Presenting the Mths Lectures! Your best bet for Qunt... MATHS LECTURE # 0 Number Systems & Working With Numbers System of numbers.3 0.6 π With the help of tree digrm, numbers cn be clssified s follows 3 0 3 Nturl numbers (positive integers) Complex Numbers These re the counting numbers used to count physicl quntities. e.g.,, 3,..., 05,..., 36, 5957,..., 37950046,.... The set of nturl numbers is denoted by N = {,, 3, 4, 5, 6, 7... } Rel Imginry Whole numbers Rtionl Irrtionl (, 3 etc.) The numbers 0,,, 3, 4,... re whole numbers. Integer rction Algebric (Roots of poly) Trnscendentl (e, π etc.) The set of ll non negtive integers (i.e. zero + nturl numbers) is sid to be the set of Whole Numbers nd is denoted s W = {0,,, 3,... }. Proper Improper (/3, /5, 3/8) (4/3, 5/, 6/5 etc.) Mixed 4 5 ( /, 3 /,... etc.) Integers PT Eduction, All rights reserved. Whole Number (0,,,...) Prime (, 3, 5, 7,...) Rel numbers Nturl Negtive odd (±, ±3, ±5...) Composite (4, 6, 8,...) Perfect (6, 8,...) Even (multiples of ) Rel numbers re those which cn represent ctul physicl quntities e.g. temperture, length, height etc. Rel numbers cn lso be defined s numbers tht cn be represented on the number line. The set of ll nturl numbers (positive, zero, negtive) re together known s integers. The set of integers is denoted s I where I = {0, ±, ±, ±3,... }. Zero nd positive integers re clled s non negtive integers. Rtionl numbers p nd q (q 0) re integers. Then p q is known s rtionl number. Thus the set Q of the rtionl numbers is given by R S T p Q = : q p, q I nd q 0 U V W Regd. Office: Indore PT centres spred cross Indi ~ Estblished 993 Our motto Kr Ke Dikhyenge is delivered through our unique Technology Driven Process Engine (TDpro engine). Emil: pthq@pteduction.com Web: www.pteduction.com, www.fcebook.com/pteduction IC : PTtkmml0 () of (8)
Nturlly, frctions such s 7 9, 3 6, re clled rtionl 5 numbers. This definition lso emphsises tht ny integer cn lso be rtionl number since p = p/, p I. Any positive rtionl number p/q, fter ctul division, if necessry cn be expressed s, p r = m + where m is non negtive integer nd 0 r < q q q or exmple, 4 5 () of (8) 3 3 0 0 = 8 + ; = 0 + ; 0 = = 0 +. 5 5 5 or the deciml representtion of frction p/q, we hve merely to consider the deciml form of frction r/q which we usully write to the right of the deciml point. Consider some frctions given below. () / = 0.5 () 3/5 = 0.6 (3) /4 = 0.5 (4) /5 = 0. (5) /8 = 0.5 (6) /6 = 0.666... (7) 5/ = 0.4545... (8) /3 = 0.33... (9) 7/ = 0.583333... Note tht the dots... represent endless recurrence of digits. Exmples (), (), (3), (4) nd (5) suggest tht we hve deciml form of the terminting type. While exmples (6), (7), (8) nd (9) tell us tht we hve deciml form of the nonterminting type. In cse of nonterminting type we hve deciml frctions hving n infinite number of digits. Some deciml frctions from this group hve digits repeting infinitely. They re clled repeting or recurring decimls. In endless recurring or infinite repeting deciml frctions we cn see tht when p is ctully divided by q the possible reminders re,, 3,..., q. So one of them hs to repet itself in q steps. Therefter the erlier numerl or group of numerls must repet itself. Note () All the rtionl numbers thus cn be represented s finite deciml (terminting type) or s recurring deciml. () The recurring digits from the recurring group re indicted by putting dot bove the first nd lst of them or br bove the recurring group. Recurring Decimls: If in deciml frction, figure or set of figures is repeted continully, then such number is clled recurring deciml. If single figure is repeted, it is shown by putting dot on it. But if set of figures is repeted, we express it either by putting one dot t the strting digit nd one dot t the lst digit of the repeting digits or by plcing br or vinculum on the repeting digit(s). (i) /3 = 0.6666... = 0. 6 = 0. 6 (ii) /7 = 3.48574857 = 3. 4857 = 3. 4857 (iii) 95/6 = 5.83333... = 5. 83 = 5. 83 Pure Recurring Decimls: A deciml in which ll the figures fter the deciml point repet is clled pure recurring deciml. Ex. 0. 6, 3. 4857 etc. Mixed Recurring Decimls: A deciml in which some figures do not repet nd some of them repet is clled mixed recurring deciml. Ex. 5. 8 3 etc. Conversion of Pure Recurring Deciml into frction Rule: Write the recurring figures only once in the numertor nd tke s mny nines in the denomintor s the number of repeting figures. () 0. 6 = 6/9 = /3. () 6. 6 = 6 + 0. 6 = 6 + 6/9 = 6 + /3 = 50/3. To convert Mixed Recurring Deciml into frction Rule: In the numertor, write the difference between the number formed by ll the digits fter deciml point (tking repeted digits only once) nd tht formed by the digits which re not repeted. In the denomintor, write the number formed by s mny nines s there re repeting digits followed by s mny zeroes s in the number of non repeting digits. () 0. 7 = (7 )/90 = 6/90 = 8/45. () 0. 5 4 (3). 53 6 A quick summry () 0. 345 = 345/99999 = (54 )/9900 = 69/550. = + (536 53)/900 = 6 300. () 0. 345 = (345 )/99990 (3) 0. 345 = (345 )/99900 (4) 0. 345 = (345 3)/99000 (5) 0. 345 = (345 34)/90000 (6) 0.345 = 345/00000 E. Let D be deciml of the form, D = 0...., where digits & lie between 0 nd 9. Then which of the following numbers necessrily produces n integer, when multiplied by D? () 8 () 08 (3) 98 (4) 08 IC : PTtkmml0
Sol. It is recurring deciml nd cn be written s D = 0.. To convert this to frction, we cn write it s /99. Thus when the number is multiplied by 99 or multiple of it, we shll necessrily get n integer. Of the given options, only (3) is multiple of 99, hence Ans.(3) Irrtionl numbers Ech non terminting recurring deciml is rtionl number. Thus the number which is non terminting non recurring deciml or more simply the number which cn not be written s frction (i.e. in the form p/q), is clled n irrtionl number. Importnt rules to remember. Addition of two odd numbers gives n even number.. Addition of two even numbers gives n even number. 3. Addition of n odd nd n even number gives n odd number. 4. Subtrction of two odd numbers gives n even number. 5. Subtrction of two even numbers gives n even number. 6. Subtrction of n even nd n odd number gives n odd number. 7. Multipliction of two odd numbers gives n odd number. 8. Multipliction of two even numbers gives n even number. 9. Multipliction of n odd nd n even number gives n even number. E.g. =.44356373095... π = 3.459653589793... log = 0.300999566398... etc. E. Identify the following numbers s rtionl, rel, even, odd, prime, composite, nturl nd irrtionl., 5, 7, 4, 4/3,, 6, 3, 3, 5 Prime numbers Sol. A positive integer which is not equl to nd is divisible by itself nd only is clled prime number. Ex., 3, 5, 7,, 3, 7, 9 etc. Thus, for the prime number 3 there re no fctors besides 3 nd. Composite numbers A positive integer which is greter thn nd is not prime is clled composite number. Thus, composite numbers will necessrily hve fctors other thn nd itself. Imp. Ex. 4, 6, 8, 9, 0,, 4, 5, 6 etc. is neither prime nor composite number. Odd numbers The integers which re not divisible by re clled odd numbers. E.g., 3, 5, 7, 9... Odd numbers re expressed in the form (n + ) where n is ny integer other thn zero (not necessrily prime). Thus,, 3, 9, +7 etc. re ll odd numbers. Even numbers The integers which re divisible by re even numbers. E.g. 0,, 4, 6, 8, 0... Even numbers re expressed in the form n where n is ny integer. Thus, 4, 6, +48 etc. re ll even numbers. Imp. The only even prime number is. Rtionl Rel Even Odd Prime Composite Nturl Irrtionl 5 7 4 4/3 6 3 3 5 The number zero It is symbolised by 0, lso clled cipher or nought which cn be interpreted in the following wys. A crdinl number of n empty (null) set.. A plce holder in rithmeticl computtion. 3. The 'Identity Element' with respect to ddition. i.e. + 0 = 0 + = Multiplying nd dividing by zero. Zero times ny number is equl to zero 0 7 = 7 0 = 0. 0 ( 5) = ( 5) 0 = 0. 8 7 5 0 ( 4) = 0. 7 3. Zero divided by ny non zero number is zero. 0 5 = 0, 0 7 = 0, 0 ( ) = 0 etc. 3. Zero is the only number which cnnot be divisor i.e. division by zero is undefined. IC : PTtkmml0 (3) of (8)
Note () Division by zero is undefined, thus not permitted t ll. () 0/0 is one of the 'indeterminte' forms. Complex numbers A number of the form x + iy, where y 0, is clled complex number, where x nd y re rel numbers nd i = ( ) is the imginry unit clled iot. In the complex number x + iy, x is clled the rel prt nd y is clled the imginry prt. If y = 0, then x + iy becomes rel number x. or exmple, ( 55). Since this cnnot be found in rel terms, hence it is clled complex number. The bsolute vlue of R is defined to be equl to. Thus =. or exmple () If =, then ( ) = =. () If = 0, then ( 0 ) = 0 = 0. We hve some simple but importnt results.. b = b. 3. b = b Let us recll some nottions which re used for certin specific sets. We list them below s 4. b = b N : The set of ll nturl numbers (i.e. positive integers). This is the set. {,, 3,..., n,...} I : The set of ll integers i.e. {..., 3,,, 0,,, 3...} W : The set of ll whole numbers i.e. {0,,, 3,...} Q : The set of rtionl numbers. R : The set of rel numbers. C : The set of ll complex numbers. Properties of rel numbers It must be noted tht ll the properties of rtionl numbers re true for rel numbers lso. (i) (ii) Commuttive property of ddition If nd b re rel numbers, then + b = b +. Associtive property of ddition If, b, c re rel numbers, then ( + b) + c = + (b + c). (iii) Commuttive property of multipliction If nd b re two rel numbers, then b = b. (iv) Associtive property of multipliction If, b, c re rel numbers, then ( b) c = (b c). (v) Distributive lw If, b, c re rel numbers, then (b + c) = b + c. 5. + b + b 6. b b 7. + b b E3. Clculte 999 999 + 999. Sol. 999 999 + 999 = 999 999 + 999 = 999 (999 + )...(Distributive Lw) = 999 (000) = 999000. E4. Simplify 9 54 + 3 58. Sol. 9 54 + 3 58 = 9 54 + 3 9 = 9( 7 + 3 )...(Distributive lw) = 9 (7 + 3)...(Distributive lw) = 58 50 = 900. Absolute vlue or modulus of rel number The bsolute vlue of ny number '' is denoted by. Definition i.e. = R S T, > 0 0, = 0, < 0 e.g. = since = > 0 0 = 0 since = 0 nd = ( ) =, since = < 0. (4) of (8) IC : PTtkmml0
Mini Revision Test # 0 DIRECTIONS: Put tick mrk ginst ech of the correct sttement.. All frctions re not rtionl numbers.. If x nd y re ny two integers, then x y number. is rtionl 3. If x be ny integer, then the rtionl number x is the sme s the integer x. 4. If x nd y re both positive integers, then the rtionl numbers x y nd x y re both negtive. 5. A rtionl number p/q is sid to be in stndrd form if q is positive integer nd the integers p nd q hve no common divisor other thn. 6. 0 50 = 3 65. 7. The rtionl number 7/3 lies to the left of zero on the number line. 8. The rtionl numbers nd re on the opposite 3 sides of zero on the number line. 9. 3 4 <. 7 35 0. If x, y nd z be three rtionl numbers such tht x < y nd y < z, then z < x. or Q. to Q.5: Answer the following questions.. Which rtionl number is the negtive of itself?. Wht is the reltion between two rtionl numbers x nd y to be reciprocl to ech other? 3. Which rtionl number hs no reciprocl? 4. Stte the property used in H I K + G + J = 3 4 H G I K + J + 3 5. The sum of two rtionl numbers is 3/4. One of them is 5 3. ind the other. 4 The concept of multiples nd fctors If X, Y nd Z re three nturl numbers nd X Y = Z, then X nd Y re clled the fctors of Z. Z is sid to be divisible by X nd Y. Z is sid to be multiple of X nd Y. or Exmple The set of positive integers which re fctors of 8 is {,, 3, 6, 9, 8}. Proper fctors A fctor of number other thn nd the number itself is clled proper fctor. Tking the previous exmple, the set of proper fctors of 8 is {, 3, 6, 9}. Tests for divisibility A number is divisible by if its unit s digit is even or zero, e.g. 68, 434, 56 etc. A number is divisible by 3 if the sum of its digits is divisible by 3, e.g. 96, 86, 99 etc. A number is divisible by 4 if the number formed by the lst two right hnd digits is divisible by 4, e.g. 6, 38, 44 etc. A number is divisible by 5 if its unit s digit is either five or zero, e.g. 535, 3970, 45 etc. A number is divisible by 6 if it is divis ible by nd 3, e.g. 484, 84966, 0368 etc. Divisibility by 7 No test upto three digits. The rule which holds good for numbers with more thn 3 digits is s follows. () (b) (c) Group the numbers in three from the right hnd side. Add the odd groups nd even groups seprtely. The difference of the odd nd even groups should be divisible by 7. Ex. Tke number 86. The groups re, 8, 6 Sum of odd groups = + 6 = 7 Sum of even groups = 8 Difference = 595 which is divisible by 7. Hence, the number is divisible by 7. A number is divisible by 8 if the number formed by the lst three right hnd digits is divisible by 8, e.g. 04, 688, 559 etc. A number is divisible by 9 if the sum of its digits is divisible by 9, e.g. 89, 59, 888993 etc. IC : PTtkmml0 (5) of (8)
A number is divisible by 0 if its unit s digit is zero. (6) of (8) e.g. 00, 580, 99990 etc. A number is divisible by when the difference between the sums of digits in the odd nd even plces is either zero or multiple of. Ex. 659989, 099989, 7645 etc. or the number 659989 Sum of the digits t even plces = + 9 + 8 = 8. Sum of the digits t odd plces = 6 + 5 + 9 + 9 = 9. Hence 9 8 =. A number is divisible by if it is divisible by 3 nd 4. e.g. 740, 7068 etc. Divisibility by 3 The rule is sme s tht of 7 with 3 replcing 7 in the divisibility check. Ex. Test the divisibility of the following numbers by 3. (i) 909987 Ans. (i) divisible. (ii) 4766983 Ans. (ii) divisible. A number is divisible by 4 if it is divisible by nd 7 e.g. 566, 354 etc. A number is divisible by 5 if it is divisible by 3 nd 5 e.g. 4745, 8970 etc. A number is divisible by 6 if the number formed by the lst four right hnd digits is divisible by 6. e.g. 579, 579568 etc. A number is divisible by 8 if it is divisible by 9 nd hs its lst digit even. e.g. 598, 73556 etc. A number is divisible by 5 if the number formed by the lst two right hnd digits is divisible by 5. e.g. 05, 3475, 55550 etc. A number is divisible by 5, if the number formed by the lst three right hnd digits is divisible by 5. e.g. 5, 450, 6375 etc. Tip. All these rules must be memorised nd prctised by the students on regulr bsis. A sincere student cn reduce clcultion time by 50%, if ll these rules re put in prctice. Imp. When number with even number of digits is dded to its reverse, the sum is lwys divisible by. e.g. 34 + 43 = 3773 which is divisible by. If x is prime number, then for ny whole number, ( X ) is divisible by x e.g. Let x = 3 nd = 5. Then ccording to our rule 5 3 5 should be divisible by 3. Now (5 3 5) = 0 which is divisible by 3. E5. If n is positive integer (> ), then prove tht n 3 n is divisible by 6. Sol. n 3 n = n(n ) = n(n ) ( n + ). As (n ), n, (n + ) re three consecutive integers with n greter thn, then it should contin fctor of nd 3. Hence, it is lwys divisible by 6. E6. ind P & Q if it is known tht the number 8563P45Q is divisible by 88. Sol. 8563P45Q is divisible by 88 nd 88 = 8 x. Therefore the number should be divisible by 8 nd both. or 8, the lst three digits should be divisible by 8. The only possible 3 digit number hving 4 nd 5 t the hundred's nd the ten's plce is 456. Hence, Q cn hve 6 s the only possible vlue. or ( + 5 + 3 + 4 + 6) (8 + 6 + P + 5) = 0 0 9 P = 0 P =. Mini Revision Test # 0 DIRECTIONS: Answer the following questions.. If 34y5 is divisible by 3, then the lest vlue of y is.... If 34b is divisible by 3 nd 4 both, then, the lest vlue of + b is... 3. If 85y4 is divisible by 6, then, the lest vlue of y is... 4. If 344x is divisible by 8, then, the gretest vlue of x is... 5. If 5g is divisible by 9, then, the lest vlue of g is... Mrk True/lse for ech of these sttements 6. 4765683 is divisible by 3. 7. 579548 is divisible by 6. 8. 74474 is divisible by 8. 9. 3b + b3 is divisible by for ll the positive vlues of nd b. 0. If p is prime number, then for ny whole number, ( p ) is divisible by. + Some fundmentl rules +(+) = + +( ) = (+) = ( ) = + + b c = + (b c) b c = (b + c) b + c = (b c) (+) (+b) = +b IC : PTtkmml0
( ) ( b) = +b ( ) (+b) = b (+) ( b) = b b b = = b = b b = b b g b b = = b b g = b b = b = + b b b = b = b b ( b) = (b ) ( + b) = ( b) ( b) = ( + b) (b ) (c b) = ( b) (b c) Expnsion of (x + ) n for ny positive integer n The following formule re very hndy tools while solving ny type of mthemticl problems. Memorise ech by hert. Imp. ( + b) = + b + b ( b) = b + b ( + b) + ( b) = ( + b ) ( + b) ( b) = 4b b = ( + b) ( b) + b = ( + b) b ( + b) 3 = 3 + 3 b + 3b + b 3 ( b) 3 = 3 3 b + 3b b 3 ( + b) 3 + ( b) 3 = ( 3 + 3b ) ( + b) 3 ( b) 3 = (3 b + b 3 ) 3 + b 3 = ( + b) ( b + b ) 3 b 3 = ( b) ( + b + b ) H G H G H G I K I K I K + J = + + J = + H I K + J G J = 4 or determining the coefficients of the terms in the expnsion of (x + ) n, for ny positive integer n, we cn use the pscl s tringle, which is explined below. Pscl s tringle Power n = n = n = 3 3 3 n = 4 4 6 4 n = 5 5 0 0 5 n = 6 6 5 0 5 6 The tringle is built s shown. Imp. e.g. or n = 4, coefficient 6 = 3 + 3, coefficient 4 = + 3. or n = 6, coefficient 6 = + 5, coefficient 5 = 5 + 0, coefficient 0 = 0 + 0. Coefficients. or the expnsion of (x + ) n, the coefficients re positive ll through.. or the expnsion of (x ) n, the coefficients of the terms re lterntively positive nd negtive with the first term positive while the numericl vlues of the coefficients re the sme s tht of (x + ) n. Squres The second power of number is clled the squre of tht number. In other words the squre of number is the product of the number with the number itself. The following tble gives the squres of ll nturl numbers from to 30. Number Squre Number Squre 6 56 4 7 89 3 9 8 34 4 6 9 36 5 5 0 400 6 36 44 7 49 484 8 64 3 59 9 8 4 576 0 00 5 65 6 676 44 7 79 3 69 8 784 4 96 9 84 5 5 30 900 rom the tble it is cler tht nturl numbers like, 4, 9, 6, 5,..., 79, 784, 84, 900,... re squres of nturl numbers but the numbers 40, 60 nd 80 re not. IC : PTtkmml0 (7) of (8)
A given number is perfect squre, if it is expressed s product of pirs of equl fctors.. A nturl number hving, 3, 7 or 8 in the unit s plce is never perfect squre (or squred number). 7, 3, 8, re not perfect squres.. The squre of n even number is lwys n even number. (8) of (8) = 4, 6 = 36, 0 = 00, = 44. 3. The squre of n odd number is lwys n odd number. 3 = 9, 7 = 49, 3 = 69, 5 = 5. 4. The number of zeroes t the end of perfect squre is never odd. 00, 400, 3600, 640000 re perfect squres nd 000, 4000, 6400000 re not perfect squres. 5. The squre of nturl number n is equl to the sum of the first n odd numbers. = = sum of the first odd number. = + 3 = sum of the first odd numbers. 3 = + 3 + 5 = sum of the first 3 odd numbers. 6. or every nturl number n, (n + ) n = (n + + n) (n + n) = (n + ) + n 4 3 = (3 + ) + 3 = 7. 6 5 = (5 + ) + 5 = 3. 7. A perfect squre (other thn ) is either multiple of 3 or exceeds multiple of 3 by. 49 = (7) = 3 6 +, 69 = (3) = 3 56 +. 8. A perfect squre (other thn ) is either multiple of 4 or exceeds multiple of 4 by. 8 = (9) = 4 0 +. 44 = () = 4 0 +. Squre roots We know tht 6 is the squre of 4. It cn lso be stted in other words tht 4 is the squre root of 6. Similrly, 5 is the squre root of 5 nd 6 is the squre root of 36 etc. We use the rdicl sign for the positive squre root. Thus 6 = 4, 5 = 5, 8 = 9 etc. We lso know tht 4 4 = 6, 9 9 = 8, 5 5 = 5. i.e. 4 is lso squre root of 6, 9 is lso squre root of 8 nd 5 is lso squre root of 5. It shows tht every number hs two squre roots, one positive nd the other negtive. Thus, Squre root of 6 = ±4. Squre root of 5 = ±5. Squre root of 8 = ±9. Note The symbol stnds for positive squre root s stted erlier. When we wnt to know both the squre roots, we put ± sign before the symbol. Thus ± 6 = ±4 but 6 = 4. Methods of finding squre root There re two methods for clculting the squre root of numbers (i) (ii) Prime fctoristion method Long division method The first method is used only when the given number is smll whole number wheres the second method cn be used for ny number. By ctoristion In this method, we (i) brek up the number into its prime fctors, (ii) mke the pirs of similr fctors nd (iii) tke one number from ech pir nd then multiply them. E7. ind the squre root of 444. Sol. 444 7 9 36 9 9 We hve, 444 = 9 9 444 = 9 = 38. By long division method The exmple given will illustrte the use of this method. Let s find 96 by this method. In this method, we (i) (ii) Divide the number into pirs of two digits beginning with the unit s digit. Think of whole number whose squre is either or just less thn. Obviously it is 3. Tke 3 s the divisor. Squre it nd put it below. Write 3 in the nswer portion. (iii) ind the first reminder nd bring down the next pir of digits i.e., the dividend is now 396. (iv) Use twice of 3 i.e., 6 s the tens digit of the next tril divisor. (v) Now we hve to think of number which used s unit with 6 will, fter multipliction with itself, be either 396 or just less thn 396. On tril (i.e., 39 6 = 6 +...), it is 6. (vi) The next divisor is 66 which when multiplied by 6 will give 396. (vii) Put 396 below 396 nd subtrct. The reminder is 0. (viii) Put 6 on the right of 3 in the nswer portion. 96 = 36. IC : PTtkmml0
E8. ind the squre root of 664 by long division method. 0 8 Sol. 6 6 4 08 0664 664 0 664 = 08. E3. A piece of lnd is in the form of isosceles right tringle. If the length of the longest side of the lnd is 98. m, find the perimeter of the lnd correct upto two deciml plces. Sol. Let the equl sides of the tringulr lnd be x m long ech. Then, x + x = (98.) 98. m As you cn see, the procedure is very long nd hence lot of prctice nd speed will hve to be gined to get commnd over this method. i.e. x = (98.) = 9643.4 i.e. x = 48.6. E9. ind the squre root of +. i.e. x = 48. 6 = 69. 44. The perimeter of the lnd, therefore, equls = 69.44 + 69.44 + 98. = 37.08 m. Sol. + = e + je + j = + e je j e + j = +. E0. ind the squre root of the following nd leve it in the product form 0 3 6 5 7 8. 0 6 8 Sol. ( 3 5 7 ) 5 3 4 6 = ( ) ( 3 ) 5 ( 7 ) ( ) = 5 3 3 5 7 4 6. E. ind the squre root of (/4) (/49) (5/). Sol. 4 5 = = =. 49 7 5 7 5 70 E. ind the squre root of 6500. 6 Sol. ( 6500 ) = 5 = 5 3 = 5 5 5 = 50. Appliction of Squres nd Squre roots We shll now tke up some problems wherein we need to find the squres nd the squre roots of numbers. E4. By wht lest number should we multiply 9900 so tht it becomes perfect squre? Sol. 9900 4950 3 475 3 85 5 75 5 55 9900 = 3 3 3 5 35 After mking pirs of similr fctors we find tht does not mke pir. 9900, if multiplied by, will become perfect squre. E5. Wht lest number should be subtrcted from 5634 so tht the resulting number becomes perfect squre? Sol. 75 7 5634 49 45 734 75 9 9 is to be subtrcted. IC : PTtkmml0 (9) of (8)
E6.The re of squre field is 00 sq. m. ind its side. Sol. Are = (side) = 00 sq m. Side = 00 m = 0 m. E7.A generl, trying to rrnge his men numbering 76674 into perfect squre formtion, found tht there were men less. How mny men were there in the front row? Sol. The number of men in the front row = (76674 + ) Cubes 76676. 5 6 5 7 6 6 7 6 5 0 66 04 046 676 676 0 56 men were there in the front row. rom the tble we get the following properties of cubes of numbers. (i) (ii) (iii) (iv) Cubes of ll odd nturl numbers re odd. Cubes of ll even nturl numbers re even. The cube of nturl number which is multiple of 3 is multiple of 7. The cube of nturl number which is of the form 3n + (e.g., 4, 7, 0,...) is lso number of the form 3n +. (v) The cube of nturl number which is of the form 3n + (e.g., 5, 8,,...) is lso number of the form 3n +. Cube roots We hve seen bove tht 5 is the cube of 5. It cn be stted in other words tht 5 is the cube root of 5. Similrly, from the tble given bove we cn sy tht 8 is the cube root of 5 nd 0 is the cube root of 000. The symbol used for cube root is 3 3. Thus 79 mens cube 3 root of 79 nd 64 mens cube root of 64. 3 is clled rdicl, 79 is clled rdicnd nd 3 is clled index. Rule for finding cube root We resolve the given number into prime fctors nd tke the product of prime fctors choosing one out of three of the ech type of prime fctors. We know tht 3 3 = 3 3 3, 5 3 = 5 5 5, 7 3 = 7 7 7, 3 =. Here, 3 is clled the third power of. The third power of is lso clled the cube of. The following tble gives the cubes of ll nturl numbers from to 30. (0) of (8) Number Cube Number Cube 6 4096 8 7 493 3 7 8 583 4 64 9 6859 5 5 0 8000 6 6 96 7 343 0648 8 5 3 67 9 79 4 384 0 000 5 565 33 6 7576 78 7 9683 3 97 8 95 4 744 9 4389 5 3375 30 7000 E8.ind the cube root of 64. 64 Sol. 3 6 8 4 We hve, 64 = 3 So, ( 64 ) = = 4 E9.ind the cube root of 3375. Sol. 3 3375 3 5 3 375 5 5 5 5 5 5 We hve, 3375 = 3 3 3 5 5 5 3 ( 3375 ) = 3 5 = 5. IC : PTtkmml0
E0. By wht lest number should we multiply 500 so tht it becomes perfect cube? Mini Revision Test # 03 Sol. 500 50 5 5 5 5 5 5 500 = 5 5 5 After mking pirs of similr fctors we find tht does not mke group of 3. 500, if multiplied by, will become perfect cube. E. By wht lest number should we divide 500 so tht it becomes perfect cube? Sol. 500 50 5 5 5 5 5 5 500 = 5 5 5 After mking pirs of similr fctors we find tht does not mke group of 3. 500 should be divided by i.e. by 4, to mke it perfect cube. Power cycles, 4, 8, 6, 3, 64, 8, 56, 5,... The series is nothing but powers of in scending order. If you see it crefully we see tht there is repetition of the unit digit fter regulr intervls or unit digits lwys hve definite pttern. The unit digit lwys follows cycle which is termed s power cycle. or exmple, power cycle of hs frequency of 4 where s 5 hs frequency of. Let us exmine the power cycle of ech number. : every time the number in the unit digit will lwys be. DIRECTIONS: Answer the following questions.. Without ctul squring, find the vlue of 589 588.. Without ctul dding, find the sum of + 3 + 5 + 7 + 9 + + 3 + 5 + 7 + 9 + + 3. 3. Is the number 4096 perfect squre? 4. ind the positive squre root of 45. 5. ind the positive squre root of 3/98. 6. Simplify : 565 + 44. 565 44 7. By wht smllest number 600 should be multiplied so tht the product becomes perfect cube? 8. By wht smllest number 048 should be divided so tht the quotient becomes perfect cube? 9. ind the cube root of 5.65. 0. ind the cube root of 0 7. ce vlue & Plce vlue Let's consider number like 598. Ech of the digits like 5,,, 9 nd 8 re t different positions or plce. The digit 8 (fce vlue) is t the unit's position nd hs plce vlue of. Similrly digit 9 (fce vlue) is t the ten's position nd hs plce vlue of 0. Hence, the given number cn be written s 5 0 4 + 0 3 + 0 + 9 0 + 8 0 0 If we tke term from the bove expression, sy 5 0 4, the first prt (5) represents the fce vlue of the number nd the second prt (0 4 ) represents the plce vlue of the number. 0 4 0 3 0 0 0 0 :, 4, 8, 6. After this, repetition of cycle gin strts. 3 : 3, 9, 7, 4 : 4, 6 5 : 5 6 : 6 7 : 7, 9, 3, 8 : 8, 4,, 6 9 : 9, Let us sy if question is sked wht is the digit t units plce in 7 7. As 7 hs power cycle with frequency 4, when we divide 7 by 4, the reminder is 3 nd the third power of 7 hs unit digit s 3, which is the nswer. Ten thousnd Thousnds Hundreds Tens Units 5 9 8 IC : PTtkmml0 () of (8)
SOLVED EXAMPLES LECTURE # 0 The following questions will help you get thorough hold on the bsic fundmentls of the chpter nd will lso help you develop your concepts. The following questions shll be covered in the clss by the techer. It is lso dvised tht sincere student should solve ech of the given questions t lest twice to develop the required expertise. DIRECTIONS: or ech of the following questions, plese give the complete solution.. The lrgest nturl number by which the product of three consecutive even nturl numbers is lwys divisible, is. There is one number which is formed by writing one digit 6 times (e.g.,, 444444 etc.). Such number is lwys divisible by. 76 + 40 =? 3. If 85 = 35, then the vlue of. The gretest prime number which exctly divides 86, is 3. How mny prime numbers re of the form 0n +, where n is whole number such tht n 9? 4. The difference between the gretest number nd the smllest number of 5 digits using 0,,, 3, 4 using ll but once, is ( 85 + 8. 5 +. 85 + 0. 085 ) is 4... 0 is equl to 5. Wht is the squre root of H G I 4 K J H G I 49 K J H G I K 5 J? 6 5. If 3 =.73, then the vlue of is 6. A certin number of persons gree to py s mny rupees s the n umber of pers ons. Th e mon ey r eceived is Rs.58449. How mny persons re there? 6. 0. 8. 0484. 0064 6. 5 =? 7. Wht is the smllest number by which 504 should be multiplied, so tht it becomes perfect squre? 7. Express the following numbers in deciml number form. () 39/4 () 3/6 (3) 3/ (4) 37/5 8. If 4 8 is multiple of 9, then the digit represented by is 9. When n is divided by 4, the reminder is 3. Wht is the reminder when n is divided by 4? 8. The sum of three numbers is 3. If the first number be twice the second nd third number is one third of the first, then the second number is 9. Wht should be multiplied with 507 so tht the product is? 0. Wht lest number must be dded to 056 to get number exctly divisible by 3? 0. By wht lest number should we multiply 8400, so tht it becomes perfect squre? () of (8) IC : PTtkmml0
Answers Mini Revision Tests Success is where preprtion nd opportunity meet... Answers to Mini Revision Test # 0. lse. lse 3. True 4. True 5. True 6. True 7. lse 8. True 9. lse 0. lse. Zero. x.y = or x = /y 3. Zero 4. Associtive Lw 5. 9/ Answers to Mini Revision Test # 0. Zero. 3 3. 4. 8 5. 0 6. True 7. lse 8. True 9. True 0. lse Answers to Mini Revision Test # 03. 77. 44 3. Yes 4. 65 5. 4/7 6. 6/9 7. 5 8. 4 9..5 0. 3 Solutions Solved Exmples Lecture # 0. Since is divisible by ech one of 7, nd 3, so numbers like 666666 etc. re lso divisible by 3, 7, nd 3 s 666666 = 6().. 40 = 7 343 = 7 7 7 7 = c 7 7 h = 49. 76 + 40 = 76 + 49 = 5 = 5. 3. 85 + 8. 5 + 85. + 0.085 85 85 = 85 + + + 00 0000 85 85 = 85 + + + 0 00 H G 35 35 35 = 35 + + + 0 00 000 I K J 85 000000 85 000 = (35 + 3.5 +.35 +.35) = 49.985. 4... 0 = = 00 00 00 H I K = G J =. 0 0 00 5. = 3 = 3 =.73 = 3.464. 6. Sum of deciml plces in the numertor nd the denomintor being the sme, we my remove the deciml plces. Given Exp. = 8 484 9 99 = = = 0.99. 64 65 8 5 00 7. 39 4 =.65, 3 = 0.7, 3 6 = 0.875, 37 = 5.48. 5 8. Replce by k. Sum of digits = (4 + + k + 8) = (4 + k). Now, the lest vlue of k for which (4 + k) is divisible by 9 is k = 4. 9. When n is divided by 4, let the quotient = k & the reminder = 3. n = 4k + 3 n = 8k + 6 = (8k + 4) + = 4(k + ) +. Thus, when n is divided by 4, the quotient is (k + ) nd the reminder is. 0. 8400 400 5 00 5 40 84 4 3 7 7 8400 = 5 5 3 7 8400, if multiplied by 3 7 = will become perfect squre.. Let n, (n + ), (n + 4) be 3 consecutive even nturl numbers. The lest vlue of n is. or this vlue of n, we hve n(n + ) (n + 4) = 4 6 = 48. The required number is 48. IC : PTtkmml0 (3) of (8)
. 43 is divisible by 7. So, it is not prime. Clerly, 59 is prime number nd divides 86. 3. Putting n =,, 3,..., 9 we find tht the requisite prime numbers re, 3, 4, 6, 7. They re 5 in number. 4. Required difference = (430 034) = 3976. 5. 5 4 4 = = = 4 49 6 7 5 7 5 35. 6. Number of persons Rupees = 58449. Here the number of persons = Rupees given Number of persons = 58449 = 607. 7. 504 = 3 3 7. The smllest number to be multiplied to mke it perfect squre = 7 = 4. 8. Let the numbers be x, y, z. Then, x = y nd z = 3 x. So, the numbers re x, x nd x 3 x + x + x = 3 x = 7. 3 So, second number = 9. Let 507 X = X = = 73. 507 x = 36. 0. On dividing 056 by 3, the reminder =. The number to be dded = (3 ) =. Short cut: 59 is the squre of 3 nd 058 is its double nd hence is required. (4) of (8) IC : PTtkmml0