# R S Aggarwal Quantitative Aptitude Chapter Wise PDF Free Download-Chapter 1st-Number System

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R S Aggarwal Quantitative Aptitude Chapter Wise PDF Free Download-Chapter 1st-Number System

NUMBERS System R S Aggarwal Quantitative Aptitude Chapter Wise

IMPORTANT FACTS AND FORMULA R S Aggarwal Quantitative Aptitude Chapter Wise Free Download

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I..Numeral : In Hindu Arabic system, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 called digits to represent any number.

A group of digits, denoting a number is called a numeral.

We represent a number, say 689745132 as shown below :

 Ten Crores (108) Crores(107) Ten Lacs (Millions) (106) Lacs(105) Ten Thousands (104) Thousands (103) Hundreds (102) Tens(101) Units(100) 6 8 9 7 4 5 1 3 2

We read it as : ‘Sixty-eight crores, ninety-seven lacs, forty-five thousand, one hundred and thirty-two’.

II  Place Value or Local Value of a Digit in a Numeral :

In the above numeral :

Place value of 2 is (2 x 1) = 2; Place value of 3 is (3 x 10) = 30;

Place value of 1 is (1 x 100) = 100 and so on.

Place value of 6 is 6 x 108 = 600000000

III.Face Value : The face value of a digit in a numeral is the value of the   digit itself at whatever place it may be. In the above numeral, the face value of 2 is 2; the face value of 3 is 3 and so on.

IV.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 whole
numbers. 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.

1. Even Numbers : A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8, 10, etc.
2. Odd Numbers : A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7, 9, 11, etc.
3. 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.

1. 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,

V.TESTS OF DIVISIBILITY R S Aggarwal Quantitative Aptitude Chapter Wise Series

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.

1. 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.

1. 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.

1. 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.
2. 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.

1. 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.

1. 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.

1. 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.

1. 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)

(8 + 7 + 3 + 4) – (1 + 2 + 8) = 11, which is divisible by 11.

1. 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.

1. 11. Divisibility By 14 : A number is divisible by 14, if it is divisible by 2 as well as 7.
2. Divisibility By 15 : A number is divisible by 15, if it is divisible by both 3 and 5.
3. 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.

1. Divisibility By 24 : A given number is divisible by 24, if it is divisible by both3 and 8.
2. Divisibility By 40 : A given number is divisible by 40, if it is divisible by both

5  and 8.

1. 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 (4×6) = 24, since

4  and 6 are not co-primes.

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VI   MULTIPLICATION BY SHORT CUT METHODS  R S Aggarwal Quantitative Aptitude Chapter Wise Series

1. Multiplication By Distributive Law :

(i) a x (b + c) = a x b + a x c    (ii) ax(b-c) = a x b-a x c.

Ex.   (i) 567958 x 99999 = 567958 x (100000 – 1)

= 567958 x 100000 – 567958 x 1 = (56795800000 – 567958) = 56795232042. (ii) 978 x 184 + 978 x 816 = 978 x (184 + 816) = 978 x 1000 = 978000.

1. Multiplication of a Number By 5n : Put n zeros to the right of the multiplicand and divide the number so formed by 2n

Ex. 975436 x 625 = 975436 x 54= 9754360000 =   609647600

16

VII.   BASIC FORMULAE

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

VIII.  DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM-R S Aggarwal Quantitative Aptitude Chapter Wise PDF Free Download

If we divide a given number by another number, then :

Dividend = (Divisor x Quotient) + Remainder

1. IX. {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.

1. 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 :R S Aggarwal Quantitative Aptitude Chapter Wise Series

(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

1. 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

sum of the n terms, Sn=   a(1-rn)

(1-r)

# SOLVED EXAMPLES from R S Aggarwal Quantitative Aptitude Chapter Wise Series

Ex. 1. Simplify :   (i) 8888 + 888 + 88 + 8

(ii) 11992 – 7823 – 456

Sol.   i )  8888                       ii) 11992 – 7823 – 456 = 11992 – (7823 + 456)

888                                                            = 11992 – 8279 = 3713-

88                                  7823                         11992

+         8                               +   456                      –   8279

9872                                  8279                          3713

Ex. 2, What value will replace the question mark in each of the following equations ?

(i) ? – 1936248 = 1635773            (ii) 8597 – ? = 7429 – 4358

Sol.  (i) Let x  – 1936248=1635773.Then, x = 1635773 + 1936248=3572021.                 (ii) Let 8597 – x = 7429 – 4358.

Then, x = (8597 + 4358) – 7429 = 12955 – 7429 = 5526.

Ex. 3. What could be the maximum value of Q in the following equation?          5P9 + 3R7 + 2Q8 = 1114

Sol. We may analyse the given equation as shown :                     1   2

Clearly, 2 + P + R + Q = ll.                                                           5   P  9

So, the maximum value of Q can be                                                3   R  7

(11 – 2) i.e., 9 (when P = 0, R = 0);                                               2   Q  8

11  1   4

Ex. 4. Simplify : (i) 5793405 x 9999  (ii) 839478 x 625

Sol.

i)5793405×9999=5793405(10000-1)=57934050000-5793405=57928256595.b

1. ii) 839478 x 625 = 839478 x 54 = 8394780000 = 524673750.

16

## Ex. 5. Evaluate : (i) 986 x 237 + 986 x 863    (ii) 983 x 207 – 983 x 107

Sol.

(i) 986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.

(ii) 983 x 207 – 983 x 107 = 983 x (207 – 107) = 983 x 100 = 98300.

Ex. 6. Simplify : (i) 1605 x 1605    ii) 1398 x 1398

Sol.

1. i) 1605 x 1605 = (1605)2 = (1600 + 5)2 = (1600)2 + (5)2 + 2 x 1600 x 5

= 2560000 + 25 + 16000 = 2576025.

(ii) 1398 x 1398 – (1398)2 = (1400 – 2)2= (1400)2 + (2)2 – 2 x 1400 x 2

=1960000 + 4 – 5600 = 1954404.

Ex. 7. Evaluate : (313 x 313 + 287 x 287).

Sol.

(a2 + b2) = 1/2 [(a + b)2 + (a- b)2]

(313)2 + (287)2 = 1/2 [(313 + 287)2 + (313 – 287)2] = ½[(600)2 + (26)2]

= 1/2 (360000 + 676) = 180338.

Ex. 8. Which of the following are prime numbers ?

(i) 241           (ii) 337         (Hi) 391           (iv) 571

Sol.

(i)         Clearly, 16 > Ö241. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13.

241 is not divisible by any one of them.

241 is a prime number.

(ii)        Clearly, 19>Ö337. Prime numbers less than 19 are 2, 3, 5, 7, 11,13,17.

337 is not divisible by any one of them.

337 is a prime number.

(iii)       Clearly, 20 > Ö39l”. Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19.

We find that 391 is divisible by 17.

391 is not prime.

(iv)       Clearly, 24 > Ö57T. Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23.

571 is not divisible by any one of them.

571 is a prime number.

Ex. 9. Find the unit’s digit in the product (2467)163 x (341)72.

Sol. Clearly, unit’s digit in the given product = unit’s digit in 7153 x 172.

Now, 74 gives unit digit 1.

7152  gives unit digit 1,

\ 7153  gives unit digit (l x 7) = 7. Also, 172 gives unit digit 1.

Hence, unit’s digit in the product = (7 x 1) = 7.

Ex. 10. Find the unit’s digit in (264)102 + (264)103

Sol. Required unit’s digit = unit’s digit in (4)102 + (4)103.

Now, 42  gives unit digit 6.

\(4)102 gives unjt digit 6.

\(4)103 gives unit digit of the product (6 x 4) i.e., 4.

Hence, unit’s digit in (264)m + (264)103 = unit’s digit in (6 + 4) = 0.

Ex. 11. Find the total number of prime factors in the expression (4)11 x (7)5 x (11)2.

Sol. (4)11x (7)5 x (11)2 = (2 x 2)11 x (7)5 x (11)2 = 211 x 211 x75x 112 = 222 x 75 x112

Total number of prime factors = (22 + 5 + 2) = 29.

Ex.12. Simplify :    (i) 896 x 896 – 204 x 204

(ii) 387 x 387 + 114 x 114 + 2 x 387 x 114

(iii) 81 X 81 + 68 X 68-2 x 81 X 68.

Sol.

(i)  Given exp  =  (896)2 – (204)2 = (896 + 204) (896 – 204) = 1100 x 692 = 761200.

(ii) Given exp  = (387)2+ (114)2+ (2 x 387x 114)

= a2 + b2 + 2ab,  where a = 387,b=114

= (a+b)2 = (387 + 114 )2 = (501)2 = 251001.

(iii) Given exp = (81)2 + (68)2 – 2x 81 x 68 = a2 + b2 – 2ab,Where a =81,b=68

=  (a-b)2 = (81 –68)2 = (13)2 = 169.

Ex.13. Which of the following numbers is divisible by 3 ?

(i) 541326                                 (ii) 5967013

Sol.

(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.

Hence, 541326 is divisible by 3.

(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.

Hence, 5967013 is not divisible by 3.

Ex.14.What least value must be assigned to * so that the number 197*5462 is r 9 ?

Sol.

Let the missing digit be x.

Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +»2) = (34 + x).

For (34 + x) to be divisible by 9, x must be replaced by 2 .

Hence, the digit in place of * must be 2.

Ex. 15. Which of the following numbers is divisible by 4 ?

(i) 67920594                    (ii) 618703572

Sol.

(i) The number formed by the last two digits in the given number is 94, which is not divisible by 4.

Hence, 67920594 is not divisible by 4.

(ii) The number formed by the last two digits in the given number is 72, which is divisible by 4.

Hence, 618703572 is divisible by 4.

Ex. 16. Which digits should come in place of * and $if the number 62684*$ is divisible by both 8 and 5 ?

Sol.

Since the given number is divisible by 5, so 0 or 5 must come in place of $. But, a number ending with 5 is never divisible by 8. So, 0 will replace$.

Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4.

Hence, digits in place of * and \$ are 4 and 0 respectively.

Ex. 17. Show that 4832718 is divisible by 11.

Sol.    (Sum of digits at odd places) – (Sum of digits at even places)

= (8 + 7 + 3 + 4) – (1 + 2 + 8) = 11, which is divisible by 11.

Hence, 4832718 is divisible by 11.

Ex. 18. Is 52563744 divisible by 24 ?

Sol.  24 = 3 x 8, where 3 and 8 are co-primes.

The sum of the digits in the given number is 36, which is divisible by 3. So, the                       given number is divisible by 3.

The number formed by the last 3 digits of the given number is 744, which is  divisible by 8. So, the given number is divisible by 8.

Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes.

So, it is divisible by 3 x 8, i.e., 24.

Ex. 19. What least number must be added to 3000 to obtain a number exactly divisible by 19 ?

Sol. On dividing 3000 by 19, we get 17 as remainder.

\Number to be added = (19 – 17) = 2.

Ex. 20. What least number must be subtracted from 2000 to get a number exactly divisible by 17 ?

Sol. On dividing 2000 by 17, we get 11 as remainder.

\Required number to be subtracted = 11.

Ex. 21. Find the number which is nearest to 3105 and is exactly divisible by 21.

Sol. On dividing 3105 by 21, we get 18 as remainder.

\Number to be added to 3105 = (21 – 18) – 3.

Hence, required number = 3105 + 3 = 3108.

Ex. 22. Find the smallest number of 6 digits which is exactly divisible by 111.

Sol. Smallest number of 6 digits is 100000.

On dividing 100000 by 111, we get 100 as remainder.

\Number to be added = (111 – 100) – 11.

Hence, required number = 100011.-

Ex. 23. On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37. Find the divisor.

Dividend – Remainder      15968-37

Sol.     Divisor = ————————– = ————- = 179.

.Quotient                    89

Ex. 24. A number when divided by 342 gives a remainder 47. When the same number ift divided by 19, what would be the remainder ?

Sol.   On dividing the given number by 342, let k be the quotient and 47 as remainder.

Then, number – 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.

\The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.

Ex. 25. A number being successively divided by 3, 5 and 8 leaves remainders 1, 4

and 7 respectively. Find the respective remainders if the order of divisors be reversed,

Sol.

 3 X 5 y – 1 8 z – 4 1 – 7

\z = (8 x 1 + 7) = 15; y = {5z + 4) = (5 x 15 + 4) = 79; x = (3y + 1) = (3 x 79 + 1) = 238.

Now,

 8 238 5 29 – 6 3 5 – 4 1 – 9,

\Respective remainders are 6, 4, 2.

Ex. 26. Find the remainder when 231 is divided by 5.

Sol.    210 = 1024. Unit digit of 210 x 210 x 210 is 4 [as 4 x 4 x 4 gives unit digit 4].

\Unit digit of 231 is 8.

Now, 8 when divided by 5, gives 3 as remainder.

Hence, 231 when divided by 5, gives 3 as remainder.

Ex. 27. How many numbers between 11 and 90 are divisible by 7 ?

Sol.  The required numbers are 14, 21, 28, 35, …. 77, 84.

This is an A.P. with a = 14 and d = (21 – 14) = 7.

Let it contain n terms.

Then, Tn = 84   =>  a + (n – 1) d = 84

=>   14 + (n – 1) x 7 = 84   or n = 11.

\Required number of terms = 11.

Ex. 28. Find the sum of all odd numbers upto 100.

Sol. The given numbers are 1, 3, 5, 7, …, 99.

This is an A.P. with a = 1 and d = 2.

Let it contain n terms. Then,

1 + (n – 1) x 2 = 99 or n = 50.

\Required sum = n (first term + last term)

2

= 50 (1 + 99) = 2500.

2

Ex. 29. Find the sum of all 2 digit numbers divisible by 3.

Sol. All 2 digit numbers divisible by 3 are :

12, 51, 18, 21, …, 99.

This is an A.P. with a = 12 and d = 3.

Let it contain n terms. Then,

12 + (n – 1) x 3 = 99 or n = 30.

\Required sum = 30 x (12+99) = 1665.

2

Ex.30.How many terms are there in 2,4,8,16……1024?

Sol.Clearly 2,4,8,16……..1024 form a GP. With a=2 and r = 4/2 =2.

Let the number of terms be n  . Then

2 x 2n-1 =1024 or 2n-1 =512 = 29.

\n-1=9 or n=10.

Ex. 31. 2 + 22 + 23 + … + 28 = ?

Sol.    Given series is a G.P. with a = 2, r = 2 and n = 8.

\sum =  a(rn-1) = 2 x (28 –1) = (2 x 255) =510

(r-1)        (2-1)

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