{"id":5890,"date":"2025-06-09T06:35:02","date_gmt":"2025-06-09T06:35:02","guid":{"rendered":"https:\/\/thecompanyboy.com\/?p=5890"},"modified":"2025-06-09T06:35:02","modified_gmt":"2025-06-09T06:35:02","slug":"rs-aggarwal-quantitative-aptitude-pdf-numbers-download","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/drive\/rs-aggarwal-quantitative-aptitude-pdf-numbers-download\/","title":{"rendered":"RS Aggarwal Quantitative Aptitude PDF Free download: Numbers"},"content":{"rendered":"<ol>\n<li><strong> NUMBERS<\/strong><\/li>\n<\/ol>\n<p><strong>IMPORTANT FACTS AND FORMULAE<\/strong><\/p>\n<p><strong>I..Numeral <\/strong>: In Hindu Arabic system, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 called <strong><em>digits <\/em><\/strong>to represent any number.<\/p>\n<p>\u00a0A group of digits, denoting a number is called a <strong><em>numeral.<\/em><\/strong><\/p>\n<p><strong><em>\u00a0<\/em><\/strong>We represent a number, say 689745132 as shown below :<\/p>\n<table>\n<tbody>\n<tr>\n<td width=\"53\">\n<p>Ten Crores (10<sup>8<\/sup>)<\/p>\n<\/td>\n<td width=\"46\">\n<p>Crores(10<sup>7<\/sup>)<\/p>\n<\/td>\n<td width=\"65\">\n<p>Ten Lacs (Millions) (10<sup>6<\/sup>)<\/p>\n<\/td>\n<td width=\"48\">\n<p>Lacs(10<sup>5<\/sup>)<\/p>\n<\/td>\n<td width=\"51\">\n<p>Ten Thousands (10<sup>4<\/sup>)<\/p>\n<\/td>\n<td width=\"47\">\n<p>Thousands (10<sup>3<\/sup>)<\/p>\n<\/td>\n<td width=\"48\">\n<p>Hundreds (10<sup>2<\/sup>)<\/p>\n<\/td>\n<td width=\"32\">\n<p>Tens(10<sup>1<\/sup>)<\/p>\n<\/td>\n<td width=\"32\">\n<p>Units(10<sup>0<\/sup>)<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td width=\"53\">\n<p><strong>6<\/strong><\/p>\n<\/td>\n<td width=\"46\">\n<p><strong>8<\/strong><\/p>\n<\/td>\n<td width=\"65\">\n<p><strong>9<\/strong><\/p>\n<\/td>\n<td width=\"48\">\n<p><strong>7<\/strong><\/p>\n<\/td>\n<td width=\"51\">\n<p><strong>4<\/strong><\/p>\n<\/td>\n<td width=\"47\">\n<p><strong>5<\/strong><\/p>\n<\/td>\n<td width=\"48\">\n<p><strong>1<\/strong><\/p>\n<\/td>\n<td width=\"32\">\n<p><strong>3<\/strong><\/p>\n<\/td>\n<td width=\"32\">\n<p><strong>2<\/strong><\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We read it as : &#8216;Sixty-eight crores, ninety-seven lacs, forty-five thousand, one hundred and thirty-two&#8217;.<\/p>\n<p><strong>II\u00a0 Place<\/strong><strong> Value or Local Value of a Digit in a Numeral <\/strong>:<\/p>\n<p>In the above numeral :<\/p>\n<p>Place value of 2 is (2 x 1) = 2; Place value of 3 is (3 x 10) = 30;<\/p>\n<p>Place value of 1 is (1 x 100) = 100 and so on.<\/p>\n<p>Place value of 6 is 6 x 10<sup>8<\/sup> = 600000000\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p><strong>III.Face Value <\/strong>: The <strong><em>face<\/em><\/strong><em> <strong>value <\/strong><\/em>of a digit in a numeral is the value of the\u00a0\u00a0 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.<\/p>\n<p><strong>IV.TYPES OF NUMBERS<\/strong><\/p>\n<p><strong>1.Natural Numbers <\/strong>: Counting numbers 1, 2, 3, 4, 5,&#8230;.. are called <strong><em>natural<\/em><\/strong><\/p>\n<p><em>numbers.<\/em><\/p>\n<p><strong>2<\/strong><em>.<\/em><strong>Whole Numbers <\/strong>: All counting numbers together with zero form the set of <strong><em>whole<br \/>\n<\/em><\/strong><strong><em>numbers<\/em><\/strong><em>. <\/em>Thus,<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <em>(i)<\/em> 0 is the only whole number which is not a natural number.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <em>(ii) <\/em>Every natural number is a whole number.<\/p>\n<p><strong>3<\/strong>.<strong>Integers <\/strong>: All natural numbers, 0 and negatives of counting numbers <em>i.e.,<br \/>\n<\/em>{\u2026, -3,-2,-1, 0, 1, 2, 3,\u2026..} together form the set of integers.<\/p>\n<p><em>(i) <\/em><strong>Positive Integers <\/strong>: {1, 2, 3, 4, \u2026..} is the set of all positive integers.<\/p>\n<p><em>(ii) <\/em><strong>Negative Integers <\/strong>: <em>{- <\/em>1, &#8211; 2, &#8211; 3,\u2026..} is the set of all negative integers.<\/p>\n<p><em>(iii) <\/em><strong>Non-Positive and Non-Negative Integers <\/strong>: 0 is neither positive nor<\/p>\n<p>negative. So, {0, 1, 2, 3,\u2026.} represents the set of non-negative integers, while<\/p>\n<p>{0, -1,-2,-3,\u2026..} represents the set of non-positive integers.<\/p>\n<ol start=\"4\">\n<li><strong> Even Numbers :<\/strong> A number divisible by 2 is called an even number, e.g., 2, 4, 6, 8, 10, etc.<\/li>\n<li><strong> Odd Numbers : <\/strong>A number not divisible by 2 is called an odd number. e.g., 1, 3, 5, 7, 9, 11, etc.<\/li>\n<li><strong> Prime Numbers :<\/strong> A number greater than 1 is called a prime number, if it has exactly two factors, namely 1 and the number itself.<\/li>\n<\/ol>\n<p>Prime numbers upto 100 are : 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43,<\/p>\n<p>47,\u00a0 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.<\/p>\n<p>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 :<\/p>\n<p>Find a whole number nearly greater than the square root of p. Let k &gt; *jp. Test whether p is divisible by any prime number less than k. If yes, then p is not prime. Otherwise, p is prime.<\/p>\n<p>e.g,,We have to find whether 191 is a prime number or not. Now, 14 &gt; V191.<\/p>\n<p>Prime numbers less than 14 are 2, 3, 5, 7, 11, 13.<\/p>\n<p>191 is not divisible by any of them. So, 191 is a prime number.<\/p>\n<p><strong>7.Composite Numbers : <\/strong>Numbers greater than 1 which are not prime, are known as composite numbers, e.g., 4, 6, 8, 9, 10, 12.<\/p>\n<p><strong>Note : <\/strong>\u00a0\u00a0\u00a0(i) 1 is neither prime nor composite.<\/p>\n<p>(ii) 2 is the only even number which is prime.<\/p>\n<p>(iii) There are 25 prime numbers between 1 and 100.<\/p>\n<ol start=\"8\">\n<li><strong> Co-primes :<\/strong> 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,<\/li>\n<\/ol>\n<p><strong>V.TESTS OF DIVISIBILITY<\/strong><\/p>\n<ol>\n<li><strong> Divisibility By 2 :<\/strong> A number is divisible by 2, if its unit&#8217;s digit is any of 0, 2, 4, 6, 8.<\/li>\n<\/ol>\n<p>Ex. 84932 is divisible by 2, while 65935 is not.<\/p>\n<ol start=\"2\">\n<li><strong> Divisibility By 3 :<\/strong> A number is divisible by 3, if the sum of its digits is divisible by 3.<\/li>\n<\/ol>\n<p><strong>Ex.<\/strong>592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which is divisible by 3.<\/p>\n<p>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.<\/p>\n<ol start=\"3\">\n<li><strong> Divisibility By 4 :<\/strong> A number is divisible by 4, if the number formed by the last two digits is divisible by 4.<\/li>\n<\/ol>\n<p>Ex. 892648 is divisible by 4, since the number formed by the last two digits is<\/p>\n<p>48,\u00a0 which is divisible by 4.<\/p>\n<p>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.<\/p>\n<ol start=\"4\">\n<li><strong> Divisibility By 5 :<\/strong> A number is divisible by 5, if its unit&#8217;s digit is either 0 or 5. Thus, 20820 and 50345 are divisible by 5, while 30934 and 40946 are not.<\/li>\n<li><strong> Divisibility By 6 :<\/strong> A number is divisible by 6, if it is divisible by both 2 and 3. Ex. The number 35256 is clearly divisible by 2.<\/li>\n<\/ol>\n<p>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.<\/p>\n<ol start=\"6\">\n<li><strong> Divisibility By 8 : <\/strong>A number is divisible by 8, if the number formed by the last<\/li>\n<\/ol>\n<p>three digits of the given number is divisible by 8.<\/p>\n<p>Ex. 953360 is divisible by 8, since the number formed by last three digits is 360, which is divisible by 8.<\/p>\n<p>But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8.<\/p>\n<ol start=\"7\">\n<li><strong> Divisibility By 9 <\/strong>: A number is divisible by 9, if the sum of its digits is divisible<\/li>\n<\/ol>\n<p>by 9.<\/p>\n<p>Ex. 60732 is divisible by 9, since sum of digits * (6 + 0 + 7 + 3 + 2) = 18, which is divisible by 9.<\/p>\n<p>But, 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is not divisible by 9.<\/p>\n<ol start=\"8\">\n<li><strong> Divisibility By 10 :<\/strong> A number is divisible by 10, if it ends with 0.<\/li>\n<\/ol>\n<p>Ex. 96410, 10480 are divisible by 10, while 96375 is not.<\/p>\n<ol start=\"9\">\n<li><strong> Divisibility By 11 :<\/strong> 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.<\/li>\n<\/ol>\n<p><strong>Ex.<\/strong> The number 4832718 is divisible by 11, since :<\/p>\n<p>(sum of digits at odd places) &#8211; (sum of digits at even places)<\/p>\n<p>(8 + 7 + 3 + 4) &#8211; (1 + 2 + 8) = 11, which is divisible by 11.<\/p>\n<ol start=\"10\">\n<li>Divisibility By 12 ; A number is divisible by 12, if it is divisible by both 4 and<\/li>\n<\/ol>\n<p>3.<\/p>\n<p><strong>Ex.<\/strong> Consider the number 34632.<\/p>\n<p>(i) The number formed by last two digits is 32, which is divisible by 4,<\/p>\n<p>(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.<\/p>\n<ol start=\"11\">\n<li><strong>11<\/strong>. Divisibility By 14 : A number is divisible by 14, if it is divisible by 2 as well as 7.<\/li>\n<li>Divisibility By 15 : A number is divisible by 15, if it is divisible by both 3 and 5.<\/li>\n<li>Divisibility By 16 : A number is divisible by 16, if the number formed by the last4 digits is divisible by 16.<\/li>\n<\/ol>\n<p><strong>Ex.<\/strong>7957536 is divisible by 16, since the number formed by the last four digits is 7536, which is divisible by 16.<\/p>\n<ol start=\"14\">\n<li><strong> Divisibility By 24 : <\/strong>A given number is divisible by 24, if it is divisible by both3 and 8.<\/li>\n<li><strong> Divisibility By 40 : <\/strong>A given number is divisible by 40, if it is divisible by both<\/li>\n<\/ol>\n<p>5\u00a0 and 8.<\/p>\n<ol start=\"16\">\n<li><strong> Divisibility By 80 : <\/strong>A given number is divisible by 80, if it is divisible by both 5 and 16.<\/li>\n<\/ol>\n<p><strong>Note :<\/strong> 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.<\/p>\n<p>If p arid q are not co-primes, then the given number need not be divisible by pq,<\/p>\n<p>even when it is divisible by both p and q.<\/p>\n<p><strong>Ex.<\/strong> 36 is divisible by both 4 and 6, but it is not divisible by (4&#215;6) = 24, since<\/p>\n<p>4\u00a0 and 6 are not co-primes.<\/p>\n<p><strong>VI\u00a0\u00a0 MULTIPLICATION BY SHORT CUT METHODS<\/strong><\/p>\n<ol>\n<li><strong> Multiplication By Distributive Law :<\/strong><\/li>\n<\/ol>\n<p>(i) a x (b + c) = a x b + a x c\u00a0\u00a0\u00a0 (ii) ax(b-c) = a x b-a x c.<\/p>\n<p><strong>Ex.<\/strong>\u00a0\u00a0 (i) 567958 x 99999 = 567958 x (100000 &#8211; 1)<\/p>\n<p>= 567958 x 100000 &#8211; 567958 x 1 = (56795800000 &#8211; 567958) = 56795232042. (ii) 978 x 184 + 978 x 816 = 978 x (184 + 816) = 978 x 1000 = 978000.<\/p>\n<ol start=\"2\">\n<li><strong> Multiplication of a Number By 5<sup>n<\/sup> : <\/strong>Put n zeros to the right of the multiplicand and divide the number so formed by 2<sup>n<\/sup><\/li>\n<\/ol>\n<p>Ex. 975436 x 625 = 975436 x 5<sup>4<\/sup>= 9754360000 =\u00a0\u00a0 609647600<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16<\/p>\n<p><strong>VII.\u00a0\u00a0 BASIC FORMULAE<\/strong><\/p>\n<ol>\n<li>(a + b)<sup>2<\/sup> = a<sup>2 <\/sup>+ b<sup>2<\/sup> + 2ab 2. (a &#8211; b)<sup>2 <\/sup>= a<sup>2 <\/sup>+ b<sup>2<\/sup> &#8211; 2ab<\/li>\n<li>(a + b)<sup>2<\/sup> &#8211; (a &#8211; b)<sup>2<\/sup> = 4ab 4. (a + b)<sup>2 <\/sup>+ (a &#8211; b)<sup>2<\/sup> = 2 (a<sup>2<\/sup> + b<sup>2<\/sup>)<\/li>\n<li>(a<sup>2 <\/sup>&#8211; b<sup>2<\/sup>) = (a + b) (a &#8211; b)<\/li>\n<li>(a + b + c)<sup>2<\/sup> = a<sup>2<\/sup> + b<sup>2<\/sup> + c<sup>2<\/sup> + 2 (ab + bc + ca)<\/li>\n<li>(a<sup>3<\/sup> + b<sup>3<\/sup>) = (a +b) (a<sup>2<\/sup> &#8211; ab + b<sup>2<\/sup>) 8. (a<sup>3<\/sup> &#8211; b<sup>3<\/sup>) = (a &#8211; b) (a<sup>2<\/sup> + ab + b<sup>2<\/sup>)<\/li>\n<li>(a<sup>3<\/sup> + b<sup>3<\/sup> + c<sup>3<\/sup> -3abc) = (a + b + c) (a<sup>2<\/sup> + b<sup>2<\/sup> + c<sup>2<\/sup> &#8211; ab &#8211; bc &#8211; ca)<\/li>\n<li>If a + b + c = 0, then a<sup>3<\/sup> + b<sup>3 <\/sup>+ c<sup>3<\/sup> = 3abc.<\/li>\n<\/ol>\n<p><strong>\u00a0<\/strong><strong>VIII.\u00a0 DIVISION ALGORITHM OR EUCLIDEAN ALGORITHM <\/strong><\/p>\n<p>If we divide a given number by another number, then :<\/p>\n<p><strong>Dividend = (Divisor x Quotient) + Remainder<\/strong><\/p>\n<ol>\n<li><strong>IX<\/strong>. {i) (x<sup>n <\/sup>&#8211; a<sup>n<\/sup> ) is divisible by (x &#8211; a) for all values of n.<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) (x<sup>n<\/sup> &#8211; a<sup>n<\/sup>) is divisible by (x + a) for all even values of n.<\/p>\n<p>\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(iii) (x<sup>n<\/sup> + a<sup>n<\/sup>) is divisible by (x + a) for all odd values of n.<\/p>\n<p><strong>\u00a0<\/strong><strong>PROGRESSION<\/strong><\/p>\n<p>A succession of numbers formed and arranged in a definite order according to certain definite rule, is called a progression.<\/p>\n<ol>\n<li><strong> Arithmetic Progression (A.P.) : <\/strong>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 <em>common difference<\/em> of the A.P.<\/li>\n<\/ol>\n<p>An A.P. with first term a and common difference d is given by a, (a + d), (a + 2d),(a + 3d),&#8230;..<\/p>\n<p>The <strong>nth term of this A.P. is given by T<sub>n<\/sub> =a (n &#8211; 1) d.<\/strong><\/p>\n<p><strong>The sum of n terms of this A.P.<\/strong><\/p>\n<p>S<sub>n<\/sub> = n\/2 [2a + (n &#8211; 1) d] = n\/2\u00a0\u00a0 (first term + last term).<\/p>\n<p><strong>SOME IMPORTANT RESULTS :<\/strong><\/p>\n<p><strong>\u00a0<\/strong>\u00a0(i) (1 + 2 + 3 +\u2026. + n) =n(n+1)\/2<\/p>\n<p>(ii) (l<sup>2<\/sup> + 2<sup>2<\/sup> + 3<sup>2<\/sup> + &#8230; + n<sup>2<\/sup>) = n (n+1)(2n+1)\/6<\/p>\n<p>(iii)\u00a0 (1<sup>3 <\/sup>+ 2<sup>3<\/sup> + 3<sup>3 <\/sup>+ &#8230; + n<sup>3<\/sup>) =n<sup>2<\/sup>(n+1)<sup>2<\/sup><\/p>\n<ol start=\"2\">\n<li><strong> Geometrical Progression (G.P.) :<\/strong> A progression of numbers in which every term bears a constant ratio with its preceding term, is called a geometrical progression.<\/li>\n<\/ol>\n<p>The constant ratio is called the common ratio of the G.P. A G.P. with first term a and common ratio r is :<\/p>\n<p>a, ar, ar<sup>2<\/sup>,<\/p>\n<p>In this G.P. T<sub>n<\/sub> = ar<sup>n-1<\/sup><\/p>\n<p>sum of the n terms, Sn=\u00a0\u00a0 <u>a(1-r<sup>n<\/sup>)<\/u><\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (1-r)<\/p>\n<h1><strong>SOLVED EXAMPLES<\/strong><\/h1>\n<p><strong>Ex. 1. Simplify :\u00a0\u00a0 (i) 8888 + 888 + 88 + 8 <\/strong>\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <strong>(ii) 11992 &#8211; 7823 &#8211; 456 <\/strong>\u00a0<\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0 i )\u00a0 8888\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ii) 11992 &#8211; 7823 &#8211; 456 = 11992 &#8211; (7823 + 456)<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 888\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0= 11992 &#8211; 8279 = 3713-<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 88\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7823\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 11992<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <u>+\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 8\u00a0\u00a0\u00a0\u00a0 <\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+ <u>\u00a0\u00a0456 <\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<u>&#8211;\u00a0\u00a0 8279<\/u><\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <u>\u00a0\u00a0\u00a0\u00a0\u00a09872\u00a0\u00a0\u00a0 <\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<u>\u00a0\u00a08279<\/u>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <u>\u00a0\u00a0\u00a0\u00a03713<\/u><\/p>\n<p><strong>Ex. 2, What value will replace the question mark in each of the following equations ?<\/strong><\/p>\n<p><strong>(i) ? &#8211; 1936248 = 1635773\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) 8597 &#8211; ? = 7429 &#8211; 4358<\/strong><\/p>\n<p><strong>Sol.<\/strong>\u00a0 (i) Let x\u00a0 &#8211; 1936248=1635773.Then, x = 1635773 + 1936248=3572021.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) Let 8597 &#8211; x = 7429 &#8211; 4358.<\/p>\n<p>Then, x = (8597 + 4358) &#8211; 7429 = 12955 &#8211; 7429 = 5526.<\/p>\n<p><strong>\u00a0Ex. 3. What could be the maximum value of Q in the following equation?\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5P9 + 3R7 + 2Q8 = 1114 <\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p><strong>Sol.<\/strong> We may analyse the given equation as shown :\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>Clearly, 2 + P + R + Q = ll.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 5\u00a0\u00a0 P\u00a0 9<\/p>\n<p>So, the maximum value of Q can be\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 3\u00a0\u00a0 R\u00a0 7<\/p>\n<p>(11 &#8211; 2) i.e., 9 (when P = 0, R = 0);\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <u>2\u00a0\u00a0 Q\u00a0 8<\/u><\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <u>11\u00a0 1\u00a0\u00a0 4 <\/u><\/p>\n<p><strong>Ex. 4. Simplify : (i) 5793405 x 9999\u00a0 (ii) 839478 x 625<\/strong><\/p>\n<p><strong>Sol.<\/strong><\/p>\n<p>i)5793405&#215;9999=5793405(10000-1)=57934050000-5793405=57928256595.b<\/p>\n<ol start=\"524673750\">\n<li>ii) 839478 x 625 = 839478 x 5<sup>4<\/sup> = <u>8394780000 <\/u>= 524673750.<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16<\/p>\n<h2><strong>Ex. 5. Evaluate : (i) 986 x 237 + 986 x 863\u00a0\u00a0\u00a0 (ii) 983 x 207 &#8211; 983 x 107 <\/strong><\/h2>\n<p><strong>Sol.<\/strong><\/p>\n<p>(i) 986 x 137 + 986 x 863 = 986 x (137 + 863) = 986 x 1000 = 986000.<\/p>\n<p>(ii) 983 x 207 &#8211; 983 x 107 = 983 x (207 &#8211; 107) = 983 x 100 = 98300.<\/p>\n<p><strong>Ex. 6. Simplify : (i) 1605 x 1605\u00a0\u00a0\u00a0 ii) 1398 x 1398 <\/strong><\/p>\n<p><strong>Sol.<\/strong><\/p>\n<ol>\n<li>i) 1605 x 1605 = (1605)<sup>2<\/sup> = (1600 + 5)<sup>2<\/sup> = (1600)<sup>2<\/sup> + (5)<sup>2<\/sup> + 2 x 1600 x 5<\/li>\n<\/ol>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 2560000 + 25 + 16000 = 2576025.<\/p>\n<p>(ii) 1398 x 1398 &#8211; (1398)<sup>2<\/sup> = (1400 &#8211; 2)<sup>2<\/sup>= (1400)<sup>2<\/sup> + (2)<sup>2<\/sup> &#8211; 2 x 1400 x 2<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =1960000 + 4 &#8211; 5600 = 1954404.<\/p>\n<p><strong>Ex. 7. Evaluate : (313 x 313 + 287 x 287).<\/strong><\/p>\n<p><strong>Sol.<\/strong><\/p>\n<p>\u00a0(a<sup>2<\/sup> + b<sup>2<\/sup>) = 1\/2 [(a + b)<sup>2<\/sup> + (a- b)<sup>2<\/sup>]<\/p>\n<p>(313)<sup>2<\/sup> + (287)<sup>2<\/sup> = 1\/2 [(313 + 287)<sup>2<\/sup> + (313 &#8211; 287)<sup>2<\/sup>] = \u00bd[(600)<sup>2<\/sup> + (26)<sup>2<\/sup>]<\/p>\n<p>= 1\/2 (360000 + 676) = 180338.<\/p>\n<p><strong>Ex. 8. Which of the following are prime numbers ?<\/strong><\/p>\n<p><strong>(i) 241\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) 337\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (Hi) 391\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (iv) 571<\/strong><\/p>\n<p><strong>Sol.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/strong>\u00a0\u00a0<\/p>\n<p>(i)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Clearly, 16 &gt; \u00d6241. Prime numbers less than 16 are 2, 3, 5, 7, 11, 13.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 241 is not divisible by any one of them.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 241 is a prime number.<\/p>\n<p>(ii)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Clearly, 19&gt;\u00d6337. Prime numbers less than 19 are 2, 3, 5, 7, 11,13,17.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 337 is not divisible by any one of them.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 337 is a prime number.<\/p>\n<p>(iii)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Clearly, 20 &gt; \u00d639l&#8221;. Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17, 19.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 We find that 391 is divisible by 17.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 391 is not prime.<\/p>\n<p>(iv)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Clearly, 24 &gt; \u00d657T. Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 571 is not divisible by any one of them.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 571 is a prime number.<\/p>\n<p>\u00a0<strong>Ex. 9. Find the unit&#8217;s digit in the product (2467)163 x (341)72. <\/strong><\/p>\n<p><strong>Sol.<\/strong> Clearly, unit&#8217;s digit in the given product = unit&#8217;s digit in 7<sup>153<\/sup> x 1<sup>72<\/sup>.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now, 74 gives unit digit 1.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 7<sup>152<\/sup>\u00a0 gives unit digit 1,<\/p>\n<p>\u00a0 \\ 7<sup>153<\/sup>\u00a0 gives unit digit (l x 7) = 7. Also, 1<sup>72<\/sup> gives unit digit 1.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, unit&#8217;s digit in the product = (7 x 1) = 7.<\/p>\n<p><strong>Ex. 10. Find the unit&#8217;s digit in (264)<sup>102<\/sup> + (264)<sup>103<\/sup><\/strong><\/p>\n<p><strong>Sol.<\/strong> Required unit&#8217;s digit = unit&#8217;s digit in (4)<sup>102<\/sup> + (4)<sup>103.<\/sup><\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now, 4<sup>2<\/sup>\u00a0 gives unit digit 6.<\/p>\n<p>\u00a0\u00a0 \\(4)<sup>102 <\/sup>gives unjt digit 6.<\/p>\n<p>\u00a0\u00a0 \\(4)103 gives unit digit of the product (6 x 4) i.e., 4.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, unit&#8217;s digit in (264)m + (264)103 = unit&#8217;s digit in (6 + 4) = 0.<\/p>\n<p><strong>\u00a0Ex. 11. Find the total number of prime factors in the expression (4)<sup>11<\/sup> x (7)<sup>5<\/sup> x (11)<sup>2<\/sup>.<\/strong><\/p>\n<p><strong>Sol.<\/strong> (4)<sup>11<\/sup>x (7)<sup>5<\/sup> x (11)<sup>2<\/sup> = (2 x 2)<sup>11<\/sup> x (7)<sup>5<\/sup> x (11)<sup>2<\/sup> = 2<sup>11<\/sup> x 2<sup>11<\/sup> x7<sup>5<\/sup>x 11<sup>2<\/sup> = 2<sup>22<\/sup> x 7<sup>5<\/sup> x11<sup>2<\/sup><\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Total number of prime factors = (22 + 5 + 2) = 29.<\/p>\n<p><strong>Ex.12. Simplify :\u00a0\u00a0\u00a0 (i) 896 x 896 &#8211; 204 x 204<\/strong><\/p>\n<p><strong>\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(ii) 387 x 387 + 114 x 114 + 2 x 387 x 114 <\/strong><\/p>\n<p><strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (iii) 81 X 81 + 68 X 68-2 x 81 X 68.<\/strong><\/p>\n<p><strong>Sol.<\/strong><\/p>\n<p>(i)\u00a0 Given exp\u00a0 =\u00a0 (896)<sup>2<\/sup> &#8211; (204)<sup>2<\/sup> = (896 + 204) (896 &#8211; 204) = 1100 x 692 = 761200.<\/p>\n<p>(ii) Given exp\u00a0 = (387)<sup>2<\/sup>+ (114)<sup>2<\/sup>+ (2 x 387x 114)<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = a<sup>2<\/sup> + b<sup>2<\/sup> + 2ab,\u00a0 where a = 387,b=114<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= (a+b)<sup>2<\/sup> = (387 + 114 )<sup>2 <\/sup>= (501)<sup>2 <\/sup>= 251001.<\/p>\n<p>(iii) Given exp = (81)<sup>2 <\/sup>+ (68)<sup>2 <\/sup>\u2013 2x 81 x 68 = a<sup>2<\/sup> + b<sup>2<\/sup> \u2013 2ab,Where a =81,b=68<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =\u00a0 (a-b)<sup>2 <\/sup>= (81 \u201368)<sup>2 <\/sup>= (13)<sup>2 = 169.<\/sup><\/p>\n<p><sup>\u00a0<\/sup>Ex.13. Which of the following numbers is divisible by 3 ?<\/p>\n<p>(i) 541326\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (ii) 5967013<\/p>\n<p>Sol.<\/p>\n<p>(i) Sum of digits in 541326 = (5 + 4 + 1 + 3 + 2 + 6) = 21, which is divisible by 3.<\/p>\n<p>Hence, 541326 is divisible by 3.<\/p>\n<p>(ii) Sum of digits in 5967013 =(5+9 + 6 + 7 + 0+1 +3) = 31, which is not divisible by 3.<\/p>\n<p>Hence, 5967013 is not divisible by 3.<\/p>\n<p><strong>Ex.14.What least value must be assigned to * so that the number 197*5462 is r 9 ?<\/strong><\/p>\n<p><strong>Sol.<\/strong><\/p>\n<p>Let the missing digit be x.<\/p>\n<p>Sum of digits = (1 + 9 + 7 + x + 5 + 4 + 6 +\u00bb2) = (34 + x).<\/p>\n<p>For (34 + x) to be divisible by 9, x must be replaced by 2 .<\/p>\n<p>Hence, the digit in place of * must be 2.<\/p>\n<p><strong>Ex. 15. Which of the following numbers is divisible by 4 ? <\/strong><\/p>\n<p><strong>(i) 67920594\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(ii) 618703572<\/strong><\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>(i) The number formed by the last two digits in the given number is 94, which is not divisible by 4.<\/p>\n<p>Hence, 67920594 is not divisible by 4.<\/p>\n<p>(ii) The number formed by the last two digits in the given number is 72, which is divisible by 4.<\/p>\n<p>Hence, 618703572 is divisible by 4.<\/p>\n<p><strong>Ex. 16. Which digits should come in place of * and $ if the number 62684*$ is divisible by both 8 and 5 ?<\/strong><\/p>\n<p><strong>Sol.\u00a0\u00a0\u00a0 <\/strong>\u00a0<\/p>\n<p>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 $.<\/p>\n<p>Now, the number formed by the last three digits is 4*0, which becomes divisible by 8, if * is replaced by 4.<\/p>\n<p>Hence, digits in place of * and $ are 4 and 0 respectively.<\/p>\n<p><strong>Ex. 17. Show that 4832718 is divisible by 11. <\/strong><\/p>\n<p><strong>Sol.\u00a0 <\/strong>\u00a0\u00a0(Sum of digits at odd places) &#8211; (Sum of digits at even places)<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = (8 + 7 + 3 + 4) &#8211; (1 + 2 + 8) = 11, which is divisible by 11.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, 4832718 is divisible by 11.<\/p>\n<p><strong>Ex. 18. Is 52563744 divisible by 24 ? <\/strong><\/p>\n<p><strong>Sol.<\/strong>\u00a0 24 = 3 x 8, where 3 and 8 are co-primes.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 The sum of the digits in the given number is 36, which is divisible by 3. So, the\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 given number is divisible by 3.<\/p>\n<p>\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0The number formed by the last 3 digits of the given number is 744, which is\u00a0 divisible by 8. So, the given number is divisible by 8.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes.<\/p>\n<p>So, it is divisible by 3 x 8, i.e., 24.<\/p>\n<p><strong>Ex. 19. What least number must be added to 3000 to obtain a number exactly divisible by 19 ?<\/strong><\/p>\n<p><strong>Sol. <\/strong>On dividing 3000 by 19, we get 17 as remainder.<\/p>\n<p>\u00a0\u00a0\u00a0 \\Number to be added = (19 &#8211; 17) = 2.<\/p>\n<p><strong>Ex. 20. What least number must be subtracted from 2000 to get a number exactly divisible by 17 ?<\/strong><\/p>\n<p><strong>Sol. <\/strong>On dividing 2000 by 17, we get 11 as remainder.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0 \\Required number to be subtracted = 11.<\/p>\n<p>Ex. 21. Find the number which is nearest to 3105 and is exactly divisible by 21.<\/p>\n<p><strong>Sol. <\/strong>On dividing 3105 by 21, we get 18 as remainder.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0 \\Number to be added to 3105 = (21 &#8211; 18) &#8211; 3.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, required number = 3105 + 3 = 3108.<\/p>\n<p><strong>Ex. 22. Find the smallest number of 6 digits which is exactly divisible by 111. <\/strong><\/p>\n<p><strong>Sol.<\/strong> Smallest number of 6 digits is 100000.<\/p>\n<p>\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0On dividing 100000 by 111, we get 100 as remainder.<\/p>\n<p>\u00a0\u00a0\u00a0 \\Number to be added = (111 &#8211; 100) &#8211; 11.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, required number = 100011.-<\/p>\n<p>Ex. 23. On dividing 15968 by a certain number, the quotient is 89 and the remainder is 37. Find the divisor.<\/p>\n<p><strong>\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>\u00a0Dividend &#8211; Remainder\u00a0\u00a0\u00a0\u00a0\u00a0 15968-37\u00a0\u00a0\u00a0\u00a0\u00a0<\/p>\n<p><strong>Sol.<\/strong>\u00a0\u00a0\u00a0\u00a0 Divisor = &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211; = &#8212;&#8212;&#8212;&#8212;- = 179.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 .Quotient\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 89<\/p>\n<p><strong>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 ?<\/strong><\/p>\n<p><strong>Sol.\u00a0\u00a0 <\/strong>On dividing the given number by 342, let k be the quotient and 47 as remainder.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, number \u2013 342k + 47 = (19 x 18k + 19 x 2 + 9) = 19 (18k + 2) + 9.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\\The given number when divided by 19, gives (18k + 2) as quotient and 9 as remainder.<\/p>\n<p><strong>Ex. 25. A number being successively divided by 3, 5 and 8 leaves remainders 1, 4<\/strong><\/p>\n<p><strong>and 7 respectively. Find the respective remainders if the order of divisors be reversed,<\/strong><\/p>\n<p>Sol.<\/p>\n<table>\n<tbody>\n<tr>\n<td width=\"23\">\n<p>3<\/p>\n<\/td>\n<td width=\"22\">\n<p>X<\/p>\n<\/td>\n<td width=\"26\">\n<p>&nbsp;<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td width=\"23\">\n<p>5<\/p>\n<\/td>\n<td width=\"22\">\n<p>y<\/p>\n<\/td>\n<td width=\"26\">\n<p>&#8211; 1<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td width=\"23\">\n<p>8<\/p>\n<\/td>\n<td width=\"22\">\n<p>z<\/p>\n<\/td>\n<td width=\"26\">\n<p>&#8211; 4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td width=\"23\">\n<p>&nbsp;<\/p>\n<\/td>\n<td width=\"22\">\n<p>1<\/p>\n<\/td>\n<td width=\"26\">\n<p>&#8211; 7<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\\z = (8 x 1 + 7) = 15; y = {5z + 4) = (5 x 15 + 4) = 79; x = (3y + 1) = (3 x 79 + 1) = 238.<\/p>\n<p>Now,<\/p>\n<table>\n<tbody>\n<tr>\n<td width=\"21\">\n<p>8<\/p>\n<\/td>\n<td width=\"36\">\n<p>238<\/p>\n<\/td>\n<td width=\"31\">\n<p>&nbsp;<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td width=\"21\">\n<p>5<\/p>\n<\/td>\n<td width=\"36\">\n<p>29<\/p>\n<\/td>\n<td width=\"31\">\n<p>&#8211; 6<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td width=\"21\">\n<p>3<\/p>\n<\/td>\n<td width=\"36\">\n<p>5<\/p>\n<\/td>\n<td width=\"31\">\n<p>&#8211; 4<\/p>\n<\/td>\n<\/tr>\n<tr>\n<td width=\"21\">\n<p>&nbsp;<\/p>\n<\/td>\n<td width=\"36\">\n<p>1<\/p>\n<\/td>\n<td width=\"31\">\n<p>&#8211; 9,<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>\\Respective remainders are 6, 4, 2.<\/p>\n<p>Ex. 26. Find the remainder when 2<sup>31<\/sup> is divided by 5.<\/p>\n<p><strong>\u00a0Sol.\u00a0\u00a0 <\/strong>\u00a02<sup>10<\/sup> = 1024. Unit digit of 2<sup>10<\/sup> x 2<sup>10 <\/sup>x 2<sup>10<\/sup> is 4 [as 4 x 4 x 4 gives unit digit 4].<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\Unit digit of 231 is 8.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Now, 8 when divided by 5, gives 3 as remainder.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Hence, 231 when divided by 5, gives 3 as remainder.<\/p>\n<p><strong>Ex. 27. How many numbers between 11 and 90 are divisible by 7 ? <\/strong><\/p>\n<p><strong>Sol. <\/strong>\u00a0The required numbers are 14, 21, 28, 35, &#8230;. 77, 84.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 This is an A.P. with a = 14 and d = (21 &#8211; 14) = 7.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Let it contain n terms.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Then, T<sub>n<\/sub> = 84\u00a0\u00a0 =&gt;\u00a0 a + (n &#8211; 1) d = 84<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 =&gt;\u00a0\u00a0 14 + (n &#8211; 1) x 7 = 84\u00a0\u00a0 or n = 11.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0 \\Required number of terms = 11.<\/p>\n<p>Ex. 28. Find the sum of all odd numbers upto 100.<\/p>\n<p>Sol. The given numbers are 1, 3, 5, 7, &#8230;, 99.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 This is an A.P. with a = 1 and d = 2.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Let it contain n terms. Then,<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1 + (n &#8211; 1) x 2 = 99 or n = 50.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0 \\Required sum = <u>n<\/u> (first term + last term)<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = <u>50 <\/u>(1 + 99) = 2500.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02<\/p>\n<p>Ex. 29. Find the sum of all 2 digit numbers divisible by 3.<\/p>\n<p><strong>\u00a0Sol.<\/strong> All 2 digit numbers divisible by 3 are :<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12, 51, 18, 21, &#8230;, 99.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 This is an A.P. with a = 12 and d = 3.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Let it contain n terms. Then,<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12 + (n &#8211; 1) x 3 = 99 or n = 30.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0 \\Required sum = <u>30<\/u> x (12+99) = 1665.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2<\/p>\n<p><strong>\u00a0<\/strong><strong>Ex.30.How many terms are there in 2,4,8,16\u2026\u20261024?<\/strong><\/p>\n<p><strong>Sol.<\/strong>Clearly 2,4,8,16\u2026\u2026..1024 form a GP. With a=2 and r = 4\/2 =2.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Let the number of terms be n\u00a0 . Then<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2 x 2<sup>n-1<\/sup> =1024 or 2<sup>n-1 =512 = 2<\/sup>9.<\/p>\n<p>\u00a0\u00a0\u00a0 \\n-1=9 or n=10.<\/p>\n<p><strong>Ex. 31. 2 + 2<sup>2<\/sup> + 2<sup>3<\/sup> + &#8230; + 2<sup>8<\/sup> = ?<\/strong><\/p>\n<p><strong>\u00a0Sol.<\/strong>\u00a0\u00a0\u00a0 Given series is a G.P. with a = 2, r = 2 and n = 8.<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 \\sum =\u00a0 <u>a(r<sup>n<\/sup>-1)<\/u> = <u>2 x (2<sup>8 <\/sup>\u20131) <\/u>= (2 x 255) =510<\/p>\n<p>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (r-1)\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0(2-1)<\/p>\n<p data-start=\"0\" data-end=\"284\">Accessing copyrighted materials, such as the &#8220;Quantitative Aptitude&#8221; book by R.S. 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Ensure that any digital copy you access is from a legitimate source to respect copyright laws.<\/li>\n<\/ul>\n<p data-start=\"1055\" data-end=\"1289\">Investing in official and authorized study materials not only ensures you receive high-quality and accurate content but also supports the authors and publishers who have dedicated their efforts to creating these educational resources.<\/p>\n<h3 data-start=\"1055\" data-end=\"1289\"><a href=\"https:\/\/eltsindia.com\/StudyMaterialFiles\/ea007acc-bc55-4e17-8470-0d9e85313252quantitative-aptitude-for-competitive-examinations-by-rs-aggarwal-reprint-2017.pdf\" target=\"_blank\" rel=\"noopener\">RS Aggarwal Quantitative Aptitude PDF Free download: Numbers<\/a><\/h3>\n","protected":false},"excerpt":{"rendered":"<p>NUMBERS IMPORTANT FACTS AND FORMULAE 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. \u00a0A group of digits, denoting a number is called a numeral. \u00a0We represent a number, say 689745132 as shown below : Ten Crores (108) [&hellip;]<\/p>\n","protected":false},"author":41,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[127],"tags":[2233,1923],"class_list":["post-5890","post","type-post","status-publish","format-standard","hentry","category-rs-aggarwal-quantitative-aptitude-pdf","tag-numbers","tag-rs-aggarwal-quantitative-aptitude"],"_links":{"self":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5890","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/users\/41"}],"replies":[{"embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/comments?post=5890"}],"version-history":[{"count":0,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/posts\/5890\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/media?parent=5890"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/categories?post=5890"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.reilsolar.com\/drive\/wp-json\/wp\/v2\/tags?post=5890"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}