Chapter 01 : Number Theory > Topic 1 : Types of Numbers
The number theory or number systems happens to be the back bone for CAT preparation.
Number systems not only form the basis of most calculations and other systems in
mathematics, but also it forms a major percentage of the CAT quantitative section. The
reason for that is the ability of examiner to formulate tough conceptual questions and
puzzles from this section. In number systems there are hundreds of concepts and
variations, along with various logics attached to them, which makes this seemingly easy
looking topic most complex in preparation for the CAT examination. The students while
going through these topics should be careful in capturing the concept correctly, as it’s not
the speed but the concept that will solve the question here. The correct understanding of
concept is the only way to solve complex questions based on this section.
Real numbers: The numbers that can represent physical quantities in a complete manner.
All real numbers can be measured and can be represented on a number line. They are of
two types:
Rational numbers: A number that can be represented in the form p/q where p and q are
integers and q is not zero. Example: 2/3, 1/10, 8/3 etc. They can be finite decimal
numbers, whole numbers, integers, fractions.
Irrational numbers: A number that cannot be represented in the form p/q where p and q
are integers and q is not zero. An infinite non recurring decimal is an irrational number.
Example: √2, √5 , √7 and Π(pie)=3.1416.
The rational numbers are classified into Integers and fractions
Integers: The set of numbers on the number line, with the natural numbers, zero and the
negative numbers are called integers, I = {…..-3, -2, -1, 0, 1, 2, 3…….}
Fractions:
A fraction denotes part or parts of an integer. For example 1/6, which can represent 1/6th
part of the whole, the type of fractions are:
1. Common fractions: The fractions where the denominator is not 10 or a multiple of it.
Example: 2/3, 4/5 etc.
2. Decimal fractions: The fractions where the denominator is 10 or a multiple of 10.
Example 7/10, 9/100 etc.
3. Proper fractions: The fractions where the numerator is less than the denominator.
Example ¾, 2/5 etc. its value is always less than 1.
4. Improper fractions: The fractions where the numerator is greater than or equal to the
denominator. Example 4/3, 5/3 etc. Its value is always greater than or equal to 1.
5. Compound fraction: A fraction of a fraction is called a compound fraction
Example 3/5 of 7/9 = 3/5 x 7/9 = 21/45
6. Complex fractions: The combination of fractions is called a complex fraction.
Example (3/5)/ (2/9)
7. Mixed fractions: A fraction which consists of two parts, an integer and a fraction.
Example 3 ½, 6 ¾
Example: Express 27/8 as a mixed fraction
Ans. Divide the numerator by denominator; note the multiplier, whatever remainder is
left divide it with the original denominator. For 27/8, 24/8 = 3, and remainder left is 3,
therefore 3 3/8 is the mixed fraction
Example: Express 35 7/17as an improper fraction.
Ans. Here we need to multiply the denominator with the non-fraction part and add it to
numerator and using same denominator.
For 35 7/17= = 602/17
The integers are classified into negative numbers and whole numbers
Negative numbers: All the negative numbers on the number line, {…..-3, -2, -1}
Whole numbers: The set of all positive numbers and 0 are called whole numbers, W =
{0, 1, 2, 3, 4…….}.
Natural numbers: The counting numbers 1, 2, 3, 4, 5……. are known as natural
numbers, N = {1, 2, 3, 4, 5…..}. The natural numbers along with zero make the set of the
whole numbers.
Even numbers: The numbers divisible by 2 are even numbers. e.g., 2, 4, 6,8,10 etc. Even
numbers can be expressed in the form 2n where n is an integer other than 0.
Odd numbers: The numbers not divisible by 2 are odd numbers. e.g. 1, 3, 5, 7, 9 etc.
Odd numbers are expressible in the form (2n + 1) where n is an integer other than 0.
Composite numbers: A composite number has other factors besides itself and unity .e.g.
8, 72, 39 etc. A real natural number that is not a prime number is a composite number.
Prime numbers: The numbers that has no other factors besides itself and unity is a prime
number. Example: 2, 23,5,7,11,13 etc. Here are some properties of prime numbers:
• The only even prime number is 2
• 1 is neither a prime nor a composite number
• If p is a prime number then for any whole number a, ap – a is divisible by p.
• 2,3,5,7,11,13,17,19,23,29 are first ten prime numbers (should be remembered)
• Two numbers are supposed to be co-prime of their HCF is 1, e.g. 3 & 5, 14 & 29 etc.
• A number is divisible by ab only when that number is divisible by each one of a and b,
where a and b are co prime.
• To find a prime number, check the rough square root of the given number and divide
the number by all the prime number lower than the estimated square root
• All prime numbers can be expressed in the form 6n-1 or 6n+1, but all numbers that can
be expressed in this form are not prime
Example: If a, a + 2 and a + 4 are prime numbers, then the number of possible solutions
for a is: (CAT 2003)
(a) 1 (b) 2
(c) 3 (d) more than 3
Ans. (a) a, a + 2, a + 4 are prime numbers. The number fits is 1, 3, 5 and 3, 5, 7 but post
this nothing will fit. Now 1, 3, 5 are not prime numbers as 1 is not prime number.
So, only one possibility is there 3, 5, 7 for a = 3.
Prime Factors: The composite numbers express in factors, wherein all the factors are
prime. To get prime factors we divide number by prime numbers till the remainder is a
prime number. All composite numbers can be expressed as prime factors, for example
prime factors of 150 are 2,3,5,5.
A composite number can be uniquely expressed as a product of prime factors.
e.g. 12 = 2 x 6 = 2 x 2 x 3 = 22 x 31
20 = 4 x 5 = 2 x 2 x 5 = 22 x 51 etc
Note : The number of divisors of a given number N ( including one and the number itself
) where N = am x bn x cp ……. Where a, b, c are prime numbers
are = ( 1 + m ) ( 1 + n ) ( 1 + p ) …………..
e.g. 90 = 2 x 3 x 3 x 3 x 5 = 21 x 32 x 51
Hence here a = 2 b = 3 c = 5
m=1n=2p=1
then the number of divisors are = ( 1 + m ) ( 1 + n ) ( 1 + p ) = 2 x 3 x 2 = 12
the factors of 90 = 1 , 2 , 3 , 5 , 6 , 9 , 10 , 15 , 18 , 30 , 45 , 90 = 12
the sum of divisors of given number N is ( am+1 – 1 ) ( bn+1 - 1 ) ( cp+1 – 1 ) ……..
___________________________________
( a – 1 ) ( b – 1 ) ( c – 1 ) ……….
Perfect number: If the sum of the divisor of N excluding N itself is equal to N , then N is
called a perfect number. e.g. 6, 28, 496
Finding a perfect number through Euclid’s method
Euclid's method makes use of the powers of 2, which are numbers obtained by
multiplying by 2 by itself over and over again, which are 1, 2, 4, 8, 16, 32, 64, 128….
Note that the sum of the two numbers in this series (in ascending order) is equal to the
third number minus 1:
1+2 = 3 = 4 - 1,
1+2+4 = 7 = 8 - 1,
STARTING FROM THE NUMBER 1, IF YOU ADD THE POWERS OF 2 AND IF
THE SUM IS A PRIME NUMBER, THEN YOU GET A PERFECT NUMBER BY
MULTIPLYING THIS SUM TO THE LAST POWER OF 2.
If you add 1+2, the sum is 3, which is a prime number. Therefore 3 x 2 = 6 is a perfect
number.
If you add 1+2+4, the sum is 7, a prime number. Therefore 7 x 4 = 28 is a perfect
number.
If you add 1+2+4+8, the sum 15 is not a prime number, so you can't use Euclid's method
here.
If you add 1+2+4+8+16, the sum is 31, a prime number. Therefore 31 x 16 = 496 is
another perfect number.
Absolute value of a number:
The absolute value of a number a is | a | and is always positive.
Fibonacci numbers: The Fibonacci numbers is a sequence where
X (n+2) = X (n+1) + X (n), X (1) = 1, X (2) = 1
Example1,1,2,3,5,8,13,21,34,55,89,144.., it can be clearly seen that any number in the
series is the addition of the last two numbers, other than the first two numbers
Example: The price of pens has increased over the years. Each year for the last 7 years
the price has increased, and the new price is the sum of the prices for the two previous
years. Last year a pen cost 60 rupees. How much does a pen cost today? How much did a
pen cost 7 years ago?
Let this year price be x. Last year it was 60, so the previous year it must have been x-60,
continuing this process backwards gives us a sequence of expressions:
x, 60, x-60, 120-x, 2x-180, 300-3x, 5x-480, 780-8x, 13x-1260
All of these increases must be positive as every year prize has gone up. That gives us a
sequence of inequalities, each of which can be solved to find a range for x:
x>0
60 > 0
x-60 > 0 x > 60
120-x > 0 x < 120
2x-180 > 0 x > 90
300-3x > 0 x < 100
5x-480 > 0 x > 96
780-8x > 0 x < 97.5
13x-1260 > 0 x > 96.92
Looking at this, we can say
96.92 < x < 97.5
The whole number value x can have is 97, with which we get
x = 97
60 = 60
x-60 = 37
120-x = 23
2x-180 = 14
300-3x = 9
5x-480 = 5
780-8x = 4
13x-1260 = 1
Seven years ago, the price was 4 rupees
In the CAT/MCQ format, where you have the four answers, you can check it by working
forward and seeing if the results are correct. You can try putting the given answers for
original price and see which one fits in the equation.
Golden ratio: The golden ratio is a special number approximately equal to:
1.6180339887498948482...
Golden ratio = (1 + √ 5)/2
To find the golden ratio, we define the golden ratio as the ratio between x and y if
xy
--- = ----y x+y
Let's say x is 1. Then we have 1/y = y/(y+1). If we solve this equation to find y, we'll find
that it is the value given above, about 1.618
A golden rectangle is a rectangle in which the ratio of the length to the width is the
golden ratio.
The concepts like Fibonacci and golden ratio are reference concepts, students are advised
not to cram them but just understand the concepts as they are.
Chapter 01 : Number Theory > Topic 2: Basic Arithmatic Operations
Addition, subtraction, multiplication and division are the four basic mathematical
operations. We have not gone into details of these concepts as they are very basic; we
have added some formulae wherever required. Students preparing for CAT are expected
to know the basic arithmetic.
Addition: Addition is used to find the total as a single number of two or more given
numbers. The number obtained is called the sum of two numbers.
Subtraction: Subtraction is the quantity left when a smaller number is taken from a
greater one. The number obtained is called the difference of two numbers. If a smaller
number is subtracted from a greater number, the difference is positive; if a greater
numbers is subtracted from a smaller number the result is negative.
Multiplication: Multiplication is the short method of finding the sum of given number of
repetitions of the same number. The resultant sum of the repetition is called the product.
If one factor is zero then the product is zero. If same factors are multiplied, they can be
represented as power or the exponent for example 3 x 3 x 3 = 33
Some short methods in multiplication :
1. multiplication by 11 , 101 , 1001 etc
Rule: add 1, 2, 3 zeroes respectively to the multiplicand and add the multiplicand to the
resulting number.
Ex 5023 x 11 = 50230 + 5023 = 55253
i. 5023 x 1001 = 5023000 + 5023 + 5028023
2. Multiplication by 5
Rule: annex a zero to the right of the multiplicand and then divide it by 2
Ex 89356 x 5 = 893560/2 = 446780
3. Multiplication by 25
Rule: annex two zeroes the right of the multiplicand and then divide it by 4
Ex 890023 x 25 = 89002300/4 = 22250575
4. Multiplication by 125
Rule: annex 3 zeroes to the right of the multiplicand and then divide it by 8
5. Multiplication by a number wholly made of nines, i.e. 9 , 99 , 999 etc
Rule: place as many zeroes to the right of the multiplicand as there are nines in the
multiplier and from the result subtract the multiplicand.
Ex: 895023 x 999 = 895023000 - 895023 = 894127977.
6. Power Patterns: see the table below and notice the pattern of last digits of powers:
• Pattern of 3: 3, 9, 7, 1 – repeat every four powers
• Pattern of 4: 4,6 – repeat every two powers
• Pattern of 7: 7, 9, 3, 1 – repeat every four powers
• Pattern of 8: 8, 4, 2, 6 – repeat every four powers
• Pattern of 9: 9,1 – repeat every two powers
Application:
Since we have seen the cyclicity of 2,3,7,8 is 4, if we want to find the last digit of any
power of these numbers of numbers with last digit as 2,3,7,8 (like 12, 13, 27) can be
calculated by finding out remainder of the power divided by four. The last digit of the
remainder power will be the last digit of given number.
Examples
Last digit of 232, since 2 has cyclicity of 4, 32/4 has remainder = 0, so the last digit will
be same as of 20 or 24, which is 6
Last digit of 325, since 3 has cyclicity of 4, 25/4 has remainder = 1, so the last digit will
be same as 31, which is 3
Example: What will be the unit’s digit in 12896?
Ans. 12896 = (12824) (12824) (12824) (12824)
Since we know multiple of 4 of power of 8, last digit is 6
Last digit = 6 × 6 × 6 × 6 = 6
Example: What is the last digit of 22^33^44^55^66^77
Ans. 22^33^44^55^66^77
It can be evaluated by just considering 2 instead of 22 and neglecting higher powers. Any
power of 33 × 4n = 3^4n ends in 1 ... that is ... it is of the form 5n + 1 thus 2^(5n + 1) as
cyclicity of 2 is 5 .....We will get the last digit as 2 × 1 = 2
Last digit of 66^77 = 6
Last digit of 55^66^77 = 5
Last digit of 44^55^66^77 = 44^(something)25 is same as 44^1 = 4
Last digit of 33^44^55^66^77 = 1
Last digit of 22^33^44^55^66^77 =2
Example: What is the digit in the unit’s place of 251? (CAT 1998)
(a) 2 (b) 8
(c) 1 (d) 4
Ans. (b) The cycle of powers of two is 2,4,8,6 as last digit and repeat. As per that a power
of 52(multiple of 4) has last digit of 6, there fore one behind 51 should have last digit of
8.
Division: Division is the method of finding how many times one number called the
divisor is contained in another number called dividend. The number of times is called the
quotient. The number left after the operation is called the remainder.
(Divisor * quotient) + Remainder = dividend
The number of divisors (including 1 and itself) of a given number N where
N = Am * Bn * Co … where A, B, C are prime numbers are (1+m)(1+n)(1+o)…
Example 2: 90 = 2 * 32 * 5, Here a,b,c are 2,3,5 and m,n,o are 1,2,1. So number of
divisors are 2*3*2 = 12, which actually are 1,2,3,5,6,9,10,15,18,30,45,90
Here the sum of the divisors is given by
(a(m+1) – 1)/(a -1) * (b(n+1) – 1)/(b -1) * (c(o+1) – 1)/(c -1) * ….
Taking values from the previous example
(22 – 1)/1 * (33 – 1)/2 * (52 – 1)/4 = 234
Tests for divisibility:
1. A number is divisible by 2 if its unit’s digit is even or zero
2. A number is divisible by 3 if the sum of its digit is divisible by 3.
3. A number is divisible by 4 when the number formed by last two right hand digits is
divisible by 4.
4. A number is divisible by 5 if its unit’s digit is 5 or zero
5. A number is divisible by 6 if it’s divisible by 2 and 3 both.
6. Divisibility by 7 has two ways:
Take the last digit, double it, and subtract it from the rest of the number; if the answer is
divisible by 7 (including 0), then the number is also. This method uses the fact that 7
divides 2*10 + 1 = 21. Start with the numeral for the number you want to test. Chop off
the last digit, double it, and subtract that from the rest of the number. Continue this until
you get a one-digit number. The result is 7, 0, or -7, if and only if the original number is a
multiple of 7.
Example 3:
123471023473
--> 12347102347 - 2*3 = 12347102341
--> 1234710234 - 2*1 = 1234710232
--> 123471023 - 2*2 = 123471019
--> 12347101 - 2*9 = 12347083
--> 1234708 - 2*3 = 1234702
--> 123470 - 2*2 = 123466
--> 12346 - 2*6 = 12334
--> 1233 - 2*4 = 1225
--> 122 - 2*5 = 112
--> 11 - 2*2 = 7.
This rule holds good for numbers with more than 3 digits is as follows:
Group the numbers in three from unit digit.
add the odd groups and even groups separately
the difference of the odd and even should be divisible by 7
e.g. 85437954 the groups are 85, 437, 954
Sum of odd groups = 954 + 85 = 1039
Sum of even groups = 437
Difference = 602 which is divisible by 7
7. A number is divisible by 8 if the number formed by the last three right hand digits is
divisible by 8.
8. A number is divisible by 9 if the sum of its digits is divisible by 9.
9. A number is divisible by 10 if its unit’s digit is zero.
10. To check the divisibility by 11, take the test, alternately add and subtract the digits
from left to right. If the result (including 0) is divisible by 11, the number is also.
Example: to see whether 365167484 is divisible by 11, start by subtracting: 3-6+5-1+67+4-8+4 = 0; therefore 365167484 is divisible by 11
11. A number is divisible by 12 if it’s divisible by 3 and 4 both.
12. A number is divisible by 13 if it fits the following rule:
Delete the last digit from the number, and then subtract 9 times the deleted digit from the
remaining number. If what is left is divisible by 13, then so is the original number.
Example: 676, 67 – 6*9 = 13, which is divisible by 13 and so is 676
13. A number is divisible by 15 when it is divisible by 3 and 5 both. E.g. 930
14. A number is divisible by 25 if the number formed by the last two right hand digits is
divisible by 25. e.g. 1025, 3475, 55550 etc.
15. A number is divisible by 125 if the number formed by the last three right hand digits
is divisible by 125. e.g. 2125, 4250, 6375 etc.
Example: Which of following numbers are divisible by 12?
(a) 188078 (b) 12496
(c) 3961815 (d) 13685
(e) 28008
Ans. Divisibility rule of 12, number has got to be divisible by 3 and 4
188078, sum of digits = 42, divisible by 3, last two digits not divisible by 4, rejected.
12496, sum of digits = 22, not divisible by 3, rejected
3961815, sum of digits = 33, divisible by 3, last two digits not divisible by 4, rejected.
13685, sum of digits = 23, not divisible by 3, rejected.
28008, sum of digits = 18, divisible by 3, last two digits divisible by 4, it is divisible by
12.
Example: What least number must be added to 127561 so that it is exactly divisible by
28?
Ans. Least number to be added plus the remainder when divided by the given number
should give the divisor. Here when we divide 127561 by 28, quotient is4555 and
remainder is 21, so 21 + least number = 28, least number = 7
Example: Find the value of ‘a’ and ‘b’ if the seven digit number ‘267a34b’ is divisible
by 72.
Ans. For a number to be divisible by 72, it should be divisible by 8 and 9.
Applying rule for 8: number formed by last 3-digits should be divisible by 8.
34b should be divisible by 8, hence
b = 4.
For divisibility by 9: digit sum should by divisible by 9.
Digit sum = 22 + a + b = 26 + a
Hence, a = 1
Chapter 01 : Number Theory > Topic 3: Properties of Numbers
1. The numbers, which give a perfect square on adding as well as subtracting its reverse,
are rare and hence termed as Rare Numbers.
If X is a positive integer and X1 is the integer obtained from X by writing its decimal
digits in reverse order, then X + X1 and X - X1 both are perfect square then X is termed
as Rare Number. For example:
For X=65, X1=56
X+X1 = 65+56 = 121 = 112
X-X1 = 65 - 56 = 9 = 32
So 65 is a Rare Number.
2. When n is odd, n( n2 – 1 ) is divisible by 2 For e.g. n = 9 then n(n2-1) = 9(92 – 1) =
720 is divisible by 24
3. If n is odd, 2n + 1 is divisible by 3, e.g. n=5, 25+1 =33, which is divisible by 3
And if n is even, 2n – 1 is divisible by 3, e.g. n=6, 26-1 =63, which is divisible by 3
4. If n is prime, then n (n4-1) is divisible by 30, e.g. n=3, 3(34-1) = 240, which is
divisible by 30
5. If n is odd, 22n + 1 is divisible by 5, e.g. n=5, 22*5+1 =1025, which is divisible by 5
And if n is even, 22n – 1 is divisible by 5, e.g. n=6, 22*6-1 = 4095, which is divisible by
5
6. If n is odd, 52n + 1 is divisible by 13, e.g. n=3, 52*3+1 =15626, which is divisible by
13
And if n is even, 52n – 1 is divisible by 13, e.g. n=4, 52*4-1 =390624, which is divisible
by 13
7. The number of divisors of a given number N ( including 1 and the number itself )
where N = ambncp where a,b,c are prime numbers , are ( 1 + m ) ( 1 + n ) ( 1 + p ).
8. xn + yn = ( x + y ) ( xn-1 – xn-2 y + …. + yn-1 ), xn + yn is divisible by x + y when n
is odd.
9. xn - yn = ( x + y ) ( xn-1 – xn-2 y + …. - yn-1 ) when n is even, so xn - yn is divisible
by (x+y)
10. xn - yn = ( x - y ) ( xn-1 + xn-2 y + …. + yn-1 ) when n is either odd or even, so xn yn is divisible by (x-y)
Example: The remainder, when ( 1523 + 2323 ) is divided by 19, is : (CAT 2004)
(a) 4 (b) 15
(c) 0 (d) 18
Ans. (c) an + bn is always divisible by a + b when n is odd.
Thus 1523 + 2323 is always divisible by 15 + 23 = 38. As 38 is a multiple of 19, 1523 +
2323 is divisible by 19. So we get a remainder of 0.
Square of a number: If a number is multiplied by itself, it is called the square of that
number. Example 4 × 4 = 16 is square of 4.
Important Properties:
1. A square cannot end in odd number of zeroes.
2. The square of an odd number is odd and that of an even number is even.
3. Every square number is a multiple of 3 or exceeds a multiple of 3 by unity.
4. Every square number is a multiple of 4 or exceeds a multiple of 4 by unity.
5. If a square number ends in 9, the preceding digit is even.
Square root: The square root of a number is the number, whose square is the given
number.
Example is 4, as 4 × 4 = 16
Methods for finding square roots:
1. Factorization: Resolve the number into prime factors and deduce if there are numbers
which are repeating themselves (square of numbers).
Example: Find
Here 2601 = 32 × 172 = = 3 × 17 = 68
2. Approximation: The approximation method is the simplest method to find the square
root of a number, but as the name suggests it is an approximate method.
This method is best explained with an example. Suppose you want to find the square root
of , you know the square root of 100 is 10 and 121 is 11, now 104 lies between 100 and
121. Difference is 21, and number is 4 more than the lower number which is 100.
Therefore we can say the square root is
10 (of 100) + 4/21 = 10.19
Cube of a number: when a number is multiplied three times with the same number, it is
called the cube of a number.
Example: 4 × 4 × 4 = 64
Cube root: The cube root of a number is the number, which if multiplied three times by
same number gives the given number.
Example: = 4. It is represented by or with the power of 1/3, example (64)1/3 = (43)1/3 =
4.
To find the cube root of a number you have to find prime factors of the numbers, and
deduce if in those numbers if a number is repeated thrice.
Example: = 3 × 17 = 51
Complex numbers
Complex numbers are numbers with square root of a negative number. They were created
as there is no root of a negative number, by assuming i (called iota) = , it was possible to
do arithmetic operations on these numbers. A complex number is represented by (a + bi),
where a and b are real numbers.
Since i = , i2 = –1, i3 = –1 × i = –i
and i4 = (i2)2 = (–1)2 = 1
Just like surds, to rationalize complex numbers, the rationalizing factor or conjugates are
used like (a + ib) and (a – ib) are relative conjugates.
HCF AND LCM
HCF: HCF is the Highest Common Factor or Greatest Common Divisor (GCD).
Actually GCD explains it well, that is the greatest division that divides given set of
numbers. Example: HCF of 10, 15 and 30 = 5 and HCF of 15, 30 and 45 is 15. It is
obvious to see in each case 5 and 15 are the highest numbers which can divide the three
numbers.
To find the HCF of given numbers, resolve the numbers into their prime factors and then
pick the common term from them and multiplying them will give you the HCF. The HCF
is 1 when no common prime factors are there, as 1 is the only number which divides the
two and is the highest.
Example : Find the HCF of 24, 48, 102
Prime factors 2 × 2 × 2 × 3, 2 × 2 × 2 × 2 × 3, 17 × 3 × 2
Common numbers = 2 × 3 = 6, therefore 6 is the HCF
LCM: LCM is Least common multiple. It represents the smallest number which is
divisible by all of the given numbers.
Example: LCM of 3, 4 and 5 = 60, as it is smallest number divisible by them.
To find the LCM, resolve all the numbers into their prime factors, take the ones which are
common and the ones which are left (uncommon) and multiplying them will give you the
LCM of the number.
Example : Find the LCM of 24, 48, 102
Prime factors 2 × 2 × 2 × 3, 2 × 2 × 2 × 2 × 3, 17 × 3 × 2
Common numbers = 2 × 3
Numbers left = 2 × 2 × 2 × 2 × 2 × 17
LCM = common numbers x numbers left
= 2 × 3 × 2 × 2 × 2 × 2 × 2 × 17 = 3264
Important Note:
1. For two numbers, LCM × HCF = product of two numbers
Example: LCM of 4, 5 = 20 and HCF is 1, 20 × 1 = 4 × 5, 20 = 20
2. HCF of fractions =
Example: Find HCF of and .
Here HCF of 3, 4 is 1
And LCM of 5, 10 is 10
HCF of and =
3. LCM of fractions =
Example: Find LCM of and .
Here LCM of 3, 4 is 12
And HCF of 5, 10 is 5
LCM of and =
4. HCF and LCM of decimals: To calculate the HCF and LCM of decimals, remove the
decimals and convert them into non-decimals, by multiplying with 10 or 100 or…. Post
that calculate the HCF and LCM in regular fashion and once you have the regular HCF
and LCM, convert that number into decimal by dividing it with power of 10 with which
you multiplied earlier.
Example: Find the HCF and LCM of 0.6, 0.9, 1.5
Convert the numbers by multiplying them with 10, therefore numbers are 6, 9, 15
HCF of 6, 9, and 15 is 3; dividing by 10 it is 0.3
LCM of 6, 9, and 15 is 90; dividing by 10 it is 9
For large numbers the way to find the HCF is using Euclid’s Algorithm, but it works for
two numbers only. From the larger number, subtract the biggest multiple of the smaller
number without getting a negative answer. Replace the larger number with the answer
and repeat this until the last number is zero, and the HCF is the next-to-last number
computed.
Example: What is the HCF of 347236 and 297228?
347236 – 1 × 297228 = 50008
297228 – 5 × 50008 = 47188
50008 – 1 × 47188 = 2820
47188 – 16 × 2820 = 2068
2820 – 1 × 2068 = 752
2068 – 2 × 752 = 564
752 – 1 × 564 = 188
564 – 3 × 188 = 0
The HCF here is 188. Once we have the HCF, the LCM is the product of the two
numbers divided by the HCF as per the properties mention on LCM and HCF. Therefore
LCM of 347236 and 297228 = 347236 × 297228/188 = 549090936.
Example: Find the LCM of 32 and 40 if their HCF is 8.
LCM × HCF = Product of two numbers
LCM = 32 × = 160
Example: One ice cream truck visits Rahul’s neighborhood every 4 days and another ice
cream truck visits her neighborhood every 5 days. If both trucks visited today, when is
the next time both trucks will visit on the same day?
Since both the trucks visited today, next they will visit together on the LCM of 4 and 5,
which is 20, so on the 20th day from now
Example: A number when divided by 7 and 11 gives 4 as remainder in each case. Find
the largest 3-digit number of such type.
Ans. General form of such a number = (LCM of 7 and 11) k + 4
= 77 k + 4 where k is any whole number
Now largest 3-digit number of the form (77k+4) is for k = 12
i.e. 77×12 + 4 = 928.
Example: LCM and HCF of 2 numbers are 248 and 8 respectively. Find the numbers.
Ans. Since HCF = 8,
So, let the two numbers be ‘8a’ and ‘8b’, where a and b are co-prime to each other.
So, LCM of (8a and 8b) = 8 × a × b
8ab = 248
ab = 31
Since a and b are co-primes, the possible values for a and b can be 1 and 31.
So the numbers are 8 × 1 and 8 × 31 i.e. 8 and 248.
Remainder Theorem
We have to take care about the following things while finding remainder:
1. 'x' when divided by 'x' the remainder will be 0.
2. 'x + a' when divided by 'x' the remainder will be a.
3. 'x - a' when divided by 'x' the remainder will be ‘-a' i.e. '-a + x'.
4. (mx + a)^n when divided by 'x' the remainder will be a^n.
5. (mx - a)^n when divided by 'x' the remainder will be (-a)^n.
Here we will always express numerator in terms of denominator.
For example
If we want to find remainder when 2^80 is divided by 5.
Here, we will express such that the base of 2^80 get near to 5.
We can write 2^80 as 4^40
Now 4 can be written as (5-1)
So 2^80 = 4^40 = (5-1)^40
Now when (5-1)^40 is divided by 5, the remainder will be (-1)^40 = 1
Example: What is the remainder when 1044 × 1047 is divided by 33?
Ans. Here when 33 divide 1044, remainder is 21.
And when 33 divide 1047 is 24
Now the rule is the remainders multiplication has to be divided by divisor to get eh
remainder
Therefore 21 × 24 = 504. Dividing by 33 remainder is 9, which is the answer.
Example: What is the reminder when 91 + 92 + 93 + ...... + 99 is divided by 6?
Ans. It is obvious that 6 is leaving remainder 3 with powers of 9, which is actually a
property, but students can check by dividing first 2–3 numbers in the series, therefore
total remainder is 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3 = 27, when 27 is divided by 6,
remainder is 3, which is the answer.
To find power of any number in the factorial of any number.
The power of 'a' in n! Can be obtained by adding the quotients obtained by dividing 'n' by
a, a2,a3 … till we get zero as the quotients.
For example power of 2 in 36! Will be
36/(2^1) + 36/(2^2) + 36/(2^3) + …...
= 18 + 9 + 4 + 2 + 1 + 0
= 34
Example: What maximum power of 6 would perfectly divide 50!?
Ans. It will be the number of 6’s in 50!, as 6 = 2 × 3, it will be equal to number of these
pairs in 50!.
No. of 2’s (leave decimals)
= 50/2 + 50/22 + 50/23 + 50/24 + 50/25
Onwards powers of 2 will exceed 50,
= 25 + 12 + 6 + 3 + 1 = 47
No. of 3’s (leave decimals)
= 50/3 + 50/32 + 50/33
Onwards powers of 3 will exceed 50,
= 16 + 5 + 1 = 22
Since 3’s are 22, there will be 22 pairs; maximum power of 6 is 22.
Example. How many zeros will be there in the value of 25!?
Ans. The number of zeros are number’s of (5 × 2)’s
No. of 5’s (leave decimals)
= 25/5 + 25/52 onwards powers of 5 will exceed 25,
= 5 + 1 = 6, as twos are plenty, but 5’s are 6 therefore number of zeros are 6
Last Two Digits
For finding the last 2-digits in any calculation like (ab)^n
Where 'ab' is a 2-digit number and n is any number, we will first find out the unit's digit
and then the ten's digit.
We will convert all the calculations in which we have unit digit of the base as 1.
1. So let’s start with those calculations which have 1 as unit digit in base
Now if the base has 1 as the unit digit then we can say that the unit digit of the result will
also be 1.
Now tens digit will be the product of tens digit of base and unit digit of power.
Consider:
(ab)^xyz
Now, if b=1 then we can say that unit digit of result will be 1,
and tens digit = unit digit of (a * z)
Example: (31)^142
Unit digit =1
Tens digit = 3 * 2 = 6
So, the last two digits = 61
Example: (81)^236
Unit digit =1
Tens digit = 8 * 6 => 8
So, the last two digits = 81
2. Now if we have 3, 7 or 9 as the unit digit in the base then we can square the base once
or twice to obtain 1 as the unit digit in the base.
Example: (39)^132
(39)^132 = ((39)^2)^66
= (21)^66 (as square of 39 has 21 as last 2 digits)
= 21 (unit = 1 and tens = 2 * 6 = 2)
Example: (67)^148
(57)^144 = ((67)^2)^74
= (89)^74
= ((89)^2)^37
= (21)^37
= 41
3. Now for those numbers which have base as even numbers.
For this we have to find the last 2 digits of any power of 2.
Things to remember for this method:
a. Any odd power of 24 will have 24 as last 2 digits.
b. Any even power of 24 will have 76 as last 2 digits.
c. (2)^10 = 1024, or we can say 24 as last 2 digits.
Example: Now let us find last two digits of (2)^135
(2)^135 = ((2)^10)^13 * 2^5
= 24^13 * 2^5
= (24^odd) * (2^5)
= 24 * 32
= 68
Example: Find last two digits of 38 ^ 96
38 = 2 * 19
So (38)^96 = (2^96) * (19^96)
= ((2^10)^9) * (2^6) * ((19^2)^48)
= (24^9) * (2^6) * (61^48)
= 24 * 64 * 81
= 36 * 81
= 16
Miscellaneous Examples
Example: How many different positive integers exist between 106 and 107, the sum of
whose digits is equal to 2?
Ans. Between 10 and 100, that is 101 and 102, we have 2 numbers, 11 and 20.
Similarly, between 100 and 1000, that is 102 and 103, we have 3 numbers, 101, 110 and
200. Extrapolating the trend, between 106 and 107, one will have 7 integers whose sum
will be equal to 2.
Example: How many strokes on computer keyboard are needed to type numbers from 1
to 999?
Ans. 1 to 9 needs 9 strokes
10 to 99 needs 180 (90 × 2) strokes
100 to 999 needs 2700 (900 × 3) strokes
Total = 2892 strokes
Example: A certain number when successively divided by 8 and 11 leaves remainders of
3 and 7 respectively. What will be remainder when the number is divided by the product
of 8 and 11?
Ans. Solution to this question comes by a rule although there is a longish method using
the choices also. Here divisor 1 = 8, divisor 2 = 11, remainder 1 = 3, remainder 2 = 7, by
rule it is equal to divisor 1 × remainder 2 + remainder 1 = 8 × 7 + 3 = 59.
Example: The number of positive integers not greater than 100, which are not divisible
by 2, 3 or 5 is: (CAT 1993)
(a) 26 (b) 18
(c) 31 (d) none of these
Ans. (a) There are 50 odd numbers less than 100 which are not divisible by 2.
Out of these 50 there are 17 numbers which are divisible by 3. [Total numbers divisible
by 3 from 1 to 99 are 33, out of last number divisible is 99(33 × 3) is an odd number, so
odd numbers divisible are 17] Out of remaining there are 7 numbers that are divisible by
5. [Total numbers divisible by 5 from 1 to 99 are 19, out of which 9 are even (10, 20, 90),
and three odd numbers are divisible by 3(15, 45, 75), therefore 7 are left]
Hence numbers which are not divisible by 2, 3, 5 = (50 – 17 – 7) = 26
Example: Let x < 0.50, 0 < y <1, z > 1. Given a set of numbers; the middle number,
when they are arranged in ascending order, is called the median. So the median of
number x, y and z would be: (CAT 1993)
(a) Less than 1 (b) between 0 and 1
(c) Greater than 1 (d) cannot say
Ans. (b) The median is the middle number in ascending order, given x < 0.50, 0 < y < 1,
z > 1, the numbers could be arranged as (x, y, z) or (y, x, z). Since x and y both are under
1, hence median will also lie between 0 and 1.
Chapter 02 : Linear Equation> Topic 1 : Variable
A variable is a symbol that represents a number. Usually we use lower case letters such
as x, y, n, t, etc for variables. For example, we might say that s stands for the side-length
of a square. We now treat s as if it were a number we could use. The perimeter of the
square is given by 4s. The area of the square is given by s2.
When working with variables, it can be helpful to use a letter that will remind you of
what the variable stands for: Let n be the number of people in a movie theatre; Let t be
the time it takes to travel somewhere; Let d be the distance from my house to the park.
• Variables can be operated upon in a manner similar to real numbers. Thus you can say
that the H.C.F of x2, x3 and x5 is x2 and that the reciprocal of 2y is 1/2y.
• Remember that strictly speaking, H.C.F of x2, x3 and x5 is x2 only when x is a natural
number or at best a positive real number. By extending the definition of H.C.F to
negative numbers, we may go wrong. For example, it is wrong to say -16 is the L.C.M of
– 8, - 4 and – 2 because – 16 is a common multiple and is smaller than – 8.
Expressions
An expression is a mathematical statement that is a combination of numbers, variables, or
both.
The following are examples of expressions:
2x + 7
2xy+5
2+6x (4-2)
z/3 x (8-z)
Example. Anil weighs 70 kilograms, and Anita weighs k kilograms. What is the
expression that gives their combined weight?
Sol: The combined weight in kilograms of these two people is the sum of their weights,
which is 70 + k.
Care should be taken while putting together numbers and variables since they represent
physical values.
While adding or subtracting them, we need to check that they have the same units.
If in the above problem, Anita’s weight was given in grams then the expression for the
weight in kilograms would be 70+ k/1000. We could also express the combined weight in
grams as 70 x 1000 + k.
To evaluate an expression at some number means replace a variable in an expression with
the number, and simplify the expression.
E.g.: To evaluate the expression (4 z + 12) when z = 15, we replace each occurrence of z
with the number 15, and simplify using the usual BODMAS rules.
4z + 12 becomes 4 × 15 + 12 = 60 + 12 = 72
Evaluation can be done for expressions with multiply variables as well. In this case the
values of each of these variables should be given for which the expression is to be
evaluated.
E.g.: To evaluate x + 3y + xz when x = 1, y =2 and z =3, we replace the variables by their
values to get 1 + 3 × 2 + 1 × 3 = 10.
An expression having only one term is called a monomial. So x, v, 2xyz, ab, 9abx2y,
…etc. are all monomials. The constant that appears in any term other than the algebraic
variable is called the ‘coefficient’ (1,2, 9, …, in the examples given).
An expression having 2 terms is called binomials, so ab+abc is a binomial.
An expression having 2 or more than two terms e.g. X + Y + 5Z, 2X2 + 4YZ3 + 3XZ2,
…, etc. is called polynomials. The point to note is that the terms are separated by an
addition or subtraction sign.
So, how we differentiate between x, 2xyz and 9abx2y?? These all are Monomials.
We define degree of an algebraic expression. The DEGREE of an expression is equal to
the maximum values of the sum of the powers of the variables in any term. You should
note two things in the definition, sum of powers and maximum sum in any term. In the
above examples of monomials, x has 1 degree, 2xyz is 1+1+1=3 degrees (1 each for
x,y,z) and 9abx2y has 5 degrees (1each for a,b,y and 2 for x).
Maximum sum in any term applies for binomials or polynomials. So for Polynomial, 2X2
+ 4YZ3 + 3XZ2, we have three different terms. Term 1 has a degree of 2, term 2 has a
degree of 4 and third term has a degree of 3. As the maximum value is 4, we say that the
degree of polynomial is 4.
Expressions of Degree 1 are called LINEAR expressions, of Degree 2 are called
QUADRATIC expressions and of Degree 3 are called CUBIC expressions. (There are
names for expressions with degree 4 ,5 and so on but let us leave them for
mathematicians)
3. Equations
An equation is simply a statement that two numbers or expressions are equal. Equations
are useful for relating variables and numbers. Many word problems can easily be written
down as equations. There are some simple rules for simplifying equations.
The following are examples of equations:
2=2
x=7
7=x
t+3=8
w + 4 = 12 – w
3 Í (d + 4) – 11 = 321 – 23.
Example: 20 added to my age in years, y, is equal to four times my age, minus 10.
Sol: The first expression stands for “20 added to my age in years”, which is y + 20
This is equal to the second expression for “four times my age, minus 10”, which is 4 y –
10
Setting these two expressions equal to one another gives us the equation y + 20 = 4 y –
10.
Example: When 17 is subtracted from 5 times a certain number, the result is 73. What is
the number?
Sol: 5x – 17 = 73,
x = 18
Solution of an Equation
When an equation has a variable, the solution to the equation is the value of the variable
that makes the equation true.
E.g.: We say y = 3 is a solution to the equation 4* y + 7 = 19, for replacing each
occurrence of y with 3 gives us
4 × 3 + 7 = 19
So, 12 + 7 = 19 so, 19 = 19, which is true.
Y = 10 is NOT a solution to the equation 4 × y + 7 = 19. When we replace each y with
10, we get
4 × 10 + 7 = 19
So 40 + 7 = 19 or, 47 = 19, which is not true.
• It is not necessary that every equation should have a solution. For instance x = x + 3 has
no solution.
An equation can have multiple solutions, the number of solutions being finite or infinite.
z2 = 4 has both z = 2 and z = -2 solutions. 0/y = 0 has infinite solutions for y since the
equation is true for any real value of y other than 0.
• If we have equations with multiple variables, the solution consists of values for each of
the variables for which the equation is true.
A solution of the equation x + 2y = 3 is x = 1 and y = 1; x = 2 and y = ½ is another. As
you can see, there are infinite solutions for the equation.
Simplifying Equations
To find the solution for an equation, we can use the basic rules of simplifying equations.
These are as follows:
• You may evaluate any parentheses, exponents, multiplications, divisions, additions, and
subtractions in the usual order of operations (BODMAS).
• You may combine like terms. This means adding or subtracting variables of the same
kind. The expression 2x + 4x simplifies to 6x. The expression 13 – 7 + 3 simplifies to 9.
• You may add any value to both sides of the equation.
• You may subtract any value from both sides of the equation. This is equivalent to
adding a negative value to each side of the equation.
• You may multiply both sides of the equation by any real number.
• You may divide both sides of the equation by any number except 0.
Example. To simplify an equation like 2x – 12 + 20 = 36, we do the following in
accordance with the above rules.
Sol: 2x + (-12) + 20 = 36
2x + 8 = 36
Add the two like terms (the integers)
2x + 8 - 8 = 36-8
Subtract 8 from each side of the equation, (or add a –8 to each side).
So. 2x = 28
2.x. ½ = 28 x ½ Divide both sides by 2 (or multiply by ½)
Or, x = 14
• Remember that dividing both sides by zero is not allowed. We commonly strike off
multiplicative terms from both sides (“cancelling”).
Such cancelling is not allowed if the term being struck off is zero. Thus, if you are
cancelling a variable you need to be sure that it cannot take the value of 0.
To illustrate, let’s take x = 2.
This can also be written as: 3x – 2x = 6 – 4
Rearranging the terms, 2x – 4 = 3x – 6 or, 2 (x – 2) = 3 (x – 2)
Cancelling (x – 2), we get 2 = 3.
• The flaw obviously is that when we are cancelling x- 2, we are actually performing a
division by x - 2.
But since x = 2, x - 2 is zero and so we cannot divide by it.
Example. A car travels down the freeway at 55 kilometers per hour. Write an expression
for the distance the car will have travelled after h hours?
Sol: Distance equals speed time’s time, so the distance travelled is equal to 55 h.
Example. There are 2000 litters of water in a swimming pool. Water is filling the pool at
the rate of 100 liters per minute.
Write an expression for the amount of water, in liters, in the swimming pool after m
minutes?
Sol: The amount of water added to the pool after m minutes will be 100 liters per minute
time m, or 100 m.
Since we started with 2000 liters of water in the pool, we add this to the amount of water
added to the pool to get the expression 100 m + 2000.
Example. Evaluate the expression (1 + z) × 2 + 12/3 - z when z = 4?
Sol: We replace each occurrence of z with the number 4, and simplify using the usual
rules: parentheses first, then exponents, multiplication and division, then addition and
subtraction.
(1 + z) 2 + 12/3 - z becomes (1 + 4) 2 + 12/3 - 4 = 5 × 2 + 12/3 - 4 = 10 + 4 – 4 = 10.
After understanding the basics of equations, let us understand Linear equations. Linear
equations are equations with one degree.
If we look at the different solutions, they may appear in three forms.
1. UNIQUE: There is only one set of values for the variables that satisfy the equation.
2. INFINITE: The number of sets of values of the variable that satisfy the equation is
infinite.
3. NO SOLUTION: There is no set of values of the variables that satisfy the given
equation.
For instance, take the case of the equation we have considered X + Y =4
If we try to solve this equation, we would find that the following values of X and Y
would satisfy the equation.
(X, Y) = (1, 3) or (1.1, 2.9) or (2.5, 1.5) or … so on. If we try to list down all the possible
values, possible set, there are infinite possibilities. So the number of solutions for (X, Y)
is infinite.
Linear equation in 1-variable
These are the equations having only one unknown variable.
General form is:
Ax + B = 0
Where A and B are constants.
For finding unique value of variable ‘x’ we need only one equation.
Linear equation in 2-variable
These are the equations having two unknown variables.
General form is:
Ax + By + C = 0
Where A, B and C are constants.
For finding unique value of variables ‘x’ and ‘y’ we need two independent and consistent
equations.
Linear equations in 3-variables.
These are the equations having three unknown variable.
General form is:
Ax + By + Cz + D = 0
Where A, B, C and D are constants.
For finding unique value of variables ‘x’, ‘y’ and ‘z’ we need three independent and
consistent equations.
Linear combination of equations or Dependent equations
Lets suppose there are ‘n’ number of linear equations: l1, l2, l3 ……ln
And there are ‘n’ constants k1, k2, k3 ……….. kn
Then these equations will be said to be in linear combination if:
K1 l1 + k2 l2 + k3 l3 ……. Kn-1 ln-1 = kn ln
So we can say that, if any equation can be written as the linear combination of some other
equations then these equations are dependent on each other.
Graphical Approach
If all these points are represented as coordinates on a graph and joined together, they
form a straight line. The graph would look as below.
On the contrary, an equation or a system of equations having a UNIQUE solution has
only one set of the variables that satisfy the equation.
An equation or a system of equations has a unique solution if and only if the number of
variable is equal to or less than the number of independent and consistent equations.
Slightly confused!!!! Let us break the above definition to make more sense.
Independent
An equation is not independent if it is derived as a combination of any of the other
equations in the system of equations.
For example, suppose there are three equations.
X + Y = 4 …(i)
2X + 3Y = 5 …(ii)
4X + 5Y = 13 …(iii)
The equation (iii) is not independent of the other two since it can be formed by
2(i) + (ii)
But in this case as we have only two variables, we may still get Unique solution for X
and Y.
Chapter 02 : Linear Equation> Topic 2: Inconsistance
Consider the two equations.
X+Y=4
2X + 2Y = 7 or X + Y = 3.5
The second equation establishes a different relationship between X and Y as compared to
the first equation. So they are said to be Inconsistance. Such a set of two equations do not
have any solution.
So to get a unique solution, the number of equations should be equal to or more than no.
of variables, these equations should be independent to each other and should be
consistent with each other.
Let us take an example of set of equations that follow all the above restrictions.
x + 2y = 4 and 2x + 3y = 6
If you solve these two equations, we get the values of x = 0 and y = 2.
Graphically,
Fro graph also, the intersection points of two lines is, (0, 2) which is the unique solution.
Corollary: (For two variable linear equations)
For Unique Solution, the two lines would intersect each other at one point. If the two
lines are parallel to each other we have no solution and if the two lines coincide, we shall
have infinite solutions.
Number of solutions:
We can find the possible number of solutions without solving the algebraic equations or
drawing the graphs. If the general representation of linear equation is,
ax + by = M
and cx + dy = N
where a and c are coefficient of x, b and d are coefficients of y and M and b/dN are
constants, then for Unique solution, a/c
for No solution, a/c = b/d M/N
for infinite solutions, a/c = b/d = M/N
Methods to solve the Linear Equations
1. Substitution Method
In this method, we find from one of the equations the value of one variable in terms of
the other and substitute it in the second equation and thus find the value of the other
variable.
Example: Solve x + y = 5… (i.)
3x + 2y = 12… (ii.)
Sol: We have x+ y = 5 or x = 5-y
Substituting in 2nd equation
3(5-y) + 2y = 12
15-y = 12
Or y = 3
x = 5-y = 5-3 = 2
Thus required solution is x = 2, y = 3
2. Elimination Method
The principle of this method consists of multiplying the coefficients of the equations by
suitable numbers such that the coefficients of the one variable may become the some in
both the equations. Then, by adding or subtracting, we get the value of one of the
variables.
Then by substituting this value of this variable in either of the equation, we find the value
of the other variable or unknown quantity.
Example: Solve 2x + 3y = 12
5x + 7y = 29
Sol: We have 2x + 3y = 12….... (i)
5x + 7y = 29….... (ii)
Here we see that if we multiply (i.) by 5 and (ii.) by 2 the coefficient of x in both
equations would become the same
5(2x + 3y = 12)
Or 10x + 15y = 60.....… (iii)
2(5x + 7y = 29)
Or 10x + 14y = 58…..... (iv)
Subtracting (iv) from (iii)
15y – 14y = 60 – 58
Or y = 2
Substituting value of y in (1)
2x + 3 x 2 = 12
Or 2x = 12 – 6
Or 2x = 6
Or x = 3
Hence solution is x = 3, y = 2
Example: What is the solution of the following simultaneous equations?
x + y + z = 6, x + 2y + 3z = 14 and x + 3y + z = 10
Sol: x + y + z = 6 … (i)
x + 2y + 3z = 14 … (ii)
x + 3y + z = 10 … (iii)
From (i), we get z = 6 – x – y
Substitute it in (ii) and (iii).
We have from (ii)
x + 2y + 18 – 3x –3y = 14 or 2x + y = 4
Similarly, from (iii)
x + 3y + 6 – x – y = 10 or 2y = 4 or y = 2
Solving, we get y = 2, x = 1, z = 3.
Example: Find for what value of K would there be a unique solution for the given set of
equations. 2x – 3y = 1 and Kx + 5y = 7
Sol: If two equations ax + by = M and cx + dy = N have a unique solution, then a/c b/d.
So in the above problem,
2/K-3/5
Solving, K –10/3.
Example: Find the value of K for which there is no solution for the given set of
equations.
2x – Ky = -3 and
3x + 2y = 1
Sol: If two equations ax + by = M and cx + dy = N have no solution, then
a/c = b/d M/N
So,
2/3=-k/2-3
Solving k = -(4/3) or K-3
So, k = -(4/3)
Example: Find the value of K for which there are infinite solutions for the given set of
equations. 5x + 2y = K and 10x + 4y = 3
Sol: If two equations ax + by = M and cx + dy =N have infinite
solutions, then a/c = b/d = M/N
So, for the two sets of equation to have infinite solutions, we have
5/10 = 2/4 = k/3
Hence, k = 3/2
Example: If the numerator and the denominator of a certain fraction are both increased
by 1 each, then the fraction will be equal to ½. Instead, if the numerator and the
denominator are increased by 4, fraction equal to 2/3. Find the fraction.
Sol: Let the fraction be x/y where x is the numerator and y is the denominator. When
both numerator and denominator are increased by 1 each, we have
(x + 1)/(y + 1) = ½
2(x + 1) = y + 1 => 2x – y = -1 ……(1)
When both numerator and denominator are increased by 4 each, we have
(x + 4)/(y + 4) = 2/3
3(x + 4) = 2(y +4) => 3x – 2y = -4 …….(2)
Solving the two equations (1) and (2), we get x = 2 and y = 5. Hence the fraction is 2/5
Example: Find the values of x and y following equations: 6/(x + y) + 5/(x – y) = 7, 12/(x
+ y) – 3/(x – y) = 1
Sol: Substitute p = 1/(x + y) and q = 1(x – y) in the given equations. Then we get
6p + 5q = 7 -------- (1)
12 p – 3q =1 --------- (2)
Now these are two equations in two unknowns that we are familiar with.
Solving these two equations, we get p = 1/3 and q = 1
Since p = 1/(x + y) and q = 1/(x – y) we get 1/(x + y) = 1/3 and 1/(x – y) = 1
x + y = 3 and x – y = 1. By solving these two equations,
we get x = 2 and y = 1
Example: The sum of the two digits of a two-digit number is 8 and if the digits are
reversed, the resulting number is 36 lower than the original number. Find the original
number.
Sol: Let the two digits of the given number be x and y such that then given number is 10x
+ y.
Since the sum of the digits is 8, we have x + y = 8 ------ (1)
When the digits are reversed, y becomes the tens digit and x, the units digit.
The number then becomes (10y + x)
So (10x+y) - (10y + x) = 36
or, 9x – 9y = 36 or x – y = 4
Solving equations (1) and (2) we get x = 6 and y = 2 and thus the number is 62.
Example: Four years from now, father’s age will be four times son’s age. Nine years
from now, father’s age will be three times son’s age. When are their present ages?
Sol: Let f be father’s present age and s be the son’s present age.
Four years from now, father will be f + 4 and son will be s + 4 years old.
f + 4 = 4(s + 4)
f – 4s = 12 ------(1)
Nine years from now, father will be f + 9 and son will be s + 9 years old.
f + 9 = 3(s + 9) => f – 3s = 18 ----- (2)
We now have two equations in two unknowns. Solving these two equations, we get f =
36 and s = 6
Father’s age is 36 years and son’s age is 6 years.
Inequalities
We just learnt about equations. Now instead of equating variables with a constant, if we
use < or > sign to have a range of solutions, we are dealing with inequalities.
So, If x = 4, then x can take only one value.
But if x > 4, then x can take infinite vales all of which are greater than 4.
All these conditions that deal with > or < situations come under the category of
Inequalities.
Graphs are used extensively to solve questions on inequalities.
Basic rules of inequalities
1. If X > Y, then –X < - Y.
2. If X> Y and A > B, then X + A > Y + B, but it is not necessary that X – A > Y – B.
3. If x/a > y/b, then bx > ay, only if a, b > 0. This is an important rule and you may often
have to use it in solving data sufficiency problems.
Example: Solve the following in-equations:
(i) 0 < -x/3 < 1 (ii) 6≤ -3(2x – 4) < 12
(iii) -3 ≤ 4 – 7x < 18 (iv) -2 < 1 -3x < 7
(v) -7 < 2x – 3 < 7 (vi) -12 < 3x – 5 ≤ - 4.
Sol: (i) We have, 0 < -x/3 < 1 => 0 > x > -3
-3 < x < 0
(Sign of inequality reverses on multiplying with negative number).
Hence, the interval (-3, 0) is the solution set of the given system of inequations.
(ii) We have, 6≤ -3(2x – 4) < 12
6/-3 ≥ -3(2x – 4)/-3 > 12/ -3
-2 ≥ 2x – 4 > - 4
-2 + 4 ≥ 2x – 4 + 4 > - 4 + 4
2 ≥ 2x > 0
1≥x>0
0<x≤1
Hence, the interval (0, 1) is the solution set of the given system of inequations.
(iii) We have, -3 ≤ 4 – 7x < 18
-3 -4 ≤ 4 – 7x – 4 < 18 – 4
-7 ≤ - 7x < 14
-7/-7 ≥ -7x/-7 > 14/-7 => 1 ≥ x > -2
-2 < x ≤ 1
Hence, the interval (-2, 1) is the solution set of the given system of inequations.
(iv) We have, -2 < 1 – 3x < 7
-2 -1 < 1 – 3x – 1 , 7 -1
-3 < - 3x < 6
-3/-3 > -3x/-3 > 6/-3
1 > x > -2
-2 < x < 1
Hence, the interval (-2, 1) is the solution set of the given system of inequations.
(v) We have, -7 < 2x – 3< 7
-7 + 3 < 2x – 3 + 3 < 7 + 3
-4 < x < 10
-4/2 < 2x/2 < 10/2
-2 < x < 5
Hence, the interval (-2, 5) is the solution set of the given system of inequations.
(vi) We have, -12 < 3x – 5 ≤ - 4
-12 + 5 < 3x – 5 + 5 ≤ -4 + 5
-7 < 3x ≤ 1
-7/3 < x ≤ 1/3
Hence, the interval (-7/3, 1/3) is the solution set of the given system of inequations.
Example: Draw the graph of the solution set of the 6,inequations 2x + 3y
x + 4y 4, x 0 and y 0.
Sol: First we draw the graph of 2x + 3y 6.
Consider the line 2x + 3y = 6.
Now, 2x + 3y = 6 => x/3 + y/2 = 1.
This line meets the axes at A (3,0) and B(0,2). Plot these points on a
graph paper and join them by hick line AB. Consider the point (0,0).
Clearly, it does not lie on 2x + 3y = 6.
Clearly, (0,0) does not satisfy 2x + 3y =6.
Therefore, the solution set of 2x + 3y =6 is represented by the line
AB and the part of the plane separated by AB, not containing (0,0).
Next, we draw the graph of x + 4y =4.
Consider the line x + 4y = 4.
Now, x + 4y = 4 => x/4 + y/1 = 1.
This line meets the axes at C (4,0) and D(0,1).
Plot these points on the same graph paper as above and join them by
thick line CD.
Consider the point (0,0). It does not lie on the line x + 4y = 4. Clearly, (0,0) 4.satisfies
the inequation x + 4y
Therefore, the solution set of x + 4y =4 is line CD and that part of
the plane separated by CD which contains (0,0).
Clearly, x = 0 consists of y-axis and the plane on the right side of y-axis. And, y = 0
consists of x-axis and the plane above x-axis.
Hence, the shaded region represents the solution set of the given system of inequations.
Note: Just a small note. The above question can never be asked in CAT. Nobody can
expect you to draw a graph and solve. So why did we mentioned it. Just for your
understanding in general. So understand what is equality and that is it.
Chapter 03 : Ratio and Proportion > Ratio
Ratio and proportion is the continuation of the basic concepts for the entrance tests. The
application of the topics is far and wide, a direct application which will be covered here is
“Allegation and Mixtures”.
Ratio:
The comparison of two quantities of the same unit is the ratio of one quantity to another.
The ratio of A and B is written as A: B or A / B, where A is called the antecedent and B
the consequent.
Example: The ratio of 10 kg to 20 kg is 10:20 or 10/20, which is 1:2 or ½, where 1 is
called the antecedent and 2 the consequent.
Properties of Ratio
1. a : b = ma : mb, where m is a constant
2. a : b : c = A : B : C is equivalent to a / A = b /B = c /C, this is an important property
and has to be used in ratio of three things.
3. If a / b = c / d , then
(a+b) / b=(c+d) / d – Componendo
Example: ½ = 2/4, so (1+2)/2 = (2+4)/4 => 3/2 = 6/4 => 3/2 = 3/2
(a-b) / b = (c-d) / d – Dividendo
Example: 10/4 = 20/8, so (10-4)/4 = (20-8)/8 => 6/4 = 12/8 => 3/2 = 3/2
(a+b) / (a-b) = (c+d) / (c-d) – Componendo and Dividendo
Example: 10/4 = 20/8, so (10+4)/(10-4) = (20+8)/(20-8) => 14/6 = 28/12 => 7/3 = 7/3
Application: These properties have to be used with quick mental calculations; one has to
see a ratio and quickly get to results with mental calculations.
Example: 10/4 = 20/8, should quickly tell us that 14/4(7/2) = 28/8(7/2), 6/4(3/2) =
12/8(3/2) and 14/6(7/3) = 28/12(7/3)
4. If a / b = c / d =e / f = ……, then (a+c+e+…..)/(b+d+f+….) = each of the individual
ratio
Example: ½ = 2/4 = 4/8, there fore (1+2+4)/(2+4+8) = 7/14 = ½
5. If A > B then (A+C)/ (B+C) < A/B Where A, B and C are natural numbers
Example: 3>2, then (3+4)/ (2+4) = 7/6 (1.16)
6. If A < B then (A+C)/ (B+C) > A/B Where A, B and C are natural numbers.
Chapter 03 : Ratio and Proportion > Proportion
Proportion is an expression in which two ratios are equal, that is A/B = C/D. Here AD =
BC
If four numbers a, b, c and d are in proportion then we can say that:
a/b=c/d
Example: The ratio of 10 kg to 20 kg is 10:20 or 10/20 or ½, and the ratio of 30 kg to 60
kg is 30:60, or 30/60 or ½, so 10/20=30/60, they are in proportion
If a: b=b: c=c: d then a, b, c, d are in continued proportion. Here a/b = b/c = c/d. Also a,
b, c, d are in geometric progression
Example: 2, 4, 8, 16 or 3, 9, 27, 81, Here 2/4 = 4/8 = 8/16, and 3/9 = 9/27 = 27/81. So
they are in continued proportion and also in geometric proportion.
Continued Proportion
If three numbers a, b and c are in continued proportion, then:
We can say that a, b, b and c are in proportion.
i.e. a / b = b / c
b2 = a c
Here we can say that
a is called first proportion, c is called third proportion and b is called mean proportion.
Types of proportion
1. DIRECT PROPORTION:
If X is directly proportional to Y, that means any increase or decrease in any of two
quantities will have proportionate effect on the other quantity. If X increases then Y will
also increase and vice-versa.
When X is directly proportional to Y, it is written as X α Y, to bring in an equality sign,
you have to introduce a constant, say k. so X = k Y. From here X/Y is a constant, so X/Y
= k.
2. INVERSE PROPORTION:
If X is inversely proportional to Y, that means any increase or decrease in any of two
quantities will have inverse proportionate effect on the other quantity. This means if X
increases, then Y decreases and if X decreases the Y increases and vice-versa for Y.
When X is inversely proportional to Y, it is written as X α 1/Y, to bring in an equality
sign, you have to introduce a constant, say k. so X = k/Y. From here XY is a constant, so
XY = k.
Both these proportions have wide applications in many subjects, especially sciences and
economics, where many factors are directly and inversely proportional to each other.
Applications of Ratio and Proportion
Partnership
Partnership, as the name suggests, its more than one person investing in something to
increase resources, reduce risk etc.
Important point: The partners are liable for income/expenditure/profit/loss as per their
percentage holding of the business or their partnership percentage. This is defined by
their capital/work/other things contribution in terms of time. The partnership terms can of
many types, in questions you may encounter partners putting monies for different time
periods, which can be calculated by multiplying the money with the time in months the
money has been invested for.
Example: Ram partnered with Rohan and Ravi (equal partners) in a business and added
capital of Rs. 10000 for six months in a year. The capital for those six months became Rs.
25000 for the year, if the end of year profit is 2500, what is Ram’s net profit?
Ram’s capital = 10000 for six months
Total capital = 25000 for six months
Ravi and Rohan’s capital = 25000 – 10000 = 15000 for 12 months
Ravi’s Capital = 7500 for 12 months Rohan’s capital = 7500 for 12 months
Ram’s share = (10000 x 6) / (10000 x 6) + (7500 x 12) + (7500 x 12)
= 60000/24000 = ¼
Ram’s Profit = ¼ x 2500 = Rs. 625
Mixtures and Allegation
“Mixtures and allegations” is about mixing different objects in order to get desired
levels/percentage/concentration of different objects. Ratio and proportion has direct
application in “mixtures and allegations”. The concept of weighted average is also used
to solve mixture questions, so it is suggested that students brush up the “Averages”
chapter before starting off with this chapter.
Example: If 100 ml water is mixed 1000 ml of milk, what is the ratio of the mixture
solution?
Using basic percentage, total solution = 1100 ml, so 100/1100 = 1/11
Water: milk = 1:10,
This can also be read as milk solution 10:11, where if milk is 10, water is 1 and total
solution is 11.
This lingo will be used extensively in this chapter; it may be for milk, alcohol, spirit etc.
Example: Two bottles contain mixture of milk and water in the ratio of 8:3 and 5:1. In
what ratio must liquid be drawn from each bottle to give a mixture in the ratio of 4:1?
Here using basic logic, suppose we take x units from bottle one and y units from bottle
two to make a mixture of 4:1, bottle with ratio 8:3 will contribute 8/11x milk and 3/11x
water, similarly other bottle will contribute 5/6y milk and 1/6y water
Total milk = 8/11x + 5/6y
Total Water = 3/11x + 1/6y
Now total milk/ total water = 4/1, so [8/11x + 5/6y]/ [3/11x + 1/6y] = 4/1
Solving we get x/y = 11/24, so x:y = 11:24
We will study a “Rule of allegation” to solve the questions on mixing products like tea
etc where different varieties(with different costs) are mixed to get desired Varity, but
students should know that formula is just an aid, another way of solving the question.
Students should know how to solve the question without the formula. The allegation uses
simple logic, but some students become slave of the allegation formula, which may lead
to lot of errors.
Rule of allegation
This rule helps us in solving questions where two varieties (of different prices) are mixed
to get a new variety with a new Average price.
Quantity of cheaper = Price of Dearer - Average price
--------------------------- ----------------------------------------Quantity of dearer Average price – Price of cheaper
Example: In what ratio should tea at the rate Rs. 40/kg be mixed with tea at the rate Rs.
27/kg, so that mixture may cost Rs 30/kg?
Using the above formula
Quantity of cheaper / quantity of dearer = (40 – 30)/ (30-27) = 10/3
So, the two should be mixed in the ratio 10:3.
In case of liquids, there is another formula which can be used in various questions.
In a vessel containing x litre of a one liquid (say pure milk), if y litres is withdrawn and
replaced by another liquid (say pure water), and this is repeated n times, then:
Milk left in vessel after nth operation = x [1-y/x] n
Example: In a vessel full of 5 litres of milk, five times 250 ml is taken and replaced with
water, what is the concentration of milk left in the vessel
As per the formula, Milk left in vessel after nth operation = x [1-y/x] n
Milk left = 5(1-0.25/5)5 (remember to covert ml into litres)
Solving, Milk left = 3.86 litres
The concentration is 3.86/5 x 100 = 77.3 %
Miscellaneous Examples:
Q1. If two numbers are in the ratio of 10:3 and 2 is subtracted from each, the resulting
numbers are in the ratio 9:2. Find the numbers
Ans1. Since the ratio is 10:3, let the numbers be 10X and 3X
Therefore (10X -2) / (3X-2) = 9/2
7X = 14, X = 2
Therefore numbers are 20 and 6
Q2. A common foodstuff is found to contain 2.5% iron. The serving size is 90.0 grams. If
the recommended daily allowance is 18 gm of iron, how many servings would a person
have to eat to get 100% of the daily allowance of iron?
Ans2. Iron percentage = 2.5/100 x 90 = 2.25 grams
Total grams required = 18
Servings required = 18/2.25 = 8 servings in a day
Q3. A substance is 99% water. Some water evaporates, leaving a substance that is 98%
water. How much of the water evaporated?
Ans3. The substance has 99% water and 1 % other substance
Let the amount of water be X and the other substance be Y
Now X/(X+Y) = 99/100, Y = X/99
After some evaporation suppose water left = Z
Therefore Z/ (Z+Y) = 98/100, Y = 2Z/98
Equating Y, X/99 = 2Z/98
Z = 0.495X
Therefore Current water level is 0.495 for earlier water level, so 50.5 % water has been
evaporated.
Q4. If I clean a 3200 square foot building five nights per week for a sum of Rs. 575 per
month, what is the cost per square foot?
Ans4. Total square foot = 3200
Total sum = Rs. 575
Cost per square foot = 575/3200 = Rs. 0.18 per square foot
Q5. A and B started a business by investing Rs 50000 and Rs 25000. What is the share of
each if yearly profit is Rs 2000?
Ans5. Total investment = 50000 + 25000 = Rs.75000
A’s share = 50000/75000 x 2000 = Rs.1333.34
B’s share = 25000/75000 x 2000 = Rs. 666.66
Q6. The salary of Ravi, Ajay and Bhuvan is Rs 350000. If they spend 70%, 75%, and
80% of their salaries respectively, their savings are in ratio of 15:10:25. Find their
salaries.
Ans 6. Total Salary of the three = 350000
Ravi’s spent = 70%, therefore Ravi’s saving = 30%
Ajay’s spent = 75%, therefore Ajay’s saving = 25%
Bhuvan’s spent = 80%, therefore Bhuvan’s saving = 20%
30 % of Ravi’s Salary: 25 % of Ajay’s Salary: 20 % of Bhuvan’s Salary = 15:10:25
30/100 R: 25/100 A: 20/100 B = 15:10:25
30R:25A:20B = 15:10:25
From here 30R/25A = 15/10, R/A = 375/300 = 17/12 = 34/24
Also 25A/20B = 10/25, A/B = 8/25 = 24/75
Now R: A: B = 34:24:75
Ravi’s Salary = 34/133 x 350000 = 89474
Ajay’s Salary = 24/133 x 350000 = 63158
Bhuvan’s Salary = 75/133 x 350000 = 197368
Q7. Divide Rs 435 among A, B and C so that if Rs 9, Rs 4 Rs 2 be subtracted from their
respective shares, the shares left may be in the ratio 6:4:5.
Ans7. Here the ratio of shares is given 3:4:5
The total is 435, and 9+4+2 = 15 needs to be subtracted from it
= 435 – 15 = 420
Now diving 420 in ratio of 6:4:5
A’s Share = 6/15 x 420 = 168, 168 + 9 = 177
B’s Share = 4/15 x 420 = 112, 112 + 4 = 116
C’s Share = 5/15 x 420 = 140, 140 + 2 = 142
Q8. A, B and C partnered in a business. A contributed Rs 12000, B Rs 10000 and C Rs
8000 and their profit was Rs. 2400. What is the share of each?
Ans8. A’s Contribution = 12000
B’s Contribution = 10000
C’s Contribution = 8000
Total = 30000
A’s share = 12000/30000 x 2400 = Rs 960
B’s share = 10000/30000 x 2400 = Rs 800
C’s share = 8000/30000 x 2400 = Rs 640
Q9. A, B and C partnered in a business for a year. A contributed Rs 12000 for 6 months,
B Rs 10000 for 8 months and C Rs 8000 for the entire year and their profit was Rs. 2728.
What is the share of each?
Ans9. A’s Contribution = 12000 for 6 months = 12000 x 6 = 72000
B’s Contribution = 10000 for 8 months = 10000 x 8 = 80000
C’s Contribution = 8000 for 12 months = 8000 x 12 = 96000
Total = 248000
A’s share = 72000/248000 x 2728 = Rs 792
B’s share = 80000/248000 x 2728 = Rs 880
C’s share = 96000/248000 x 2728 = Rs 1056
Q10. Ravi and Mayank enter into a partnership by investing Rs. 7000 and Rs. 3000
respectively. At the end of one year, they divided their profits such that 1/3 of the profit is
divided equally for the efforts they have put into the business and the remaining amount
of profit is divided in the ratio of the investments they made in the business. If Ravi
received Rs. 8000 more than Mayank did, what was the profit made by their business in
that year?
Ans10. Let the profit made during the year be X
The Profit to be divided equally = X/3
The profit to be divided as per contribution = 2X/3
The diving ratio is 7000:3000 = 7:3
Ravi shall get more profit in 2X/3, by 70-30 = 40% more
Therefore 40/100 x 2X/3 = 8000
The profit, X = Rs. 30000
Chapter 04 : Percentages, Profit and Loss> Percentages
Percent (%)
Percent means for every hundred or per hundred. The numerator of “per hundred” is
called the rate percent. Example 12/100 can be called as 12%, and 12% is the rate
percent, and 12 is the rate. The other way to look at it is if some makes a profit of 20%,
then one has gained 20/100 of the value invested.
Important notes
1. To express percent as a fraction divide it by 100,
12 % =
2. To express a fraction as a percent multiply it by 100,
½=
3. Increase/Decrease Percent
=
Here Increase/Decrease = (Final value – initial value)
Important relations in percentage
1. If the price of a commodity increases by r%, then percentage reduction in
consumption, so as not to increase expenditure is
Example: If the cost of petrol increases by 40%, by what percent the person should
reduce his consumption considering expenditure on petrol remains the same.
Increase in price = 40%, by the formula, decrease in consumption is = = 28.57%
2. If the price of a commodity decreases by r%, then increase in consumption, so as not to
decrease expenditure is
Example: If the cost of petrol decreases by 10 %, by what percent can a person increase
his consumption considering expenditure on petrol remains the same?
Decrease in price = 10%, by the formula, increase in consumption is = 11.11%
3. If A’s income is r% more than B’s then B’s income is % less than A’s.
Example: If A’s income is 20% more than B’s, then what percent is B’s income lesser
than A?
A’s income more than B = 20%, by the formula, B’s income is = × 100 = 16.66 % less of
A
4. If A’s income is r % less than B’s then B’s income is
% more than A’s
Example: If A’s income is 20 % less than B’s, then what percent is B’s income more
than A?
A’s income less than B = 20%, by the formula, B’s income is = × 100 = 25% more of A
5. If the present population of a town is p and let there be an increase of X % per annum.
Then:
(i) Population after n years =
(ii) Population n years ago =
This is the compound interest formula, which we will study in detail later. If the decrease
or depreciation is r%, then population or value of a machine (after depreciation) after n
years
=
CALCULATIONS IN PERCENTAGES
Let’s start with a number A (= 1A)
1. A increased by 10% would become A + 0.1A = 1.1A
2. A decreased by 10% would become A – 0.1A = 0.9 A
3. A increased by 200% would become A + 2A = 3A
4. A decreased by 50 % would become = 0.5A
5. Use decimal fractions while adding and subtracting and normal fractions while
multiplying.
SOLVED EXAMPLES
Q1. Express as percentages: .
Ans. To convert to percent, multiply each by 100
× 100 = 100/3 %, do the others similarly.
Q2. If a pipe, A is 30 meters and 45% longer than another pipe, B find the length of
the pipe B.
Ans. A = 30m, A = B + B => A = B => A = 1.45B Therefore B = = = 20.68.
Q3. On my sister’s 15th birthday, she was 159 cm in height, having grown 6% since
the year before. How tall was she the previous year?
Ans. Height this year = 159 cm, growth = 6 %, let last year height be A
Now A = 159 A = = 150
Last year height = 150
Q4. Arun spent 25 % of his pocket money, and has Rs 125 left. How much had he at
first?
Ans. Pocket money spent = 25%, left = 75%, Let original pocket money be A
Therefore A = 125 A = 125 ×
(By now students should be able to do this in single step), A = 166.66
Q5. If the cost of electricity increases by 30%, by what percent one should reduce
his spend in order that spent on electricity stays the same?
Ans. As per the formula, × 100 The reduced percentage spent = × 100 = 23.07%
Q6. If the price of petrol increases by 25% and Rajesh intends to spend only 15%
more on petrol, by how much % should he reduce the quantity of petrol that he
buys?
Ans. Let the initial cost of 1 litre be A
Cost after increase = 1.25A (25% increase)
Let Rajesh’s initially buy be ‘B’ litres of petrol
Initial spent = AB
Increase in spent = 15%,
Current spent = 1.15 AB (15% increase)
Let the number of litres he is buying now is C.
Therefore 1.25AC = 1.15 AB, 1.25 C = 1.15 B, C = 0.92B, which means that current
consumption is 92% of earlier consumption, therefore Rajesh has reduced his
consumption by 8 %
Q7. In an election, Congress secured 10% of the total votes more than BJP (consider
only two parties in the election and everyone voting). If BJP got 126000 votes, by
how many votes did it lose the election?
Ans. Let congress secured X % of the total votes, therefore BJP had secured (X – 10) %
of votes, being a two party election:
X + X – 10 = 100
2X = 110
X = 55
Therefore Congress has 55% of vote and BJP has 45%, since BJP got 126000 votes
of total votes = 126000
Total votes = 280000
Congress votes = 55/100 × 280000 = 154000
Difference = 28000, which is the victory margin.
Q8. If the population is1500000 and the expected birth rate is 50%, while the
expected death rate is 31%, What will be the net change in the in the population at
the end of the one year.
Ans. The current population is 1500000
Number of births will be 50/100 × 1500000 = 750000
Number of deaths was 31/100 × 1500000 = 465000
Net change = 750000 – 465000 = 285000
Q9. What is the % change in the area of a square (which will become rectangle) if its
length side is increased by 10% and its width side is decreased by 10%?
Ans. In these types of problems, assume a percent (100) base and then move forward.
Let the side of square be 100
So length side will become
1.1(10% increase) × 100 = 110
And the width side will become
0.9(10% decrease) × 100 = 90
Old Area = 100 × 100 = 10000
New Area = 110 × 90 = 9900
Difference = 10000 – 9900 = 100
Difference percent = 100/10000 × 100
= 1% decrease in the area
Q10. Ram obtains 40 % of the marks in a paper of 200 marks. Shyam is ahead of
Ram by 25 % of Ram’s marks, while Bhuvan is ahead of Shyam by one ninth of his
own marks. How many marks does Bhuvan get?
Ans. Ram’s marks = 40/100 × 200 = 80
Shyam’s Marks = 1.25 (25% ahead) × 80 = 100
Let Bhuvan’s Marks be A, therefore
A = 100 (Shyam’s marks) + 1/9A
8/9 A = 100, A = 112.5
After grasping the concept of percentages, let us move to its biggest application,
problems of profit and losses.
Chapter 5: Calendar
The year consists of 365 days, 5 hours, 48 minutes (52 weeks and 1 odd day). An extra
day is added once in every fourth year which was called the leap year, which has 366
days (52 weeks and 2 odd days).
To find the day of any given date of the year, you need to understand the calendar
calculations:
1. First thing to remember, first January 1 AD was Monday therefore, we must count
days from Sunday. This means the 0th day was Sunday, so the 7th day was Sunday again
and so on and so forth.
2. The day gets repeated after every seventh day (concept of a week), if today is Monday,
then 28th day from now will also be Monday as it a multiple of 7 (28/7 = 4, so four
weeks). Here the 30 day will be calculated by 30/7, which is 4 weeks and 2 days, these
two days are called odd days. With starting day as Monday and two odd days, the day
will be Wednesday; this point is the most critical in calendars. The other of looking at it
is since the 28th day is Monday, so the 30th day will be Wednesday. But you have to
understand and use the concept of odd days as the question may be about thousands of
years.
3. In a normal year there are 365 days so 52 weeks and 1 odd day, in a leap year there are
366 days so 52 weeks and 2 odd days.
4. In 100 years there are 24 leap years and 76 normal years, so the number of odd days
are 24(2) + 76 = 124, which is 17 weeks + 5 odd days, so 100 years have 5 odd days.
5. In 200 years the number of odd days is twice the number in 100 years which is 10,
which is one week and 3 odd days, so 200 years have 3 odd days. In 300 years, the
number of odd days is 15, which is two weeks and 1 odd day, so 300 years have one odd
day.
6. 400 year is a leap year; similarly the multiples of 400 are also leap years.
7. In 400 years, the number of odd days become 20 + 1(from the leap year), so total days
are 21, which is three weeks and 0 odd days. In 400 years there are 0 odd days
Example1: What was the day on 25th January, 1975?
Counting the years 1600 + 300 + 74
In 1600 years, there are zero odd days
In 300 years, there is one odd day
In 74 years, there are 18 leap years and 56 normal years, so the odd days are:
18(2) + 56(1) = 36 + 56 = 92,
Which is 13 weeks and 1 odd day
In 25 days of January, 1975, there are 3 weeks and 4 odd days
Total odd days = 0 + 1 + 1 + 4, six odd days, so it was a Saturday.
Q1. What day of the week was 20th June 1837?
Ans. 20th June, 1837 means 1836 complete years + first 5 months of the year 1837 + 20
days of June
1600 years give no odd days
200 years give 3 odd days
36 years give (36 + 9 ) or 3 odd days
Thus 1836 years give 6 odd days
From 1st January to 20th June there are 3 odd days
Odd days: January 3, Feb. 0, March 3, April 2, May 3, June 6 = 17
Thus the total number of odd days = 6 + 3 or 2 odd days
This means that the 20th of June fell on 2nd day commencing from Monday.
The required day was Tuesday.
Q2. How many times does the 29th day of the month occur in 400 consecutive years?
Ans. In 400 consecutive years there are 97 leap years. Hence, in 400 consecutive years
February has the 29th day 97 times and the remaining eleven months have the 29th day
400 × 11 or 4400 times.
Thus the 29th day of the month occurs
= 4400 + 97 = 4497 times.
Q3. Today is 3rd November. The day of the week is Monday. This is a leap year. What
will be the day of the week on this date after 3 years?
Ans. This is a leap year. So, none of the next 3 years will be leap years. Each year will
give one odd day so the day of the week will be 3 odd days beyond Monday i.e. it will be
Thursday.
Q4. December 9, 2001 is Sunday. What was the day on December 9, 1971? (CAT)
a) Thursday (b) Wednesday
(c) Saturday (d) Sunday
Ans. Total number of days = 30 × 365 + 8 days from leap years = 10958
Thus number of weeks = = 1565
Hence December 9, 1971 must have been Tuesday.
Q5. The calendar of year 1982 is same as which year
Ans. We need to have 0 odd days, counting from 1982,
Therefore 1988 could be the year with the same calendar as 1982, buts it’s a leap year
and 1982 is not. Therefore next is 1993, where it fits, so calendar of 1982 is same as 1993
Q6. If today is Monday, what will be the day 350 days from now?
Ans. 350 days, 350/7 = 50, no odd days, so it will be a Monday.
Q7. If today is Monday, what will be the day one year and 50 days from now?
Ans. Cannot be determined, as we don’t know if it is a leap year or not.
Q8. The calendar for the year 1984 is same as which upcoming year.
Ans. Follow the process as done in Question 5, remember it has to be a leap year, since
1984 is a leap year, answer is 2012
Chapter 6: Averages
An Average is a measure that tells us information about a group of things rather than data
about individuals.
Mean, median, and mode are all types of averages, although the mean is the most
common type of average and usually refers to the arithmetic mean. There are other kinds
of means that are more difficult.
The arithmetic mean is a simple type of average. Suppose you want to know what the
numerical average is in your math class. Let’s say your grades so far are 80, 90, 92, and
78 on the four quizzes you have had. To find your quiz average, add up the four grades:
80 + 90 + 92 + 78 = 340
Then divide that answer by the number of grades that you started with, four: 340/4 = 85.
So, your quiz average is 85. Whenever you want to find a mean, just add up all the
numbers and divide by the number of numbers you started with.
But sometimes the arithmetic mean doesn’t give you all the information you want.
Suppose you are looking for a job. You interview with a company that has ten
employees, and the interviewer tells you that the average salary is Rs. 20000 per month.
That’s a lot of money! But that’s not what you would be making.
For this particular company, you would make half of that. Each employee makes
Rs.10000 per month, except for the owner, who makes Rs.110000 per month. Nine
employees make Rs.10000, so adding those up is 9 x 10000 =90000. Then the owner
makes Rs.110000, so the total is Rs.110000 + Rs.90000 = Rs.200000. Divide by the total
number of employees, ten, and we have Rs.20000/10 = Rs.2000. Because the owner
makes so much more than everyone else, her salary “pulls” the average up.
A better question to ask is, “What is the median salary?” The median is the number in the
middle, when the numbers are listed in order. For example, suppose you wanted to find
the median of the numbers 6, 4, 67, 23, 6, 98, 8, 16, 37. First, list them in order: 4, 6, 6, 8,
16, 23, 37, 67, 98. Now, which one is in the middle? Well, there are nine numbers, so the
middle one is the fifth, which is 16, so 16 is the median.
Now, what about when there is an even number of numbers? Look at the quiz grade
example again: 90, 80, 92, 78. First list the numbers in order: 78, 80, 90, 92. The two
middle ones are 80 and 90. So do we have two medians? No, we find the mean of those
two: 80 + 90 = 170, and 170/2 = 85. So 85 is the median.
Now look at those salaries again. To find the median salary, we look at the salaries in
order: 10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000, 10000, 110000. This is
an even number of salaries, so we look at the middle two. They are both 10000, so the
median is Rs.10000. That’s much better at telling you how much you’ll make if you
accept the job.
But the median doesn’t always give you the best information either. Suppose you
interview with a company that has 10 general employees, 7 assistants, 3 managers, and 1
owner. For this company, the mean salary could be Rs.5000, and the median Rs. 4000.
But you are applying for the position of general employee, whose starting salary is Rs.
1000.
The mode is the type of average you want to know in this situation. The mode is the
number that occurs most frequently. In the example for median, 6 would be the mode
because it occurs twice, while the other numbers each occur once. In our employee
example, the mode is Rs. 10000 because that number occurs ten times, which is more
than the number of times any other number occurs.
The most common value found in a group consisting of several readings is called as the
‘mode’. A group of observation can have more than one modal value.
± Mean, median and mode are all good types of averages, and each works best in
different types of situations.
In the following example, we will measure the central tendency (that is mean; median,
and mode values) of the given group of readings; and later we also will try to understand
how these measures of central tendency may change when we add more values to the
group of readings.
Example:
A person wrote down the time he spent on the exercises in a week. These readings are as
follows.
Day
Exercise time
Monday
20 min.
Tuesday
30 min.
Wednesday
Thursday
Friday
20 min.
35 min.
25 min.
The graph of the data is as shown below.
GRAPH
1. FNow we calculate the mean.
Mean = Sum of the readings/Number of readings = 20 + 30 +20 +35 +25/5
Therefore Mean = 130/5 = 26 minutes
Now we calculate the median, after arranging the given set of numbers in the increasing
order we get,
Exercise time: 20, 20, 25, 30, and 35
The median (middle) value is 25 minutes.
Now we calculate the modal value. To calculate the mode we need to see how many
times each number is repeating. And then the mode is the number that is repeating for the
maximum.
Number 20 is repeating 2 times. Numbers 25, 30 and 35 are repeating only once.
Therefore the mode is 20 minutes.
• ± the mean, median and mode must have the same unit that is associated with the given
set of readings.
Let’s say the person adds Saturday in the exercise schedule. And on Saturday 30 minutes
are spent on the exercise. To calculate the new mean value, add up all the readings and
divide the sum by total number of readings.
Mean = 20 + 30 +20 +35 +20 + 30/6 = 160/6 = 26.66 minutes
Or, what we can do is multiply the original mean by the original number of readings to
get the sum of the original readings.
26 x 5 = 130
We add the new value of 30 to this to get 160 and divide it by 6 to get the new mean. To
calculate the new median, we rearrange the numbers in the increasing order. Therefore
we now get,
Exercise time: 20, 20, 25, 30, 30, and 35
However in this case we do not have one single middle value. Instead we have two
numbers in the middle that are 25 and 30. In this case we take the average of the two
middle numbers to find the median. Therefore average of 25 and 30 is (25+30)/2 = 27.5
minutes.
Therefore the new median value is 27.5 minutes.
To calculate the new mode, we observe that numbers 20 and 30 minutes repeat 2 times.
Numbers 25 and 35 minutes repeat only once. Therefore this set of readings has two
mode values and the two modes are 20 and 30 minutes.
Example: Find the measures of central tendency for the following readings.
-2, 350%, 12, (-3)2, -80%
Solution: To compute the mean, median, or mode all the readings must be presented in
the same number form.
Therefore we will need to convert the percentages into decimal form.
Therefore after converting the given set of readings into same number form, the new set
has the following values.
-2, 3.5, 12, 9, -0.8
Now, mean = -2 + 3.5 +12 +9 − 0.8/5 = 21.7/5 = 4.34
Now median of the readings is the middle value of the set after arranging the numbers in
the increasing order, which is
-2, -0.8, 3.5, 9, 12
Since none of the numbers repeats more than once, the above set of readings has no
mode.
The concept of averages can be applied to problems in slightly different way.
Example: During the week, John watches TV every night after he finishes his
homework. The following table shows how much time he spent last week watching
television.
Day
Entertainment time
Monday
1 hour
Tuesday
30 min.
Wednesday
30 min.
Thursday
45 min.
Friday
2 hours
John decided to include his TV time for Saturday and Sunday as well. When he did and
recalculated his mean, mode and median he discovered that only the mean had changed.
What can we say about the TV time for Saturday and Sunday that would produce this
situation?
Solution: To proceed with the numbers, they must be converted to the same units. After
rearranging the numbers to get the increasing order for entertainment time, we have: 30
min, 30 min, 45min, 60min, 120 min
Therefore, mean = 285/5 = 57 min and median = 45 min.
Since number 30 repeats 2 times and other numbers repeat only once, we can say that
mode = 30 min.
We need to find additional possible values for Saturday and Sunday that will not change
the median and mode.
Let us look at the number sequence.
30, 30, 45, 60, 120
If 45 should remain as the median (in the middle), then if we write a number to its left,
we must write a number to its right side also. And if number 30 should remain as the
mode, then the two more numbers, representing Saturday and Sunday, that we will
include into to the data set must be different from 45, 60 and 120.
Therefore we can say if 45 minutes should remain the median value for the entire week,
then if John spends more than 45 minutes but neither 1 hour nor 2 hours watching TV on
Saturday, then on Sunday he must spend less than 45 minutes watching television. (The
amount of time spent on TV on Saturday and Sunday can be swapped).
Average properties
1. If each number in a set of data points is increased by a certain number x, then the
average will be increase by x, similarly for incase of decrease it will decrease by x, and in
case of multiplication it will be multiplied by x.
Example: In a class of ten students, where the average weight is 50 kgs, if weight of each
student is increased by 5 kg what will be the new average?
Number of students = 10, average = 50,
Total weight of class = 500 (50 × 10)
In case each student’s weight is increased by 5 kg, excess weight = 50 (5 × 10)
Total weight of the class will become = 500 + 50 = 550
Average weight = = 55, so average is also increased by 5 kgs
2. The average speed in case of a journey from X to Y at speed of A m/sec and returning
back to X at a speed of B m/sec, is m/sec
Example: Sunil travels from Delhi to Noida at the speed of 40 km/hr and returns at the
speed of 50 km/hr. What is the average speed of the journey?
Using the formula,
=
= = 44.44 Km/hr
Always remember the units, in what units you have started and in what units you are
ending.
Alternative way is by the basic principle of mechanics,
Speed =
Which means Time =
And Average Speed =
Suppose distance from Delhi to Noida is A, therefore
Time while going, T1 =
Time while coming T2 =
Total time = =
Total Distance = A + A = 2A
Average speed = =
=
=
= 44.44 Km/hr
Here are some concepts of various kinds of mean, students are requested to go through
the concepts and understand the applications rather that cramming the formulas. There is
no direct question expected in CAT on basis of these formulas, But application may be
used widely in many concepts especially data interpretation.
Arithmetic mean: As the definition given above, The AM of n numbers X1, X2, X3 ……
Xn is denoted by X and calculated as
X=
X=
The S is called sigma, which represents summation, from values 1 to n.
For many values it is difficult to calculate mean using this formula. There is an easier
formula for that using two concepts, Assumed mean (a) and Deviation (d) from the
Assumed mean
X=
Here di = Xi – a
Example: AM of 20, 30, 40, 52, 96, 34, 36. Let the assumed mean be 40
Xi di
20 20 – 40 = – 20
30 30 – 40 = – 10
40 40 – 40 = 0
52 52 – 40 = 12
96 96 – 40 = 56
34 34 – 40 = – 6
36 36 – 40 = – 4
S di = 28
Mean, X = 40 + = 40 + = 40 + 4 = 44
Weighted AM: If there are two values, a and b, have weights (or frequency) of x and y,
then the weighted arithmetic mean is given by
WAM =
What is “weight” in weighted mean?
The weight is the relative strength of a data point, lets take respective an example:
Strength: Suppose there is a paper with four sections of 100 marks each – Maths,
English, Social studies and History. Now there are different weightages assigned to each
that is Maths – 40%, English – 20%, Social studies – 20% and History – 20%. If a
student gets following marks (Maths –60, English – 60, Social studies – 50 and History –
50) in different subjects, what are the net marks out of 100?
Here to get net marks out of 100, the respective marks in subjects have to be treated with
its weights as per following table:
Subject Marks Net marks
(marks*weight)
Maths 60 0.4(40%) 24
English 60 0.2(20%) 12
Social studies 50 0.2(20%) 10
History 50 0.2(20%) 10
Total (out of 100) – Mean Marks 56
Let’s also understand what is frequency?
Frequency: The concept of frequency is not the same as weight but can be treated in the
same manner. Suppose the marks in English for a class of 10 people are 50, 60, 50, 51,
65, 60, 50, 36, 60 and 65. Here 50 have occurred 3 times, 60 has occurred three times etc,
so the times a data point occurs in a set of data points is called its frequency. Now for the
above give data points, the following is the frequency and mean calculation:
Marks Frequency Total
50 3 150
60 3 180
51 1 51
65 2 130
36 1 36
Total 10 547
So the mean here is = 54.7
If Weights W1, W2 …..Wn are assigned to the values X1, X2 ….. Xn, then the Weighted
AM
WAM =
Example: WAM of 10, 15, 20, 25 with frequency 1, 2, 3, 5
Xi Wi Wi Xi
10 1 10
15 2 30
20 3 60
25 5 125
10 225
S Wi = 10 S Xi Wi = 225
WAM = = 22.5
Mean of means: If the means (M1, M2...) are given along with numbers (N1, N2...), then
the mean of means is given by using the Weighted Arithmetic Mean formula, where
weights are replaced by respective number:
Mean of means =
Example: The mean marks scored by 60 boys are 80 and those scored by 100 girls are 60
Mean = = = 67.5 Marks
Geometric mean: The other mean is the geometric mean, which is given by nth root of
the product of n numbers.
GM of A1 , A2 , A3 , ... An =
Example: GM of 2 and 8 is = = 4
Properties
1. The AM of given set of data points is always equal to or greater than their GM, that is
AM =GM
Example: Find the Geometric mean and Arithmetic mean of 2, 8
Arithmetic Mean = = 5
Geometric Mean = = = 4
2. If the geometric mean of one group of “a” numbers is x and that of another group of
“b” numbers is y, then the geometric mean of the combined groups is:
Harmonic mean: The Harmonic mean is give by
For two numbers a and b Harmonic mean is given by
HM =
==
For N numbers (a, b, c … n),
HM =
Harmonic mean is used to calculate average speed, with different speeds for same
distance.
Example: Ajay travels from Delhi to Shimla at the speed of 50 km/hr and returns at the
speed of 40 km/hr, what is the average speed of the journey?
Using the formula of Harmonic Mean,
=
= = 44.44 Km/hr
Weighted Harmonic Mean: The Weighted Harmonic mean is give by
For two numbers a and b with weights x and y it is given by
Weighted HM =
This is used to calculate the average speed when different distances are covered at
different speeds.
Example: A man covers his first 6 km at an average speed of 10 m/s, another 4 km at 8
m/s and the last 3 km at 5 m/s.
Average speed = weighted HM =
Mode: It is the number that occurs most frequently in a given set of numbers.
Example: In set of numbers, 1, 2, 3, 4, 5, 5, 4, 3, 2, 1 and 5, the number 5 occurred the
maximum number of times, so the mode is 5.
Median: It is the middle value of a set of data points arranged in an ascending or
descending order.
If the number of data points is odd, the median is value.
If the number of data points is even, the median is give by the average of the two middle
values, suppose the medium values are a and b, here median is given by .
Here
a = and b = + 1
Example: Median of numbers 25, 22, 30, 29, 36
First the values need to be arranged in the ascending order 22, 25, 29, 30, 36
Median is the value Þ 3rd value, which is 30
Example: Median of numbers 25, 22, 30, 29, 36, 38
First the values need to be arranged in the ascending order 22, 25, 29, 30, 36, 38
The two middle values are = 3 and + 1 = 4, the 3rd and 4th values are 29 and 30
Thus median = = = 28.
Chapter 7: Simple and Compound Interest
The lending and borrowing of money has been happening since thousands of years. Any
sum of money, borrowed for a certain period, will invite an extra cost to be paid on the
money borrowed; this extra cost at a fixed rate is called interest. The money borrowed is
called principal. The sum of interest and principal is called the amount. The time for
which money is borrowed is called period.
Amount = Principal + Interest
The interest paid per hundred (or percent) for a year is called the rate percent per annum.
The rate of interest is almost always taken as per annum, in calculations we will always
consider it per annum unless indicated.
The interest is of two types, one is simple, the other is compound:
Simple interest
It is the interest paid as it falls due, at the end of decided period (yearly, half yearly or
quarterly), the principal is said to be lent or borrowed at simple interest.
Simple Interest, SI = PRT / 100
Here P = principal, R = rate per annum, T = time in years
Therefore Amount, A = P + PRT/100 = P [1 + ( RT / 100 )]
If T is given in months, since rate is per annum, the time has to be converted in years, so
the period in months has to be divided by 12. if T = 2 months = 2/12 years)
Example 1: Find the amount on S.I. when Rs 4000 is lent at 5 % p.a. for 5 years.
By the formula, A = P (1 + RT/100) = 4000( 1 + 5 x 5/100 ) = Rs 5000
Compound Interest
The compound interest is essentially interest over interest. The interest due is added to
the principal and that becomes the new principal for the interest to be levied. This method
of interest calculation is called compound interest, this can be for any period (yearly, half
yearly or quarterly) and will be called “Period compounded” like Yearly compounded or
quarterly compounded and so on.
First period’s principal + first period’s interest = second period’s principal
Compound interest = principal {1 + Rate/100}time - Principal
CI = P { 1 + R/100 }T – P
Here Amount = principal {1 + Rate/100 }time
Example 2: Find the compound interest on Rs 4500 for 3 years at 6 % per annum
Using the formula, A = P (1 + R/100)T = 4500(1 + 6/100)3 = 4500 (1.06)3 = 5360
Compound interest = 5360 – 4500 = Rs 860
THE RULE OF 72
The rule of 72 is a quick way to show how long it will take to double your money under
The equation for the rule of 72 is:
Number of years for money to double = (72/Annual Interest Rate) interest rate
At 8% interest, it will take 72/8 = 9 years for your money to double.
Here are more examples:
At 6%, it will take 12 years ( 72/6 = 12)
At 12%, it will take 6 years ( 72/12 = 6)
The rule of 72 is a short cut to estimate the magic of compound interest that makes your
money grow.
• Remember that the rule of 72 is an approximation and its accuracy reduces as the
interest rate becomes high.
Important notes
1. In case interest is paid half yearly, then the interest is divided by 2, and used as (R/2) in
the formula and the time is multiplied by 2, and used as 2T in the formula, given by A =
P [ 1 + ( R / 200 ) ]2T
Example 3: Find the compound interest on Rs 5000 for 3 years at 6 % per annum
compounded half yearly.
Using the formula, A = P [ 1 + ( R / 200 ) ]2T
= 5000(1 + 6/200)3x2
= 5000 (1.03)6 = 5971
Compound interest = 5971 – 5000 = Rs 971
2. In case interest is paid quarterly, then the interest is divided by 4, and used as (R/4) in
the formula and the time is multiplied by 4, and used as 4T in the formula, given by A =
P [ 1 + ( R / 400 ) ]4T payable quarterly (rate = R/4, time = 4T)
Example 4: Find the compound interest on Rs 5000 for 3 years at 6 % per annum
compounded quarterly.
Using the formula, A = P [ 1 + ( R / 200 ) ]2T
= 5000(1 + 6/400)3x4
= 5000 (1.015)12 = 5978
Compound interest = 5978 – 5000 = Rs 978
3. In case the rates are different(R1, R2, R3….) for different years, the amount is given
by P{1 + R1/100}{1 + R2/100}{1 + R3/100}
Example 5: Find the compound interest on Rs 5000 for 3 years at 6 % per annum for first
year, 7% for the second year and 8% for the third year
Using the formula, P{1 + R1/100}{1 + R2/100}{1 + R3/100}
= 5000(1 + 6/100) (1 + 8/100) (1 + 9/100)
= 6125
Compound interest = 6125 – 5000 = Rs 1125
4. For population increase the formula to be used is P {1 + R/100 }T, and for decrease P
{ 1 - R/100 }T. It can also be used for depreciation factor.
Example 6: The death rate of a town with population of 100000 is 5 %, considering there
are no new births, what is the population of town in next three years?
Using the formula, P { 1 - R/100 }T
= 100000(1-5/100)3
= 100000(0.857) = 85738
5. In case the period is a fraction like 3 and ½ years, or a and b/c years, then the amount
should be calculated by this formula
A = P { 1 + R/100 }3{1+(1/2 x R)/100}
Or A = P { 1 + R/100 }a{1+(b/c x R)/100}
Example 7: The birth rate of a town with population of 100000 is 5 %, considering there
are no deaths in the town, what is the population of town in next three years and fours
months?
Three years and four months mean 3 1/4
Using the formula, A = P { 1 + R/100 }a{1+(b/c x R)/100}
= 100000(1+5/100)3(1+ ¼ x 5/100)
= 100000(1.157) (1.012)
= 117210 will be the population
6. The SI and CI earned during the first period remains the same.
Example 8: The compound interest on a certain sum of money in 2 years is 210 and the
simple interest on the same amount is 200, what are the principle and the rate of interest
Since SI and CI for first year is the same, and SI for each year is the same, so SI for the
first year = 200/2 = 100, CI for year I = 100, that means CI for the year II = 210 – 100 =
110. Here the excess of interest over year I = 10. Since the excess of interest in CI is
interest over first years interest, assuming I is the interest, I/100 x 100 = 10, so I = 10, and
the principal is obviously 1000.(calculate yourself)
Example 9: A sum of money placed at Compound Interest doubles in every 5
years, then in how many years it will become 16 times?
Now, it is given that the principle gets doubled in every 5 years.
So, if we start from initial amount P, then in first 5 years it will become 2P.
In the next 5 years 2P will become 4P, next 5 years 4P will become 8P and finally
in next 5 years 8P will become 16P.
So, it will take (5+5+5+5) = 20 years.
Net present value (NPV)
Money received or paid today is not the same as money received or paid after a period.
This is because the money has an opportunity cost of interest in the same period. What it
simply means is that you can earn interest on money if you have it now, and if you get the
money later, you loose the opportunity to make interest on that. For example, if the going
interest rate in the market is 10%, and someone has to pay me Rs. 1000, and he pays after
an year, so he should pay, 1100 (100 has the interest), Here 1100 is called the future
value and 1000 is called the present value.
Here the Future value (FV) = Present value (PV) {1 + Rate/100 }time, which is the basic
formula for amount in the case for compound interest, this is the formula to be used for
calculating present value. From here
PV = FV / {1 + Rate/100 }time
This is the same formula as of the compound interest; herein we are calculating principal
from the amount, which’s it!
Practical applications of the NPV
1. Installment schemes
Today there are all kinds of loans and financing of various products right from two
wheelers to houses. When a loan is taken, customer generally pays a monthly installment,
his dues reduced by that amount and the interest is charged only on the balance amount
which is known as reducing balance. Also there are many other concepts like floating
rates etc, but they are out of purview of CAT. Here is the monthly installment formula for
a fixed rate of interest (fixed means which does not change over time, floating means
which changes with market conditions):
Monthly Installment, M = [A/(1-B)] x P
Here A = R/1200 (where R is the rate of interest)
B = [1/(1+A)]T (Where T is time in months)
And P is the principal amount that is the amount of loan taken
The installment can be calculated with this formula by using concept of NPV also. This
formula is derived from there only. You can find this formula in Microsoft excel also
under PMT in the formula section. But these annuity formula questions will not be asked
in CAT.
Example 10: If Ram takes a home loan of 500000 for 3 years at the rate of 7.5%, what
will be his monthly installment?
T = 12 x 3 = 36 months
R = 7.5%
P = 500000
Using the formula, M = [A/(1-B)] x P
A = 7.5/1200 = 0.00625
B = [1/(1+A)]T = [1/(1+0.00625)]36
M = [0.00625 / {1 - [1/(1+0.00625)]36}] x 500000
M = 15553
Chapter 10 : Indices and Surds
Indices
If any number ‘a’ is multiplied 5 times, we say it as ‘ a raise to the power 5’ or a5.
Here ‘a’ is expressed as an exponent, where ‘a’ is the base and
5 is called the power or index of ‘a’.
So, we can write ‘a’ as any exponent ‘n’ which will be written as ‘an‘.
This can be written as ‘a to the power n’.
Example: 45 = 4 × 4 × 4 × 4 × 4 = 1024
Rules of Indices
1. am × an = am+n
2. am / an = am-n
3. (am)n = amn
4. a-m = 1 / am
5. (ab)m = am bm
6. a0 = 1
7. a1 = a
Note:
1. If given that (am) = (an)
Case 1: if a = 1 or 0 then we can not comment on ‘m’ and ‘n’.
Case 2: if a = -1, then we can say either both m and n are even or they are odd.
Case 3: in other cases we can say that m = n.
2. If given that (am) = (bm)
Case 1: if m = 0, then we can not comment on ‘a’ and ‘b’.
Case 2: if m is odd, then a = b.
Case 3: if m is even, then a = b or a = -b.
SURDS
The numbers of the type ‘ a + √b’ are called surds,
Where ‘a’ and ‘b’ are rational numbers.
If there is a surd of the form a + √b, then a - √b is called the conjugate of this surd.
The product of a surd and its conjugate is always a rational number.
Rationalization of a Surd
When we have a surd in the denominator of any calculation, then it is difficult to perform
certain operations on that number. It is always good to have the rational number in the
denominator.
So, in order to remove a surd we use a technique of rationalization in which of we have a
surd of the form a + √b in the denominator, then we multiply and divide that number with
the conjugate of that surd.
Doing this we can remove a surd from the denominator.
Comparison of surds and indices
Surds and indices can be compared among themselves in various ways.
1. By making the bases same.
2. By making the powers same.
3. In case of surds, they can be compared by squaring them.
4. Or surds can also be compared using rationalization method.
Example: Find which one is greater.
1. 299 or 834 834 = 2102 So 834 is greater.
2. 375 or 750
375 = 33×25 = 2725
750 = 72×25= 4925 So, 750is greater.
3. (√7 + √19) or (√5 + √21)
Squaring both of them, we get,
(√7 + √19)2 = 7 + 19 + 2√133 = 26 + 2√133
(√5 + √21)2 = 26 + 2√105 So, now we can say (√7 + √19) is greater.
4. (√23 - √17) or (√29 - √23) Rationalizing both of them,
(√23 - √17) = 6 / (√23 + √17)
(√29 - √23) = 6 / (√29 + √23)
Clearly, (√23 - √17) is greater.
Chapter 11 : Progression
A succession of numbers formed and arranged in a definite order according to a certain
definite rule is called a sequence or progression.
The number occurring at the nth place of a sequence is called its nth term or the general
term, to be denoted by tn.
A sequence is said to be finite or infinite according as the number of distinct terms in it is
finite or infinite.
We would learn about three types of progressions: Arithmetic, Geometric and Harmonic.
Arithmetic progression
It is a sequence in which each term, except the first one, differs from its preceding term
by a constant, called the common difference.
So, for examples 1, 6, 11, 16, …., so on and 1, – 1 , – 3, – 5, …., are A.P.
In the examples above, the difference between any two successive numbers is equal to 5
and – 2 respectively. This difference is called the Common Difference.
The general form of expressing this series is a, a + d, a + 2d, a + 3d, … so on.
The standard notations are as follows.
a = The first term, d = Common difference, Tn = The nth term
l = The last term, Sn = Sum of n terms,
1. Tn = a + (n – 1)d
2. Sn =
In the formula above, would give the average of the terms of the progression. So the sum
of the series is average of all the terms multiplied by number of terms.
Substituting, l = a + (n – 1)d
We also have Sn = n/2 [2a + (n – 1)d]
If to each term of an A.P. a fixed non-zero number is added, then the resulting
progression is also an A.P.
If each term of a given A.P. is multiplied or divided by a given non-zero fixed number k,
then the resulting progression is an A.P.
Arithmetic mean
If a and b are any two numbers, n is called the arithmetic mean of a and b and is given by
n=.
Arithmetic mean can also be found if there are more than two terms. For instance, the
arithmetic mean of a, b, c and d is equal to (a + b + c + d)/4.
In problems, Three numbers in an AP should be taken as a – d, a, a + d.
Four numbers in an AP should be taken as a – 3d, a – d, a + d, a + 3d.
This would make sum of three numbers in one variable so that we can solve it easily.
Chapter 12 : Quadratic and higher order equations
Chapter 13 : Function
Functions are relationships defined between a set of variables. The functional
operator works on the domain and maps it onto the range.
Take an example f(x) = 2x + 4 for all x (- R.
This is a function that maps any value of x, say 2 to a value y = f(x) = 2x + 4 = 8.
For f to be a functional operator, it maps one value of x to one value of y.
For any input, the output is defined by the relationship it stands for.
For instance, if f(x) = 3x2 + 4x + 2, then f(1) = 3(1)2 + 4(1) + 2 = 9
The input is 1 and the output is 9.
Similarly, if f(x) = max (x3 , x2, x) for x = ½ , f(1/2) = max(1/8, ¼, ½) = ½. The
functions can be defined as per the user. It need not necessarily be just an
algebraic expression.
Even or Odd Functins
Chapter 14 : Logarithms
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Chapter 15 : Binomial Theorem
Chapter 16 : Plain Geometry
Chapter 18 : Solid Geometry
Chapter 19 : PERMUTATION AND COMBINATION
Chapter 20 : PROBABILITY
Chapter 21 : SET THEORY
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