maths formulae

[1]

Arithmetical Formulae

Natural Numbers : Counting numbers 1, 2, 3, 4, 5, . . . are known as natural numbers. The set of allnatural numbers can be represented by N = [1, 2, 3, 4, 5, ...]Whole Numbers : Numbers 0, 1, 2, 3, 4, 5, . . . are known as whole numbers (if we include 0 amongthe natural numbers). The set of whole numbers can be represented by N = [0, 1, 2, 3, 4, 5, ...]. Everynatural number is a whole number, but 0 is a whole number which is not a natural number.Integers : All counting numbers and their negatives, and also zero (0) are known as integers. The setof integers can be represented by I = {..., –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, ...}.Positive Integers : All natural numbers are positive integers. Thus, the set I+ = {1, 2, 3, 4, 5, ...} is theset of all positive integers.Negative Integers : The set I– = {–1, –2, –3, –4, –5, ...} is the set of all negative integers.Rational Numbers : The number of the form x/y, where x and y are integers and y ¹ 0 are calledrational numbers. Thus 2/5, 3/7, 4/9, 5/3, 7,4, –5/7, –3/8, –2/7, 0/1, etc. are rational numbers. The setof all rational numbers is denoted by Q. Thus, Q ={n : n = x/y, x,y Î I, y ¹ 0}.Irrational Numbers : Those numbers which when expressed in decimal form are neither terminatingnor repeating decimals are called irrational numbers. For example, Ö2, Ö3, Ö5, p. [Note : the exact value

of p is not 22/7. 22/7 is a rational number but p is irrational. When we take the value of p as 22/7 or

3.14, both the values are only approximate values of p].

Odd Numbers : All those numbers which are not exactly divisible by 2 are called odd numbers.Even Numbers : All those numbers which are exactly divisible by 2 are called even numbers.When n is an odd number n(n2 - 1) is divisible by 24; 2n + 1 is divisible by 3; 22n + 1 is divisible by 5;52n + 1 is divisible by 13. When n is an even number 2n - 1 is divisible by 3; 22n - 1 is divisible by 5; 52n

- 1 is divisible by 13.Odd number raised to even or odd number shall remain odd. For example, 32 = 9, 33 = 27, 34 = 81.Similarly even number raised to even or odd number shall remain even. For example, 42 = 16, 43 = 64,44 = 256.Prime Numbers : Any natural number other than 1 if it is divisible by 1 and itself only is called aprime number. For example, 2, 3, 5, 7, 11 are prime numbers.Composite Numbers : Natural numbers greater than 1 which are not prime, are called composite

numbers. For example, 4, 6, 8, 9, 12, 15 are composite numbers.The number 1 is neither a prime number nor a composite number, and 2 is the only even number

that is prime.The 25 numbers, viz., 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83,

89, 97 are the prime numbers between 1 and 100.Two numbers which have only 1 as the common factor are called co-primes, i.e., relatively prime to

each other. For example, 3 and 5 are co-primes.If p is prime p(p4 - 1) is divisible by 30. If p is prime greater than 3, p2 - 1 is divisible by 24. If p is a

prime number then for any whole number n, np - n is divisible by p, i.e., n3 - n is divisible by 3, n7 - nis divisible by 7, n13 - n is divisible by 13, n17 - n is divisible by 17, and so on.

Multiplication made easy :Any number to be multiplied by 9, 99, 999, etc. can be solved easily by using 10n - 1, and adding as

many zeros (0) to the right of multiplicant as there are nines in the multiplier, and from the resultsubtract the multiplicant to get the answer.

Any number to be multiplied by 11, 101, 1001, etc. can be solved easily by using 10n + 1, and addingas many zeros (0) to the right of multiplicant as there are nines in the multiplier, and to the result addthe multiplicant to get the answer.

Any number to be multiplied by 5, 25, 125, 625, etc. i.e., by a number which is some power of5,place as many zeros (0) to the right of the multiplicant as is the power of 5 in the multiplier, thendivide the number so obtained by 2 raised to the same power as is the power of 5.

Any number to be multiplied by 15, 25, 35, etc. double the multiplier and then multiply the multiplicantby this new number and divide the product by 2 to get the required answer.

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Squares :

The square of an odd number will be odd and the square of an even number will be even. No squarenumber will end with an odd number of zeros.

The numbers ending with 0, 1, 5, or 6 will always give the same number in the unit digit. For example,

102 = 100; 112 = 121, 252 = 625, 162 = 256.

The square of a number shall never end with 2, 3, 7 or 8.

The square of any number with unit digit 2 or 8 shall always have 4 in the unit digit. For example,12 × 12 = 144; 18 × 18 = 324; 22 × 22 = 484; 28 × 28 = 784.



Every square is either a multiple of 3 or exceeds a multiple of 3 by unity; or a multiple of 4 or exceedsa multiple of 4 by unity. For example, 36 = 3 × 12; 49 = (4 × 12) + 1; 64 = (3 × 21) + 1; 81 = 3 × 27;144 = 4 × 36.

To square a number in which every digit is 1 count the number of digits in the given number and writethe numbers in ascending order followed soon in descending order.up to one.Thus, 112 = 121; 1112 = 12321; 11112 = 1234321; 222 = 22 (11)2 = 4 × 121 = 484;2222 = 22 (111)2 = 4 × 12321 = 49284; 22222 = 22 (1111)2 = 4 × 1234321 = 4937284;332 = 32 (11)2 = 9 × 121 = 1089; 3332 = 32 (111)2 = 9 × 12321 = 110889;33332 = 32 (1111)2 = 9 × 1234321 = 11108889.To square a number which is nearer to 10x : x2 = (x2 - y2) + y2 = (x + y) (x - y) + y2

For example : 9982 = (998 + 2) (998 - 2) + 22 = 996000 + 4 = 996004; 10042 = (1004 - 4) (1004 + 4) + 42 = 1008000 + 16 = 10008016To square a number which is ending with 25, multiply the number in the hundred’s place with thenumber in hundred’s placesuffixed by 5, and attach 625 to the resultant value.For example : 3252 = 3 × 35 = 105, hence the square of 325 is 105625. 5252 = 5 × 55 = 275, hence the square of 525 is 275625.

Cube of a number consisting of nines only : Say there are n nines in the number. Write (n - 1) ninesfollowed by 7, then (n - 1) zeroes follwed by 2 and finally n nines. Thus 993 = 970299; 9993 =997002999; 99993 = 999700029999;

Fourth power of a number consisting of nines only : Say there are n nines in the number. Write (n - 1)nines followed by 6, then (n - 1) zeroes follwed by 5, then (n - 1) nines follwed by 6 and finally(n - 1) zeroes follwed by 1. Thus 994 = 96059601; 9994 = 996005996001; 99994 = 9996000599960001.

Perfect square : The square of a natural number is known as perfect square.For example, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ... are the squares of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ... andso on.

Division made easy :A number is divisible by 2, if its unit’s place digit is 0, 2, 4, 6, or 8.

A number is divisible by 3, if the sum of its digits is divisible by 3.For example, 1782 is divisible by 3 since the sum of its digits is 18 which is divisible by 3.

A number is divisible by 4, if the number formed by its last two digits is divisible by 4.For example, 3462 is divisible by 4 since the sum of its last two digits is 8 which is divisible by 4.

A number is divisible by 5, if the unit’s digit in the number is either 0 or 5.

A number is divisible by 6, if the number is even and the sum of its digits is divisible by 3.For example, 5442 is divisible by 6 since it is an even number and the sum of the digits is 15 which isdivisible by 3.

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A number is divisible by 7, if the unit digit of the number is doubled and then it is subtracted from thenumber obtained after omitting the unit digit, and the remainder is divisible by 7, the given number isdivisible by 7.For example, 203 is divisible by 7 since doubling the unit digit we get 6 and when this is subtractedfrom 20 we get 14 which is divisible by 7.

A number is divisible by 8, if the number formed by its last three digits is divisible by 8.For example, 65792 is divisible by 8 since the number formed by its last 3 digits is divisible by 8.

A number is divisible by 9, if the sum of its digits is divisible by 9.For example, 41094 is divisible by 9 since the number formed by its digits is divisible by 9.

A number is divisible by 10, if it ends in 0.

A number is divisible by 11, if, starting from the RHS, (Sum of its digits at the odd place) – (Sum of itsdigits at even place) is equal to 0 or divisible by 11.For example, 79387 is divisible by 11 since the difference between the sum of its digits at odd placesand the sum of its digits in even places is 0.

A number is divisible by 12, if it is divisible by both 3 and 4.For example, 5508 is divisible by 12 since the number is divisible by both 3 and 4.

A number is divisible by 13, if the number of tens added to 4 times the number of units is divisible by 13.For example, 16224 is divisible by 13 since the number of tens 1622 added to 4 times the unit (4 × 4)= 1622 + 16 = 1638, and this number is divisible by 13.

A number is divisible by 15, if the sum of the digits of the number is divisible by 3 and the digit in itsunit place is either 0 or 5.For example, 86940 is divisible by 15 since the sum of the digits of this number (27) is divisible by 3and the digit in its unit place is 0.

A number is divisible by 16, if the number formed by the last four digits of the given number isdivisible by 16.For example, 5703482 is divisible by 16 since the number formed by the last four digits (3824) isdivisible by 16.

A number is divisible by 17, if the number of tens added to 12 times the number unit is divisible by 17.For example, 833 is divisible by 17 since the number of tens (83) added to 12 times unit (36) gives 119which is divisible by 17.

A number is divisible by 19, if the number of tens added to twice the number of unit is divisible by 19.For example, 1083 is divisible by 19 since the number of tens (108) added to twice the unit (6) gives114 which is divisible by 19.

A number is divisible by 29, if the number of tens added to thrice the number of unit is divisible by 29.For example, 1073 is divisible by 29 since the number of tens (107) added to thrice the unit (3) gives116 which is divisible by 29.Results on Division : Dividend = Quotient × Divisor + Remainder

HCF and LCM of Numbers :A number which divides a given number exactly is called a factor or divisor of the given number andthe given number is the multiple of the factors. For example, 45 is exactly divisible by 3, 5, 9 and 15;Here 3, 5, 9 and 15 are the factors of 45 and 45 is the multiple of these factors.

Common Factor : A common factor of two or more numbers is a number which divides each of themexactly.

Highest Common Factor : Highest common factor (HCF) of two or more numbers is the greatestnumber that divides each one of them exactly. HCF is also called Greatest Common Measure (GCM)or Greatest Common Divisor (GCD). For example, 12 = 2 × 2 × 3 = 6 × 2; 18 = 2 × 3 × 3 = 6 × 3;24 = 2 × 2 × 2 × 3 = 6 × 4; 36 = 2 × 2 × 3 × 3 = 6 × 6; 48 = 24 = 2 × 2 × 2× 2 × 3 = 6 × 8; Here we canfind that the highest common factor of 12, 18, 24, 36 and 48 is 6.

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Common Multiple : A common multiple of two or more numbers is a number which divides each ofthem exactly.

Least Common Multiple : The Least Common Multiple (LCM) of two or more given numbers is thelowest or least number which is exactly divisible by each of them. For example, Consider the twonumbers 18 and 24. Multiples of 18 = 36, 54, 72, 90, 108, 126, 144, ...; and the multiples of 24 = 48, 72,96, 120, 144; common multiples of the two numbers 18 and 24 are 72 and 144, and the least commonmultiple of 18 and 24 is 72.

HCF of the numbers × LCM of the numbers = Product of two numbers. The greatest number that willdivide a, b and c leaving remainders p, q and r will be HCF of (a - p) , (b - q) and (c - r).

Square Root and Cube Root :Square root of a given number is that number which when multiplied by itself is equal to the givennumber. To find square root one may adopt two distinct methods as follows :

Factorisation method : Find the prime factors of the given number; group the factors in pairs; take onenumber from each pair of factors and then multiply them together. The product thus obtained will bethe square root of the given number.For example, 3136 = 2 × 2 × 4× 4 × 7 × 7, therefore, Ö3136 = 2 × 4 × 7 (taking one number from each

pair of factor) = 56.

Division method : Normally when the number is large and the factors cannot be easily determined themethod of division can be adopted.For example, consider the number 5475600. Separate the digits of this numberinto periods of two beginning from the right side. Thus the given number canbe separated as 5 47 56 00.

Find a number whose square may be equal to or less than the first period.Here the number is 2 whose square is 4 which is less than the first period 5.The remainder will be 1. Bring down the next period, i.e.,47 and double thequotient (4) and take it as divisor. Now 3 affixed to the divisor makes it 43 andmultiplied by 3 gives 129, leaving a remainder of 18. Now bring the nextperiod (56) and write it next to 18 , thus the number will be 1856.

Now add the divisor (43 + 3) = 46, and multiply it by a number that can give next quotient. Here it is464 and multiplied by 4 shall yield 1856. The last period of couple of zeroes can be taken as 0. Thus,the square root of 5475600 will be 2340.

Cube root of a given number is that number which when raised to the third power produces the givennumber. To find cube root one may resort to method of factorisation. In this method write the givennumber as product of prime factors. Take the product of prime numbers, choosing one out of three ofeach type.For example, consider the number 614125. This can be factorised as 5 × 5 × 5 × 17 × 17 × 17. Thus theof

3Ö614125 will be 5 × 17 = 85.

Tips to find cube roots :If the cube ends in 0, then its cube root also ends in 0.If the cube ends in 1, then its cube root also ends in 1.If the cube ends in 2, then its cube root also ends in 8If the cube ends in 3, then its cube root also ends in 7.If the cube ends in 4, then its cube root also ends in 4.If the cube ends in 5, then its cube root also ends in 5.If the cube ends in 6, then its cube root also ends in 6.If the cube ends in 7, then its cube root also ends in 3.If the cube ends in 8, then its cube root also ends in 2.If the cube ends in 9, then its cube root also ends in 9.13 = 1; 23 = 8; 33 = 27; 43 = 64; 53 = 125; 63 = 216; 73 = 343; 83 = 512; 93 = 729; 103 = 1000.

2 3 4 0

2 5 47 56 00

4

43 147

129

464 1856

1856

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Simple Arithmetical Operatioms :

To simplify the expressions formulated according to the statements of the problems relating to practicallife. The order of various operations in exercises involving brackets and fractions must be performedstrictly according to the order of the letters of the word BODMAS.

Each letter of the word BODMAS stands as follows : B stands for brackets. There are 4 brackets, viz.,bar

_, ( ), { }, [ ]; These brackets are to be removed strictly in the order

_, ( ), { }, [ ]; following brackets

it is O that stands for of; following O it is D that stands for division; following division it is M that standsfor multiplication; following multiplication it is A that stands for addition; following addition it is S thatstands for subtraction.

Algebraic Formulae :

(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

(a + b)2 + (a – b)2= 2(a2 + b2)

(a + b)2 – (a – b)2 = 4ab

a2 – b2 = (a + b) (a – b)

(a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab (a + b)

(a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab (a – b)

a3 + b3 = (a + b) (a2 – ab + b2)

a3 – b3 = (a – b) (a2 + ab + b2)

a3 + b3 + c3 – 3abc = (a + b + c)a2 + b2 + c2 – ab – bc – ca

a4 – b4 = (a2 + b2) (a + b) (a – b)

xn + yn = (x + y) (xn-1 – xn-2y + . . . + yn-1)

xn + yn is divisible by (x + y) when n is odd.

xn – yn is divisible by (x + y) when n is even.

xn – yn is always divisible by (x – y) whether n is even or odd.

Surds and Indices :

an is called a surd if n is a fraction and an is called the index if n is an integer; a is called the base.

am × an = am+n am ÷ an = am–n (am)n = (an)m = amn

am ÷ b–n = am × bn (a ÷ b)m/n = (b ÷ a)m/n (nÖa)n = anÖa.nÖb = nÖab (Öa + Öb)2 = a + b + 2Öab (Öa – Öb)2 = a + b – 2Öab

a + Öb = c + Öd Þ a = c and b = d

Percentage :

To express x% as a fraction, we have x% = x ÷ 100

To express a / b as a per cent, we have a ÷ b = (a ÷ b × 100) %

If A is R% more than B, then B is less than A by {100R ÷ [100 + R] } %

If the price of a commodity increases by R%, then the reduction in consumption, not to increase theexpenditure is {100R ÷ [100 + R] } %

If ‘A’ is R% less than ‘B’, then ‘B’ is more than ‘A’ by {100R ÷ [100 – R] } %

If the price of a commodity decreases by R%, then the increase in consumption, not to increase theexpenditure is {100R ÷ [100 – R] } %

If the population of a town is P in a year, then its population after N years is P (1 + R/100)N

If the population of a town is P in a year, then its population N years ago is P ÷ [(1 + R/100)N]

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Average :

Sum of quantities ÷ Number of quantities = Average

Sum of quantities ÷ Average = Number of quantities

Average × Number of quantities = Sum of quantities

Average of two or more groups taken together, say n1 and n

2 are two quantities and their average is p

and q respectively, then the combined average of all of them put together is (n1p + n

2q) ÷ (n

1 + n

2).

When the average of n1 quantities is p and the average of n

2 quantities is q then the average of the

remaining quantities in the group is (n1p – n

2q) ÷ (n

1 – n

2).

When the average of n quantities is p and if one of the given quantities whose value is a is replaced bya new quantity of value b, the average becomes q, then b = a + n(q – p).

When the average of n quantities is p and if one of the given quantities is removed, the averagebecomes q. The value of the removed quantity will be n(p – q) + q.

When the average of n quantities is q and if one new quantity p is added, the average becomes q. Thevalue of the new quantity will be n(q – p) + q.

The average of first n natural numbers is (n + 1) ÷ 2.

The average of square of natural numbers till n is (n + 1) (2n + 1) ÷ 6.

The average of cubes of natural numbers till n is n(n + 1)2 ÷ 4.

The average of even numbers from 1 to n is (last even number + 2) ÷ 2.

The average of odd numbers from 1 n is (last odd number + 1) ÷ 2.

The average of n consecutive numbers, consecutive even numbers or consecutive odd numbers isalways the middle number, if n is odd.

The average of n consecutive numbers, consecutive even numbers or consecutive odd numbers isalways the middle two numbers, if n is even.

The average of first n consecutive even numbers is (n + 1).

The average of first n consecutive odd numbers is n.

The average of squares of first n consecutive even numbers is [2(n + 1)(2n + 1)] ÷ 3.

The average of squares of first n consecutive even numbers till n is (n + 1)(n + 2)] ÷ 3.

The average of squares of first n consecutive odd numbers till n is n(n + 2) ÷ 3.

When the average of n consecutive numbers is p, then the difference between the smallest and thelargest number is 2(n – 1).

Weighted Average : When some items are more important than others proper weightage is given tothe various items. If the values x

1, x

2, x

3, . . . , x

n are assigned weights w

1, w

2, w

3, . . . , w

n respectively,

then the weighted average is calculated as (w1x

1 + w

2x

2 + w

3x

3 + . . . + w

nx

n) ÷ (w

1 + w

2 + w

3 + . . . + w

n)

The term weight stands for the relative importance that is attached to the different values.

Ratio & Proportion :

Ratio : A ratio is a comparison of two quantities by division. If x and x are two numbers, the ratio ofx to y is x/y and is denoted by x : y.

The ratio of the squares of two numbers is called the duplicate ratio of the two numbers, e.g., 72 : 92

or 49 : 81 is the duplicate ratio of 7 : 9.

The ratio of the cubes of two numbers is called the triplicate ratio of the two numbers, e.g., 23 : 33 or

8 : 27 is the triplicate ratio of 2 : 3.

The ratio of the square roots of two numbers is called the sub-duplicate ratio of the two numbers,e.g., 7 : 9 is the sub-duplicate ratio of 49 : 81.

The ratio of the cube roots of two numbers is called the sub-triplicate ratio of the two numbers, e.g.,2 : 3 is the sub-triplicate ratio of 8 : 27.

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If the antecedent and the consequent of a ratio interchange their places, the new ratio is called thereciprocal ratio or inverse ratio of the first, e.g., if p : q be the given ratio, then 1/p : 1/q or q/p isits reciprocal or inverse ratio.The ratio of the product of the antecedents to that of the consequents of two or more given ratios iscalled compound ratio, e.g., m : n and p : q are two given ratios then mp : nq is the compound ratio

of the given ratios, for example, 2/3, 3/4, 4/5, 5/7 are the given ratios, then the comppound ratio willbe 2 × 3 × 4 × 5 ÷ 3 × 4 × 5 × 7, i.e., 2/7. .

Proportion : The equality of two ratios is called a proportion. If a : b = c : d, we write a : b : : c : d andwe say that a, b, c, d are in proportion.

In a proportion, the first and fourth terms are known as extremes, while the second and third areknown as means. If four quantities are in proportion then, Product of extremes = Product of means.

Third proportional : When m : n : : n : q, then q is called the third proportional of m and n. Thisimplies, q = n2/m, and this is the third proportional of m and n.

Fourth proportional : When p : q : : r : x, then x is called the fourth proportional of p, q and r. Thisimplies, p/q = r/x, or x = qr/p.

When m : x : : x : n, x is known as the mean proportional or second proportional of m and n. Thisimplies, m/x = x/n, or x2 = mn or x = Ömn.

Order of ratio : To determine which of the two given ratios m : n and x : y is greater or smaller, wecompare m × x and n × y. For n > 0 and y > 0. (i) If m × y > n × x, then m/n > x/y. (ii) If m × y < n× x, then m/n < x/y. (iii) If m × y = n × x, then m/n = x/y.

Profit & Loss :

Cost Price : The price of an article at which it has been purchased.

C. P. = {100 ÷ (100 + Gain%)} × S. P. or {100 ÷ (100 – Loss%)} × S. P.

Selling Price : The price of an article at which the article is sold.

S. P. = {(100 + Gain%) ÷ 100} × C. P. or {(100 – Loss%) ÷ 100} × C. P.

Profit : If the selling price is more than the cost price the transaction gives a gain or profit.

Profit percentage : (Profit × 100) ÷ Cost Price

Loss : If the cost price of an article is more than its selling price the transaction ends up at loss.

Loss percentage : (Loss × 100) ÷ Cost Price

Marked Price : Marked price or list price is the price on the label of a product.

S. P. = Marked Price – Percentage discount on marked Price.

Discount : It is the reduction made on the marked price of an article.

Discount per cent = (Marked Price – Selling Price) ÷ Marked Price × 100.

If the C. P. of x article is equal to the S. P. of y articles then gain or loss % = [(x – y) ÷ y] × 100

When X sells an article to Y at a loss or gain of m% and Y sells it to Z at a loss or gain of n%, and if Zpays Rs. a for the article the original C. P. = 1002Z ÷ [(100 + m) (100 + n)].

When X sells an article to Y at a loss or gain of m% and Y sells it to Z at a loss or gain of n%, then theresultant loss or profit per cent is given by m + n + (mn/100).

Partnership :

If two partners X and Y have Rs. P1 and Rs. P

2 as their shares in capital for the same period of time and

the total profit accrued is Rs. P, then the shares of the two partners in the profit will be as follows :X = Rs. (P

1 × P) ÷ (P

1 + P

2) and Y = Rs. (P

2 × P) ÷ (P

1 + P

2).

If three partners X, Y and Z have Rs. P1, Rs. P

2 and P

3 as their shares in capital for the same period of

time and the total profit accrued is Rs. P, then the shares of the three partners in the profit will be :X = Rs. (P

1 × P) ÷ (P

1 + P

2 + P

3), Y = Rs. (P

2 × P) ÷ (P

1 + P

2 + P

3) and Z = Rs. (P

3 × P) ÷ (P

1 + P

2 + P

3).

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If two partners X and Y have Rs. P1 and Rs. P

2 as their shares in capital for the period of time T

1 and T

2

and the total profit accrued is Rs. P, then the shares of the two partners in the profit will be :

X = Rs. (P1 × T

1 × P) ÷ (P

1T

1 + P

2T

2) and Y = Rs. (P

2 × T

2× P) ÷ (P

1T

1 + P

2T

2).

If three partners X, Y and Z have Rs. P1, Rs. P

2 and P

3 as their shares in capital for the period of time

T1, T

2 and T

3 and the total profit accrued is Rs. P, then the shares of the three partners in the profit will

be as follows : X = Rs. (P1 × T

1 × P) ÷ (P

1T

1 + P

2T

2 + P

3T

3) Y = Rs. (P

2 × T

2 × P) ÷ (P

1T

1 + P

2T

2 + P

3T

3) and

Z = Rs. (P3 × T

3 × P) ÷ (P

1T

1 + P

2T

2 + P

3T

3).

If the capital of two partners X and Y be Rs. P1 and Rs. P

2 for the period of time T

1 and T

2 respectively

then profit of X : profit of Y = Rs. (P1 × T

1) : Rs. (P

2 × T

2).

If the capital of three partners X, Y and Z be Rs. P1, Rs. P

2 and Rs. P

3 for the period of time T

1, T

2 and

T3 respectively then profit of X : profit of Y : profit of Z = Rs. (P

1 × T

1) : Rs. (P

2 × T

2) : Rs. (P

3 × T

3).

If the capital of three partners X, Y and Z be Rs. I1, Rs. I

2 and Rs. I

3 and their profits are in the ratio P

1

: P2 : P

3 , then the ratio of the time period of investment of the three partners will be I

1/P

1 : I

2/P

2 : I

3/P

3.

If the profits of three partners X, Y and Z be Rs. P1, Rs. P

2 and Rs. P

3 and the ratio of the time period of

investment of the three partners be T1, T

2 and T

3 , then the ratio of the capital invested by the three

partners will be : P1/T

1 : P

2/T

2 : P

3/T

3.

Note : The same formula in each case holds good for loss also.

Time & Work :

If A can do a piece of work in n days, then A’s 1 day’s work = 1/n.

If A’s 1 day’s work = 1/n, then A can finish the work in n days.

If A is thrice as good a workman as B, then:

Ratio of work done by A and B = 3 : 1,

Ratio of times taken by A & B to finish a work = 1 : 3

If A does a work in x days and B does the same work in y days, then both of them working togetherwill do the same work in xy ÷ (x + y) days.

If A, B and C can do one work in x, y and z days respectively, then the three of them working togetherwill do the same work in xyz ÷ (xy + yz + zx) days.

If A and B working together are able to do one work in x days, and if A alone can do the same workin y days, then B alone can do the same work in xy ÷ (y – x) days.

If among three workers A, B and C, if A and B working together are able to do one work in x days, ifB and C working together are able to do one work in y days, and if C and A working together are ableto do one work in z days, then all the three workers A, B and C working together can do the samework in 2xyz ÷ (xy + yz + zx) days. A alone can do the work in 2xyz ÷ (xy + yz – zx) days; B alone cando the work in 2xyz ÷ (zx + yz – xy) days; and C alone can do the work in 2xyz ÷ (xy + zx – yz) days.

If x men and y women can do a piece of work in d days then m men and n women can do the samework in dxy ÷ (xn + ym) days.

Suppose two groups of people are involved in a job with same efficiency. In one M1 persons can do W

1

works in T1 time and in the other M

2 persons can do W

2 works in T

2 time. The relationship between the

two groups can be given by M1W

1T

1 : M

2W

2T

2.

Pipes and Cisterns :

When an inlet pipe takes n hours to fill one empty tank to the brim, then in 1 hour 1/n part of the tankwill be filled.

When an outlet pipe takes m hours to completely drain one full tank, then in 1 hour 1/m part of thetank will be emptied.

When both the pipes are open simultaneously 1/n – 1/m part of the tank will be filled in 1 hour.

[9]

Three pipes P1, P

2 and P

3 take t

1, t

2 and t

3 hours to fill a cistern respectively, when opened one at a time.

If all the three pipes P1, P

2 and P

3 are opened simultaneously they will take (t

1 × t

2 × t

3) ÷ (t

1t2 × t

2t3 × t

3t1)

hours to fill the cistern.

Two pipes P1 and P

2 can fill a tank in x hours and y hours respectively and another pipe P

3 is for

emptying the tank.

When all the pipes are open together the tank fills in t hours. In this instance the capacity of pipe P3 to

empty the full tank will be xyt ÷ (xt + yt – xy) hours.

Time Speed and Distance :

Speed = Distance travelled ÷ Time taken

Distance = Speed × Time

Time = Distance ÷ Speed

One kmph = 1000 metres ÷ 3600 seconds = 5/18 m. per sec. 1 m. per sec. = 18/5 kmph.

Suppose a man covers a distance at x kmph and an equal distance at y kmph, then average speedduring his whole journey is [2xy ÷ (x + y)] kmph

Suppose a man covers a distance d1 km. at s

1 kmph and then d

2 km. at s

2 kmph, then his average speed

during the whole journey will be : [s1s2 (d

1 + d

2)] ÷ (s

1d

2 + s

2d

1) kmph.

Suppose a man X goes from A to B at s1 kmph, and returns from B to A at s2 kmph, then the averagespeed during the whole journey will be : 2 s

1s2 ÷ (s

1 + s

2).

A man goes certain distance X to Y at a speed of s1 kmph and returns from Y to X at a speed of s

2 kmph.

If the total time taken for the journey both ways was t hours, the distance between A and B will be :

t × [s1s2 ÷ (s

1 + s

2)].

If two cars A and B start at the same time from two points x and y towards each other and after crossingthey take T

1 and T

2 hours to reach y and x respectively, then Speed of A : Speed of B = ÖT

2 : ÖT

1.

Trains :

Lengths of trains are x km and y km, moving at u kmph and v kmph (where, u > v) in the samedirection, then the time taken by the overtaking train to cross the slower train is [(x + y) ÷ (u – v)] hoursand time taken to cross each other is [(x + y) ÷ (u + v)] hours.

If a train overtakes a pole or a man or a tree, then the distance covered in overtaking is equal to thelength of the train.

If a train overtakes a bridge or a tunnel or a platform, then the distance covered in overtaking is equalto the sum of the two lengths, i.e., the length of the train and the length of the bridge/tunnel/platform.

Boats and Streams :

If the speed of a boat in still water is u km/hr and the speed of the stream is v hm/hr,then, Speed downstream = (u + v) km/hr.and Speed upstream = (u –v) km/hr.

If the speed downstream is a km/hr and the speed upstream is b km/hr, then:

Speed in still water = ½ (a + b) km/hr.

Rate of stream = ½ (a – b) km/hr.

A boatman rows a certain distance downstream in h1 hours and returns the same distance upstream in

h2 hours, when the speed of the running water is s kmph.

Thus, the speed of the boatman in still water will be : s [(h2 + h

1) ÷ (h

2 - h

1)].

A boatman rows a boat in still water x kmph. In a stream with water current w kmph he takes t hoursto go upstream from A to B and return downstream from B to A.Hence the distance between A and B will be : [t(x2 - w2)] ÷ 2x km. .

[10]

Races and Games of Skill :

X beating Y by t seconds means X finishes the race t seconds before Y finishes.

X gives Y a start of t seconds means X starts t seconds after Y starts from the same point.

X gives Y a start of l metres means while X starts from the starting point Y starts l metres ahead of thestarting point, though both start at the same time.

In a game of 100 X can give Y 30 points means while X scores 100 points Y scores 100 – 30 = 70 points.

If a boy A can run x metre race in t1 seconds and another boy B can run the same distance in t

2 seconds,

where t1 < t

2, then A will beat B by a distance of x ÷ t

2 × (t

2 – t

1) metres.

Alligation or Mixture :

Alligation : It is the rule that enables us to find the ratio in which two or more ingredients at

the given price must be mixed to produce a mixture at a given price.

Mean Price : The cost price of a quantity of the mixture is called the mean price.

Rule of Alligation : If two ingredients are mixed, then it is calculated as :(Cheaper quantity) : (Dearer quantity) = (CP of Dearer – Mean Price) : (Mean Price – CP of Cheaper).

Problems on Ages :

If the age of X was n1 times the age of Y, t years ago, and at present age of X is n

2 times the age

of Y, then present ge of X will be : [(n1 – 1) ÷ (n

1 – n

2)] × n

2t years; and the present age of Y will

be : [(n1 – 1) ÷ (n

1 – n

2)] × t years.

If the present age of X is n1 times the present age of Y, and t years hence the age of X will be n

2

times the age of Y, then the present age of X will be : [(n2 – 1) ÷ (n

1 – n

2)] × n

2t years, and the

present age of Y will be [(n2 – 1) ÷ (n

1 – n

2)] × t years.

If the sum of the present ages of X and Y is T years, and if p years hence, if the age of X would ben times the age of Y, then the present age of X will be : [Tn + p(n – 1)] ÷ (n + 1) years; and thepresent age of Y will be : [T – p(n – 1)] ÷ (n + 1) years.

If the ratio of ages of X and Y is m : n and after t years it will become p : q, then the present age ofX will be : [mt (p – q)] ÷ (mq – np) years; and the present age of Y will be : [nt (p – q)] ÷ (mq – np)years.

Simple Interest :

Principal = P, Rate = R% per annum, T is the number of years, Interest = I, and Amount = A..

Simple Interest (S.I.) = ( P × R × T ) ÷ 100 P = ( 100 × S.I. ) ÷ ( R × T )

R = ( 100 × S.I. ) ÷ ( P × T ) T = ( 100 × S.I. ) ÷ ( P × R )

Compound Interest :

Principal = P, Rate = r % per annum, T is the number of years, Interest = I, and Amount = A.

Amount (A) = P [ 1 + r/100 ]T

Compound Interest (C.I.) = A – P, i.e., P[(1 + r/100)T – 1]

Rate of Interest R = [(A/P)1/T – 1]

When interest is compounded Half-yearly : Amount = P [ 1 + r/200 ]T

When interest is compounded Half-yearly: Compound Interest (C.I.) = A – P, i.e., P[(1 + r/200)T – 1]When interest is compounded Quarterly: Amount = P [ 1 + r/400 ]T

When interest is compounded Quarterly: Compound Interest (C.I.) = A – P, i.e., P[(1 + r/400)T – 1]

When the rate of interest differs for different years say R1, R

2, R

3 for first, second and third years

respectively, then Amount (A) = P (1 + R1/100) (1 + R

2/100) (1 + R

3/100)

When a certain sum of money at compound interest amounts to Rs. a in X years and to Rs. b in Y years,then the rate of interest (R) per annum will be : [(b/a)1/Y–X – 1] × 100 %

[11]

Logarithms :

Logarithm is the exponent or power to which a stated number called the base, is raised to yield aspecific number.If for a positive real number (a ¹ 1), am = b, then the index m is called the logarithm of b to the basea. This is written as : log

ab = m

The exponential form of 24 = 16 can be written in logarithmic form as log216 = 4.

Product Formula : The logarithm of the product of two numbers is equal to the sum of their logarithms,i.e., log

a(mn) = log

am + log

an.

Quotient Formula : The logarithm of the quotient of two numbers is equal to the difference of theirlogarithms, i.e., log

a(m/n) = log

am – log

an.

Power Formula : The logarithm of a number raised to a power is equal to the power multiplied bylogarithm of that number, i.e., log

a(mn) = n log

am, where a, m are positive and a ¹ 1.

Base Changing Formula : logn

m = logam ÷ log

an, hence log

n

m = logm / logn, where a, m, n are positiveand a ¹ 1, n ¹ 1.Reciprocal Relation : log

ba × log

ab = 1, where a, b are positive and not equal to 1.

logba = 1/log

ab.

alog

ax = x, where a and x are positive and a ¹ 1.

If a > 1 and x > 1, then logax > 0.

If 0 < a < 1 and 0 < x < 1, then logax > 0.

If 0 < a < 1 and x > 1, then logax < 0.

If a > 1 and 0 < x < 1, then logax < 0.

For a >1, a ¹ 1, c > 0 ab > c Û logac < b.

For 0 < a < 1, a ¹ 1, c > 0 ab > c Û logac > b.

Logarithm of 1 to any base is 0, i.e., loga I = 0, where a > 0, a ¹ 1.

Logarithm of any two numbers to the same base is 1, i.e., loga a = 1, where a > 0, a ¹ 1.

Stocks and Shares

Brokerage : The broker’s charge is called brokerage.

When stock is purchased, brokerage is added to the cost price.

When the stock is sold, brokerage is subtracted from the selling price.

The selling price of a Rs. 100 stock is said to be : (i) at par, if S.P. is Rs. 100 exactly; (ii) above par (orat premium), if S.P. is more than Rs. 100; (iii) below par (or at discount), if S.P. is less than Rs. 100.

By a Rs. 800, 9% stock at 95, we mean a stock whose face value is Rs. 800, annual interest is 9% of theface value and the market price of a Rs. 100 stock is Rs. 95.True Discount

Say, a man has to pay Rs. 156 after 4 years and the rate of interest is 14% per annum.

Clearly, Rs. 100 at 14% will amount to Rs. 156 in 4 years. So, the payment of Rs. 100 now will clearoff the debt of Rs. 156 due 4 years hence. We say that : Sum due = Rs. 156 due 4 years hence; Present

Worth (P.W.) = Rs. 100; and True Discount (T.D.) = Rs. (156 - 100) = (Sum due) – (P.W.)

T.D. = Interest on P.W.

Amount = (P.W.) + (T.D.)

Interest is reckoned on R.W. and true discount is reckoned on the amount. Suppose rate = R% perannum and time = T years. Then,

(i) P.W. = (100 × Amount) ÷ [100 + (R × T)] = (100 × T.D.) ÷ (R × T);

(ii) T.D. = (P.W. × R × T) ÷ 100;

(iii) Sum = (S.I. × T.D.) ÷ (S.I. – T.D.);

(iv) (S.I.) – (T.D.) = S.I. on T.D.;

(v) When the sum is put at compound interest, then P.W. = Amount ÷ (1 + R/100)T

[12]

Banker’s Discount

Banker’s Discount (B.D.) is the S.I. on the face value for the period from the date on which the bill wasdiscounted and the legally due date.

Banker’s Gain (B.G.) = (B.D.) – (T.D.) for the unexpired time

When the date of the bill is not given, grace days are not to be added

B.D. = S.I. on bill for unexpired time = (B.D.) – (T.D.) = S.I. on T.D. = (Amount × Rate × Time) × 100

T.D. = (Amount × Rate × Time) ÷ [100 + (Rate × Time)] = (B.G. × 100) ÷ (Rate × Time)

Amount = (B.D. × T.D.) ÷ (B.D. – T.D.)

Clocks and Calendars

Clocks :

The hour hand in a clock moves through 30° in one hour, i.e., it moves through 0.5° in one minute.

The minute hand in a clock moves through 360° in one hour, i.e., it moves through 6° in one minute.Thus in every minute the minute hand gains 5.5° more than the hour hand.

Both hands – the hour hand and the minute hand – coincide once in every hour.

Both hands – the hour hand and the minute hand – are straight (at 180°) once in every hour, in thisposition the hands will be 30 minutes apart.

Both hands – the hour hand and the minute hand – are right angles (at 90°) twice in every hour, in thisposition the hands will be 15 minutes apart.

Calendars :

An ordinary year has 365 days.

A leap year – a year divisible by four (except century) – has 366 days. For example, 2004, 2008, 2012are leap years, whereas 1998, 2002, 2006 and 2010 are not leap years.

The centuries which are divisible by 400 are leap years, while those which are not divisible by 400 arenot leap years. For example, 1200, 1600, 2000 are leap years, whereas 1700, 1800 and 1900 are not leapyears.

A century has 24 leap years and 76 ordinary years.

In 100 years there will be 5 odd days (because 76 ordinary years will give 76 odd days and 24 leap

years will give 48 odd days, which add up to 124 odd days or 17 weeks and 5 odd days). In 200 yearsthere will be 3 odd days. In 300 years there will be 1 odd day. In 400 years there will be 0 odd day.

An ordinary year of 365 days has 52 weeks and 1 odd day.

A leap year of 366 days has 52 weeks and 2 odd days.

In an ordinary year the month of February will have no odd day, but in a leap year it will have 1 oddday.

The first day of a century must be a Monday, Tuesday, Thursday or Saturday.

The last day of a century can never be a Tuesday, Thursday or Saturday.

Number of odd days in first century = 5 days, hence the last day of the first century is Friday.

Number of odd days in two centuries = 3 days, hence the last day will be Wednesday.

Number of odd days in three centuries = 1 day, hence the last day will be Monday.

Number of odd days in four centuries = 0 day, hence the last day will be Sunday.

{Since the order is kept in continous successive cycles, the last day of a century cannot be Tuesday,

Thursday or Saturday, hence the last day has to be either Sunday, Monday, Wednesday or Friday. This

suggests that the first day of a century must be either Monday, Tuesday, Thursday or Saturday].

To find the day of the week on a particular date when no reference day is given, count the number ofodd days on the given date. For 0 odd day it will be Sunday, for 1 odd day it will be Monday, for 2 odddays it will be Tuesday, for 3 odd days it will be Wednesday, for four odd days it will be Thursday, for5 odd days it will be Friday, and for 6 odd days it will be Saturday.

[13]

Progressions

We have three particular types of sequences, viz., arithmetic sequence, geometric sequence and harmonic

sequence.A sequence is a function whose domain is the set N of natural numbers and range, a subset of realnumbers or complex numbers. Sequence which has a subset of real numbers as its range is called a realsequence.If the terms of a sequence follow certain pattern, then the sequence is called a progression.By adding or subtracting the terms of a sequence, we obtain a series.

Arithmetic Progression :

A sequence whose terms decrease or increase by a fixed number is known as arithmetic progression

(A.P.); the fixed number is called the common difference of the A.P.

A sequence x1, x

2, x

3, x

4, x

5, . . . will be in A.P. if x

2 – x

1 = x

3 – x

2 = x

4 – x

3 = x

5 – x

4 = . . ., i.e., x

n – x

n-1

= constant, for n ³ 2.

Three numbers x, y, z are in A.P. if and only if y – z = z – x, i.e., if and only if x + z = 2y.

Suppose p is the first term and q is the common difference of an A.P., then the A.P. will be : p, p + q, p+ 2q, p + 3q, p + 4q, . . .

The first term t1 is p = p + (1 – 1)q;

The second term t2 is p + q = p + (2 – 1)q;

The third term t3 is p + 2q = p + (3 – 1)q;

The fourth term t4 is p + 3q = p + (4 – 1)q;

The fifth term t5 is p + 4q = p + (5 – 1)q;

The nth term tn is p + (n – 1)q;

Therefore, the formula, tn is p + (n – 1)q gives the general term of an arithmetic progression.

Any three numbers in an A.P. can be taken as p – q, p, p + q.

Any four numbers in an A.P. can be taken as p – 3q, p – q, p + q., p + 3q.

Any five numbers in an A.P. can be taken as p – 2q, p – q, p, p + q, p + 2q.

Geometric Progression :

A sequence (finite or infinite) of non-zero numbers, in which every term, except the first one, bears aconstant ratio with its preceding term is called a geometric progression (G.P.).

In a G.P., any term may be obtained by multiplying the preceding term by the common ratio of the G.P.In general, in geometric progression, the first term is denoted by a and the common ratio by r.

Suppose a be the first term and r (¹ 0) be the common ratio of a G.P., and x1, x

2, x

3, x

4, x

5, . . ., x

n denote

first, second, third, fourth, fifth, . . ., nth terms respectively, then we get, x1 = x

2r, x

2 = x

3r, x

3 = x

4r,

x4 = x

5r, . . ., x

n = x

n–1r.

On multiplying these we get, x2x

3x

4x

5. . .x

n = x

1x

2x

3x

4x

5. . .x

n–1rn–1 Þ x

n = x

1rn–1; but x1 = a.

Therefore, the formula, xn = arn–1 gives the general term of a geometric progression.

Harmonic Progression :

A sequence of non-zero numbers a1, a

2, a

3, a

4, a

5, . . . is said to be in harmonic progression (H.P.) if the

sequence 1/a1, 1/a

2, 1/a

3, 1/a

4, 1/a

5, . . . is in arithmetic progression.

nth term of harmonic progression = 1/nth term of the corresponding arithmetic progression.

No term of harmonic progression can be 0.

Three numbers x, y, z are in H.P. if and only if 1/x, 1/y, 1/z are in A.P., i.e., 1/x + 1/z = 2 × 1/y ory = 2xz ÷ (x + z).

Reciprocals of terms of H.P. are in A.P., hence in case of H.P. properties of A.P. can be used.

Harmonic mean : A number H is said to be the single harmonic mean between two given numbersx and y, if x, H, y are in H.P.

[14]

Linear Equations

Linear equation in one variable : It is an equation of the type ax + b = 0 or ax = c, where, a, b, c areconstants (real numbers), a ¹ 0 and x is an unknown variable. Thus, an equation, mx + n = 0, can besolved as x = n/m, and – (n/m) can be called as the root of the linear equation x = n/m.

Linear equation in two variables : It is an equation of the type px + qy + r = 0, or px + qy = n, wherep, q, r and n are constants, p ¹ 0 and q ¹ 0.

Solving two simultaneous linear equations : (i) Find the value of one variable, say x in terms of theother, i.e., y, from either of the two equations. (ii) Substitute the value of x thus obtained in the otherequation, therby getting an equation with only one variable y; (iii) Solve this equation for y; (iv)

substitute the value of y thus obtained in (i) and find the value of x.

Method of elimination : (i) Multiply both the equations by such numbers so as to make the coefficientsof one of the two unknowns numerically the same; (ii) Add or subtract the two equations to get anequation comprising only one unknown variable, and solve the resultant equation to get the value ofthe unknown variable; (iii) Substitute the value of this unknown variable in either of the two givenequations, and solve the equation thus obtained to find the other unknown variable.

Quadratic Equations

An algebraic expression of the form ax2 + bx + c = 0, where a ¹ 0, b, c Î R is called a quadraticequation. The numbers a, b, c are known as the coefficients of the quadratic equation and the expressionb2 – 4ac is known as its discriminant. The discriminant of a quadratic equation is generally denoted byD or D.

Roots of the quadratic equation : The root of a quadratic equation ax2 + bx + c = 0 is a, a real

number or a complex number, such that aa2 + ba + c = 0.

The roots of the quadratic equation ax2 + bx + c = 0 are given by : x = (–b ± Öb2 – 4ac) ÷ 2a

Nature of the roots of the quadratic equation : (i) If D < 0, then the roots a, b are imaginary;(ii) If D > 0, then the roots a, b are real and distinct; (iii) If D = 0, then the roots a, b are real and equal.

Roots are rational Û D is a perfect square, and roots are irrational Û D is positive but not a perfectsquare.

If the roots are equal in magnitude but opposite in sign, then b = 0, ac > 0.

If the roots are reciprocal to each other, then c = a.

If a and c are of opposite signs, the roots must also be of opposite signs.

The imaginary roots occur in conjugate pair, thus, if ax2 + bx + c = 0, where a, b, c Î R has one root x+ i y, then the other root will be x – i y.

The irrational roots occur in conjugate pair if the coefficients are rational, thus, if ax2 + bx + c = 0,where a, b, c are rational, has one root x + Öy, then the other root will not be x – Öy.

if ax2 + bx + c = 0 is satisfied by more than two values, it is an identity and a = b = c = 0.

If a and b are the roots of ax2 + bx + c = 0, then the sum of the roots will be : a + b = –b/a = coefficientof x ÷ coefficient of x2.

If a and b are the roots of ax2 + bx + c = 0, then the product of the roots will be : ab = c/a = constantterm ÷ coefficient of x2.

Symmetric functions of the roots :

a2 + b2 = (a + b)2 – 2ab; a3 + b3 = (a + b)3 – 3ab (a + b);

a4 + b4 = (a3 + b3) + (a + b) – ab (a2 + b2); a5 + b5 = (a3 + b3) + (a2 + b2) – a2b2 (a + b);

| a – b | = Ö(a + b)2 – 4ab; a2 – b2 = (a + b) (a – b);

a3 – b3 = (a – b) [(a + b)2 – ab; a4 – b4 = (a + b) (a – b) (a2 + b2).

[15]

Polynomial

A function p(x) of the form p(x) = a0 + a

1x + a

2x2 + a

3x3 + . . . + a

nxn, where, a

0, a

1, a

2, a

3, . . ., a

n are

real numbers, an ¹ 0 and n is a non-negative integer is called a polynomial in x over reals.

The real numbers a0, a

1, a

2, a

3, . . ., a

n are known as the coefficients of the polynomial.

If a0, a

1, a

2, a

3, . . ., a

n are all integers, it will be considered as polynomial over integers.

If a0, a

1, a

2, a

3, . . ., a

n are rational numbers, it will be considered as polynomial over rationals.

A polynomial having only one term is known as a monomial.

A polynomial having two terms is known as a binomial.

A polynomial having three terms is known as a trinomial.

The exponent in the term with the highest power is called the degree of the polynomial.

Suppose p (x) is a polynomial of degree n > 0. If p (a) = 0 for a real number a, then (x – a) is a factorof p (x). Conversely, if (x – a) is a factor of p (x) then, p (a) = 0. This is known as Factor Theorem.

Suppose p (x) is any polynomial of degree ³ 1 and a is any number. If p (x) is divided by x – a, theremainder will be p (a). This is known as Remainder Theorem.

Mensuration

Area and Perimeter

Area of Square = (side)2 = ½ (diagonal)2; Perimeter of a square = 4 × side.

Area of a Rectangle = (length × breadth); Perimeter of a rectangle= 2 (length + breadth)

Area of 4 walls of a room = 2 (length + breadth) × height

Area of a Quadrilateral = 0.5 × one diagonal × sum of perpendiculars to it from opposite vertices, thisis denoted as 0.5d(p

1 + p

2).

Area of a Parallelogram = base × corresponding altitude.

Area of a Rhombus = ½ × (Product of diagonals) = ½ (d1 × d

2) = base × height = a × h. Perimeter of a

rhombus = 4 × sides = 2 Ö(d1

2 + d2

2). Side of a rhombus = ½ Ö(d1

2 + d2

2).

Area of a Trapezium = ½ × sum of parallel sides × perpendicular distance between the parallel sides,i.e., ½ (a + b) × h.

Area of a regular hexagon = [(3Ö3) ÷ 2] × (side)2

Area of a regular octagon = 2(Ö2 + 1] × (side)2

Area of a Triangle = ½ (base × height) or Ös (s – a) (s – b) (s – c), where a, b, c are the sides of thetriangle and s is the semi-perimeter of the triangle, i.e., 0.5 (a + b + c). Thus, perimeter of a triangle isa + b + c = 2s.

Area of an isoceles triangle = b/4 Ö4a2 – b2; perimeter of an isoceles triangle = (2a + b); height of anisoceles triangle = 0.5 Ö4a2 – b2.

Area of a circle = pr2, i.e., pd2/4, i.e., c2/4p, i.e., ½(c × r ), where, r = radius, d = diameter and c =circumference of the circle; Circumference of the circle = 2pr.

Volume and Surface Area

Volume of a Cube = (side)3 cubic units; Surface Area of a Cube = 6s2 sq. units.

Volume of a Cuboid = (l × b × h) cubic units; Surface Area of a Cuboid = 2 (lb + bh + hl) sq. units.

Diagonal of a Cuboid = Öl2 + b2 + h2

Volume of a Cylinder = Area of the base × height = pr2h cubic units.

Area of the curved surface of a Cylinder = circumference of the base × height = 2prh sq. units.

Area of the total surface of a Cylinder = Area of the curved surface + Area of the 2 circular ends =2prh + 2pr2 = 2pr (h + r) sq. units.

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Volume of a right circular Cone = 1/3 Area of the base × height = 1/3pr2h cubic units.

Curved surface area of a Cone = prl = pr Öh2 + r2 sq. units.

Surface area of a Cone = Area of the base + Area of the curved surface = pr2 + prl = pr(r + l) sq. units.

Slant height of a cone = l = Öh2 + r2.

Frustum of a Cone : Given R = radius base of the frustum, r = radius of the top of the frustum,h is the height of the frustum, and l is the slant height of the frustum.

Slant height = Öh2 + (R + r)2 sq. units.

Curved surface area of frustum = p (R + r) l sq. units.

Total surface area of frustum = p [(R2 + r2) + l (R + r)] sq. units

Volume of the frustum = (ph/3) (R2 + r2 + Rr) cubic units.

Volume of a Sphere : 4/3 pr3 cu. units.

Surface area of Sphere = 4pr2 sq. units.

Volume of a Hemisphere : 2/3 pr3 cu. units.

Total Surface area of Hemisphere = 3pr2 sq. units.

Area of the curved Surface of Hemisphere = 2pr2 sq. units.

Volume of a Prism : Area of the base × height cubic units.

Lateral surface area of a Prism : Perimeter of the base × height sq. units.

Total surface area of a Prism : 2 × area of the base + lateral surface area sq. units.

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