Quantitative Aptitude


1.     NUMBERS

TYPES OF NUMBERS
1.Natural Numbers : Counting numbers 1, 2, 3, 4, 5,..... are called natural numbers.
2.Whole Numbers : All counting numbers together with zero form the set of wholenumbers. Thus,
(i) 0 is the only whole number which is not a natural number.
(ii) Every natural number is a whole number.
3.Integers : All natural numbers, 0 and negatives of counting numbers i.e.,{…, - 3 , - 2 , - 1 , 0, 1, 2, 3,…..} together form the set of integers.
(i) Positive Integers : {1, 2, 3, 4, …..} is the set of all positive integers.
(ii) Negative Integers : {- 1, - 2, - 3,…..} is the set of all negative integers.
(iii) Non-Positive and Non-Negative Integers : 0 is neither positive nor negative. So, {0, 1, 2, 3,….} represents the set of non-negative integers, while{0, - 1 , - 2 , - 3 ,…..} represents the set of non-positive integers.
4. Even Numbers : A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8,10, etc.
5. Odd Numbers : A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7,9, 11, etc.
6. Prime Numbers : A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself.
Prime numbers upto 100 are : 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.
Prime numbers Greater than 100 : Let p be a given number greater than 100. To find out whether it is prime or not, we use the following method :
Find a whole number nearly greater than the square root of p. Let k > *jp. Test whether p
is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is
prime.
e.g,,We have to find whether 191 is a prime number or not. Now, 14 > V191.
Prime numbers less than 14 are 2, 3, 5, 7, 11, 13.
191 is not divisible by any of them. So, 191 is a prime number.
7.Composite Numbers : Numbers greater than 1 which are not prime, are known as
composite numbers, e.g., 4, 6, 8, 9, 10, 12.
Note : (i) 1 is neither prime nor composite.
(ii) 2 is the only even number which is prime.
(iii) There are 25 prime numbers between 1 and 100.
8. Co-primes : Two numbers a and b are said to be co-primes, if their H.C.F. is 1. e.g.,
(2, 3), (4, 5), (7, 9), (8, 11), etc. are co-primes.

TESTS OF DIVISIBILITY
1. Divisibility By 2 : A number is divisible by 2, if its unit's digit is any of 0, 2, 4, 6, 8.
Ex. 84932 is divisible by 2, while 65935 is not.
2. Divisibility By 3 : A number is divisible by 3, if the sum of its digits is divisible by 3.
Ex.592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which
is divisible by 3.
But, 864329 is not divisible by 3, since sum of its digits =(8 + 6 + 4 + 3 + 2 + 9) = 32,
which is not divisible by 3.
3. Divisibility By 4 : A number is divisible by 4, if the number formed by the last two digits is divisible by 4.
Ex. 892648 is divisible by 4, since the number formed by the last two digits is 48, which is divisible by 4.
But, 749282 is not divisible by 4, since the number formed by the last tv/o digits is 82,which is not divisible by 4.
4. Divisibility By 5 : A number is divisible by 5, if its unit's digit is either 0 or 5. Thus,20820 and 50345 are divisible by 5, while 30934 and 40946 are not.
5. Divisibility By 6 : A number is divisible by 6, if it is divisible by both 2 and 3. Ex. The number 35256 is clearly divisible by 2.
Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, which is divisible by 3. Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6.
6. Divisibility By 8 : A number is divisible by 8, if the number formed by the last three digits of the given number is divisible by 8. Ex. 953360 is divisible by 8, since the number formed by last three digits is 360, which is divisible by 8.
But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8.
7. Divisibility By 9 : A number is divisible by 9, if the sum of its digits is divisible by 9 Ex. 60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which is divisible by 9.
But, 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is not divisible by 9.
8. Divisibility By 10 : A number is divisible by 10, if it ends with 0.
Ex. 96410, 10480 are divisible by 10, while 96375 is not.
9. Divisibility By 11 : A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11.
Ex. The number 4832718 is divisible by 11, since :
(sum of digits at odd places) - (sum of digits at even places) ie (8 + 7 + 3 + 4) - (1 + 2 + 8) = 11, which is divisible by 11.
10. Divisibility By 12 ; A number is divisible by 12, if it is divisible by both 4 and 3.
Ex. Consider the number 34632.
(i) The number formed by last two digits is 32, which is divisible by 4,
(ii) Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is divisible by 3. Thus, 34632 is divisible by 4 as well as 3. Hence, 34632 is divisible by 12.
11. Divisibility By 14 : A number is divisible by 14, if it is divisible by 2 as well as 7.
12. Divisibility By 15 : A number is divisible by 15, if it is divisible by both 3 and 5.
13. Divisibility By 16 : A number is divisible by 16, if the number formed by the last4 digits is divisible by 16.
Ex.7957536 is divisible by 16, since the number formed by the last four digits is 7536, which is divisible by 16.
14. Divisibility By 24 : A given number is divisible by 24, if it is divisible by both3 and 8.
15. Divisibility By 40 : A given number is divisible by 40, if it is divisible by both 5 and 8.
16. Divisibility By 80 : A given number is divisible by 80, if it is divisible by both 5 and 16.
Note : If a number is divisible by p as well as q, where p and q are co-primes, then the given number is divisible by pq.
If p arid q are not co-primes, then the given number need not be divisible by pq, even when it is divisible by both p and q.
Ex. 36 is divisible by both 4 and 6, but it is not divisible by (4x6) = 24, since 4 and 6 are not co-primes.


BASIC FORMULAE
1. (a + b)2 = a2 + b2 + 2ab
2. (a - b)2 = a2 + b2 - 2ab
3. (a + b)2 - (a - b)2 = 4ab
4. (a + b)2 + (a - b)2 = 2 (a2 + b2)
5. (a2 - b2) = (a + b) (a - b)
6. (a + b + c)2 = a2 + b2 + c2 + 2 (ab + bc + ca)
7. (a3 + b3) = (a +b) (a2 - ab + b2)
8. (a3 - b3) = (a - b) (a2 + ab + b2)
9. (a3 + b3 + c3 -3abc) = (a + b + c) (a2 + b2 + c2 - ab - bc - ca)
10. If a + b + c = 0, then a3 + b3 + c3 = 3abc.

DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM
If we divide a given number by another number, then : Dividend = (Divisor x Quotient) + Remainder
{i) (xn - an ) is divisible by (x - a) for all values of n.
(ii) (xn - an) is divisible by (x + a) for all even values of n.
(iii) (xn + an) is divisible by (x + a) for all odd values of n.

PROGRESSION
A succession of numbers formed and arranged in a definite order according to certain definite rule, is called a progression.
1. Arithmetic Progression (A.P.) : If each term of a progression differs from its preceding term by a constant, then such a progression is called an arithmetical progression. This constant difference is called the common difference of the A.P.
An A.P. with first term a and common difference d is given by a, (a + d), (a + 2d), (a +3d)...
The nth term of this A.P. is given by Tn =a (n - 1) d.
The sum of n terms of this A.P.
Sn = n/2 [2a + (n - 1) d] = n/2 (first term + last term).
SOME IMPORTANT RESULTS :
(i) (1 + 2 + 3 +…. + n) =n(n+1)/2
(ii) (l2 + 22 + 32 + ... + n2) = n (n+1)(2n+1)/6
(iii) (13 + 23 + 33 + ... + n3) =n2(n+1)2
2. Geometrical Progression (G.P.) : A progression of numbers in which every term bears a constant ratio with its preceding term, is called a geometrical progression.
The constant ratio is called the common ratio of the G.P. A G.P. with first term a and common ratio r is : a, ar, ar2,
In this G.P. Tn = arn-1

 2 H.C.F. AND L.C.M. OF NUMBERS
I. Factors and Multiples : If a number a divides another number b exactly, we say that a is a factor of b. In this case, b is called a multiple of a.
II. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.): The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly.
There are two methods of finding the H.C.F. of a given set of numbers :

1. Factorization Method : Express each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.
2. Division Method: Suppose we have to find the H.C.F. of two given numbers. Divide the larger number by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is the required H.C.F.
Finding the H.C.F. of more than two numbers: Suppose we have to find the H.C.F. of three numbers. Then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given numbers. Similarly, the H.C.F. of more than three numbers may be obtained.
III. Least Common Multiple (L.C.M.) : The least number which is exactly divisible by each one of the given numbers is called their L.C.M.
1. Factorization Method of Finding L.C.M.: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors,
2. Common Division Method {Short-cut Method) of Finding L.C.M.:
Arrange the given numbers in a row in any order. Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. Repeat the above process till no two of the numbers are divisible by the same number except 1. The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers,
IV. Product of two numbers =Product of their H.C.F. and L.C.M.
V. Co-primes: Two numbers are said to be co-primes if their H.C.F. is 1.
VI. H.C.F. and L.C.M. of Fractions:
1.H C F= H.C.F. of Numerators         2.L C M = L.C.M of Numerators
     L.C.M. of Denominators                     H.C.F. of Denominators
VII. H.C.F. and L.C.M. of Decimal Fractions: In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.
VIII. Comparison of Fractions: Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M. as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

3. DECIMAL FRACTIONS
I. Decimal Fractions : Fractions in which denominators are powers of 10 are known as decimal
fractions.
Thus, 1/10=1 tenth=.1;1/100=1 hundredth =.01;
99/100=99 hundreths=.99; 7/1000=7 thousandths=.007, etc
II. Conversion of a Decimal into Vulgar Fraction: Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point. Now, remove the decimal point and reduce the fraction to its lowest terms.
Thus,  0.25=25/100=1/4; 2.008=2008/1000=251/125.
III. 1. Annexing zeros to the extreme right of a decimal fraction does not change its value
Thus, 0.8 = 0.80 = 0.800, etc.
2. If numerator and denominator of a fraction contain the same number of decimal places, then we remove the decimal sign.
Thus, 1.84/2.99 = 184/299 = 8/13; 0.365/0.584 = 365/584=5


IV. Operations on Decimal Fractions :
1. Addition and Subtraction of Decimal Fractions: The given numbers are so placed under each other that the decimal points lie in one column. The numbers so arranged can now be added or subtracted in the usual way.
2. Multiplication of a Decimal Fraction By a Power of 10 : Shift the decimal point to the right by as many places as is the power of 10.
Thus, 5.9632 x 100 = 596.32; 0.073 x 10000 = 0.0730 x 10000 = 730.
3.Multiplication of Decimal Fractions : Multiply the given numbers considering them without the decimal point. Now, in the product, the decimal point is marked off to obtain as many places of decimal as is the sum of the number of decimal places in the given numbers.
Suppose we have to find the product (.2 x .02 x .002). Now, 2x2x2 = 8. Sum of
decimal places = (1 + 2 + 3) = 6. .2 x .02 x .002 = .000008.
4.Dividing a Decimal Fraction By a Counting Number : Divide the given number without considering the decimal point, by the given counting number. Now, in the quotient, put the decimal point to give as many places of decimal as there are in the dividend. Suppose we have to find the quotient (0.0204 + 17). Now, 204 / 17 = 12. Dividend contains 4 places of decimal. So, 0.0204 + 17 = 0.0012.
5. Dividing a Decimal Fraction By a Decimal Fraction : Multiply both the dividend and the
divisor by a suitable power of 10 to make divisor a whole number. Now, proceed as above.
Thus, 0.00066/0.11 = (0.00066*100)/(0.11*100) = (0.066/11) = 0.006V
V. Comparison of Fractions : Suppose some fractions are to be arranged in ascending or descending order of magnitude. Then, convert each one of the given fractions in the decimal form,
and arrange them accordingly. Suppose, we have to arrange the fractions 3/5, 6/7 and 7/9 in descending order. now, 3/5=0.6,6/7 = 0.857,7/9 = 0.777....
since 0.857>0.777...>0.6, so 6/7>7/9>3/5
VI. Recurring Decimal : If in a decimal fraction, a figure or a set of figures is repeated
continuously, then such a number is called a recurring decimal.
In a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it.
If a set of figures is repeated, it is expressed by putting a bar on the set
                                        _                                                     ______
Thus 1/3 = 0.3333….= 0.3; 22 /7 = 3.142857142857.....= 3.142857
Pure Recurring Decimal: A decimal fraction in which all the figures after the decimal point
are repeated, is called a pure recurring decimal.
Converting a Pure Recurring Decimal Into Vulgar Fraction : Write the repeated figures
only once in the numerator and take as many nines in the denominator as is the number of
repeating figures.
thus ,0.5 = 5/9; 0.53 = 53/59 ;0.067 = 67/999;etc...
Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some
of them are repeated, is called a mixed recurring decimal.
e.g., 0.17333.= 0.173.
Converting a Mixed Recurring Decimal Into Vulgar Fraction : In the numerator, take the
difference between the number formed by all the digits after decimal point (taking repeated
digits only once) and that formed by the digits which are not repeated, In the denominator,
take the number formed by as many nines as there are repeating digits followed by as many
zeros as is the number of non-repeating digits.
Thus 0.16 = (16-1) / 90 = 15/19 = 1/6;
____
0.2273 = (2273 – 22)/9900 = 2251/9900
VII. Some Basic Formulae :
1. (a + b)(a- b) = (a2 - b2).
2. (a + b)2 = (a2 + b2 + 2ab).
3. (a - b)2 = (a2 + b2 - 2ab).
4. (a + b+c)2 = a2 + b2 + c2+2(ab+bc+ca)
5. (a3 + b3) = (a + b) (a2 - ab + b2)
6. (a3 - b3) = (a - b) (a2 + ab + b2).
7. (a3 + b3 + c3 - 3abc) = (a + b + c) (a2 + b2 + c2-ab-bc-ca)
8. When a + b + c = 0, then a3 + b3+ c3 = 3abc

4. SIMPLIFICATION
I. ’BODMAS’Rule: This rule depicts the correct sequence in which the operations are to be
executed, so as to find out the value of a given expression.
Here, ‘B’ stands for ’bracket’ ,’O’for ‘of’ , ‘D’ for’ division’ and ‘M’ for ‘multiplication’, ‘A’
for ‘addition’ and ‘S’ for ‘subtraction’.
Thus, in simplifying an expression, first of all the brackets must be removed, strictly in the order(), {} and [].
After removing the brackets, we must use the following operations strictly in the order:
(1)of (2)division (3) multiplication (4)addition (5)subtraction.
II. Modulus of a real number : Modulus of a real number a is defined as
|a| = {a, if a>0
         -a, if a<0
Thus, |5|=5 and |-5|=-(-5) =5.
III. Virnaculum (or bar): When an expression contains Virnaculum, before applying the ‘BODMAS’ rule, we simplify the expression under the Virnaculum.
5. SQUARE ROOTS AND CUBE ROOTS
Square Root: If x2 = y, we say that the square root of y is x and we write, √y = x.
Thus, √4 = 2, √9 = 3, √196 = 14.
Cube Root: The cube root of a given number x is the number whose cube is x. We denote the cube
root of x by 3√x.
Thus, 3√8 = 3√2 x 2 x 2 = 2, 3√343 = 3√7 x 7 x 7 = 7 etc.
Note:
1.√xy = √x * √y 2. √(x/y) = √x / √y = (√x / √y) * (√y / √y) = √xy / y
 6. SURDS AND INDICES
1. LAWS OF INDICES:
(i) am x an = am + n
(ii) am / an = am-n
(iii) (am)n = amn
(iv) (ab)n = anbn
(v) ( a/ b )n = ( an / bn )
(vi) a0 = 1

2. SURDS: Let a be a rational number and n be a positive integer such that a1/n = nsqrt(a)
is irrational. Then nsqrt(a) is called a surd of order n.
3. LAWS OF SURDS:
(i) n√a = a1/2
(ii) n √ab = n √a * n √b
(iii) n √a/b = n √a / n √b
(iv) (n √a)n = a
(v) m√(n√(a)) = mn√(a)

7. PERCENTAGE
1. Concept of Percentage : By a certain percent ,we mean that many hundredths. Thus x percent means x hundredths, written as x%.
To express x% as a fraction : We have , x% = x/100.
Thus, 20% =20/100 =1/5; 48% =48/100 =12/25, etc.
To express a/b as a percent : We have, a/b =((a/b)*100)%.
Thus, ¼ =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%.
2. If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is
[R/(100+R))*100]%.
If the price of the commodity decreases by R%, then the increase in consumption so as to decrease the expenditure is
[(R/(100-R)*100]%.
3. Results on Population : Let the population of the town be P now and suppose it increases at the rate of R% per annum, then :
1. Population after n years = P [1+(R/100)]^n.
2. Population n years ago = P /[1+(R/100)]^n.
4. Results on Depreciation : Let the present value of a machine be P. Suppose it depreciates at the rate R% per annum. Then,
1. Value of the machine after n years = P[1-(R/100)]n.
2. Value of the machine n years ago = P/[1-(R/100)]n.
5. If A is R% more than B, then B is less than A by
[(R/(100+R))*100]%.
If A is R% less than B , then B is more than A by
[(R/(100-R))*100]%.
8. PROFIT AND LOSS
Cost Price: The price at which article is purchased. Abbrevated as C.P.
Selling Price: The price at which article is sold. Abbrevated as S.P.
Profit or Gain: if Sp is greater than Cp, the selling price is said to have profit or gain.
Loss: if sp is less than cp, the seller is said to incured a loss.
formula
1.gain=(sp)-(cp).
2.loss=(cp)-(sp).
3.loss or gain is always reckoned on cp
4. gain %={gain*100}/cp.
5.loss%={loss*100}/cp.
6.sp={(100+gain%) /100}*cp.
7.sp={(100-loss%)/100}*cp.
8.CP={100/(100+gain%)} *sp
9.cp=100/(100-loss%)}*sp
10.if the article is sold at a gain of say 35%, then sp =135% of cp
11.if a article is sold at a loss of say 35%. Then sp=65% of cp.
12.when a person sells two items, One at a gain of x% and other at a loss of x%. then the seller always incures a loss given:
{loss%=(comon loss and gain ) 2}/10. =(x/10) 2
13.if the trader professes to sell his goods at cp but uses false weights, then      GAIN=[ERROR/(TRUE VALUE)-(ERROR)*100]%

9. RATIO AND PROPORTION
I. RATIO: The ratio of two quantities a and b in the same units, is the fraction a/b and we write it as a:b.
In the ratio a:b, we call a as the first term or antecedent and b, the second term or consequent.
Ex. The ratio 5: 9 represents 5/9 with antecedent = 5, consequent = 9.
Rule: The multiplication or division of each term of a ratio by the same non-zero number does not affect the ratio.
Ex. 4: 5 = 8: 10 = 12: 15 etc. Also, 4: 6 = 2: 3.
2. PROPORTION: The equality of two ratios is called proportion.
If a: b = c: d, we write, a: b:: c : d and we say that a, b, c, d are in proportion . Here a and d are called extremes, while b and c are called mean terms.
       Product of means = Product of extremes.
Thus, a: b:: c : d <=> (b x c) = (a x d).
3. (i) Fourth Proportional: If a : b = c: d, then d is called the fourth proportional to a, b, c.
(ii) Third Proportional: If a: b = b: c, then c is called the third proportional to a and b.
(iii) Mean Proportional: Mean proportional between a and b is square root of ab
4. (i) COMPARISON OF RATIOS:
We say that (a: b) > (c: d) <=> (a/b)>(c /d).
(ii) COMPOUNDED RATIO:
The compounded ratio of the ratios (a: b), (c: d), (e : f) is (ace: bdf)
5. (i) Duplicate ratio of (a : b) is (a2 : b2).
(ii) Sub-duplicate ratio of (a : b) is (a : b).
(iii)Triplicate ratio of (a : b) is (a3 : b3).
(iv) Sub-triplicate ratio of (a : b) is (a : b ).
(v) If (a/b)=(c/d), then ((a+b)/(a-b))=((c+d)/(c-d)) (Componendo and dividendo)
6. VARIATION:
(i) We say that x is directly proportional to y, if x = ky for some constant k and we write, x y.
(ii) We say that x is inversely proportional to y, if xy = k for some constant k and we write, x∞(1/y)
10. PARTNERSHIP
1. Partnership: When two or more than two persons run a business jointly, they are called partners and the deal is known as partnership.
2. Ratio of Division of Gains:
i) When investments of all the partners are for the same time, the gain or loss is distributed a among the partners in the ratio of their investments.
Suppose A and B invest Rs. x and Rs. y respectively for a year in a business, then at the end of the year:
(A’s share of profit) : (B's share of profit) = x : y.
ii) When investments are for different time periods, then equivalent capitals are calculated for a unit of time by taking (capital x number of units of time). Now, gain or loss is divided in the ratio of these capitals.
Suppose A invests Rs. x for p months and B invests Rs. y for q months, then
(A’s share of profit) : (B's share of profit) = xp : yq.
3. Working and Sleeping Partners: A partner who manages the business is known. As a
working partner and the one who simply invests the money is a sleeping partner.
11. CHAIN RULE
1. Direct Proportion: Two quantities are said to be directly proportional, if on the increase (or decrease) of the one, the other increases (or decreases) to the same
Ex. 1. Cost is directly proportional to the number of articles. (More Articles, More Cost)
Ex. 2. Work done is directly proportional to the number of men working on it (More Men, More Work)
2. Indirect Proportion: Two quantities are said to be indirectly proportional,if on the increase of the one, the other decreases to the same extent and vice-versa.
Ex. 1. The time taken by a car in covering a certain distance is inversely proportional to the speed of the car.(More speed, Less is the time taken to cover a distance)
Ex. 2. Time taken to finish a work is inversely proportional to the num of persons working at it.
(More persons, Less is the time taken to finish a job)
Remark: In solving questions by chain rule, we compare every item with the term to be found out.
12. TIME AND WORK
1. If A can do a piece of work in n days, then A's 1 day's work = (1/n).
2. If A’s 1 day's work = (1/n),then A can finish the work in n days.
3. 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 and B to finish a work = 1 : 3.
13. PIPES AND CISTERNS
1. Inlet: A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an
inlet.
Outlet: A pipe connected with a tank or a cistern or a reservoir, emptying it, is known as an outlet.
2.  (i) If a pipe can fill a tank in x hours, then : part filled in 1 hour = 1/x
     (ii) If a pipe can empty a full tank in y hours, then : part emptied in 1 hour = 1/y
     (iii) If a pipe can .fill a tank in x hours and another pipe can empty the full tank in y hours
      (where y> x), then on opening both the pipes, the net part filled in 1 hour = (1/x)-(1/y)
(iv) If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours (where
x > y), then on opening both the pipes, the net part emptied in 1 hour = (1/y)-(1/x)

14. TIME AND DISTANCE
                  Distance                Distance
1. Speed = Time ,         Time= Speed ,     Distance = (Speed * Time)
2. x km / hr = x * 5/18
3. x m/sec = (x * 18/5) km /hr
4. If the ratio of the speeds of A and B is a:b , then the ratio of the times taken by them to cover the same distance is 1/a: 1/b or b:a.
5. Suppose a man covers a certain distance at x km/ hr and an equal distance at y km / hr .
Then , the average speed during the whole journey is 2xy/(x+y) km/ hr.
15. PROBLEMS ON TRAINS
1. a km/hr= (a* 5/18) m/s.
2. a m / s = (a*18/5) km/hr.
3 Time taken by a train of length 1 metres to pass a pole or a standing man or a signal post is equal to the time taken by the train to cover 1 metres.
4. Time taken by a train of length 1 metres to pass a stationary object of length b
metres is the time taken by the train to cover (1 + b) metres.
5. Suppose two trains or two bodies are moving in the same direction at u m / s and
v m/s, where u > v, then their relatives speed = (u - v) m / s.
6. Suppose two trains or two bodies are moving in opposite directions at u m / s
and v m/s, then their relative speed is = (u + v) m/s.
7. If two trains of length a metres and b metres are moving in opposite directions at
u m / s and v m/s, then time taken by the trains to cross each other = (a + b)/(u+v)
sec.
8.If two trains of length a metres and b metres are moving in the same direction
at u m / s and v m / s, then the time taken by the faster train to cross the slower
train = (a+b)/(u-v) sec.
9. If two trains (or bodies) start at the same time from points A and B towards each
other and after crossing they take a and b sec in reaching B and A respectively,
then
(A's speed) : (B’s speed) = (b1/2: a1/2).
16.BOATS AND STREAMS
1.In water ,the direction along the stream is called downstream and ,the direction against
the stream is called upstream.
2.If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then:
speed downstream=(u+v)km/hr.
speed upstream=(u-v)km/hr.
3.If the speed downstream is a km/hr and the speed upstream is b km/hr,then :
speed in still water=1/2(a+b)km/hr, rate of stream=1/2(a-b)km/hr
17. SIMPLE INTEREST
1.. Principal: The money borrowed or lent out for a certain period is called the principal or the sum.
2. Interest: Extra money paid for using other's money is called interest.
3. Simple Interest (S.I.) : If the interest on a sum borrowed for a certain period is reckoned
uniformly, then it is called simple interest.
Let Principal = P, Rate = R% per annum (p.a.) and Time = T years. Then,
(i) S.I. = (P*R*T )/100
(ii) P=(100*S.I)/(R*T) ;R=(100*S.I)/(P*T) and T=(100*S.I)/(P*R)
18.COMPOUND INTEREST
Compound Interest: Sometimes it so happens that the borrower and the lender agree to fix up a certain unit of time, say yearly or half-yearly or quarterly to settle the previous account. In such cases, the amount after first unit of time becomes the principal for the second unit, the amount after second unit becomes the principal for the third unit and so on.
After a specified period, the difference between the amount and the moneyborrowed is called the Compound Interest (abbreviated as C.I.) for that period.
Let Principal = P, Rate = R% per annum, Time = n years.
I. When interest is compound Annually:
          Amount = P(1+R/100)n
II. When interest is compounded Half-yearly:
            Amount = P[1+(R/2)/100]2n
III. When interest is compounded Quarterly:
          Amount = P[ 1+(R/4)/100]4n
IV. When interest is compounded AnnuaI1y but time is in fraction, say 3(2/5) years.
            Amount = P(1+R/100)3 x (1+(2R/5)/100)
V. When Rates are different for different years, say Rl%, R2%, R3% for 1st, 2nd and 3rd year respectively.
Then,  Amount = P(1+R1/100)(1+R2/100)(1+R3/100)
VI. Present worth of Rs.x due n years hence is given by :
Present Worth = x/(1+(R/100))n
19 .AREA
FUNDEMENTAL CONCEPTS
I.RESULTS ON TRIANGLES:
1.Sum of the angles of a triangle is 180 degrees.
2.Sum of any two sides of a triangle is greater than the third side.
3.Pythagoras theorem:
In a right angle triangle,
(Hypotenuse) 2 = (base)2 + (Height)2
4.The line joining the midpoint of a side of a triangle to the opposite vertex is called the
MEDIAN
5.The point where the three medians of a triangle meet is called CENTROID.
Centroid divides each of the medians in the ratio 2:1.
6.In an isosceles triangle, the altitude from the vertex bi-sects the base
7.The median of a triangle divides it into two triangles of the same area.
8.Area of a triangle formed by joining the midpoints of the sides of a given triangle is one-fourth of the area of the given triangle.
II.RESULTS ON QUADRILATERALS:
1. The diagonals of a parallelogram bisects each other .
2. Each diagonal of a parallelogram divides it into two triangles of the same area
3. The diagonals of a rectangle are equal and bisect each other.
4. The diagonals of a square are equal and bisect each other at right angles.
5. The diagonals of a rhombus are unequal and bisect each other at right angles.
6. A parallelogram and a rectangle on the same base and between the same parallels are equal
in area.
7. Of all the parallelograms of a given sides , the parallelogram which is a rectangle has the
greatest area.
IMPORTANT FORMULAE
I.1.Area of a rectangle=(length*breadth)
Therefore length = (area/breadth) and breadth=(area/length)
2.Perimeter of a rectangle = 2*(length+breadth)
II.Area of a square = (side) 2 =1/2(diagonal) 2
III Area of four walls of a room = 2*(length + breadth)*(height)
IV 1.Area of the triangle=1/2(base*height)
2. Area of a triangle = (s*(s-a)(s-b)(s-c)), where a,b,c are the sides of a triangle and s= ½(a+b+c)
3.Area of the equilateral triangle =(3/4)*(side) 2
4.Radius of incircle of an equilateral triangle of side a=a/23
5.Radius of circumcircle of an equilateral triangle of side a=a/√3
6.Radius of incircle of a triangle of area del and semiperimeter S=del/S
V.1.Area of the parellogram =(base *height)
2.Area of the rhombus=1/2(product of the diagonals)
3.Area of the trapezium=1/2(size of parallel sides)*distance between them
VI 1.Area of a circle =pi*r2,where r is the radius
2. Circumference of a circle = 2ΠR.
3. Length of an arc = 2ΠRθ/(360) where θ is the central angle
4. Area of a sector = (1/2) (arc x R) = pi*R2*θ/360.
VII. 1. Area of a semi-circle = (pi)*R2.
2.             Circumference of a semi-circle = (pi)*R.
20.VOLUME AND SURFACE AREA
I. CUBOID
Let length = 1, breadth = b and height = h units. Then,
1. Volume = (1 x b x h) cubic units.
2. Surface area= 2(lb + bh + lh) sq.units.
3. Diagonal.= √(l2 +b2 +h2 ) units
II. CUBE
Let each edge of a cube be of length a. Then,
1. Volume = a3 cubic units.
2. Surface area = 6a2 sq. units.
3. Diagonal = 3 a units.
III. CYLINDER
Let radius of base = r and Height (or length) = h. Then,
1. Volume = (π r2h) cubic units.
2. Curved surface area = (2πrh). units.
3. Total surface area =2πr (h+r) sq. units
IV. CONE
Let radius of base = r and Height = h. Then,
1. Slant height, l = √(h2+r2 )
2. Volume = (1/3) πr2h cubic units.
3. Curved surface area = (πrl) sq. units.
4. Total surface area = (πrl + πr2 ) sq. units.
V. SPHERE
Let the radius of the sphere be r. Then,
1. Volume = (4/3) πr3 cubic units.
2. Surface area = (4πr2) sq. units.
VI. HEMISPHERE
Let the radius of a hemisphere be r. Then,
1. Volume = (2/3) πr3 cubic units.
2. Curved surface area = (2πr2) sq. units.
3. Total surface area = (3πr2) units.
Remember: 1 litre = 1000 cm3.
21. CALENDAR
Under this heading we mainly deal with finding the day of the week on a particular
given date the process of finding it lies on obtaining the number of odd days.
Odd Days : Number of days more than the complete number of weeks in a given.
Period ., is the number of odd days during that period.
LeapYear: Every year which is divisible by 4 is called a leap year.
Thus each one of the years 1992, 1996, 2004, 2008, 2012, etc. is a leap year. Every
4th century is a leap year but no other century is a leap year. Thus each one of 400,
800, 1200,' 1600, 2000, etc. is a leap year.
None of 1900, 2010, 2020, 2100, etc. is a leap year.
An year which is not a leap year is called an ordinary year.
(I )An ordinary year has 365 days. (II) A leap year has 366 days.
Counting of Odd Days:
i)1 ordinary year = 365 days = (52 weeks + 1 day).
:. An ordinary year has 1 odd day.
ii)1 leap year = 366 days = (52 weeks + 2 days).
:. A leap year has 2 odd days.
_ iii)100 years = 76 ordinary years + 24 leap years
= [(76 x 52) weeks + 76 days) + [(24 x 52) weeks + 48 days]
= 5200 weeks + 124 days = (5217 weeks + 5 days).
:. 100 years contain 5 odd days.
200 years contain 10 and therefore 3 odd days.
300 years contain 15 and therefore 1 odd day.
400 years contain (20 + 1) and therefore 0 odd day.
Similarly, each one of 800, 1200, 1600, 2000, etc. contains 0 odd days.
Remark: (7n + m) odd days, where m < 7 is equivalent to m odd days.
Thus, 8 odd days ≡ 1 odd day etc.
22. CLOCKS
IMPORTANT FACTS
The Face or dial of a watch is a circle whose circumference is divided into 60
equal parts, called minute spaces.
A clock has two hands, the smaller one is called the hour hand or short hand
while the larger one is called the minute hand or long hand..
i) In 60 minutes, the minute hand gains 55 minutes on the hour hand.
ii) In every hour, both the hands coincide once.
iii) The hands are in the same straight line when they are coincident or opposite to
each other.
iv) When the two hands are at right angles, they are 15 minute spaces apart.
v)When the hand's are in opposite directions, they are 30 minute spaces apart.
vi)Angle traced by hour hand in 12 hrs = 360°.
vii)Angle traced by minute hand in 60 min. = 360°.
Too Fast and Too Slow: If a watch or a clock indicates 8.15, when the correct
time , 8 is said to be 15 minutes too fast.
On the other hand, if it indicates 7.45, when the correct time is 8, it is said to be
15 minutes too slow
23. TRUE DISCOUNT
Suppose 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 clear off
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;
True Discount (T.D.) = Rs. (156 - 100) = Rs. 56
(Sum due) - (P.W.).
We define : T.D. = Interest on P.W.
Amount = (P.W.) + (T.D.).
Interest is reckoned on P.W. and true discount is reckoned on the amount.


IMPORTANT FORMULAE
Let rate = R% per annum and Time = T years. Then,
1. P.W.=[100 x Amount /100 + (R x T)
=100 x T.D./ RxT
2. T.D.=[(P.W.) x R x T /100]
= [ Amount x RxT/100 + (R x T)]
3.(S.I.)*(T.D.) /(S.I.)-(T.D.)
4. (S.I.) - (T.D.) - S.I. on T.D.
5. When the sum is put at compound interest, then
P.W. = Amount/[1 +R/100]^T.




 Number Prefix Symbol
10 1 deka- da
10 2 hecto- h
10 3 kilo- k
10 6 mega- M
10 9 giga- G
10 12 tera- T
10 15 peta- P
10 18 exa- E
10 21 zeta- Z
10 24 yotta- Y

Number Prefix Symbol
10 -1 deci- d
10 -2 centi- c
10 -3 milli- m
10 -6 micro- u (greek mu)
10 -9 nano- n
10 -12 pico- p
10 -15 femto- f
10 -18 atto- a
10 -21 zepto- z
10 -24 yocto- y


Roman Numerals

I=1 V=5 X=10 L=50 C=100 D=500 M=1 000
_
V=5 000
_
X=10 000
_
L=50 000
_
C = 100 000
_
D=500 000
__
M=1 000 000




Fraction = Decimal    
1/1 = 1    
1/2 = 0.5    
1/3 = 0.3,        2/3 = 0.6  
1/4 = 0.25,      3/4 = 0.75  
1/5 = 0.2,        2/5 = 0.4,     3/5 = 0.6,    4/5 = 0.8
1/6 = 0.16,      5/6 = 0.83  
1/7 =  0.142857 ,    2/7 =  0.285714 ,    3/7 =  0.428571,     4/7 =  0.571428       5/7 =    0.714285,   6/7 =  0.857142
1/8 = 0.125 ,          3/8 = 0.375,           5/8 = 0.625,          7/8 = 0.875
1/9 = 0.1,               2/9 = 0.2 ,             4/9 = 0.4,               5/9 = 0.5
  7/9 = 0.7,             8/9 = 0.8
1/10 = 0.1,             3/10 = 0.3,            7/10 = 0.7,            9/10 = 0.9
1/11 = 0.09,           2/11 = 0.18,          3/11 = 0.27,          4/11 = 0.36   5/11 = 0.45,         6/11 = 0.54,            7/11 = 0.63             8/11 = 0.72,         9/11 = 0.81,          10/11 = 0.90
1/12 = 0.083,         5/12 = 0.416,        7/12 = 0.583,        11/12 = 0.916
1/16 = 0.0625,       3/16 = 0.1875,        5/16 = 0.3125,      7/16 = 0.4375
  11/16 = 0.6875,     13/16 = 0.8125,     15/16 = 0.9375
1/32 = 0.03125,      3/32 = 0.09375,     5/32 = 0.15625,     7/32 = 0.21875
  9/32 = 0.28125,    11/32 = 0.34375,   13/32 = 0.40625
  15/32 = 0.46875,  17/32 = 0.53125,   19/32 = 0.59375
  21/32 = 0.65625,  23/32 = 0.71875,   25/32 = 0.78125
  27/32 = 0.84375,  29/32 = 0.90625,   31/32 = 0.96875



_______________________________________________________________________________
                                                              Courtesy: Quantitative Aptitude by R.S. Agarwal

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