useful advantages of this Mathematics are;

  • Its more than 1700% times faster than General Math. Thus, it could be considered as the World’s Fastest.
    It helps a child to lose the Math fear from his mind. Usually, students are scared of doing mathematical calculation because of the logic they have to use to solve it. Thus, this ancient Maths will help to solve the problems in the easiest way.
    It helps to increase the thinking capability and intelligence, along with a sharpening of the mind.
  • Helps in increasing the speed and giving accurate answers.
  • Students feel very confident about this subject, after improving their memory power.
  • Students get interested more in numbers, they just have to apply their skills and have knowledge of tables to learn this.
  • These maths calculation are of great use while preparing for competitive exams.

Vedic Maths Sutras

As mentioned before, it consists of 16 sutras. Below are the names of the Sutras and Upa sutras with their meaning and corollary.

Name Upa Sutra Meaning Corollary
Ekadhikena Purvena Anurupyena By one more than the previous one Proportionately
Nikhilam Navatashcaramam Dashatah Sisyate Sesasamjnah All from 9 and the last from 10 The Remainder Remains Constant
Urdhva-Tiryagbyham Adyamadyenantyamantyena Vertically and crosswise First by the First and the Last by the Last
Paraavartya Yojayet Kevalaih Saptakam Gunyat Transpose and adjust For 7 the Multiplicand is 143
Shunyam Saamyasamuccaye Vestanam When the sum is the same that sum is zero By Osculation
Anurupye Shunyamanyat Yavadunam Tavadunam If one is in ratio, the other is zero Lessen by the Deficiency
Sankalana-vyavakalanabhyam Yavadunam Tavadunikritya Varga Yojayet By addition and by subtraction Whatever the Deficiency lessen by that amount and set up the Square of the Deficiency
Puranapuranabyham Antyayordashake’pi By the completion or non-completion Last Totalling 10
Chalana-Kalanabyham Antyayoreva Differences and Similarities Only the Last Terms
Yaavadunam Samuccayagunitah Whatever the extent of its deficiency The Sum of the Products
Vyashtisamanstih Lopanasthapanabhyam Part and Whole By Alternate Elimination and Retention
Shesanyankena Charamena Vilokanam The remainders by the last digit By Mere Observation
Sopaantyadvayamantyam Gunitasamuccayah Samuccayagunitah The ultimate and twice the penultimate The Product of the Sum is the Sum of the Products
Ekanyunena Purvena Dhvajanka By one less than the previous one. On the Flag
Gunitasamuchyah Dwandwa Yogiji The product of the sum is equal to the sum of the product.
Gunakasamuchyah Adyam Antyam Madhyam The factors of the sum are equal to the sum of the factors

We hope the detailed article on Vedic Maths is helpful. If you have any doubt regarding this article or Vedic Maths, drop your comments in the comment section below and we will get back to you as soon as possible.

FAQs on Vedic Maths

1. What is Vedic Mathematics?

Vedic Maths is the world’s fastest mental maths system. It helps you calculate faster.

2. How useful is Vedic Maths?

Vedic Maths improves mentality and promotes creativity. The ease and simplicity of Vedic Mathematics mean that calculations can be carried out mentally. All these features of Vedic math encourage students to be creative in doing their math.

3. What are the best ways to learn Vedic Maths?

Learn about the Vedic Maths Tricks, Primary Definitions, and the importance of it using the direct links available on our page. You can view or download and use them as a reference during your preparation.

4. Which are good books to learn Vedic Mathematics?

Aspirants can look for some of the best books to learn on Vedic Mathematics by accessing our page.

 

HCF AND LCM WORD PROBLEMS

HCF and LCM Word Problems : In this section, we will learn how to solve word problems involving highest common factor and lowest common multiple.

HCF and LCM Word Problems

Problem 1 :

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together? (excluding the one at start)

Solution :

For example, let the two bells toll after every 3 secs and 4 secs respectively.

Then the first bell tolls after every 3, 6, 9, 12 seconds…
Like this, the second bell tolls after every 4, 8, 12 seconds…

So, if the two bell toll together now, again they will toll together after 12 seconds. This 12 seconds is nothing but the L.C.M of 3 seconds and 4 seconds

The same thing happened in our problem. To find the time, when they will all toll together, we have to find the L.C.M of (2, 4, 8, 6, 10, 12).

L.C.M of (2, 4, 8, 6, 10, 12) is 120 seconds  =  2 minutes.

So, after every two minutes, all the bell will toll together.

For example, in 10 minutes, they toll together :

10/2  =  5 times

That is, after 2,4,6,8,10 minutes. It does not include the one at the start

Similarly, in 30 minutes, they toll together :

30/2  =  15 times

(excluding one at the start).

Problem 2 :

The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 8:20:00 hrs, when will they again change simultaneously ?

Solution :

For example, let the two signals change after every 3 secs and 4 secs respectively.

Then the first signal changes after 3, 6, 9, 12 seconds…

Like this, the second signal changes after 4, 8, 12 seconds…

So, if the two signals change simultaneously now, again they will change simultaneously after 12 seconds. This 12 seconds is nothing but the L.C.M of 3 seconds and 4 seconds

The same thing happened in our problem. To find the time, when they will all change simultaneously, we have to find the L.C.M of (48, 72, 108).

L.C.M of (48,72,108) is 432 seconds  =  7 min 12 sec

So, after every 7 min 12 sec, all the signals will change simultaneously.

At 8:20:00 hrs, if all the three signals change simultaneously, again they will change simultaneously after 7 min 12 sec. That is at 8:27:12 hrs.

Hence, three signals will change simultaneously at 8:27:12 seconds.

Problem 3 :

A merchant has 120 ltrs of and 180 ltrs of two kinds of oil. He wants the sell oil by filling the two kinds of oil in tins of equal volumes. What is the greatest of such a tin.

Solution :

The given two quantities 120 and 180 can be divided by 10, 20,… exactly. That is, both the kinds of oils can be sold in tins of equal volume of 10, 20,… ltrs.

But, the target of the question is, the volume of oil filled in tins must be greatest.

So, we have to find the largest number which exactly divides 120 and 180.That is nothing but the H.C.F of (120, 180)

H.C.F of (120, 180)  =  60

The 1st kind 120 ltrs is sold in 2 tins of of volume 60 ltrs in each tin.

The 2nd kind 180 ltrs is sold in 3 tins of volume 60 ltrs in each tin.

Hence, the greatest volume of each tin is 60 ltrs.

Problem 4 :

Find the least number of soldiers in a regiment such that they stand in rows of 15, 20, 25 and form a perfect square.

Solution :

To answer this question, we have to find the least number which is exactly divisible by the given numbers 15,20 and 25.That is nothing but the L.C.M of (15, 20, 25)

L.C.M of (15, 20, 25)  =  300

So, we need 300 soldiers such that they stand in rows of 15, 20 , 25.

But, it has to form a perfect square (as per the question)

To form a perfect square, we have to multiply 300 by some number such that it has to be a perfect square.

To make 300 as perfect square, we have to multiply 300 by 3.
Then, it is 900 which is a perfect square.

Hence, the least number of soldiers required is 900.

Problem 5 :

Find the least number of square tiles by which the floor of a room of dimensions 16.58 m and 8.32 m 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 the tile, we have to find the largest number which exactly divides 16.58 and 8.32. That is nothing but the H.C.F of (16.58, 8.32).

To convert meters into centimeters, we have to multiply by 100.

16.58 ⋅ 100  =  1658 cm

8.32 ⋅ 100  =  832 cm

H.C.F of (1658, 832)  =  2

Hence the side of the square tile is 2 cm

Required no. of tiles :

=  (Area of the floor) / (Area of a square tile)

=  (1658 ⋅ 832) / (2 ⋅ 2)

=  344,864

Hence, the least number of square tiles required is 344,864.

Problem 6 :

A wine seller had three types of wine. 403 liters of 1st kind, 434 liters of 2nd kind and 465 liters of 3rd kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing.

Solution :

For the least possible number of casks of equal size, the size of each cask must be of the greatest volume.

To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465. That is nothing but the H.C.F of (403, 434, 465)

The H.C.F of (403, 434, 465)  =  31 liters

Each cask must be of the volume 31 liters.

Req. No. of casks is

=  (403/31) + (434/31) + (465/31)

=  13 + 14 + 15

=  42

Hence, the least possible number of casks of equal size required is 42.

Problem 7 :

The sum of two numbers is 588 and their HCF is 49. How many such pairs of numbers can be formed ?

Solution :

Because the H.C.F is 49, the two numbers can be assumed as 49x and 49y

Their sum is 588. So, we have

49x + 49y  =  588

Divide each side 49.

x + y  =  12

We have to find the values of “x” and “y” such that their sum is 12.

The possibles pairs of values of (x, y) are

(1, 11), (2, 10), (3, 9), (4, 8), (5, 7), (6, 6)

Here, we have to check an important thing. That is, in the above pairs of values of (x, y), which are all co-primes ?

[Co-primes = Two integers are said to be co-primes or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1]

Therefore in the above pairs, (1, 11) and (5, 7) are the co-primes.

Hence, the number of pairs is 2.

Problem 8 :

The product of two numbers is 2028 and their H.C.F. is 13. Find the number of such pairs

Solution :

Since the H.C.F is 13, the two numbers could be 13x and 13y

Their product is 2028.

So, we have

(13x) ⋅ (13y)  =  2028

169xy  =  2028

Divide each side by 169.

xy  =  12

We have to find the values of “x” and “y” such that their product is 12.

The possibles pairs of values of (x, y) are

(1, 12), (2, 6), (3, 4)

Here, we have to check an important thing. That is, in the above pairs of values of (x, y), which are all co-primes?

[Co-primes = Two integers are said to be co-primes or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1]

Therefore in the above pairs, (1, 12) and (3, 4) are the co-primes.

Hence, the number of pairs is 2

Problem 9 :

Lenin is preparing dinner plates. He has 12 pieces of chicken and 16 rolls. If he wants to make all the plates identical without any food left over, what is the greatest number of plates Lenin can prepare ?

Solution :

To make all the plates identical and find the greatest number of plates, we have to find the greatest number which can divide 12 and 16 exactly.

That is nothing but H.C.F of 12 and 16.

H.C.F of (12, 16)  =  4

That is, 12 pieces of chicken would be served in 4 plates at the rate of 3 pieces per plate.

And 16 rolls would be served in 4 plates at the rate of 4 rolls per plate.

In this way, each of the 4 plates would have 3 pieces of chicken and 4 rolls. And all the 4 plates would be identical.

Hence, the greatest number of plates Lenin can prepare is 4

Problem 10 :

The drama club meets in the school auditorium every 2 days, and the choir meets there every 5 days. If the groups are both meeting in the auditorium today, then how many days from now will they next have to share the auditorium ?

Solution :

If the drama club meets today, again they will meet after 2, 4, 6, 8, 10, 12…. days.

Like this, if the choir meets today, again they will meet after 5, 10, 15, 20 …. days.

From the explanation above, If both drama club and choir meet in the auditorium today, again, they will meet after 10 days.

And also, 10 is the L.C.M of (2, 5).

Hence, both the groups will share the auditorium after ten days.

Problem 11 :

John is printing orange and green forms. He notices that 3 orange forms fit on a page, and 5 green forms fit on a page. If John wants to print the exact same number of orange and green forms, what is the minimum number of each form that he could print ?

Solution :

The condition of the question is, the number of orange forms taken must be equal to the number of green forms taken.

Let us assume that he takes 10 orange and 10 green forms.

10 green forms can be fit exactly on 2 pages at 5 forms/page. But,10 orange forms can’t be fit exactly on any number of pages.

Because, 3 orange forms can be fit exactly on a page. In 10 orange forms, 9 forms can be fit exactly on 3 pages and 1 form will be remaining.

To get the number of forms in orange and green which can be fit exactly on some number of pages, we have to find L.C.M of (3,5). That is 15.

15 orange forms can be fit exactly on 5 pages at 3 forms/page.

15 green forms can be fit exactly on 3 pages at 5 forms/page.

Hence,the smallest number of each form could be printed is 15.

Problem 12 :

Lily has collected 8 U.S. stamps and 12 international stamps. She wants to display them in identical groups of U.S. and international stamps, with no stamps left over. What is the greatest number of groups Lily can display them in ?

Solution :

To make all the groups identical and find the greatest number of groups, we have to find the greatest number which can divide 8 and 12 exactly.

That is nothing but H.C.F of 8 and 12.

H.C.F of (8, 12) = 4

That is, 8 U.S stamps can be displayed in 4 groups at 2 stamps/group.

And 12 international stamps can be displayed in 4 groups at 3 stamps/group.

In this way, each of the 4 groups would have 2 U.S stamps and 3 international stamps. And all the 4 groups would be identical.

Hence, the greatest number of groups can be made is 4

Problem 13 :

Abraham has two pieces of wire, one 6 feet long and the other 12 feet long. If he wants to cut them up to produce many pieces of wire that are all of the same length, with no wire left over, what is the greatest length, in feet, that he can make them ?

Solution :

When the two wires are cut in to small pieces, each piece must of same length and also it has to be the possible greatest length.

6 feet wire can be cut in to pieces of (2, 2, 2) or (3, 3)

12 feet wire can be cut in to pieces of (2, 2, 2, 2, 2, 2 ) or (3, 3, 3, 3)

The length of each small piece must be of possible greatest length.

To find the possible greatest length, we have to find the greatest number which can divide both 6 and 12. That is H.C.F of (6, 12).

H.C.F of (6, 12) = 6.

Hence, the greatest length of each small piece will be 6 ft.

(That is, 6 feet wire is not cut in to small pieces and it is kept as it is. Only the 12 feet wire is cut in to 2 pieces at the length of 6 feet/piece)