CHAPTER-1 Number Theory Number is a fundamental concept of mathematics. The first conceptions of numbers were acquired by man in remote antiquity. It began with the counting of people, animals and various articles. The process of counting produced the numbers one, two, three, etc. In ancient times numbers were denoted by straight-line strokes. This notation was found inconvenient to write large numbers. Therefore symbols were invented for large numbers. These symbols are known as numerals. The most popular and widely used number system is the Hindu-Arabic system that consists of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. Addition (+), Multiplication (), Subtraction (-), Division (), Involution (raising a number to a power) and Evolution (finding the root of a number) are the six fundamental arithmetic operations. Addition and subtraction, multiplication and division and involution and evolution are pair wise inverse operations. Natural Number The No’s 1, 2, 3, . . . . . are called natural numbers. They are also known as counting no positive integers. The set of natural no’s is usually denoted by N i.e. N={1, 2, 3,…} Note: 1. 1 is the least natural no. 2. There is no greatest natural no. Whole Numbers The No. 0, 1, 2, 3,. . . . . are called whole No’s. They are also known as non- negative integers. The set of whole no is usually denoted by W. W= {0, 1, 2, 3,…..} Note: ‘0’ is he only whole no which is not a natural no. Integers The natural no’s together with their negative counterparts and zero called integers. The set of integers is denoted by I or Z. Z = {…….-2, -1, 0, 1, 2,…..} 1
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CHAPTER-1
Number TheoryNumber is a fundamental concept of mathematics. The first conceptions of numbers were acquired by man in remote antiquity. It began with the counting of people, animals and various articles. The process of counting produced the numbers one, two, three, etc. In ancient times numbers were denoted by straight-line strokes. This notation was found inconvenient to write large numbers. Therefore symbols were invented for large numbers. These symbols are known as numerals. The most popular and widely used number system is the Hindu-Arabic system that consists of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.
Addition (+), Multiplication (), Subtraction (-), Division (), Involution (raising a number to a power) and Evolution (finding the root of a number) are the six fundamental arithmetic operations. Addition and subtraction, multiplication and division and involution and evolution are pair wise inverse operations.
Natural Number
The No’s 1, 2, 3, . . . . . are called natural numbers. They are also known as counting no positive integers. The set of natural no’s is usually denoted by N i.e. N={1, 2, 3,…}
Note: 1. 1 is the least natural no.2. There is no greatest natural no.
Whole Numbers
The No. 0, 1, 2, 3,. . . . . are called whole No’s. They are also known as non-negative integers. The set of whole no is usually denoted by W.
W= {0, 1, 2, 3,…..}
Note: ‘0’ is he only whole no which is not a natural no.
Integers
The natural no’s together with their negative counterparts and zero called integers. The set of integers is denoted by I or Z.
Z = {…….-2, -1, 0, 1, 2,…..}
Note: 1. The set Z = {-1, -2, -3,…..} is called the set of negative integers.
2. ‘0’ is neither a positive nor a negative integer.
Rational Numbers
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A number which can be expressed in the form P/Q, where P is an integer and Q is a natural number is called a rational number. The set of rational numbers is denoted by Q.
Example: 2, -2, -5.5, 5.5 and 10.8752 etc are all rational numbers.
Note: Every integer is as rational number, for eg if x is an integer, if can be written as x/1
Irrational Numbers
A number which cannot the expressed in form P/Q, P is an integer and Q is a natural number is called irrational number. An irrational number is also known as incommensurable quantity. The set of irrational numbers is usually noted by Q’.
Examples: 2=1.414 ………, 3=1.732 ……, 5=2.236 …., are all irrational numbers.
Note:
1. The ratio of the circumference to diameter of a circle is an irrational number and is denoted by . Its value is 3.14159….. The rational number 22/7 very often taken as an approximated value of . In fact a better approximation of is 355/113.
2. The sum 1+1/1+1/12+1/123+….. is an irrational number and is denoted by ‘e’ and its value is 2.718….
Real Numbers
The rational numbers and irrational numbers are together called real numbers. The set of real numbers is denoted by R.
The German mathematician Richard Dedekind established that “Every real number can be represented by a point on a line called the real number line and conversely every point on the real number line represents a real number”
Concepts of Decimals
Terminating Decimals
A real number is called a terminating decimal if the number of digits succeeding the decimal part is finite.
Example: 1, 2, 12, 34 and 123.456 are all terminating decimals.
Non-terminating recurring decimals
A real number is called a non-terminating recurring decimal if a certain number of continuous digits repeat infinitely after the decimal point. It is also known as a non-terminating periodic decimal.Examples: 0.333…….and 5.23467467….are non-terminating recurring decimals. They are
denoted by 0.3 and 5.23467 respectively. Note: Every non-terminating non-recurring decimal is an irrational number.
Concepts of FractionsA common fraction is a part of units or several equal parts of units. The numbers which indicated how many parts a unit is divided into is called the denominator of the fraction and the number
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indicating how many parts are taken is the numerator of the fraction. 4/7. 2/3, 5/1, are examples of common fractions.
Proper Fractions
A proper fraction is one in which the numerator is less than the denominator.
Example: 4/7, 1/3, 1/7, 1/5 etc. A proper fraction is always less than 1.
Improper Fractions
An improper fraction is one in which the numerator is greater than or equal to the denominator.
Example: 5/4. 3/2, 9/5, 7/3, etc. Improper fractions are a form of mixed fractions. An improper fraction is always greater than or equal to1.
Mixed Fraction
Fraction consisting of whole numbers and fraction are mixed fractions.
Example: 5½, 3¾ etc.
To change a mixed number into a fraction:
a. Multiply the whole number by the denominator of the fractional part.
b. Add the result to the numerator of the fractional part.
c. Divide the result by denominator of the fractional part.
Example: 5½
Step 1 52=10
Step 2 10+1=11
Step 3 11/2 Result
Complex Fractions
Fractions whose numerator and denominators are fractions are known as complex fractions.
Decimal Fractions
A set of digits after a decimal is known as a decimal fraction. For example 0.464, 0.5, 0.33, 0.49 etc. The numbers are known as decimal fraction because they can be converted to proper fractions by dividing and multiplying them by 10n (where n is the number of digits after the decimal point)
Arithmetical Operations With Fractions
Multiplying Fractions
In order to multiply fractions, multiply their numerators and divide the result by the product of their denominators.
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Examples:
1. a/b c/d = ac/bd 2. a/b c/d e/f = ace/bdf
Dividing Fractions
In order to divide two fractions, multiply one fraction by the inverted second fraction
Example: a/b d/c = a/b c/d
Equivalent Fractions
Two fractions are equivalent if on reducing both the fraction to their lowest terms they are found to be equal.
Example: 4/5 and 40/50 are equivalent fractions because 40/50 = 4/5 on being reduced to its lowest terms.
Note: The value of a fraction is not changed if the numerator and denominator are multiplied by the same number.
Example: 4/5= 4*3/5*3= 12/15
This is known as reducing the fraction to higher terms. The value of the fraction is not changed if the numerator ad denominator are both multiplied (or both divided by the same number)
Even and Odd Numbers:An integer is called an even number if it is a multiple of 2, otherwise it is called an odd number.
Thus the set of the even integers is {0, +2, +4…..} and the set of odd integers is {+1, +3, +5, ……}
Note:
1. The sum, difference and product of two even numbers is again an even number. For example 8+4=12 is even, 16-12=4 is even 64=24 is even.
2. The sum and difference of two odd numbers is an even number. For example 3+5=8 is even and 21-7=14 is even.
Prime NumbersThe integers P > 1 is called a prime number if and only if the integer has got exactly two distinct factors.
Example: 2, 3, 5, 7, 11, ………… are all prime numbers.
Note: 1. Around 2,500 years age. Euclid established that shore are an infinite number of primes.
2. The largest prime number known till date is 230,21,377-1 and has 9,09,526 digits. It was discovered very recently day a 19-year old student called Roland Clarkson from the California State University.
3. 2 is the only even prime number, for any other even number will have 2 as a factor.
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Composite NumbersAn integer C > 1 is called a composite number, if and only if it has at least three or more factors including 1 and the integer itself.
Example: 4, 9, 10……..are all composite numbers.
Twin PrimesPrime numbers differing by 2 are called twin primes.
Example: 3 and 5, 5, and 7, 11 and 13 are all twin primes.
Prime TripletA set of three prime numbers, differing by 2 in called a prime triplet.
Example: The only set of prime triplets is {3, 5, 7}.
Perfect Numbers
A number is known as a perfect number it the sum of all the possible divisors of the number (excluding the number itself) is equal to the value of the number. For example, take integer 6.
The possible divisors of 6 are 1, 2 and 3 and 1 + 2+ 3 = 6. Similarly for 28, 496, 8128 etc., the sum of the possible divisors equal the number itself.
Complex NumbersThere is no real number which when squared given a negative number. In higher mathematical calculations often we come across such situations where in we need to find the roots of negative numbers. The set of real numbers is inadequate in handling such calculations. In order that these calculations are facilitated, imaginary numbers were defined. Imaginary numbers are root of negative numbers. The number I = -1 was defined with the property that is i2 =-1.
The sum of the real number and an imaginary number constitutes a complex number. For example 2 + -5 can also be denoted an 2 i5 (where I = square root of –1).
FactorsAny natural number will have at least two factors, they are 1 and the number itself.
Only the number 1 has a single factor i.e. itself consider for example:
i. 2 Factors of this number are 1, 2.ii. 4 Factors are 1, 2, 4iii. 6 Factors are 1, 2, 3, 6iv. 28 Factors are 1, 2, 4, 7, 14, 28
MultipleWhen a given a number ‘X’ is multiplied by an integer (Y) we get the multiple of that number ‘X’. It like adding the value X, Y times say 12. It is multiple of 3, 4, also 2, 6
3 4 = 3+3+3+3=12
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X Y
2 6 =2+2+2+2+2+2=12 X Y
Highest Common Factor (HCF)
It is the highest common factor to two or more given numbers. It is also known as greatest or highest common factor and is denoted as either HCF or GCF
Example: (a) 5 is the HCF of 10, 25(b) 1 is the HCF / GCF of 7, 9, 2.
Note: The HCF of any to prime numbers is always ‘One’.
Method to find other HCF / GCF of given numbers.
i. Factorization method :
Express each number is product of primes. Now take the product of common factors, this is HCF.
Example: 30, 24, 36, 810
30 =23524 =222336 =2233810 = 333325
HCF = 2 3 = 6
ii. By division method:
Suppose two number are given, Divide the greater number by lesser number, divide the lesser by the reminder, Divide the first reminder by the second reminder and so on till there is no reminder. The last divisor is the HCF required.
Example:
a. Find HCF of 2002, 182
By Division Method:
182) 2002 (11 2002 0000
HCF = 182;
b. 189, 1197
189) 1197 (6 1134
63) 189 (3
6
189 000
HCF = 63
The HCF of 189, 1197 is a factor of 1197 – (189 6) =63
Again the common factor of 63, 189 is 63 which gives a reminder of zero after dividing it with 189 The HCF of these two numbers is 63.
Note: In case of more than two numbers, say three nos. choose any two of them and find their HCF. The HCF of these two and the third gives HCF of the third numbers.
Least Common Multiple (LCM):
Least common multiple of two or more numbers is the smallest (least) number that is exactly divisible by each of them. Eg: 24
a. The number 24 is the LCM of 2, 3, 4, 6, 8, 12…….
b. 45, is the LCM of =3, 5, 9, 15
Method to find out LCM of a given number:
i. By method of factorization:
Resolve each one of the given numbers into prime factors, them their LCM is the product of highest power of all factors, that occur in these numbers.
Ex: LCM of 40, 81, 256, 111
40 = 4 10 = 2225 = 23 5
81 = 34
111 = 3 37
LCM = 23 34 51 37 = 119880
ii. By Formula:
Product of numbers = HCF LCM
LCM= Product of Numbers/HCF
Ex. LCM 21, 54, is LCM= 21 54/3 = 378
iii. Using factors:
LCM of 15, 24, 64, 100
LCM = 222222355= 26 3 52
= 4800
Important Rule
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Let A and B be two numbers, then
Product of A and B = HCF(A, B) LCM (A, B)
HCF and LCM of fractions
1. HCL of fractions = HCF of Numerators/LCM of Denominators
2. LCM of fractions = LCM of Numerators/HCF of Denominators
Example: Find the HCF and LCM of 2/7 = 4/24
i. HCF = 2/74 = 1/140 = 1/14ii. LCM = 4/7
Concepts of Divisibility of NumbersWe will categories the Divisibility criteria in three broad types:
i. For Numbers 2 to 11ii. Using the concept of seed for bigger prime numbers (not discussed in this booklet as it is not required for Campus Placements)iii. Using co-prime concept for composite numbers(not discussed in this booklet as it is not required for Campus Placements)
Divisibility for Numbers from 2 to 11
i. Divisibility by 2: A number is said to be divisible by two if its units digit is ‘even’ or ‘zero’.
E.g.: 24, 36, 100, 10, 2 etc.
ii. Divisibility by 3: A number is said to be divisible by three, if the sum of the digits is divisible by 3.
E.g.: 27=2+7=9 is divisible by ‘3’ 27 is div. by 3339=3+3+9=15 is divisible by ‘3’ 339 is div. by 3
The method is known as “digit sum”, where in we add up the individual digits of a given umber and divide the result to check for divisibility.
iii. Divisibility by 4: If the last two digits of a given number is divisible by ‘4’ or the last two digits of the given number is ‘00’, then the number is divisible by 4.
E.g.: 256, 5120, 1336
iv. Divisibility by 5: A number is said to be divisible by if the units digits of the given number is 5 or ‘0’
E.g.: 225, 125, 100 etc
v. Divisibility by 6: A number is said to be divisible by 6 if it is divisible by 2 and 3 both. This will be further explained in the co prime concept for composite numbers.
E.g.: 12, 216, 78 etc.
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vi. Divisibility for 7: Test for divisibility of 7.
See if the number is a large number i.e. more than 6 digits then, Do the following.
E.g.: Consider this numbers: 8235437
The above number is divisible by 7 if the sum of the product of the digits of the number from left to right with 1, -2, -3, -1, 2, 3…….Successively us divisible by 7 or is equal to ‘0’.
We would discuss this again in the seed concepts of divisibility.
vii. Divisibility of Eight: A number is said to be divisible by eight if its, last three digits are divisible by eight or if the last three digits are ‘000’.
E.g.: 2512, 14256, 17200
viii. Divisibility by Nine: A number is said to be divisible by ‘9’ is the sum of the digits is divisible by ‘9’. I.e., we are again taking into account the digits um
E.g.: 43731 = 4 + 3 +7 +3 +1 = 18 is div. By 9.
ix. Divisibility by ten (10): A number is said to be divisible by 10 if it last digits is ‘0’
E.g.: 1000, 500, 100, etc.
x. Divisibility for 11: A number is said to be divisible by 11 if the difference between the sum of the digits in odd position and even position is zero or eleven or a multiple of eleven.
E.g.: 1331 = 1+3 = 43 +1 = 4 4 – 4 = 0, is divisible by 11.
9174 = 9+ 7 = 16 sum of digits of even position1+4 = 5 sum of digits of odd position 16 – 5 = 11 is divisible by 11.
Square of a numberWhen a number is multiplied by itself the number obtained is called the square of that number.
E.g.: Square of 5 = 5 5 = 25;
Square of 13 = 13 13 = 169;
Square of 9 = 9 9 = 81; etc
Perfect Square:
The square of a natural number is called perfect square.
Some properties of square number:
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i. Square of a number cannot end with odd number of zeros.
ii. Square of a number cannot end with 2, 3, 7, 8.
iii. Square of 1, 5, 6 always end with the digits 1, 5, 6 respectively.
iv. Square of an even number is always even.
v. Square of an odd numbers is always odd.
vi. Every square number is a multiple of 3, or exceeds a multiple of 3 by unity.
vii. Every square number is a multiple of 4, or exceeds a multiple of 4 by unity.
viii. Square of a negative number is always positive.
ix. To find the square of a number ending in 5, multiply the ten’s digits by next higher integer and annex 25.
E.g.: 1052 = 10 (ten’s digits) 11 (next higher integer) = 11025; 25 is annexed to get the Ans.
xi. If a square ends in 9, the proceeding digit is even.
Cube of Number
Cube a number is obtained by multiplying the number twice by itself: * *
i. Cube of a number can end in any integer from 0 to 9.
ii. Cube of a negative number will always have negative sign in the result.
Square & Cube roots
1. The number of digits in the square root of a ‘n’ digit number is
a. n+1/2 when n is odd
b. n/2 when n is even
2. The square root of any number is either positive or negative.
3. The cube root of any positive number is always positive.
Square root of a number
1. Prime factorization method
Find square root 3600 = 3600 22223355 = 2235 = 60
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2. Division method: Find the square root of 2116. Do the following steps
Square Root of a Perfect Square:Step 1: The integers are grouped in pair (2 at a time)
From the RHS of number i.e. from the units digits
We have two groups 21, 16
Step 2: The integer whose square is less than or equal total first group is written on the top and the left of the number.
Step 3: The Square of the integer is subtracted from the first group
Step 4: Now take the next pair of numbers. We get 516 as divided.
Step 5: Double the divisor (which make it 8) and add a number (in this case 6) which multiplied by 8 (in this care 86) gives a numbers less than or equal to 516. 86 6 = 516; But (here 6) in the quotient. There is no reminder. So square of 2116 is 46.
Square Root of not a Perfect Square:
a. To find the square root of 51
First find the nearest square to 51 i.e. 49. Subtract 49 from 51 = 2
Now, 49 is the reminder of 7. Now, divide the reminder 2 by two times 7, (27) to get
= 2/27 = 1/7 = 0.1412.
Add this to 7, and you have 7.14 as the square root of 51.
Step 1: 51 - 49=2
Step 2: 2/14 = 1/7 =0.14
Step 3: 7 + 0.14 = 7.14 square root
b. To find square root of 80
The nearest square number is 81.
Subtract 80 from 81. = 1
Now 81 is square of 9.
Divide the reminder 1 by two times 9.
= 1/29=1/18 = 0.055;
Subtract this from 9 = 8.95 as the square root of 80
Step 1: 80 = 81 –1
Step 2: -1/18 =-0.05
Step 3: 9 – 0.05 = 8.95
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Memorize the following
i. Square from 1 to 60, look in for some pattern.
ii. Square roots from 1 to 50; look in for some patterns
iii. Cube from 1 to 30
iv. Cube roots from 1 to 30
v. Reciprocals from 1 to 30;
Concepts of Last digit
For any no. the digit in units place is called as last digit.
Observe some important facts in the last digits and memorize till the fourth power.
* Observe that again the fifth power of the above given odd number will have the same digit as the first power.
* See that fourth power of even number is ending with ‘6’ multiplied by any even number (say x) when multiplied with this number ‘6’ will have the same last digit of (x).
Example: 6 12 = 7216 4 = 646 18 = 108
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* Also observe that the fourth power of every odd number ends in 1; when any given number is multiplied with this fourth power, the product will have the same last digit as the no. used as multiplier.
81 7 = 56781 6 = 486
* Observe that for numbers ending in 0, 1, as 5 the last digit is 0, 1, 6 and 5 respectively.
Some examples of last digit concept
1. Say the number is of the form xy.
* Observe x: Is the last digit odd or even? If it is even, any power of that should end in even only. If last digit of x is odd, any power of this should end in odd only.
* Break up ‘y’ as a multiple of 4 + odd /even depending on the example (case) i.e express y = 4g + k (k can be odd or even and will be definitely less than 4)
* Now x to power of 4g + odd /even will have the same last digit as power of last digit of ‘x k if x is even.
Example: What is the last digit in the following?
1. 2898
Step 1: x = 28 even last digit should be even.
Step 2: y = 93 = 4 23 + 1 = 92 + 1 4g + k (odd)
Step 3: 284g+k = 8k = 81 = 8 Where z = 23
The last digit is eight = 8.
2. 2334
Step 1: x = 23 =Odd Last digit should be odd
Step 2: y = 34 = 4 8 + 2 = 32 + 2 | 4g + k (even)
Step 3: 234g+k = 3k = 32 = 9
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2. Binomial expansion: How last digit is obtained?
Remember this expansion will always have (n+1) terms.
= a[1st n terms] + bn where M is the polynomial expansion Remember this for applicationslast term
In general
(a + b)n = aM + bn
(a – b)n = aM + (-b)a
Lets consider the same example as given before
1. (28)93 = (20+8)93 = 20 M + 8 93
= 8(92+1) = 8(4g+1) = 84z 81, where g=23
This will always end in ‘6’
Using the fact that any even number when multiplied when multiplied with six will give the same last digit we have, 6 81 = 8. Since we are only concerned about the last digit is 8.
2. Lets take an easy example:
27 = (0+2)7 = O M +27 where z=1= 24x1+3 = 24xg+3 = 24g 23
= 16g 8 = ---6 8= ----8
3. Lets see for an odd no.: What is the last digit
(23)34 = (20 +3)34 = 20 M + 334
= 332+2 = 348+2 = 34g+2 = ---132 (Since 34g will always end in 1)=32 = 9
The last digit is 9
4. Lets do 1 more:
(43)75 = (40+3)75 = 40 M + 375
=372+3 = 34xg+3 = ---133 = -----7
The last digit is 7.
Remember:
Even4x will always end in the digit: ‘6’Odd4x will always end in the digit: ‘1’
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Computations using Vedic MethodologyComputations form the most important aspect of problem solving in tests that disallow the use of
computation devices. All of us, over a period of time would have become conversant with small numbers
and hence computations involving small numbers are quick and accurate. As we have not developed the
same type of instincts with large numbers and fractions, it takes us more time to compute with these
numbers. Vedic Mathematics takes this fact into account and reduces large computations into small
numbers with the help of formulae so that computations can be quicker and more accurate. For example,
the usual procedure in schools and colleges in case of multiplication is to cram up multiplication tables up
to 20 and 30. But according to Vedic Mathematics, the multiplication tables are not required above 5. With
the help of simple sutras any multiplication work can be performed. We shall examine some of the simple
sutras and their application in computations.
Multiplication
Using Bases – 10, 100, etc.
Suppose we have to multiply 99 by 97.
(a) We should take, as base for our calculations, that power of 10 which is nearest to the number
to be multiplied. In this case 102 or 100 is that power.
(b) Put the numbers 99 and 97 above and below on the left hand side.
99 -197 -3
96 3
(c) Subtract each of them from the base (100) and write down the remainders (01 and 03) on the
right hand side with a connecting minus sign (-) between them, to show that the numbers to
be multiplied are both less than 100.
(d) The product will have two parts, one on the left side and one on the right. A vertical dividing
line may be drawn for the purpose of demarcation of the two parts.
(e) Now the left hand side digits (of the answer) can be arrived at in one of 4 ways.
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(i) Subtract the base 100 from the sum of the given numbers (99 and 97) i.e., 196, and put
196-100 i.e., 96 as the left hand side part of the answer. 99 + 97 - 100 = 96
(ii) Subtract the sum of the two deficiencies (01 + 03 = 04) from the base (100) and you get
the same answer (96) again. 100 - (01 + 03) = 96
(iii) Cross subtract deficiency 3 on the second row from the original number 99 in the first row.
And you get (99 - 3) i.e., 99 - 03 = 96 or
(iv) Cross subtract in the converse way i.e., 1 from 97). And you get 96 again as the left-hand
side of the portion of the required answer. 97 - 1 = 96.
(f) Now vertically multiply the two deficiencies (01 and 03). The product is 03. And this is the
right hand side portion of the answer.
(g) Thus 99 x 97 = 9603
This method holds well in all cases and is therefore capable of infinite application.
Solved Examples
1. 9 x 9 (Base: 10) 2. 999 x 996 (Base: 100)
3. 8 x 8 (Base: 10) 4. 9 x 6 (Base: 10)
8 -2 9 -18 -2 6 -4
6 4 5 4
The algebraic explanation for this is (x - a) (x - b) = x(x - a - b) + ab, where x takes the base value,
a and b are deficiencies.
In cases where the multiplication of deficit digits yields a product consisting of more than one digit,
the second digit must be carried forward to the left-hand side.
For example, 7 x 6
7 -3
6 -4
3 +
1
2
4 2
Here 3 x 4 yields 12 and 1 is carried forward to the left hand side and is added to 3 to get the product
as 42.
9 -1 999 -19 -1 996 -4
8 1 995 4
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Further Examples
1. 8 x 5 2. 93 x 95
8 -2 93 -7
5 -5 95 -5
3 0 -2 * -5 = 10 88 35
1 (Carry)
4 0
3. 89 x 95 4. 87 x 87
89 -11 87 -13
95 -5 87 -13
84 55 74 169
74+1 69
75 69
As you must have observed, we have till now considered only examples in which the multiplicands are
both on the lower side of the base. If both the multiplicands are on the higher side of the base, the
story remains the same.
Solved Examples
1. Multiply 101 with 103
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101 + 1
103 + 3
104 / 03 = 10403
Do not forget to put an extra 0 on the right hand side, since in this case our base 100 has 2 zeros.
We must have 2 digits on the right hand side.
2. Multiply 109 with 121
109 + 09
121 + 21
130 / 189 = 130 + 1 / 89 = 13189
The algebraic explanation for this is : (x+a) (x+b)= x(x+a+b) + ab
3. Multiply 1021 with 1034
1021 + 21
1034 + 34
1055 / 714 =1055714
(As base is 1000, there should be 3 digits on the right side. No carry-over is required)
4. Multiply 10134 with 10111
10134 + 134
10111 + 111
10245 /
We multiply 134 x 111 as
134 + 34
111 + 11
145 / 374 = 14874
10134 + 134
10111 + 111
10245 / 14874 = 102464874
The method is applicable for numbers as large as possible.
5. 1001 x 101
This one is not so tricky as it looks, we choose our base as 100 (remember that the differences for
both the numbers have to be calculated from the same base)
1001 + 901
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101 + 1
1002 / 901 = 1002 + 9 / 01 = 101101
Further down the line, we will see how to multiply numbers choosing different bases for them.
6. 102 x 97
How should we go about multiplying these? One of the numbers is above the base (100) while the
other one is below. Before proceeding with solving such kind of problems, we will have to learn
the concept of vinculum in Vedic Mathematics.
Concept of Vinculum
We will come across a lot of numbers with Vinculum above them in Vedic Mathematics, e.g. 102 , 10
.
The bar is the Vinculum. How do we convert between the normal numbers and Vinculum numbers,
and actually first of all, what does the Vinculum mean?
We will first see how the numbers with Vinculum above them are converted to normal numbers, can
you then figure out what does it mean?
The 2 shlokas which deals with vinculum state, “All from 9 and last from 10” and “One less from the
previous digit”.
So suppose we have to convert 210 to a normal decimal number. Then we proceed as follows:
1. “All from 9 and last from 10”: All the digits under the vinculum should be subtracted from 9,
except for the last one, which should be subtracted from 10.
So in this case, 9 should be subtracted from 9 and 1 from 10. Thus we get the last 2 digits of our
answer as: 09
2. “One less from the previous digit”: One should be subtracted from the digit preceding the
vinculum.
Since in this case it is 0, we borrow 1 from the digit to the left of it (as we do in normal mathematics)
to arrive at the first part of our answer: 209
Thus, answer is: 20909
Solved Examples
1.
Step 1 yields 88
Step 2 yields 09
So our answer is 988.
2.
Step 1 yields 54
19
Step 2 yields 22
So our answer is 2254.
Further Examples
1. 2.
Step 1 yields 1 Step 1 yields 66
Step 2 yields 0 Step 2 yields 1
Answer is 01 Answer is 166
Just try to see whether you have understood the concept of vinculum by trying out the following
examples:
1. 2. 3. 4.
Answers are:
1. 433 2. 9980001 3. 2611 4. 2257
Do you see a pattern?
So does that tell you what actually does a vinculum signify?
Let’s get back to the problem of multiplying 102 with 97. The method is still essentially what it was,
except that a vinculum comes into play.
102 + 2
97 – 3
99 /
We cross-subtract (or cross-add) as we used to do earlier for the left hand side of our product. Since for
the right side, we are multiplying + 2 with – 3, we should put – 6, our vinculum is exactly that, as you
can see from the “patterns” above.
Our method of converting vinculum numbers to normal numbers is essentially to carry out the
subtraction. Now 99 = 9894 and that is our answer.
Solved Examples
1. 103 97 2. 111 91 3. 1024 981
20
103 + 03 111 + 11 1024 + 24
97 – 03 91 - 9 981 - 19
100 / = 9991 102 / = 10101 1005/ = 1004544
4. 121 x 81 5. 9 x 11 6. 976 x 1041
121 + 21 9 – 1 976 - 24
81 – 19 11 + 1 1041 + 41
102 / 10 / = 99 1017 / = 1016016
= 102 – 3 /
= 99 / = 9801
7. 46 x 51
In this case, we can choose our base as 10 or 100, but the difference with the base in both cases will
be very large. Wouldn’t it be better if we could choose 50 as our base? We certainly can and this
brings us to our new extension of this method of multiplication.
Using Sub-Bases
1. First of all, choose the normal base, the sub-base will be calculated from that. Let’s choose, in this
case, our base to be 10.
2. Choose the sub-base, in this case, it is 50. Calculate the differences from the sub-base.
3. Multiply the number as we used to do earlier, the only extension being that the left part of the
answer has to be multiplied by . The right hand side of the answer should have
number of digits equal to number of digits in base (not sub-base), and it will be obtained by
multiplying the differences.
Thus:
46 – 4
51 + 1
47 / ( = 5, hence multiply by 5)
= 235 / = 2346
Note that we can also use 100 as our base, in which case:
46 – 4
51 + 1
21
47 / ( = ½)
= 23 / (Take on the right hand side, it becomes 50)
= 23 / 46 (since 50 – 04 = 46)
= 2346.
Solved Examples
1. 241 x 263
Choose 1000 as the base and 250 as the sub-base.
241 – 9 (Calculate the differences from the sub-base)
263 + 13
254 /
= 63 / (take on the right-hand side, it becomes 500)
= 63 / 383 (500 – 117 = 383)
= 63383
2. 313 x 283
It will be better to take 100 as the base and 300 as the sub-base in this case, instead of taking 250
as the base.
313 + 13
283 – 17
296 /
3
= 888 /
= 886 /
(As, right hand side can have only two digits, take on the left side, so it becomes 888 – 2 = 886)
22
= 88579
3. 486 x 486
Take 1000 as the base and 500 as the sub-base.
486 – 14
486 – 14
472 / 196
= 236 / 196
= 236196
4. 1024 x 1031
Take base as 1000 sub-base is also 1000.
1024 + 24
1031 + 31
1055 / 744 = 1055744
Thus we see that this method encompasses the previously encountered methods.
How did we multiply 24 and 31 (the differences) in the previous example? One method is to use the
normal mathematics, another one is to use Vedic Mathematics (take 10 as base and 30 as sub-base)
but there is a third much easier method.
24 x 31 = 24 x 30 + 24 x 1 = 720 + 24 = 744
The basic law of number theory that we use here: a (b + c) = a b + a c is, as you will
remember, called the distribution of multiplication over addition. Used properly, it is a much more
powerful tool for some kind of multiplications than both the conventional multiplication and Vedic
Mathematics.
Example:
98 x 34 = 98 x (30 + 4) = 98 x 30 + 98 x 4
Now 98 x 30 = 98 x 3 x 10, so we find 98 x 3 and then put a 0.
How do we calculate 98 x 3? One way is to multiply. Other is to realize that:
98 x 3 = (100 – 2) x 3 = 100 x 3 – 2 x 3 = 300 – 6 = 294
Similarly, 98 x 4 = 400 – 2 x 4 = 392.
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Or, since we know from the first step that 98 x 3 = 294, add 98 to it to arrive at 98 x 4.
How do we add 98 to 294? One method is to add. Other is to add 100 to 294 and then subtract 2 to
arrive at our answer 392.
Hence,
98 x 34 = 2940 + 392 = 2940 + 400 – 8 = 3332
The calculations are always much faster and accurate if we try to find some bases (multiplies of 10)
near our numbers.
We have seen till now how to multiply 2 numbers both of which are closer to some base (or sub-base).
Can we multiply two numbers one of which is closer to one base and another to another base?
Yes we can. Let’s see how we can do that with an example.
Example:
Multiply 34 with 87.
Our base for 34 is 10, and for 87 too is 10.
For 34, our sub-base is 30, for 87 our sub-base is 90.
87 – 3
34 + 4
The left hand side of the answer is calculated as:
1. Let n1 and n2 be the two numbers; d1 and d2 their differences, b1 and b2 the bases and r1 and r2 the
ratio of the sub-bases with bases.
Then left hand side = n1 x r2 + d2 x r1 = n2 x r1 + d1 x r2
Ratio is the relation which one quantity has to another of the same kind. Ratio is the comparison of two quantities.
Example: 2 kg : 3 kg, 5 men : 10 men.
Note that in both the examples the quantities we have compared have the same unit.
The ratio of two quantities l and m denoted by l : m (read as l is to m). Here l and m are called the terms of the ratio where l is called the antecedent and m is called the consequent.
Example: In the ratio 32 : 51, 32 is called the antecedent and 51 is called as the consequent.
Important points to remember1. The ratio of 2 quantities x and y denoted by x : y (read as x is to y the first term x is called the
antecedent and second term y is called the consequent.)
2. In order to find the ratio of two quantities of the same kind they must first be expressed in terms of a common unit.
For instance, the ratio of Rs. 3 and 45 Ps. is 300 : 45.
Or 20 : 3
3. Two or more ratios may be compared by reducing their equivalent fractions to a common denominator.
For instance to compare 5 : 2 and as
and as
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Since , we get 5 : 2 > 1 : 3
4. A ratio is a pure number and has no units.
5. Comparison of Ratios
a. l : m is called the ratio of greater inequality, if l > m.
b. l : m is called the ratio of lesser inequality, if l < m.
c. l : m is called the ratio of same inequality, if l = m.
Examples: 1. 3 : 2 is a ratio of greater inequality since 3 > 2
2. 1 : 6 is a ratio of lesser inequality since 1 < 6
3. 5 : 5 is a ratio of equal inequality since 5 = 5
Two ratios can l : m and n : p can be compared. The different comparisons are as given below:
a. If lp > mn then l : m > n : p
b. If lp < mn then l : m < n : p
c. If lp = mn then l : m = n : p
Examples:
1. Compare the ratios 6 : 5 and 3 : 2.
Clearly (6 x 2 = 12) < (5 x 3 = 15). Hence 6 : 5 > 3 : 2
2. Compare the ratios 4 : 7 and 7 : 2.
Clearly (4 x 2 = 8) < (7 x 7 = 49). Hence 4 : 7 < 7 : 2
3. Compare the ratios 9 : 2 and 4 : 7.
Clearly (9 x 7 = 63) > (4 x 2 = 8). Hence 9 : 2 > 4 : 7
6. Commensurable Quantities
Two quantities are said to be commensurable, if their ratio can be expressed as the ratio of two integers.
Example:
Show that and are commensurable quantities.
Consider : = : = 78 : 26 = 6 : 2. This is the ratio of 2 integers.
Hence and are commensurable numbers.
7. Some Important Results on Inequality
a. If the same positive number is added to both the terms of ratio of greater inequality then the ratio is diminished.
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Example : is ratio of greater inequality.
A positive number 4 is added to both the terms of ratio i.e.,
Since , the result is proved.
b. If the same positive number is added to both the terms of ratio of less inequality then the ratio is increased.
Example : is ratio of lesser inequality.
If 3 is added to both the terms of the ratio
Since , the result is proved.
8. Compound Ratio
If a : b and c : d are two ratios then the ratio ac : bd is called their compound ratio.
For example the compound ratio of 2 : 5 and 3 : 1 is 6 : 5.
9. Important terms and definitions on Ratio
i) Duplicate ratio of a : b is a2 : b2
ii) Triplicate ratio of a : b is a3 : b3.
iii) Sub duplicate ratio of a : b is a½ : b½.
iv) Compounded ratio of a : b and c : d is ac : bd.
v) Duplicate ratio of a : b is a2 : b2.
vi) Sub triplicate ratio of a : b is a1/3 : b1/3.
vii) Reciprocal ratio of a : b is :
ProportionLet us consider two ratios a : b and c : d. When we compare these two ratios i.e., when we compare ad and bc, if ad = bc then we say the ratios are in proportion.
Example:
Consider two ratios 2 : 3 and 4 : 6.
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a = 2, b = 3, c = 4, d = 6.
ad = 2 x 6 = 12 and bc = 3 x 4 = 12.
Since ad = bc the two ratios are in proportion and is denoted by a : b :: c : d and we read it as a is to b is to c is to d.
When a : b :: c : d then a and d are called extremes and b and c are called means.
The product of the means is equal to the product of the extremes.
Important points to remember
1. Mean Proportion
In the proportion a : b :: b : c, the term b is called the mean proportion.
Since the product of the means is equal to product of the extremes, we get
Or .
2. Third Proportional
In the proportion a : b :: b : c, the extreme c is called the third proportional.
3. Fourth Proportional
In the proportion a : b :: c : d, the extreme d is called the fourth proportion.
4. Direct Variation
One quantity A is said to vary directly as another B, when the two quantities depend upon each other in such a manner that if B changes, A changes in the same ratio. Ratio of A and B is a constant always.
Example: If a train moving at a uniform rate travels 40 miles in 60 min, it will travel 20 miles in 30 min, 80 miles in 120 min. This is expressed by telling the distance is proportional to time.
5. The symbol is used to denote variation
A B is read as A varies as B.
6. The ratio of A and B is always a constant.
A B A = constant B
i.e., = constant.
Hence, if A1, A2 are distinct values of A and B1, B2 are distinct values of B, and if
A1 B1 and A2 B2
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then, =
Example: The number of men working is directly proportional to units of work completed. More men, more work can be completed.
7. Directly Proportional Graph
If, X Y
Then, X = KY
where, K CONSTANT
The graph illustrates this relationship.
8. Inversely or Indirectly Proportional
One quantity A is said to vary inversely as another quantity B, when A varies directly as the reciprocal of B.
i.e., A A = AB = K, where K = constant.
Hence, if A1, A2 are distinct values of A and B1, B2 are distinct values of B, and if
A1 and A2
then, =
The graph illustrates this relationship.
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Worked Examples1. Two numbers are in the ratio 5 : 6. If 8 is added to each they are in the ratio 9 : 10. Find the
numbers.
Solution :
Since, the two numbers are in the ratio 5 : 6,
let us assume that the numbers are 5x & 6x.
If 8 is added to each number, we get 5x + 8 & 6x + 8 respectively.
The new ratio = .
But according to the condition given, this ratio is 9 : 10
Equating the two ratios we get
10(5x + 8) = 9(6x + 8) 50x + 80 = 54x + 72
4x = 8 x = 2
first number = 5 x 2 = 10 & second number = 6 x 2 = 12
2. Two numbers are in the ratio 3 : 5. If 6 is subtracted from each they are in the ratio 5 : 9. Find the numbers.
Solution:
Since, the two numbers are in the ratio 3 : 5,
we can assume that the numbers are 3x & 5x.
Subtracting 6 from each of them, they are in the ratio 5 : 9.
Thus should be subtracted from each term so as to change the ratio from 3 : 5 to 5 : 9.
4. What should be added to each term in the ratio 7 : 9 so that it becomes 5 : 6 ?
Solution :
Let x be added to each term then
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6(7 +x) = 5(9 +x) 42 + 6x = 45 + 5x
6x – 5x = 45 – 42 x = 3
5. The prices of two sites are in the ratio 6 : 11, 6 months later when the price of the first site was raised by Rs. 600 and the price of the second site was raised by Rs. 800, the ratio of their prices become 3 : 5. Find the original prices of the sites.
Solution :
Let the original prices of two sites be 6x & 11x respectively.
6 months later, the price of first site will be 6x + 600 & the price of second site will be 11x800.
It is given that, six months later, the ratio of two sites will be 3 : 5. Thus we have
5(6x + 600) = 3(11x + 800)
30x +3000 = 33x + 2400
600 = 3x x = 200
Thus the original price of the first site is 6 x 200 = 1,200
and the original price of the second site is 11 x 200 = 2,200
6. In a 1,200 meters race, A beat B by 200 meters and C beat B by 400 meters. If A’s speed is 400 mts/min, find C’s speed.
Solution :
Distance covered by A is 1,200 mts in the given time
Distance covered by B is 1,200 - 200 = 1000 mts in the same time
Similarly,
C’s speed = A’s speed = x 400 = 500
7. There are two clerks A and B drawing the same salary. A saves 15% of his salary while B spends 80%. Find the ratio of the savings made by the former to that made by the latter.
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Solution :
Let us assume that A and B have Rs. 100 as salary.
Then A saves Rs. 15 and B saves 100 - 80 = Rs. 20
The ratio of their savings is 15 : 20 i.e. 3 : 4.
8. If x : y : z = 4 : -3 : 2 & 2x +4y -3z = 20, find the values of x, y, z.
Solution :
Let = k (say)
x = 4k, y = -3k & z = 2k and 2x + 4y - 3z = 20
8k -12k - 6k = 20
-10k = 20 k = - 2
Hence,
x = 4(-2) = - 8
y = (-3)(-2) = 6
z = 2(-2) = - 4
9. Three numbers are in the ratio 4 : 5 : 7. The sum of their cubes is equal to 4256. Find the numbers.
Solution :
Let the first number be 4x, then the second number will be 5x & the third number will be 7x.
Further it is given that, the sum of their cubes is equal to 4256.
10. Two lecturers A and B whose salaries are Rs. 7,500 and Rs. 14,500 received an increase in their salaries. If their salaries were increased in the same ratio, what increase did B receive when A’s salary reached Rs. 9,000?
Let us assume the ratio increase of B’ salary as .
Then = (because the increase of A & B are in the same ratio)
Thus x : 14,500 :: 1,500 : 7500
7500x = 14,500 x 1,500
x = = 2,900
Thus B’s salary was increased by Rs.2,900.
Practice Exercise 1
1. Two numbers are in the ratio of 5 : 8. If 9 is added to each they will be in the ratio 8 : 11. The original number in the numerator is
1. 24 2. 15 3. 8 4. 5
2. The ages of a man and his son are in the ratio of 3 : 1. Fifteen years hence the ratio will be 2 : 1. How old is the father now?
1. 45 years 2. 30 years 3. 40 years 4. None of these
3. The ratio of the areas of two squares is 9 : 1. The ratio of their perimeters will be
1. 1 : 2 2. 3 : 1 3. 3 : 2 4. None of these
4. X and Y have their annual incomes in the ratio of 8 : 5 and their annual expenditure is in the ratio 5 : 3. If they save Rs.1400 and Rs.1000 per year respectively, find their annual incomes.
1. Rs.19200, Rs.13000 2. Rs.6400, Rs.4000
3. Rs.12000, Rs.7500 4. Rs.7200, Rs.4500
5. Find three numbers in the ratio of 3 : 2 : 5 such that the sum of their squares is equal to 1862.
6. Two numbers are in the ratio 7 : 1. The difference between them is 44. What is the sum of these two numbers?
1. -208 2. 3. 4. None of these
7. A college has 53 boys and girls in the ratio 5 : 4. If there is an increase of 20% in the number of girls, what is the new ratio of boys to girls?
1. 25 : 24 2. 235 : 570 3. 196 : 531 4. 4 : 5
8. For what value of x will the ratio (7 + x) : (12 + x) be equal to the ratio 5 : 6?
1. 20 2. 13 3. 22 4. 18
9. The ratio between the numerator and denominator of a number is 3 : 7. If the difference between the denominator and numerator is 28, what is the sum of the numerator and denominator?
1. 60 2. 70 3. 65 4. 75
10. The ages of Anil and Sagar are in the ratio of 6 : 5 and the sum of their ages is 44 years. What will be the ratio of their ages after 8 years?
1. 7 : 6 2. 3 : 4 3. 8 : 7 4. 2 : 5
11. Two numbers are in the ratio of 3 : 5. If 9 is subtracted from each, they are in the ratio of 12 : 23. What is the second number?
1. 55 2. 25 3. 40 4. 30
12. A greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 4 leaps of the hound are equal to 5 of the hare. The ratio of distance traversed by the hound and hare area :
1. 24 : 25 2. 25 : 24 3. 5 : 6 4. 6 : 7
13. The ages of A and B are in the ratio 4 : 5; the ages of B and C are in the ratio 3 : 2. The eldest of the three is
1. A 2. B 3. C 4. Can’t say.
14. A contractor undertook to build a road in 100 days. He employed 70 men. After 30 days, he found
that only of the road could be built. How many additional men should be employed to complete
the work in time ?
1. 6 2. 20 3.10 4. None of these
15. The age of a man and his son are now in the ratio of 3 : 1. Fifteen years hence the ratio will be 2 : 1. How old is the father now?
1. 45 years 2. 30 years 3. 40 years 4. None of these
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Practice Exercise 21. An alloy consists of two metals iron and copper which are in the ration 1 : 2 and another alloy
contains the same metals in the ratio 2 : 3. How many parts of each alloy must be taken to obtain a third alloy which would contain the same metals in the ratio 5 : 8 ?
1.2 : 3 2. 4 : 5 3. 1 : 1 4. 3 : 10
2. If +b = 2b find a : b.
1. 3 : 1 2. 1 : 3 3. -1 : 3 4. 1 : -3
3. If 9 men and 6 boys can do in 2 days what 5 men & 7 boys can do in 3 days, what is the ratio of work rate of a man to a boy.
1. 3 : 1 2. 2 : 1 3. 3 : 2 4. 4 : 3
4. Rama and Sita started a business with initial investments in the ratio 6 : 5. Their profits were in the ratio 16 : 5. If Rama had invested his money for 8 months, find for how long had Sita invested her money.
1. 4 months 2. 2 months 3. 6 months 4. 3 months
5. 50 lts of wine are mixed with 10% water. How much water must be added to make the mixture contain wine water in the ratio 4 : 1 ?
1.10 lts 2. 7.5 lts 3. 12 lts 4. 8 lts
6. The ratio of the ages of mother and son is 5 : 3. If the sum of their ages is 64 years, find the difference of their ages.
1. 20 years 2. 16 years 3. 25 years 4. 21 years
7. In a country, in a decade, the population increased by 15.9%. If the increase in the urban population was 18% and that in the rural population was 4%. Find the ratio of the urban to rural population at the beginning of the decade.
1. 5 : 1 2. 17 : 3 3. 10 : 3 4. 19 : 5
8. Vessels A and B contain milk and water in the ratio 4 : 5 and 5 : 1 respectively. In what proportion would quantities be taken from the vessels A and B respectively to form a mixture in which milk : water is in the ration 5 : 4.
1. 2 : 5 2. 4 : 3 3. 5 : 2 4. 3 : 4
9. A piece of wire 78 cm long is bent in the form of an isosceles triangle. If the ratio of one of the equal sides to the base is 5 : 3, the length of the base in cms is
1. 19 cm 2. 18 cm 3. 16 cm 4. 17 cm
10. One man adds 3 liters of water to 12 liters of milk and another man adds 4 liters of water to 10 liters of milk. Compare the amount of milk in the two mixtures.
1. 25 : 26 2. 28 : 25 3. 3 : 2 4. None of these
11. If A : B = 3 : 4 and B : C = 5 : 6 then A : B : C is _____
1. 15 : 20 : 24 2. 2 : 3 : 4 3. 3 : 2 : 7 4. None
12. Two numbers are in the ratio 4 : 5. If each is diminished by 6, they would be in the ratio 3 : 4. Find them.
1. 24, 30 2. 12, 15 3. 16, 19 4. 80, 100
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13. Write 6 : 5.2 as the ratio of integers
1. 5 : 4 2. 6 : 5 3. 8 : 7 4. 13 : 1
14. The incomes of A and B are in the ratio 4 : 3 and their expenditures are in the ratio 2 : 1. If each one saves Rs.1000, their incomes are
1. Rs.2000 and Rs.1500 2. Rs.3000 and Rs.2250
3. Rs.4000 and Rs.3000 4. None of these
15. Divide 27 into two parts so that 5 times the first and 11 times the second are together equal to 195
The French word ‘Cent’ or the Latin word ‘Centum’ means hundred.
Therefore the word percentage means per hundred or hundredths.
Any fraction with hundred in the denominator is called percentage.
Percent is indicated using the symbol: %
80% is ; 5% is ; 365% is 365
100
A percentage can be represented by a fraction or a decimal.
Important points to remember1. Converting Fractions, Numbers and Decimals to Percentages
5 can also be written as 500%
1.25 can also be written as 125%
3.75 can also be written as 375%
0.575 can also be written as 57.5%
2. A fraction can be converted to percentage by multiplying by 100
100 = 0.8 x100= 80%
Percentage Fraction Decimal
75% 0.75
30% 0.3
65% 0.65
23% 0.23
66% 0.66
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100 = 0.1666 x 100 = 16.66%
3. Finding a percent of a given number.
P% of A =
Example:
a) 10% of 50 = =5
b) 25% of 120= =30
4. Percentage Comparison.
Given x and a, to find x is what% of a
Multiply the ratio by 100 and add % sign, i.e.,
Example:
a) 4 is what percent of 60? 100 = 6.67%
b) 10 is what percent of 100? 100 = 10%
c) 207 is what percent of 100? 100 = 207%
5. Percent Increase and Decrease
In many cases we are interested in knowing by how much fraction of its origin value a quantity has changed. If the value is given in percentage then:
Percentage change =
6. Percentage Error
Error = Actual value - Value taken (AV>VT)
Error = Value taken - Actual value (VT>AV)
If the value of x is taken as y then error is y - x
, where (y – x) is error and x is the actual value
7. Relative Percentage
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In relative percentage, there is percentage comparison between two or more quantities.
Example:
If A’s income is 20% less than B’s, then by what percentage is B’s income more than A?
Solution:
Let B’s income be Rs.100
A’s income would be 20% less than B
Hence, A’s income = 80% of Rs.100 = Rs.80
The percentage by which B’s income would be more than A
= % = 25%
Worked Examples1. Find 5 percent of 400.
Solution:
5% = = 0.05
Hence, the required amount = 0.05 x 400 = 20
2. Find 2% of Rs.18.75 to the nearest paise.
Solution:
2% of Rs.18.75 18.75 = 0.02 x 18.75 = 0.375 = 38 paise.
3. In a school the strength of the students is 260 out of which 65% are boys. Find the number of boys and girls in the school.
Solution:
Total number of students = 260
Given that 65% of the students are boys. Hence, 35% of the students are girls.
No. of girls = 35
100 260 = 91 and, No. of boys = 260 – 91 = 169
There are 169 boys and 91 girls in the school.
4. A piece of cloth 50 meters long shrinks by 0.6%. How much did it shrink?
Solution:
0.6% of 50 meter = 0.3 meters = 30 cm
5. In a spelling test of 80 words, Ravi spelt 80% of the words correctly. How many words did he spell right?
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Solution:
The number of words in the spelling test = 80 words.
Ravi spells 80% words correctly.
No. of words spelt correctly by Ravi = 80
100 80 = 64 words.
6. A man spends 25% of his money and then Rs.75 and then 5% of the remainder. If he had Rs.3291.75 left with him, how much money did he possess initially?
Solution:
Let the money before spending 5% be Rs.100.
So, amount after spending 5% is Rs.(100 - 5) = Rs.95
Actual amount of money after spending 5% = Rs.3291.75
So, actual amount of money before spending 5% = Rs. = Rs.3465
Therefore, amount of money left before spending Rs.75 = Rs.3465 + Rs.75 = Rs.3540
If the original money is Rs.100, then the amount of money left after spending 25% of Rs.75.
Money left after spending 10% Original money
Rs.75 Rs.100
Rs.3540 Rs.4720
Therefore his original sum of money is Rs.4720.
7. Find the percentage error in writing 2.54 as 2.5.
Solution:
Actual value=2.5
Value taken=2.54
Error value=2.54-2.5=0.04
8. The price of rice is raised by 30%, by what percent should a house wife reduce consumption so as to not to increase her total expenditure on rice?
Solution:
Let her initial consumption be 100 kg at a price of Rs.10 per kg.
Hence her total expenditure = 100 10 = Rs.1000
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The new price after increase of 30% is 13 per kg
Let the consumption be X kgs
Then,
13X = 1000
Alternative Method
where X is the increase in percentage.
9. Satish’s income is 20% more than that of Sanjana’s. By what percentage is Sanjana income less than that of Satish?
Solution:
Method 1
Let Sanjana income be Rs.100
Satish’s income is
Sanjana income is less than Satish by
= %
Method 2
, where x is percentage difference
10. If the present population of a town is 20000, which is increasing at r rate of 10% every year what will be population of the town after 2 years?
Solution:
56
(Or Original price - 15% of Rs.40, 000) = 40,000 = Rs.34,000
Practice Exercise 1
1. X’s salary is half that of Y. If X got a 50% increase in his salary while Y got a 25% increase, then what is the percentage increase in the total salary received by both?
1. 50% 2. 25% 3. 33 % 4. 66.66%
2. A man spends 30% of his salary on food and 60% of the balance on clothing. If his total savings per month is Rs.265, what is his total annual income?
3. A’s income is 10% more than B’s. By what percentage is B’s income less than A’s?
1. 10% 2. 9.09% 3. 10.9% 4. 11%
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4. Kumar earns 25% more while Raman earns 75% more than their colleague Mohan. If the monthly income of Raman is Rs.1750, what is Mohan’s salary as a percentage of the combined salary of the three colleagues?
1. 50% 2. 25% 3. 33 % 4. 40%
5. A mango tree’s height is 30% less than that of a coconut tree. By what percent is the coconut tree taller than the mango tree?
1. 42.8% 2. 40% 3. 28.3% 4. 34.15%
6. What is 20% of 50%?
1. 10% 2. 40% 3. 25% 4. None of these
7. If the population of a town increases by r% every year, then the population after ‘t’ years is
1. 2. 3. 4.
8. When the price of a radio was reduced by 20% the quantity sold increased by 80%. What was the net effect on the sales revenue?
9. A man spends 40% of his salary on books. Out of this amount, 60% is spent on buying text books, 20% on novels and the remaining on journals. If 80% of the money on journals is spent on buying foreign journals and the remaining on Indian journals, what percentage of his salary is spent on Indian Journals?
1. 2.4% 2. 1.6% 3. 6.4% 4. 3.2%
10. 50% of a% of b is 75% of b% of c. What is c?
1. 0.5a 2. 0.667a 3. 1.25a 4. 1.5a
11. The drivers of 65% of BTS buses are on strike today. 4% of the remaining buses are under repair due to accidents. If the number of buses still operating is 168, what is the total number of BTS buses operating under normal circumstances?
1. 542 2. 500 3. 600 4. 550
12. In a test, a student gets 80% of what he expected. His expected score happens to be 60% more than the class average. What is his score as a percentage of the average score?
1. 115% 2. 160% 3. 144% 4. 128%
13. Two sides of a cuboid are increased by 10% and the third side is decreased by 20%. What is the percentage volume change?
14. A block of ice is to be carried through a certain distance. The ice melts by th of its
own weight for every rd of the total distance. What percent of ice remains at the end?
1. 94% 2. 72% 3. 33.33% 4. 66.66%
15. Mohan spends 50% of his monthly allowance on food, 18% on magazines, 10% on writing material and 15% on miscellaneous items. If he saves Rs.840 at the end of the year out of his allowance, what is his monthly allowance?
1. Rs.3000 2. Rs.2200 3. Rs.1000 4. Rs.750
Practice Exercise 21. Star TV is presently connected to 2 million houses in India. If 1000 new connections are
given every day, what is the annual percentage growth in star TV viewer-ship?
1. 18.25% 2. 22.15% 3. 7.07% 4. 11.11%
2. The price of wheat drops by 20%. How many kg can be bought now with the same money that was sufficient to buy 16 kg prior to price change?
1. 18 2. 20 3. 20.5 4. 17
3. In a village, 12% of goats were lost in a flood and 5% of the remainder died due to starvation. If the number of goats remaining is 8,360, what was the original number of goats before the flood?
1. 10,000 2. 8,800 3. 9,400 4. 10,240
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4. Anupama’s income is 15% greater than Sivakami’s. By what percent is Sivakami’s income less than Anupama’s?
1. 13.04% 2. 15% 3. 20% 4. 17.1%
5. A salt solution contains 5% salt and the remaining water. The total volume of the solution is 10 litres. If 2 litres of water evaporate, what percent of the remaining solution is salt?
1. 6 % 2. 5% 3. 7 % 4. 6 %
6. The price of Vadilal ice cream is 50% more than Dollops and that of Walls is 50% more than Vadilal. By what percent is Dollops cheaper than Walls?
1. 50% 2. 66.66% 3. 55.55% 4. 60%
7. The sales of a business in 1977 was 20% less than in 1976. By what percent must the sales in 1978 be increased to bring it to the level of 1976?
1. 20% 2. 25% 3. No increase4. 30%
8. The price of sugar is increased by 20% and its consumption is reduced by 20%. Find out the net increase/decrease in expenditure on sugar.
9. If the price of each cup of tea served in a restaurant is reduced by 20%, one can buy 3 more cups of tea for Rs.30. What is the price of tea per cup?
1. Rs.2.00 2. Rs.2.25 3. Rs.2.30 4. Rs.2.50
10. A company manufactures electric tube lights which are priced at Rs.25 each. The tube lights consume power @ Re.1 per hour. The company is planning to launch a new type of tube-light which, as per the tests conducted, can cut the electricity bill by 50%. What higher percentage of price can the company fix for the new tube over the old variety if the extra cost is to be recovered in just 10 hours of lighting?
1. 50% 2. 35% 3. 20% 4. 48%
11. A and B are two cinema halls in a city. A charges an entrance fee of Rs.10.80 which includes entertainment tax @ 20% on Basic fee. Basic fee at B is cheaper by 60%, tax being 25%d on basic fee. By what percentage is the entrance fee of A costlier than that of B?
1. 350% 2. 140% 3. 235% 4. 180%
12. A fruit vendor is sanctioned a loan of Rs.5000 from a bank. The loan carries an interest of 40% compounded half-yearly. The vendor intends to recover the entire annual interest within 4 months of operation. By what percent should he load the prices, if he currently has an average monthly income of Rs.4400 and has 100% margin.
1. 25% 2. 5% 3. 15% 4. 30%
13. A rectangle shaped orchard has trees planted in 12 rows and 48 columns. The adjoining land is empty and square in shape, with two sides being half and double of the existing rectangular orchard farm. How many trees can be planted in the square land if the same spacing between trees is followed?
1. 400 2. 900 3. 121 4. 576
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14. Suresh weighs 65 kg. He has an attack of viral fever and loses 1/5 of his weight. Within a few days he recovers and gains 25% weight. But when the fever relapses after two weeks, he loses 10 kg. By what percentage should he improve his weight in order to reach his normal weight?
1. 20% 2. 18.2% 3. 25% 4. 15%
15. The price of an article is cut by 20%. By what percent should it be increased to sell it at the original price?
Profit, Loss & DiscountImportant points to remember
1. Cost Price (CP) : The price at which a commodity is purchased is called the COST PRICE of the commodity.
2. Selling Price (SP): The price at which a commodity is sold is called the SELLING PRICE of the commodity.
3. The amount that is deducted from the marked price is called Retailer’s Discount.
The difference between the marked price (List price) and the discount becomes the new selling price.
The reduction made on the marked price of the article is called the Discount.
DISCOUNT = Market price – Selling price.
4. The difference of cost price and selling price gives LOSS or PROFIT.
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Profit = Selling price – Cost price
Loss = Cost price – Selling price
Cost price = Selling price – Profit
Selling Price = Cost price + Profit
Cost price = Selling price + Loss
Selling price = Cost price – Loss
If the selling price is equal to the cost price then there is neither Profit nor Loss.
5. When profit or loss is calculated for Rs.100 (or any currency of 100 denominations), we get Profit Percentage and Loss Percentage.
Discount% = 100.
6. To find Selling Price, when Cost Price and Profit /Loss Percentage are given
7. The Selling Price of 2 commodities being the same, if one of them is sold at a Loss of R% and the other sold at a Profit of R%, then there will be an overall LOSS Percentage of
i.e., and, Loss =
Worked Examples1. A man sells two cars for Rs.2,00,000 each. On one he gains 5% and on the other he
loses 5%. Find his gain or loss percent on the whole transaction.
Solution:
Loss % = = .
2. If the selling price of 10 articles is the same as the cost price of 15 articles, find the gain percent.
Solution:
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Let C.P. of each article be Re.1.
C.P. of 10 articles = Rs.10.
S.P. of 10 articles = C.P. of 15 articles = Rs.15.
Gain = .
3. By selling 100 bags, a shopkeeper gains the selling price of 25 bags. Find his gain percent.
Solution:
(S.P. of 100 bags) - (C.P. of 100 bags) = S.P. of 25 bags
S.P of 75 bags = C.P. of 100 bags
Let C.P. of each bag be Rs.1
then C.P of 75 bag = Rs.75
S.P of 75 bags = Rs.100
Gain = = 33.33%.
4. Santosh sells 2 bags at Rs.200, each he gets a profit of 15% on one bag an a loss of 15% on the other bag. What is overall profit or loss?
Solution:
CP = SP = 200
CP = 200 = Rs.173.91
CP = SP ( ) = 200( ) = 200 = 235.2
Total CP = 235.24 + 173.91 = Rs. 408.25
Total SP = 400
He has loss.
Loss% = = = 2.25%
In the previous problem % profit and % loss was 15%.
Then loss% = = = 2.25%
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Loss =
Loss =
= = 8.022
But the above is not valid when the cost price of both the articles are same.
5. A dishonest dealer prefers to sell his goods at cost price but uses a weight of 950 gms for 1 kg. Find his gain percent.
Solution:
Gain % = = .
6. By selling an article of Rs.120, one gains Rs.20. Find the gain percent.
Solution:
S. P. = Rs. 120, Gain = Rs. 20
C. P. = Rs. (120 - 20) = Rs. 100
Gain of Rs. 100 is Rs. 20
So, gain % = .
7. By selling an article for Rs.100, a man loses 20% for how much should he sell it to gain 8%?
Solution:
S.P. = Rs.100, loss = 10%
C.P = Rs. = Rs.125
Now C.P = Rs.125, Gain required, = 8%
S.P = Rs. = Rs.135.
8. A man loses Rs.50 by selling some pens at the rate of Rs.3 per piece and gains Rs.100, if he sells them at Rs.4 per piece. Find the number of pieces sold by him.
Solution:
Suppose the no. of all pieces sold is x. Then, the difference of selling prices at the new rates is Rs.150.
Rs.4x – Rs.3x = 150 x = 150
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Thus the number of pens sold = 150.
9. The percent profit made when an article is sold for Rs.50 is twice that when it is sold for Rs.45. Find the C.P of the article.
Solution:
Let C.P. be Rs.x. Then
=
50 - x = 2(45 - x) x = 40
Thus, C.P. of the article = Rs.40.
10. X spends 80% of his income. His income was increased by 10% and his expenditure was increased by 5%. By what percentage did his savings increase?
Solution:
Let the original income of X = Rs.100
Expenditure = 80% of 100 or Rs.80
His savings = Rs.100 - Rs.80 = Rs.20
X’s new expenditure = (100 + 5)% of 80 = Rs.84
Savings = Rs.110 - Rs.84 = Rs.26
Increase in savings = Rs.26 - Rs.20 = Rs.6
Percentage of increase in savings = = 30%
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Practice Exercise 11. Find the selling price if the cost price is Rs.150 and the profit is 20%.
1. 120 2. 180 3. 210 4. 100
2. Find the cost price if selling price is Rs.200 and profit is 12%.
1. Rs.175.87 2. Rs.170 3. Rs.178.57 4. Rs.185.77
3. Find the percentage profit if the cost price is Rs.175.00 and selling price is 250.00.
1. 42.8 2. 48.2 3. 48 4. 40.8
4. Find the selling price if cost price is Rs. 1,000 and percentage loss is 10.
1. 1100 2. 1010 3. 1001 4. 900
5. A sells a cycle to B at a profit of 20% and B sells it to C at a profit of 25%. If C pays Rs. 225 for the bicycle, how much did A pay for it?
1. Rs.150 2. Rs.120 3. Rs.125 4. Rs.225
6. A merchant sells two articles for Rs. 150/- each. He sells one at a loss of 10% and the other at a profit of 10%. What is his net profit or loss?
1. Loss of Rs.1.50 2. Gain of Rs.3.03 3. Loss of Rs.3.03 4. No profit, no loss
7. A fruit seller bought some oranges at 4 per rupee and an equal number at 5 per rupee. He then sold the entire quantity at 9 for two rupees. What is his percentage gain or loss?
1. 8 % 2. 2% 3. 1 % 4. No profit, no loss
8. A tape recorder is sold at Rs.935 at a profit of 10%. What would have been the actual profit or loss if it had been sold for Rs.810?
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1. Rs.55 loss 2. Rs.75 loss 3. Rs.20 gain 4. Rs.40 loss
9. A merchant sells rice and makes a profit of 6%. His cost price increases by 10% and as a result he increases his selling price also by 10%. What percent profit does he earn now?
1. 6% 2. 6.6% 3. 10% 4. None of these
10. Ram buys a Maruti car at a 20% discount and immediately sells it at 20% more than the list price. What will be his profit?
1. 40% 2. 50% 3. 60% 4. 20%
11. The printed sale price of a book is Rs.60. The trader offers two successive discounts of 20% and 30%. The net sale price is subject to a sales tax of 5%. How much does the buyer pay?
1. Rs.35.28 2. Rs.31.50 3. Rs.36.00 4. Rs.32.48
12. John sells two cameras at Rs.1000 each. On one camera he gets a profit of 10%, while on the other he incurs a loss of 10%. What is his overall profit or loss?
1. Gain of Rs.20 2. No profit, no loss 3. Loss of Rs.20.20 4. None of these
13. A retailer buys a dozen pens for Rs. 96. The wholesaler gives him a pen free on the condition that he should sell at Rs. 8 per pen. What will be the retailer’s percentage profit?
1. 8.33% 2. 10% 3. 6.66% 4. 12.25%
14. A cloth merchant sells at a profit of 10%. What is his net profit if he sells goods and bills Rs.1000 on a particular day, given that the measuring tape of the merchant measures 995 mm, instead of 1m?
1. Rs.95.45 2. Rs.82.33 3. Rs.110.25 4. Rs.78.75
15. A trader buys 78 kg of wheat for Rs.892. He sells 40% of this quantity at a loss of 20%. What should be his percentage mark up on the remaining in order that he earns an overall profit of 25%?
1. 40% 2. 55% 3. 28% 4. 45%
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Practice Exercise 21. Sheela buys eggs at Rs. X per dozen and sells them at 8 times that price per hundred.
What is her percentage profit or loss?
1. 16% profit 2. 4% profit 3. 4% loss 4. 16% loss
2. Ram buys an article at th its value and sells it at 50% more than its value. What is his
percentage profit?
1. 50% 2. 100% 3. 75% 4. 125%
3. A manufacturer’s list price for an item is 40% more than its cost. If he allows a discount of 10% on the list price, what will be his percentage profit?
1. 30% 2. 28% 3. 26% 4. 24%
4. Krishna buys three articles at Rs.800 each. He sells the first at 15% profit and the second at 20% loss. What should be his percentage profit on the third article to ensure that he makes an overall profit of 10%?
1. 30% 2. 35% 3. 40% 4. 25%
5. X buys 6 books for Rs.100 and sells 5 books for Rs.100. What is his percentage profit?
1. 30% 2. 35% 3. 40% 4. 25%
6. Shyam buys X eggs to resell them at a profit of 10% but loses 10% of the eggs due to transit damage. By how much should he mark up the selling price of the eggs in order to retain a 10% profit?
1. 30% 2. 40% 3. 33.33% 4. 22.22%
7. A man sells 2 goats at Rs.120 each. By doing so, he gains 25% on one goat and loses 25% on the other. What is his loss/gain percent if any?
1. 0 2. 6.25 3. 4.3 4. 2
8. A merchant loses 7% on a certain tea, but he uses a weight of 900 gm instead of 1 kg. Find his profit %.
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1. 3.33 2. 2.05 3. 4.67 4. 4.21
9. A sells a bicycle to B at a profit of 20% and B sells it to C at a profit of 25%. If C pays Rs.225 for the bicycle, how much did A pay for it?
1. 100 2. 125 3. 150 4. 200
10. How much percent above the C.P. should a shop keeper mark his goods so that he gains 26% after allowing a discount of 10%?
1. 40 2. 50 3. 70 4. 30
11. A trader marks his goods 25% above the C.P. and allows his customers a discount of 12%. What is his profit%?
1. 6 2. 8 3. 10 4. 9
12. A saree was priced at Rs.1,200/-. After two successive discounts, the selling price was Rs.816. If the first discount was 15%, find the rate of the second discount.
1. 10% 2. 15% 3. 20% 4. 33 %
13. By selling 25 metres of cloth at Rs.50 per metre, a merchant earns a profit equivalent to the cost price of 5 metres of cloth. The percentage profit is
1. 15% 2. 25% 3. 20% 4. 18%
14. A man buys 50 chairs for Rs.5,000/- but 20 of them are damaged. He decides to sell the
damaged chairs each at th the price of the normal one. What should be the price of a
normal chair in order that the person makes a profit of 35% for the whole lot?
1. Rs.125 2. Rs.170 3. Rs.130 4. Rs.175
15. An article was bought for Rs. 1,300 and sold immediately for Rs. 1,436.50 payable in 10 months time. Find the loss or gain accrued if the rate of interest is 10% per annum?
The word “interest” has been taken from the Latin word Interesse.
1. Interest is the money paid for the privilege of using another person’s money. When a lender lends the money to the borrower, the money that is lent is called the PRINCIPAL and the sum of money paid for the use of it is called the INTEREST.
2. In other words, the principal is the amount of money that is borrowed and interest is the proportionate sum of money paid for the use of it. The sum of the interest and principal at the end of any time is called the AMOUNT.
Interest: It is money paid for the privilege of using another person’s money.
Principal: When Lender lends the money to the borrowers, the money that is lent is called the principal.
Amount: Amount = Principal + Interest.
The sum of the interest and principal at the end of any time is called the Amount.
Simple Interest1. Simple interest is the interest calculated on the original P rincipal . It is sum of money paid to the
lender by the borrower for particular period.
2. Simple Interest on Principal “P” for “T” number of years, at then Rate of R% per annum is given by,
And the Amount,
A = P + I where I =
70
A = P +
The principal amount does not change for any time period, while calculating Simple Interest.
Compound Interest1. In compound interest calculations, the interest for each period is added to the principal
before interest is calculated for the next period. With this method the principal grows as the interest is added to it. This method is used in investments such as savings account and bonds.
2. When the compound interest is calculated Yearly:
1. Let Principal = P
Rate = R%
Time = n years
2. Let the rates of interest for 2 successive years be R1% & R2% per annum, respectively. Let A be the amount after 2 years, then
If R1%, R2% and R3% be the interest rates for 3 successive years then the amount after 3 years will be
3. If the period is not complete number of year:
If it is n years and m months and interest is R% p.a. compounded yearly, then
3. Simple Interest and Compound Interest for 1 year at a given Rate of Interest per annum are always Equal.
4. When the compound interest is calculated Half-Yearly:
Principal = P, Rate = R% per annum, Time = n years.
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Since interest is earned half yearly, the rate of interest will be for every half year, and time
will be 2n half years.
Let A be amount after n years, then
5. When the compound interest is calculated Quarterly:
Rate of interest is per quarter.
Time = 4n (quarterly)
Then the Amount,
Worked Examples
1. Mr. Christy borrowed Rs. 7,500 from a bank on 1st June 1994 at 63
4% p.a. S.I. and paid the amount
back on 3rd June 1995. Find the amount paid.
Solution:
Amount = Principal + S.I.
Principal = Rs. 7500
S.I. = , R = 27
4% p.a.
N = 12 months + 3 days 1 year
S.I. = = Rs. 506.25
Amount = Rs. 8006.25.
2. Find S.I on Rs. 8000/- at 161
2% p.a. for the period from 7th June 1997 to 8th September 1997.
Solution:
P = Rs. 8000, N = June 23 days + September 8 days + 2 months
N =( + ) years = 0.252 years
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R = % p.a.
So, I = = Rs. 332.64
3. On what capital will S.I. for 3 weeks and 4 days at 12% p.a. amount to Rs. 16?
Solution:
S.I. = ; R = 12% ; S.I. = 16
N = 3 weeks + 4 days = years
N = 0.0685 years
16 =
P = = Rs. 1946.5
Capital or Principal amount = Rs. 1946.5
4. A sum of money trebles itself in 15 years 6 months. In how many years would it double itself?
Solution:
Let the sum i.e. P = x
When the sum trebles itself, S.I. = 2x, N = 151
2 years = years
2x = x×R 31
×100 2
R = %
When the sum doubles itself, S.I. = x
x = x N N = = 7.75 years
i.e., 7 years 9 months.
5. Find the compound interest on Rs.6500 for 3 years at 10% p.a. compounded annually.
Solution:
C.I. = A - P
A = P (1 + r)n
P = 6500 A = 6500
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A = 6500(11)3 = 8651.5
C.I. = 8651.5 - 6500= Rs. 2151.5
6. At what rate percent of compound interest will Rs.970 earn Rs.250 as interest in 10 years?
Solution:
C.I = Rs. 250
P = Rs. 970
A = P + C.I = 970 + 250 = Rs. 1220
A = P (1 + r)n
1.258 = (1+r)10
1 + r =
7. A certain sum put at S.I. amounts in 4 years to Rs.780 and double the same sum amounts to Rs.2000 in 6 years. Find the sum and rate of interest.
Solution:
Amount = P +
780 = P + .......... (1)
and,
2000 = ......... (2)
Equation (2) Equation (1)
2000
R = -22
R = = 32.4% p.a.
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8. Mr. Christy and Mr. Clint lend Rs.400 for 2 years at 4% p.a. compound interest. Mr. Christy’s interest is paid half yearly and Mr. Clint’s interest is paid yearly. Find the difference between the amounts.
Solution:
Christy’s case
P = Rs. 400
n = 2 years
R = 4%
(Because interest is payable half yearly.)
A = P = 400 = 400 (1.02)4 = Rs. 432.97
Clint’s case
A = P = 400 = 400 (1.04)2 = Rs. 432.64
Difference between the amounts is Rs.0.33/-
9. Find the compound interest on Rs.5600 for 11
2 year at 10% p.a. compounded annually.
Solution:
A = 5600 = Rs.6468
C.I = 6468 - 5600 = Rs.868
10. Find the CI on Rs. 10,000 in 10 years if the rate of interest is 5% for the first 2 years, 8% for the next 3 years, 9% for the 6th and 7th years and 10% for last 3 years.
Solution:
CI = A - P
A = 10,000 = Rs. 21,965
CI = 21,965 - 10,000 = Rs. 11,965
{Note using varying rate of interest of interest formula}
Practice Exercise 1
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1. A sum of money placed at CI doubles itself in 5 years. It will amount to eight times itself in how
many years?
1. 15 years 2. 20 years 3. 33 years 4. 21 years
2. In how many years will a sum at Rs.800 at 10% per annum compounded semi-annually become
Rs.926.10?
1. 4.5 years 2. 1.5 years 3. 4 years 4. 6 years
3. A sum of money, compounded, amounts to Rs.6690 after 3 years and to Rs.10035 after 6 years.
Find the sum.
1. Rs.1110 2. Rs.5560 3. Rs.5432 4. Rs.4460
4. A certain sum amounts to Rs.7350 in 2 years and to Rs.8575 in 3 years. Find the sum and rate