Fast Track Objective Mathematics PDF

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Sales & Support Offices Agra, Ahmedabad, Bengaluru, Bhubaneswar, Chennai, Delhi(I&II),Guwahati, Haldwani, Hyderabad,Jaipur, Kolkata, Kota, Lucknow, Nagpur, Meerut & Pune ISBN : 978-93-5176-014-6 Typeset by Arihant DTP Unit at Meerut PRINTED & BOUND IN INDIA BY ARIHANT PUBLICATIONS (INDIA) LTD. For further information about the products from Arihant, log on to www.arihantbooks.com or email to [email protected] Fast Track objective Arithmetic Today, there is a plethora of books available in the market on Objective Arith- metic which seems to be complete in their way, but are still unable to fully satisfy the aspirants. LET US KNOW SOME OF THE REASONS Lack of Understanding the Basic Concepts Mostly, students face a competitive examination on the base of their knowledge about mathemat- ical rules, formulae and concepts. Inspite of having the knowledge, he lacks behind when he faces questions in the examination. Does he realise this inability? Yes, he does but feels confused and blocked when he is unable to solve them and is left with a sense of grudge that he could solve it. The only reason behind this problem is the understanding of basic concepts. If he would have been clear with them, he could solve any of the questions because as a matter of fact, every ques- tion is based on a particular concept which is just twisted in the examinations to judge the overall ability of a student. Inappropriate Use of Short Tricks This is the second biggest problem in front of the aspirants. The number of questions asked in the competitive examination is much more than the time assigned for diem. This leads die aspirants to use shortcut mediods. Although, diese mediods prove to be beneficial in some cases, but due to time management problems, he gets bound to use these methods irrationally and inappropriately. As a result, he jumbles between all the shortcuts which lead to wrong answers which could have been solved if he knew when and where to apply the shortcut methods. Inability to Distinguish Between the Applications of Formulae We all are aware of the amount of stress and pressure a competitive examination creates on the mindset of an aspirant. Succumbed to such pressure, an aspirant is unable to decide the appro- priate formula to be applied in a particular step. During the crisis of time, such confusion adds to the problems and squeezes in more time and results to an unsatisfactory score. Keeping in mind all kinds of problems faced by an aspirant in a competitive examination, we have developed diis book with profound interest in a step-wise method to encounter all your queries and worries. This book named 'FAST TRACK ARITHMETIC is wordiy to fulfill your expectations and will help you as loyal guide throughout. OUTSTANDING QUALITIES OF FAST-TRACK OBJECTIVE ARITHMETIC Use of Fundamental Formulae and Method In this book, all the fundamental formulae and methods have been presented in such a striking yet friendly and systematic manner that just going through them once will give you an effective grasp. They have been present in such a manner that they, will never let you get confused between fast track technique and basic method. Appropriate Short Cut Methods An important feature of this book is its short cut methods or tricks given in the name of "fast track formulae or techniques". Each technique is given with its basic or fundamental method. So, that a student can use these tricks according to their desire and save there precious time in exams. Division of Exercises According to the Difficulty Level Based on the standard and level of difficulty of various questions, the exercises are divided into two parts i.e., 'Base level exercise' for relatively easier questions and 'Higher skill level exercise' for difficult ques- tions. 'Multiconcept questions' which requires a use of different concepts in a single question have also been incorporated with important chapters. Special Emphasise on Geometry, Trigonometry and Mensuration Now-a-days, Questions from geometry, trigonometry and mensuration are asked in large numbers in dif- ferent exams. So, a large variety and number of questions are provided for tiiese chapters. Completely Updated with Questions from Recent Exams This book is incorporated with die questions from all the recent competitive exams, held in year 2013-14. This book is a brain child of Mr Deepesh Jain, Director, Arihant Publications (India) Limited. Richa Agar- wal, Diwakar Sharma and Shivam Mittal have given their best and sincere efforts for die completion and final presentations of die book. The entire project has been managed and supervised by Mr Mahendra Singh Rawat and Mr Amit Verma. Aas Mohammed and Pradeep are to be complemented for very apt designing to the book cover. Amit Bansal and Mayank Saini have given their expertise in the layout of the book. Everyone's contribution for tiiis book is very special and is worthy of great applause. Reader's recommendations will be highly treas- ured. With best compliments Rajesh Verma Contents 1. Number System Numerals Face Value and Place Value of the Digits in a Number Types of Numbers Operations on Numbers Divisibility Tests Unit's Place of an Expres- sion Basic Number Theory 2. Number Series Types of Series Types of Questions Asked on Number Series 3. HCF and LCM Factors and Multiples Least Common Multiple (LCM)« Highest Common Factor (HCF) Method to Calculate LCM and HCF of Fractions Fast Track Techniques to Solve the Questions Method to Solve Questions Based on Bells 4. Simple and Decimal Fractions Simple Fraction Decimal Fraction Operations on Simple Fractions Operations on Decimal Fractions Comparison of Simple Fractions Fast Track Formulae to Solve die Questions 5. Square Root and Cube Root Square Square Root Properties of Squares and Square Roots Fast Track For- mulae to Solve the Questions Fast Track Techniques to Solve the Questions Cube Cube Root Properties of Cube and Cube Roots 6. Indices and Surds Indices Surds Properties of surds Operations on Surds Fast Track Techniques to Solve the Questions 7. Simplification VBODMAS Rule Basic Formulae 8. Approximation Basic Rules to Solve the Problems by Approximation 9. Word Problems Based on Numbers Types of Word Problems Based on Numbers 10. Average Average Properties of Average Important Formulae Related to Average of Numbers Fast Track Techniques to Solve the Questions Average Speed 11. Percentage Percentage Formulae to Calculate Percentage Fast Track Techniques to Solve the Questions 12. Profit and Loss Basic Formulae Related to Profit and Loss Fast Track Techniques to Solve die Ques- tions 13. Discount Marked Price Basic Formulae Related to Discount Successive Discount Fast Track Techniques to Solve die Questions 14. Simple Interest Simple Interest (SI) Instalments Fast Track Techniques to Solve die Questions 15. Compound Interest Basic Formulae Related Compound Interest Instalments Fast Track Techniques to Solve die Questions 16. True Discount and Banker's Discount True Discount Fast Track Formulae to Solve die Questions Banker's Discount 17. Ratio and Proportion Ratio Comparison of Ratios Proportion Fast Track Techniques to Solve die Questions 18. Mixture or Alligation Mixture Rule of Mixture or Alligation Fast Track Techniques to Solve die Questions 19. Partnership Types of Partnership Types of Partners Fast Track Techniques to Solve the Questions 20. Unitary Method Direct Proportion Indirect Proportion 21. Problem Based on Ages Important Rules for Problem Based on Ages Fast Track Techniques to Solve the Ques- tions 22. Work and Time Basic Rules Related to Work and Time Fast Track Techniques to Solve the Questions 23. Work and Wages Important Points Fast Track Formulae to Solve die Questions 24. Pipes and Cisterns Important Points Fast Track Techniques to Solve the Questions 25. Speed, Time and Distance Basic Formulae Related to Speed, Time and Distance Fast Track Techniques to Solve the Questions 26. Problems Based on Trains Basic Rules Related to Problems Based on Trains Fast Track Techniques to Solve the Questions 27. Boats and Streams Basic Formulae Related to Boats and Streams Fast Track Techniques to Solve die Ques- tions 28. Races and Games of Skill Important Terms Some Facts about Race Fast Track Techniques to Solve die Ques- tions 29. Clock and Calendar Clock Important Points Related to Clock Fast Track Techniques to Solve die Questions Calender Ordinary Year Leap Year Odd Days 30. Linear Equations Linear Equations in One, Two and Three Variables Metiiods of Solving Linear Equa- tions Consistency of the System of Linear Equations 31. Quadratic Equations Important Points Related to Quadratic Equations Methods of Solving Quadratic Equations Fast Track Formulae to Solve die Questions 32. Permutations and Combinations Permutation Cases of Permutation Combination Cases of Combination Factorial Fundamental Principles of Counting Fast Track Formulae to Solve die Questions 33. Probability Terms Related to Probability Event Rules/Theorems Related to Probability. Types of Questions 34. Area and Perimeter Area Perimeter Triangle Properties of Triangle Quadrilateral Regular Polygon Circle Fast Track Techniques to Solve the Questions 35. Volume and Surface Area Volume Surface Area Cube Cuboid Cylinder Cone Sphere Prism Pyramid Fast Track Techniques to Solve the Questions 36. Geometry Point Line Angle Triangle Congruency of Triangles Similarity of Triangles Quadrilateral Polygons Circle 37. Coordinate Geometry Rectangular Coordinate Axes Quadrants Distance Formula Basic Points Related to Straight Lines 38. Trigonometry Measurement of Angles Relation between Radian and Degrees Trigonometric Ratios Trigonometric Identities Sign of Trigonometric Functions Trigonometric Ratios of Combined Angles 39. Height and Distance Line of Sight Horizontal Line Angle of Elevation Angle of Depression 40. Data Table 41. Pie Chart 42. Bar Chart Types of Bar Chart 43. Line Graph Types of Line Graph 44. Mixed Graph 45. Data Sufficiency Fast Track Practice Sets Chapter 01 Number System A system in which we study different types of numbers, their relationship and rules govern in them is called as number system. In the Hindu-Arabic system, we use the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. These symbols are called digits. Out of these ten digits, 0 is called an insignificant digit whereas the others are called significant digits. Numerals A mathematical symbol representing a number in a systematic manner is called a numeral repres- ented by a set of digits. How to Write a Number To write a number, we put digits from right to left at the places designated as units, tens, hundreds, thousands, ten thousands, lakhs, ten lakhs, crores, ten crores. Let us see how the number 308761436 is denoted It is read as Thirty crore eighty seven lakh sixty one thousand four hundred and thirty six. Face Value and Place Value of the Digits in a Number Face Value In a numeral, the face value of a digit is the value of the digit itself irrespective of its place in the numeral. For example In the numeral 486729, the face value of 8 is 8, the face value of 7 is 7, the face value of 6 is 6, the face value of 4 is 4, and so on. Place Value (or Local Value) In a numeral, the place value of a digit changes according to the change of its place. Look at the following to get the idea of place value of digits in 72843016. It is clear from the above presentation that to obtain the place value of a digit in a numeral, we multiply the digit with the value of its place in the given numeral. Types of Numbers 1. Natural Numbers Natural numbers are counting numbers. They are denoted by N. For example N = {1,2,3,...}. ♦ All natural numbers are positive. ♦ Zero is not a natural number. Therefore, 1 is the smallest natural number. 2. Whole Numbers All natural numbers and zero form the set of whole numbers. Whole numbers are denoted by W. For example W = {0,1,2,3,...} ♦ Zero is the smallest whole number. Whole numbers are also called as non-negative integers. 3. Integers Whole numbers and negative numbers form the set of integers. They are denoted by/. For example / = {...,-4,-3,-2,-1,0,1,2,3,4,...} Integers are of two types. (i) Positive Integers Natural numbers are called as positive integers. They are denoted by I. For example I+ = {1,2,3,4,...} (ii) Negative Integers Negative of natural numbers are called as negative integers. They are denoted by I~. For example I~ ={-1,-2,-3,-4,...} ♦ '0' is neither +ve nor -ve integer. 4. Even Numbers A counting number which is divisible by 2, is called an even number. For example 2, 4, 6, 8, 10, 12,... etc. ♦ The unit's place of every even number will be 0, 2, 4, 6 or 8. 5. Odd Numbers A counting number which is not divisible by 2, is known as an odd number. For example 1, 3, 5, 7, 9, 11, 13, 15, 17, 19,... etc. ♦ The unit's place of every odd number will be 1, 3, 5, 7 or 9. 6. Prime Numbers A counting number is called a prime number when it is exactly divisible by, 1 and itself. For example 2, 3, 5, 7, 11, 13,... etc. ♦ 2 is the only even number which is prime. ♦ A prime number is always greater than 1. ♦ 1 is not a prime number. Therefore, the lowest odd prime number is 3. ♦ Every prime number greater than 3 can be represented by 6n + 1, where n is integer. 7. Composite Numbers Composite numbers are non-prime natural numbers. They must have atleast one factor apart from 1 and itself. For example 4, 6, 8, 9, etc. ♦ Composite numbers can be both odd and even. ♦ 1 is neither a prime number nor composite number. 8. Coprimes Two natural numbers are said to be coprimes, if their HCF is 1. For example (7, 9), (15, 16) ♦ Coprime numbers may or may not be prime. 9. Rational Numbers A number that can be expressed as p/q is called a rational number, where p and q are inteqers and a * 0. 10. Irrational Numbers The numbers that cannot be expressed in the form of p/q are called irrational numbers, where p and q are integers and q * 0. For example -J2, V3, -Jl, VTT etc. ♦ 7C is an irrational number as 22 / 7 is not the actual value of n but it is its nearest value. ♦ Non-periodic infinite decimal fractions are called as irrational number. 11. Real Numbers Real numbers include rational and irrational numbers both, ♦ Real numbers are denoted by R. Operations on Numbers Addition When two or more numbers are combined together, then it is called addition. Addition is de- noted by ' + ' sign. For example 24 + 23 + 26 = 73 Subtraction When one or more numbers are taken out from a larger number, then it is called subtraction. Subtraction is denoted by '-' sign. For example 100-4-13=100-17=83 Division u is the dividend and d is the divisor. A number which tells how many times a divisor (d ) exists in dividend D is called the quotient Q. If dividend D is not a multiple of divisor d, then D is not exactly divisible by d and in this case remainder R is obtained. Let us see the following operation of division Multiolication When 'a! is multiplied by '£>', then 'a' is added 'b' times or 'b' is added 'a' times. It is denoted by 'x'. Let us see the following operation on Multiplication If a =2 and b = 4, then 2x4=8 or (2+ 2+ 2+ 2) =8 Here, 'a! is added 'b' times or in other words 2 is added 4 times. Similarly, 4x2=8 or (4+ 4) =8 In this case, 'b' is added 'a! times or in other words 4 is added 2 times. Divisibility Tests Divisibility by 2 When the last digit of a number is either 0 or even, then the number is divisible by 2. For example 12, 86, 472, 520, 1000 etc., are divisible by 2. Divisibility by 3 When the sum of the digits of a number is divisible by 3, then the number is divisible by 3. For example (i) 1233 1 + 2 + 3 + 3 = 9, which is divisible by 3, so 1233 must be divisible by 3. (ii) 156 1 + 5 + 6 = 12, which is divisible by 3, so 156 must be divisible by 3. Divisibility by 4 When the number made by last two-digits of a number is divisible by 4, then that particular number is divisible by 4. Apart from this, the number having two or more zeroes at the end, is also divisible by 4. For example (i) 6428 is divisible by 4 as the number made by its last two digits i.e., 28 is divisible by 4. (ii) The numbers 4300, 153000, 9530000 etc., are divisible by 4 as they have two or more zeroes at the end. Divisibility by 5 Numbers having 0 or 5 at the end are divisible by 5. For example 45, 4350, 135, 14850 etc., are divisible by 5 as they have 0 or 5 at the end. Divisibility by 6 When a number is divisible by both 3 and 2, then that particular number is divisible by 6 also. For example 18, 36, 720, 1440 etc., are divisible by 6 as they are divisible by both 3 and 2. Divisibility by 7 A number is divisible by 7 when the difference between twice the digit at ones place and the number formed by other digits is either zero or a multiple of 7. For example 658 is divisible by 7 because 65 - 2 X 8 = 65 - 16 = 49. As 49 is divisible by 7, the number 658 is also divisible by 7. Divisibility by 8 When the number made by last three digits of a number is divisible by 8, then the number is also divisible by 8. Apart from this, if the last three or more digits of a number are zeroes, then the number is divisible by 8. For example (i) 2256 As 256 (the last three digits of 2256) is divisible by 8, therefore 2256 is also divisible by 8. (ii) 4362000 As 4362000 has three zeroes at the end. Therefore it will definitely divisible by 8. Divisibility by 9 When the sum of all the digits of a number is divisible by 9, then the number is also divisible by 9. For example (i) 936819 9+3 + 6 + 8 + 1 + 9= 36 which is divisible by 9. Therefore, 936819 is also divisible by 9. (ii) 4356 4 + 3 + 5 + 6 = 18 which is divisible by 9. Therefore, 4356 is also divisible by 9. Divisibility by II When a number ends with zero, then it is divisible by 10. For example 20, 40, 150, 123450, 478970 etc., are divisible by 10 as these all end with zero. Divisibility by 1' When the sums of digits at odd and even places are equal or differ by a number divisible by 11, then the number is also divisible by 11. For example (i) 2865423 Let us see Sum of digits at odd places (A) = 2 + 6+4 + 3 = 15 Sum of digits at even places (B) = 8 + 5 + 2 = 15 =>A = B Hence, 2865423 is divisible by 11. (ii) 217382 Let us see Sum of digits at odd places (A) = 2 + 7 + 8 = 17 Sum of digits at even places (B) = 1 + 3 + 2 = 6 A- B = 17-6 = 11 Clearly, 217382 is divisible by 11. Divisibility by 12 A number which is divisible by both 4 and 3 is also divisible by 12. For example 2244 is divisible by both 3 and 4. Therefore, it is divisible by 12 also. Divisibility by 14 A number which is divisible by both 7 and 2 is also divisible by 14. For example 1232 is divisible by both 7 and 2. Therefore, it is divisible by 14 also. Divisibility by 1! A number which is divisible by both 5 and 3 is divisible by 15 also. For example 1275 is divisible by both 5 and 3. Therefore, it is divisible by 15 also. Divisibility by II A number is divisible by 16 when the number made by its last 4-digits is divisible by 16. For example 126304 is divisible by 16 as the number made by its last 4-digits i.e., 6304 is divisible by 16. Divisibility by 11 A number is divisible by 18 when it is even and divisible by 9. For example 936198 is divisible by 18 as it is even and divisible by 9. Divisibility by 25 A number is divisible by 25 when its last 2-digits are either zero or divisible by 25. For example 500, 1275, 13550 are divisible by 25 as last 2-digits of these numbers are either zero or divis- ible by 25. Divisibility by 125 A number is divisible by 125 when the number made by its last 3-digits is divisible by 125. For example 630125 is divisible by 125 as the number made by its last 3-digits are divisible by 125. To Find a Number Completely Divisible by Given Number Consider a number xr which is when divided by d, gives a quotient q and leaves a remainder r. Then, To find the number which is completely divisible by d such that remainder r is zero, follows the example given below. Ex. 1 Find the number, which on (1) addition (2) subtraction from the number 5029 is completely divisible by 17. Sol. Dividing 5029 by 17 we find. Remainder = 14 1. The minimum number on adding of which the given number is completely divisible by 17 = Divisor - Remainder = 17 - 14 = 3. 2. The minimum number on subtraction of which the given number is completely divisible by 17 = Re- mainder = 14. Unit's Place of an Expression Given expression can be of following two types 1. When Number is Given in the form of Product of Number To find the units digit in the product of two or more number we take units digit of every numbers and then multiply them. Then, the unit digit of the resultant product is the units digit of the product of original num- bers. For example 207x781x39x94 Taking units digit of every number and then multiplying them = 7x1x9x4 = 7x36 [taking units place digit] Again, taking units digit and then multiplying = 7x6=42.-. Units digit for 207 x 781x39 x 94 is 2. 2. When Number is Given in the form of Index ♦ If the unit's digit number are 0,1, 5 or 6, then the resultant unit's digit remains same. For example, (576)1151, its units digit is 6. (155)120, its unit digit is 5. (191)19, its unit digit is 1. (900)51, its unit digit is 0. ♦ If units place is 2, then the power of the number is first divided by 4 and there after represented in the form of 2. For example (572)443 (—) i.e., (2)443 = (2) ^ 4 'taking units place digit = (24)110 x23 =24 x23 [v (24)110 =24] = 16 x 8 taking units place digit Units place is 8 ♦ In the same way, if units place digit are 4 or 8, then units digit for 4 and 8=6. e.g., (124)372 taking units place digit = (4)372 = (44 )93 =44 =256 Unit's place digit = 6 ♦ If units digit is 3 or 7, then units digit for 3 and 7 = 1. For example (2467)153 taking units place = (7)153 => (74)38 xf => 74 X71 =72 x72 x7 =49 x49 x7 taking unit's place digit =9x9x7 = 81x7 taking unit's place digit = 7 Unit's digit = 7 ♦ If units place is 9 and if the power of 9 is even, then units digit will be 1 and if the power of 9 is odd, then units digit will be 9. For example (539)140 Since, power is even for unit's digit 9 Units digit =1(539)141 Since, power is odd for unit's digit 9 Units digit =9 Basic Number Theory ♦ Square of every even number is an even number while square of every odd number is an odd number. ♦ A number obtained by squaring a number does not have 2, 3 ,7 or 8 at its unit place. ♦ There are 15 prime numbers between 1 and 50 and 10 prime numbers between 50 and 100. ♦ If p divides q and r, then p divides their sum and difference also. e.g., 4 divides 12 and 20, then 20 + 12 = 32 and 20 -12 = 8 are also divisible by 4. ♦ For any natural number n, (n - n) is divisible by 6. ♦ The product of three consecutive natural numbers is always divisible by 6. ♦ (xm - a"1) is divisible by (x - a) for all values of m. ♦ {xm - cr"1) is divisible by (x + a) for even values of m. ♦ {xm + a"1) is divisible by (x + a) for odd values of m. ♦ Number of prime factors of ap bq cr ds is p + q + I + s, where a, b, c and d are prime number. Multi Concept QUESTIONS 1. If n is any odd number greater than 1, then n(n2 -1) is (a) divisible by 96 always (b) divisible by 48 always (c) divisible by 24 always (d) None of these ^ (c) Solving the question by taking two odd numbers greater than 1, i.e., 3 and 5, then n{n2 - 1) for n = 3 => 3(9-1) => 3x8 = 24 n(n2 -1)forn = 5 => 5(25-1) => 24x5 = 120 Using option we find that both the number are divisible by 24 2. 76n -66n, where n is a integer greater than 0, is divisible by (a) 13 (b) 127 (c) 559 (d) None of these * (b) 76n - 66n for n = 1, 76 - 66 => (73)2-(63)2 {a2 -b2 = (a+ b)(a-b)} => (73 - 63)(73 + 63) => (343 - 216) (343 + 216) => 127 x 559 ∴ It is clearly divisible by 127, 4. If the sum of first 11 terms of an arithmetic progression equal that of the first 19 terms, Then, what is the sum of first 30 terms? (a) 0 (b) -1 (c) 1 (d) Not unique w (a) Let the first term be a common difference of progression be d According to the question, Fast Track Practice Exercise© Base Level Questions 1. Find the place value of 4 in 46127. (a) 4 (b) 400 (c) 40000 (d) 4000 (e) None of the above 2. Find the place value of 7 in 837218. (a) 7000 (b) 7 (c) 700 (d) 70000 fej None of the above What is the place value of 6 in 65489203? [SSCLDC2010] (a) 6x105 (b) 6x104 (c) 6x107 (dj 6x10a (e) None of the above 4. Find the face value of 7 in 942756. (a) 7 (b) 700 (c) 7000 (d) 70000 5. Find the face value of 6 in 652410. [SSC LDC 2008] (a) 6x105 (b) 6x104 (c) 6000 (d) 6 6. Find the sum of the face values of 9 and 6 in 907364. [Hotel Mgm, 2007] (a) 15 (b) 20 (c) 9 (d) 18 7. Find the difference of the face values of 7 and 2 in 210978. (a) 4 b) 3 (c) 6 r^ 5 (e) None of the above 8. Find the sum of place and face values of 8 in 43836. [Hotel Mgmt 2008] (a) 88 (b) 808 (c) 880 (d) 888 (e) None of the above 9. Find the difference of place and face values of 4 in 324372. (a) 3996 (b) 3998 (c) 3398 (dj 3396 fej None of the above 10. Find the sum of place value of 6 and face value of 9 in 927653. (a) 608 (b) 508 (c) 609 (d) 507 (e) None of the above 11. Find the difference of place value of 4 and face value of 3 in 3784105. (a) 3997 (b) 1 (c) 1000 (d) 3845 (e) None of the above 12. When 121012 is divided hy 12, the remainder is [CTET 2012] (ajO (b)2 (c)3 (d)4 13. The sum of place values of 2 in 2424 is [CTET 1012] (a) 4 (b) 220 (c) 2002 (d) 2020 14. The pair of numbers which are relatively prime to each other is [CDS 2012] (a) (68, 85) (b) (65, 91) (c) (92, 85) (d) (102, 153) 15. Find the sum of 1st and 2nd prime numbers. (a) 5 (b) 3 (c) 7 (d) 2 (e) None of the above 16. Find the product of 1st natural number and 1st prime number. (a) 4 (b) 3 (c) 2 (d) 5 (e) None of the above 17. The product of 1st natural, 1st whole and the 1st prime numbers is equal to (a) 5 (b) 0 (c) 9 (d) 7 (e) None of the above 18. The product of any number and the 1st whole number is equal to (a) 0 (b) 2 (c) 1 (d) -1 (e) None of the above 19.A rational number is expressed as... where, p and q are integers and q * 0. (e) None of the above 20.2/3 is a rational number whereas V2/V3 is [CLAT2013] (a) also a rational number (b) an irrational number (c) not a number (d) a natural periodic number 21. Which of the following is a prime number? (a) 35 (b) 53 (c) 88 (d) 90 (e) None of the above 22. The number of all prime numbers less than 40 is... (a) 15 (b) 18 (c) 17 (d) 12 (e) None of the above 23. Find the quotient when 445 is divided by 5. (a) 78 (b) 48 (c) 79 fdj 89 (e) None of the above 24. Find the remainder when 54 is divided by 17. (a) 5 (b) 3 (c) 2 (d) 7 (e) None of the above 25. Find the dividend when divisor is 13, quotient is 30 and remainder is 12. (a) 402 (b) 543 (c) 436 (d) 455 (e) None of the above 26. What is the remainder in the expression 29—? 26 (a) 29 (b) 26 fc; 18 (d) 0 fej None of the above 27. Find the dividend from the expression 41±. 19 (a) 783 fbj 800 (cj 893 (dj 387 fej None of the above 28. When 1/7 of a number is subtracted from the number itself, it gives the same value as the sum of all the angles of a triangle. What is the number? [Bank PO 2010] (a) 224 (b) 210 (c) 140 (d) 350 (e) 187 29. What least number must be added to 1057 to get a number exactly divisible by 23? (a) 1 (b) 3 (c) 2 (d) 4 (e) None of the above 30. In a division sum, the divisor is ten times the quotient and five times the remainder. If the remainder is 46, then find the dividend. [General Insurance 2007] (a) 5388 (b) 5343 (c) 5336 (d) 5391 (e) None of the above 31. The product of two consecutive odd numbers is 6723. What is the greater number? [Bank Clerks 2009] (a) 89 (b) 85 (c) 91 (d) 83 (e) None of the above 32. The sum of the four consecutive even numbers is 284. What would be the smallest number? [Bank PO 2010] (a) 72 (b) 74 (c) 68 (d) 66 (e) None of the above 33. The sum of the digits of a two-digit number is 14 and the difference between the two digits of the number is 2. What is the product of the two digits of the two-digit number? [Bank Clerks 2009] (a) 56 (b) 48 (c) 45 (d) Couldn't be determined (e) None of the above 34. What number should be added to 231228 to make it exactly divisible by 33? [CDS 2012] (a) 1 (b) 2 (c) 3 (d) 4 Exercise © Higher Skill Level Questions 1. Find the sum of first 25 natural numbers. (a) 432 (b) 315 (c) 325 (d) 335 (e) None of the above 2. Find the sum of the squares of first 35 natural numbers. (a) 14910 (b) 15510 (c) 14510 (d) 16510 (e) None of the above 3. Find the sum of the cubes of first 15 natural numbers. (a) 15400 (b) 14400 (c) 16800 (d) 13300 4. Find the sum of first 37 odd numbers. [Hotel Mgmt. 2010] (a) 1369 (b) 1295 (c) 1388fdJ 1875 (e) None of the above 5. Find the sum of first 84 even numbers. [Bank Clerks 2008] (a) 7140 (b) 7540 (c) 6720 (d) 8832 (e) None of the above 6. Sum of first 15 multiples of 8 is [CUT 2013] (a) 960 (b) 660 (c) 1200 (d) 1060 7. The product of four consecutive natural numbers plus one is [CDS 2014] (a) a non-square (b) always sum of two square numbers (c) a square (d) None of the above Find the unit digit in the product of (268 x 539 x 826 x 102). [MBA 2009] (a) 5 (b) 3 (c) 4 (d) 2 (e) None of the above 9. Find the unit digit in the product of (4326 x 5321). [Hotel Mgmt. 2010] (a) 6 (b) 8 (c) 1 (d) 3 (e) None of the above 10. What is the unit digit in (6817)754 ? (a) 8 (b) 4 (c) 2 (d) 9 (e) None of the above 11. What is the unit digit in (365 x659x771)? (a) 6 (b) 4 (C) 2 (d) 1 (e) None of the above 12. Find the last two-digits of 15x37x63x51x97x17 [IBAC1O2012] (a) 35 (b) 45 (c) 55 (d) 85 13. How many rational numbers are there between 1 and 1000? [CDS 2012] (a) 998 (b) 999 (c) 1000 (d) Infinite 14. The sum of 5 consecutive even numbers A,B,C,D andE is 130. What is the product of A and E ? [Bank Clerks 2009] (a) 720 (b) 616 (c) 660 (d) 672 (e) None of the above 15. The sum of the five consecutive numbers is equal to 170. What is the product of largest and the smallest numbers? [Bank Clerks 2011] (a) 1512 (b) 1102 (c) 1152 (d) 1210 (e) None of the above 16. Which of the following numbers always divides the difference between the squares of two consecutive odd integers? [Bank Clerks 2009] (a) 7 (b) 3 (c) 8 (d) 6 (e) None of the above 17. A number divided by 56 gives 29 as remainder. If the same number is divided by 8, the remainder will be... [SSC CCL 2007] (a) 4 (b) 5 (c) 6 (d) 7 18. On dividing a certain number by 357, the remainder is 39. On dividing the same number by 17, what will be the remainder? (a) 5 (b) 3 (c) 7 (d) 6 (e) None of the above 19. A number when divided by 5, leaves 3 as remainder. What will be the remainder when the square of this number is divided by 5? (a) 3 (b) 4 (c) 5 (d) 0 (e) None of the above 20. In a question on division with zero remainder, a candidate took 12 as divisor instead of 21. The quotient obtained by him was 35. Find the correct quotient. (a) 10 (b) 12 (c) 20 (d) 15 (e) None of the above 21. A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor? [IB ACIO 2013] (a) 13 (b) 59 (c) 35 (d) 37 22. The number 58129745812974 is divisible by [CDS 2012] (a) 11 (b)9 (c) 4 (d) None of these 23. How many numbers between - Hand 11 are multiples of 2 or 3? [CDS 2012] (a) 11 (b) 14 (c) 15 (d) None of these 24. Which one of the following numbers is divisible by 11? [CDS 2013] (a) 45678940 (b) 54857266 (c) 87524398 (d) 93455120 25. When 17200 is divided by 18, find the remainder. (a) 1 (b) 4 (c) 5 (d) 3 (e) None of the above 26. What is the remainder when 41000 is divisible by 7? [CDS 2014] (a) 1 (b) 2 (c) 4 (d) None of these 27. A common factor of (4143 + 4343) and (4141+4341)is... (a) (43 - 41) (b) (4141 + 4341) (c) (4143 + 4343) (d) (41 + 43) (e) None of the above 28. The remainder when 9 + 6 is divided by 8 is [SSC CGL (Main) 2012] (a) 2 (b) 3 (c) 5 (d) 7 29. What will be the remainder when 19100 is divided by 20? [SSC CGL (Main) 2012] (a) 19 (b) 20 (c) 3 (d) 1 30. It is given that (2s2 + 1) is exactly divisible by a certain number. Which of the following is also definitely divisible by the same number? (a) (216 + 1) (b) (216-1) (c)7x213 fdj (296 + 1) (e) None of the above 31. The number (6x + &x) for natural number x is always divisible by... (a) 6 and 12 (b) 12 only (c) 6 only (d) 3 only (e) None of the above 32. 195 + 215 is divisible by [CDS 2013] (a) Only 10 (b) Only 20 (c) Both 10 and 20 (d) Neither 10 nor 20 33. If 'a' is a natural number, then the largest number dividing (a - a) is (a) A (b) 5 (c) 6 (d) 7 (e) None of the above 34. 7 - 4 is exactly divisible by which of the following number? [SSC FCI 2012] (a) 34 (b) 33 (c) 36 fdj 35 35. If N, (N + 2) and (N + 4) are prime numbers, then the number of possible solutions for N are [CDS 2013] (a) 1 (b) 2 (c) 3 (d) None of these 36. The smallest positive prime (say p) such that 2P - lis not a prime is ^DS 2013] raj 5 (b) 11 (c) 17 fdj 29 37. If b is the largest square divisor of c and a divides c, then which one of the following is correct? (where, a, b and c are integers) [CDS 2013] (a) b divides a (b) a does not divide b (c) a divides b (d) a and b are coprime 38. If re is a whole number greater than 1, then re (re - 1) is always divisible by [CDS 2014] (a) 12 (b) 24 (c) 48 (d) 60 39. What is the sum of all positive integers lying between 200 and 400 that are multiples of 7? [IB ACIO 2013] (a) 8729 (b) 8700 (c) 8428 (d) 8278 40. Consider the following statements I. To obtain prime numbers less than 121, we are to reject all the multiples of 2, 3, 5 and 7. II. Every composite number less than 121 is divisible by a prime number less than 11. Which of the statements given above is/are correct? [CDS 2013] (a) Only I (b) Only II (c) Both I and II (d) Neither I nor II 41. Consider the following statements I. 7710312401 is divisible by 11. II. 173 is a prime number. Which of the statements given above is/are correct? [CDS 2013] (a) Only I (b) Only II (c) Both I and II (d) Neither I nor II 42. If & is a positive integer, then every square integer is of the form [CDS 2013] (a) Only 4k (b) Ak or Ak + 3 (c) 4/c + 1 or 4/c + 3 (d) Ak or Ak + 1 43. Every prime number of the form 3k + 1 can be represented in the form 6m + 1 (where k, m are integers), when [CDS 2013] (a) k is odd (b) k is even (c) k can be both odd and even (d) No such form is possible Answer with Solutions Exercise© Base Level Questions 1. (C) 4 is at the place of ten thousand. /. Required place value = 4 x 10000 =40000 2. (a) 7 is at the thousand place. /. Required place value = 7 X 1000 = 7000 3. (c) 6 is at the place of crore..'. Required place value = 6 x 10000000 =6 x 107 4. (a) Face value is the value of digit itself..". Required face value of 7 = 7 5. (d) Face value is the value of digit itself. /. Required face value = 6 6. (a) The face value is the value of digit itself. So, required sum = 9 + 6 = 15 7. (d) The face value is the value of digit itself. So, required difference = 7-2 = 5 8. (6) Place value of 8 = 800 and face value of 8 = 8 ∴ Required sum = 800 + 8 = 808 9. (a) Place value of 4 = 4000 and face value of 4 = 4 ∴ Required difference = 4000 - 4 = 3996 10. (c) Place value of 6 = 600 and face value of 9 = 9 ∴ Required sum = 600 + 9 = 609 11. (a) Place value of 4 = 4000 and face value of 3 = 3.'. Required difference = 4000 - 3 = 3997 Hence, when 121012 is divided by 12, then remainder is 4. 13. (d) The sum of place values of 2 in 2424 = 2 X 1000 + 2X10 = 2000 + 20 = 2020 14. (c) 92 and 85 are coprime numbers because their HCF is 1. 15. (a) 1st prime number = 2 and 2nd prime number = 3.'. Required sum = 2+3 = 5 16. (c) 1st natural number = 1 and 1 st prime number = 2 Required product = 1x2 = 2 17. (6) 1st natural number = 1, 1 st whole number = 0 and 1 st prime number = 2 ∴ Required product = 1x0x2 = 0 18. (a) 1st whole number = 0 Clearly, when any number is multiplied with 0 (the 1st whole number), then the result is 0. 19. (d) A proven fact. 21. (6) 53 has only two factors itself and 1. Hence, it is a prime number. 22. (d) Prime numbers less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37 23. (d) Dividend = 445 and divisor = 5 /. Remainder = 3 25. (a) Given, Divisor (d) = 13, Quotient (Q) = 30 Remainder {R) = 12 and Dividend (D) = ? We know that, D=dxQ+R D = 13X30 + 12 = 390 + 12 = 402 26. (c) It is clear from the expression that remainder is 18. 27. (a) Given, Quotient (Q) = 41, Divisor (d) = 19 Remainder (.R) = 4 and Dividend (D) = ? ∴ Dividend D = dxQ + £ = 19x41 + 4 = 779 + 4 = 783 28. (6) Let the number be x. According to the question, ∴ Number to be added = (23 - 22) = 1 30.(C) Divisor = 5 X Remainder = 5 X 46 = 230 Also, 10 x Quotient = 230 Quotient = 23 We know that Dividend = (Divisor X Quotient) + Remainder ∴ Dividend = (230 x 23) + 46 = 5290 + 46 = 5336 (d) Let two consecutive odd number be (x + 1) and (x + 3). According to the question, (x + 1) (x + 3) = 6723 => x2 + 3x + x + 3 = 6723 => xz + 4x+3 = 6723 => x2+4x+3-6723 = 0 => xz + 4x - 6720 = 0 => x2 + 84x - 80x - 6720 = 0 => x (x + 84) - 80 (x + 84) = 0 => (x - 80) (x + 84) = 0 x = 80,(x*-84) Hence, the greater number = 80 + 3 = 83 32. (c) Let four consecutive even numbers are x, x + 2, x + 4 and x + 6. According to the question, x+x + 2+x + 4 + x + 6 = 284 => 4x+ 12 = 284 => 4x = 284-12 = 272 33. (6) Let be the ten's digit be x and unit's digit be y. The two-digit number = lOx + y (where, x > y) According to the question, x + y = 14...(i) and x - y = 2... (ii) Solving Eqs. (i) and (ii), we get x = 8 and y = 6.'. Required product = 8 x 6 = 48 34. (c) Given, 33)231228(7006 231 228 198 30 Now, 33 - 30 = 3.'. On adding 3 to 231228, it is completely divisible by 33. Exercise © Higher Skill Level Questions 1. (c) We know that, Sum of first n natural numbers 2. (a) We know that, Sum of the squares of first n natural 3. (6) We know that, Sum of the cubes of first n natural numbers 4. (a) We know that, 2 Sum of first n odd numbers = n Given, n = 37 ∴ Required sum = (3 if =37x37 = 1369 5. (a) We know that, Sum of first n even numbers = n {n + 1) Given, n = 84 ∴ Required sum = 84 (84 + 1) = 84 X 85= 7140 6. (a) first 15 multiple of 8 are 8, 16, 24.....120 So, 8(1,2,3,4.....15) 7. (c) Product of four consecutive numbers plus one is always a square Illustration 1 Let four consecutive numbers be 3, 4, 5 and 6. = (3 x 4 x 5 x 6) + 1 = 361 = (19)z Illustration 2 Let four consecutive numbers be 9, 10, 11 and 12. = (9 x 10 x 11 x 12) + 1 = 11881 = (109)2 8. (c) Product of unit digits = 8X9X6X2 = 864 /. Required digit = 4 9. (a) Product of unit digits =6x1=6 /. Required digit = 6 10. (d) Required digit = Unit digit in (7)754 = Unit digit in {(74)188 x 72} = Unit digit in (1 x 49) = 9 11. (6) Unit digit in 34 = 1 ∴ Unit digit in (34)16 = 1 Unit digit in 385 = 3 Unit digit in 659 = 6 Unit digit in 74 = 1 ∴ Unit digit in (74)17 =1 77l = (74jl7 x f As, Unit digit in 73 =3 Unit digit in 771 =3.". Required unit digit = Unit digit in (3x6x3) = Unit digit in 54 =4 12. (a) 15 X 37 X 63 X 51 X 97 X 17 = 255X37X63X51 X 97 = 35X21 [last two digits] = 735 = 35 [last two digits] Hence, last two digits of the product is 35. 13. {d) There can be infinite number of rational numbers between 1 and 1000. 14. (c) Let five consecutive even number be A = x,.B = x + 2, C = x+4, D = x + 6 and E = x + 8 According to the question, X+X+2+X+4+ x + 6+ x + 8 = 130 => 5x+20 = 130 => 5x = 130-20 = 110 15. (c) Let the five consecutive numbers are x, (x + 1) and (x + 2), (x + 3) and (x + 4). According to the question, x+x+l + x+2+x+3+x + 4 = 170 5x+ 10 = 170 => 5x = 160 = 32 Largest number = (x + 4) = 32 + 4 = 36 /. Required product = 32x36 = 1152 16. (c) Let the two consecutive odd numbers be (2x + 1) and (2x + 3). ∴ Difference = (2x + 3)2 - (2x + l)2 = (2x + 3 + 2x + 1) (2x + 3 - 2x- 1) = (4x + 4) X2 = 8 (x+ 1), which is exactly divisible by 8. 17. (6) Let the number be x. According to the question, x = 56A- + 29 Then, x = (8 X 7 k) + (8 X 3) + 5 = 8 X (7i + 3) + 5 Therefore, when x is divided by 8, the required remainder = 5 Dired Approach In such type of questions to get the new remainder we simply divide the first remainder by second divisor and remainder obtained is the required answer. 18. (a) Let the given number be (357Jr + 39). Then, (357* + 39) = (17 X 21k) + (17 X 2) + 5 = 17x(21i + 2) + 5.. Required remainder = 5 19. (6) Let the number be x. According to the question, x = (5k + 3) On squaring both sides, we get => x2 = (5k + 3)2 = (25£2 + 30Jc + 9) = 5(5£2 + 6k + 1) + 4.. On dividing x by 5, the remainder is 4. 20. (C) Number = 35 X 12 = 420 21. (d) Let the divisor be x and quotient be y. Then, number = xy + 24 Twice the number = 2xy + 48 Now, 2xy is completely divisible by x. On dividing 48 by x remainder is x. x = 48 -11 = 37 22. (a) We know that, a number is divisible by 11 when the difference between the sum of its digit at even places and sum of digit at odd places is either 0 or the difference is divisible by 11. So, number is 58129745812974 Sum of digits at odd places = 4 + 9 + 1 + 5+7+2 + 8 = 36 Sum of digit at even places 7+2 + 8 + 4 + 9 + 1 + 5 = 36 So, the required difference = 36 - 36 = 0 /. The number is divisible by 11. 23. (c) Method I Following are the numbers between - 11 and 11 which are multiples of 2 or 3? -10,-9,-8,-6,-4, -3,-2, 0, 2, 3, 4, 6, 8, 9, 10. /. The numbers of multiples 2 or 3, between - 11 and 11 are 15. Method II Numbers between 0 and 11 which are multiples of 2 or 3 /. Number be 15, including "0". 24. (d) we know that, if the difference between the sum of digits at even places and sum of digits at odd places is (0), then the number is divisible by 11. From options. (a) 45678940 Sum of even places =5+7+9 + 0 = 21 Sum of odd places =4 + 6 + 8 + 4 = 22 Their difference = 22-21*0 (b) 54857266 Sum of even places =4 + 5+2+6 = 17 Sum of odd places = 5+8 + 7+6 = 26 Their difference = 26-17 = 9*0 (c) 87524398 Sum of even places =7+2+3+8= 20 Sum of odd places =8 + 5 + 4 + 9= 26 Their difference = 26-20=6*0 (d) 93455120 Sum of even places =3+5+1 + 0=9 Sum of odd places =9 + 4 + 5+2 = 20 Their difference = 20-9 = 11+11 = 1 So, it is divisible by 11. 25. (a) We know that, (xm - am) is divisible by (x + a), for even values of m..: (17200 - l200) is divisible by (17 + 1). => (17200 - 1) is divisible by 18. When 17200 is divisible by 18, then the remainder is 1. 27. (d) We know that, when m is odd (xm + am) is divisible by (x + a)... Each one is divisible by (41 + 43). ∴ Common factor = (41 + 43) 28. (d) Required remainder =9 +6 = (l)19 + 6=4 [v8 = 9 - 1, so replaced by 1] 30. (d) Let 232 = x and let (232 + 1) = (x + 1) be divisible by a number n. Then, (296 + 1) = (x3 + 1) = (x + 1) (x2 - x + 1) which is clearly divisible by n as (x + 1) is divisible by n. 31. (a) (6X2 + 6x) = 6x (x + 1) which is clearly divisible by 6 and 12 as x(x + 1) is even. 32. (c) We can check divisibility of 195 + 215 by 10 by adding the unit digits of 9 and 1 which is equal to 9 + 1 = 10. So, it must be divisible by 10. Now, for divisibility by 20 we add 19 and 21 which is equal to 40. So, it is clear that it is also divisible by 20. So,195 + 215isdivisiblebybothl0and20. 33. (C) (2 - 2) = 6 is the largest natural number that divides (a - a) for every number a. 34. (6) We know that, (xn - y") is divisible by (x — y) for all n and is divisible by (x + y) for even n. :. (712 - 412) is divisible by(7 + 4) and(7 - 4) => (712 _ 412) is divisible by 11 and 3 (71Z - 412) is divisible by 33. 35. (a) When N is a natural number, then there is only one possible case that N, (N + 2), [N + 4) are prime numbers, When N = 3, then N, (N + 2), {N + 4) = 3, 5, 7 all are primes. 36. (b) Taking p = 5 2P - 1 = 25 - 1 = 31 which is prime Taking p = 11 2P - 1 = 211 - 1 = 2047 Since, 2047 is divisible by 23, so it is not prime. Thus, required least positive prime number is 11. 37. (c) Since, b is largest square divisor of c. So, c = bx (where, x is not a whole square number) Also, a divides c. So, a will divide bx. or a will divide b. (since, it cannot divide x as it is not a whole square) 38. (a) If n greater than 1, then nz(nz - 1) is always divisible by 12. Illustration 1 Put n = 2, then nz{nz - 1) = (2)2(22 - 1) = 4 X3 = 12 Illustration 2 Now, put n = 3, then n2(nz - 1) = (3)2(32 - 1) = 9 X 8 = 72 39. (a) Least number divisible by 7 and above 200 is 203. Greatest number divisible of 7 and below 400 is 399. Total numbers divisible by 7 between 200 to 400 are 29 Now, sum of n terms of AP = — (a + 1) 2 where, a = 203, 1 = 399 and n = 29 40. (c) Both the statements given are correct. As 121 is the square of 11. So, to obtain prime numbers less than 121, we reject all the multiples of prime numbers less than 11 i.e., 2, 3, 5 and 7. Similarly, every composite number less than 121 is divisible by a prime number less than 11 i.e., 2, 3, 5 or 7. 41. (c) I. Any number in order to get completely divided by 11 must have the difference between the sum of even place digits and the sum of odd place digits equal to 0 or the multiple of 11. In 7710312401, difference between sum of even place digits and the sum of odd place digit 0. So, it is divisible by 11. II. To check divisibility of 173, we can divide the number by all the prime numbers from 2 to 13. It is not divisible by 2, 3, 5, 7, 11 and 13. So, it is a prime number. Hence, both Statements I and II are correct. 42. (d) If k is a positive integer, then every square integer is of the form 4i or 4£ + 1, as every square number is either a multiple of 4 or exceeds multiple of 4 by unity. 43. (b) Every prime number of the form 3k + 1 can be represented in the form 6m + 1 only, when k is even. Chapter 2 Number Series A number series is a sequence of numbers written from left to right in a certain pattern. To solve the ques- tions on series, we have to detect/find the pattern that is followed in the series between the consecutive terms, so that the wrong/missing term can be find out. Types of Series There can be following types of series 1. Prime Number Series The number which is divisible by 1 and itself, is called a prime number. The series formed by using prime number is called prime number series. Ex. 1 Find out the next term in the series 7,11,13,17,19,.... Sol. Given series is a consecutive prime number series. Therefore, the next term will be 23. Ex. 2 Find out the next term in the series 3, 7,17, 31,.... Sol. Here, every next prime number takes place skipping one, two, three and four prime numbers, respect- ively. Hence, the required answer will be 53. Let us see 2. Addition Series The series in which next term is obtained by adding a specific number to the previous term, is known as ad- dition series. Addition series are increasing order series and difference between consecutive term is equal. Ex. 3 Find out the missing term in the series 2, 6,10,14,..., 22. Sol. Here, every next term is obtained by adding 4 to the previous term... Required term = 14 + 4 = 18 3. Difference Series Difference series are decreasing order series in which next term is obtained by subtracting a fixed/specific number from the previous term. Ex. 4 Find out the missing term in the series 108, 99, 90, 81,..., 63. Sol. Here, every next number is 9 less than the previous number. So, required number = 81 - 9 = 72 4. Multiple Series When each term of a series is obtained by multiplying a number with the previous term, then the series is called a multiplication series. Note Number which is multiplied to consecutive terms, can be fixed or variable Ex. 5 Find out the missing term in the series 4, 8,16, 32, 64,..., 256. Sol. Here, every next number is double the previous number. So, required number = 64 x 2 = 128 Ex. 6 Find out the missing term in the series 5, 20, 80, 320,..., 5120. Sol. Here, every next number is 4 times the previous number. So, required number = 320 X 4 = 1280 5. Division Series Division series are those in which the next term is obtained by dividing the previous term by a number. Note Number which divides consecutive terms can be fixed or variable Ex. 7 Find out the missing term in the series 10080,1440, 240,..., 12,4. Hence, missing term is 48. Ex. 8 Find out the missing term in the series 512, 216, 72,..., 12. Hence, missing term is 36. 6. n2 Series When a number is multiplied with itself, then it is called as square of a number and the series formed by square of numbers is called rr series. Ex. 9 Find out the missing term in the series 4,16, 36, 64,..., 144. Sol. This is a series of squares of con- secutive even numbers. Let us see 22 = 4, 42 = 16, 62 = 36, 82 = 64, 102 = 100, 122 = 144 Hence, missing term is 100. Ex. 10 Find out the missing term in the series 1, 4, 9,16, 25,..., 49. Sol. This is a series of squares of consecutive natural numbers. Let us see l2 = 1,22 = 4,32 = 9, 42 = 16, 52 = 25, 62 = 36, 72 = 49 Hence, missing term is 36. 7. (n2 + 1) Series If in a series each term is a sum of a square term and 1, then this series is called (n 2 +1) series. Ex. 11 Find out the missing term in the series 10,17, 26, 37,..., 65. Sol. Series pattern 32 + 1 = 10, 42 + 1 = 17, 52 + 1 = 26, 62 + 1 = 37, 72 + 1 = 50, 82 + 1 = 65 So, required number = 50 Ex. 12 Find out the missing term in the series 122,145,170,197,..., 257. Sol. Series pattern ll2 + 1 = 122, 122 + 1 = 145, 132 + 1 = 170, 142 + 1 = 197, 152 + 1 = 226, 162 + 1 = 257 So, correct answer is 226 8. (n2 -1) Series In a series, if each term is obtained by subtracting 1 from square of a number, then such series is known as (n2 -1) series. Ex. 13 Find out the missing term in the series 0, 3, 8,15, 24,..., 48. Sol. Series pattern l2 - 1 = 0, 22 - 1 = 3, 32 - 1 = 8, 42 - 1 = 15, 52 - 1 = 24, 62 - 1 = 35, 72 - 1 = 48 So, correct answer is 35. Ex. 14 Find out the missing term in the series 224,195,168,..., 120. Sol. Series pattern 152 - 1,142 - L 132 - 1,122 - 1, ll2 — 1 So, required number =12 - 1 = 143 9. (n2 + n) Series Those series in which each term is a sum of a number with square of that number, is called as (ir + n) series. Ex. 15 Find out the missing term in the series 12, 20, 30, 42,..., 72. Sol. Series pattern 32 + 3, 42 + 4, 52 + 5, 62 + 6, 72 + 7, 82 + 8 So, required number =7 + 7 = 56 Ex. 16 Find out the missing term in the series 420 , 930,1640,..., 3660. Sol. Series pattern 202 + 20, 302 + 30, 402 + 40, 502 + 50, 602 + 60 So, required number = 50 + 50 = 2550 10. (n2 - n) Series Series in which each term is obtained by subtracting a number from square of that number, is known as (n2 - n) series. Ex. 17 Find out the missing term in the series 42, 30,..., 12, 6. Sol. Series pattern 72 - 7, 62 - 6, 52 - 5, 42 - 4, 32 - 3 So, required number =5 - 5 = 20 Ex. 18 Find out the missing term in the series 210, 240, 272, 306,..., 380. Sol. Series pattern 152 - 15, 162 - 16, 172 - 17, 182 - 18, 192 - 19, 202 - 20 So, required number = 192 - 19 = 342 11. n3 Series A number when multiplied with itself twice, then the resulting number is called the cube of a number and series which consist of cube of different number following a specified sequence, is called as n 3 series. Ex. 19 Find out the missing term in the series 1, 8, 27,..., 125, 216. Sol. Series pattern l3, 23, 33, 43, 53, 63 So, required number = 4 = 64 Ex. 20 Find out the missing term in the series 1000,8000,27000,64000,125000,.... Sol. Series pattern 103, 203, 303, 403, 503, 603 So, required number = 603 = 216000 12. (n3 + 1) Series Those series in which each term is a sum of a cube of a number and 1, are known as (n3 +1) series. Ex. 21 Find out the missing term in the series 126, 217, 344,..., 730. Sol. Series pattern 53 + \ 63 + 1, 73 + 1, 83 + 1, 93 + 1 So, required number = 8 + 1 = 513 Ex. 22 Find out the missing term in the series 1001, 8001,..., 64001, 125001, 216001. Sol. Series pattern 103 + 1, 203 + \ 303 + 1,403 + 1, 503 + 1, 603 + 1 So, required number = 303 + 1 = 27001 13. (n3 -1) Series Series in which each term is obtained by subtracting 1 from the cube of a number, is known as (n3 -1) series. Ex. 23 Find out the missing term in the series 0, 7, 26, 63,124, 215,.... Sol. Series pattern l3 - \ 23 - 1, 33 - 1, 43 - 1,53 - \ 63 - 1, 73 - 1 So, required number = 73 - 1 = 342 Ex. 24 Find out the missing term in the series......, 7999, 26999, 63999,124999. Sol. Series pattern 103 - 1, 203 - 1, 303 - 1, 403 - \ 503 - 1 So, required number = 10 - 1 = 999 14. (n3 + n) Series When each term of a series is a sum of a number with its cube, then the series is known as (n3 + n) series. Ex. 25 Find out the missing term in the series 2,10, 30,..., 130, 222. Sol. Series pattern l3 + 1, 23 + 2, + 33 + 3, 43 + 4, 53 + 5, 63 + 6 So, required number = 4 + 4 = 68 Ex. 26 Find out the missing term in the series..., 8020, 27030, 64040. Sol. Series pattern 103 + 10, 203 + 20, 303 + 30,403 + 40 So, required number = 103 + 10 = 1010 15. (n3 - n) Series When each term of a series is obtained by subtracting a number from its cube, then series is termed as (n - n) series. Ex. 27 Find out the missing term in the series 0, 6, 24, 60,120,.... Sol. Series pattern l3 - \ 23 - 2, 33 - 3, 43 - 4, 53 - 5, 63 - 6 So, required number =6 - 6 = 210 Ex. 28 Find out the missing term in the series..., 7980, 26970, 63960,124950. Sol. Series pattern 103 - 10, 203 - 20, 303 - 30, 403 - 40, 503 - 50 So, required number = 103 - 10 = 990 16. Alternating Series In alternating series, successive terms increase and decrease alternately. The possibilities of alternating series are f It is a combination of two different series. ♦ Two different operations are performed on successive terms alternately. Ex. 29 Find the next term in the series 15,14,19,11, 23, 8,... Ex. 30 Find the next term in the series 50, 200,100,100,200,50,400,... 17. Arithmetic Progression (AP) The progression of the form a, a+d, a + 2d, a + 3d,...is known as an arithmetic progression with first term a and common difference d. Then, nth term Tn = a + (n -1) d Ex. 31 In series 359, 365, 371,..., what will be the 10th term? Sol. The given series is in the form of AP. Since, common difference i. e., d is same. Here, a = 359 and d =6 10th term = a + (n - 1) d= 359 + (10 - 1) 6= 359 + (9 x 6)= 413. 18. Geometric Progression (GP) The progression of the form a, ar, ai, ar3,... is known as a GP with first term a and common ratio = r. Then, nth term of GP Tn = ar" ~ * Ex. 32 In the series 7,14, 28, , what will be the 10th term? Sol. The given series is in the form of GP. Since, common ration i. e., r is same. Here, a = 7 and r = 2 10th term = arn " J = 7(2)(10 _1)=7x29 = 3584 Types of Questions Asked on Number Series There are mainly three types of questions which are asked from this chapter which are as follows Type O To Find the Missing Term In this type of questions a series is given in which one of the term is missing and it is required to find the missing term by detecting the pattern of series. Ex. 33 What should be the next term in the following series? 12, 24, 72,144, ? Sol. Series pattern 12 x 2 = 24; 24 x 3 = 72 ; 72 x 2 = 144; 144 x 3 = 432 Clearly, next term is 432. Ex. 34 What should come in place of question mark in the series given below? 92, ?, 46, 69, 138, 345 Clearly, 46 should come in place of question mark. Type @ To Find the Wrong Term in this type of questions, a series is given in which one of term is required to detect the pattern of series and find that wrong term. Ex. 35 In the following series, a wrong number is given. Find out that wrong number. 4, 5,10,18, 34, 59, 95 Sol. Series pattern 4 + l2 = 5; 5 + 22 = 9 b > 12, then what will be the values of a and 6? [RRB2012] (a) 12,24 (b) 24,12 (c) 24,36 (d) 36,24 29. The sum of HCF and LCM of two numbers is 403 and their LCM is 12 times their HCF. If one number is 93, then find the another number. [MBA 2007] (a) 115 (b) 122 (c) 124 (d) 138 (e) None of the above 30. The LCM of two numbers is 495 and their HCF is 5. If sum of the numbers is 100, then find the differ- ence of the numbers. [Hotel Mgmt. 2008] (a) 10 (b) 46 (c) 70 (d) 90 (e) None of the above 31. The LCM of two numbers is 20 times of their HCF and (LCM + HCF) = 2520. If one number is 480, what will be the triple of another number? (a) 1200 (b) 1500 (c) 2100 (d) 1800 (e) None of the above 32. The sum of two numbers is 1056 and their HCF is 66, find the number of such pairs. (a) 6 (b) 2 (c) 4 (d) 8 (e) None of the above 33. What is the smallest possible length that can be exactly measured by the scales of lengths 3 cm, 5 cm and 10 cm? (a) 15 cm (b) 30cm(cJ 28 cm (d) 40 cm (e) None of the above 34. What is the least number which is exactly divisible by 8, 9, 12, 15 and 18 and is also a perfect square? (a) 3600 (b) 7200 (C) 5200 (d) 6500 (e) None of the above 35. Find the greatest number of 3-digits which when divided by 6, 9, 12 leaves 3 as remainder in each case. [CBI 2008, BOI 2007] (a) 975 (b) 996 (c) 903 (d) 939 (e) None of the above 36. What will be the greatest number that divides 1356, 1868 and 2764 leaving 12 as remainder in each case? [Delhi Police 2007] (a) 64 (b) 124 (c) 156 (d) 260 37. Find the greatest number that divides 130, 305 and 245 leaving remainders 6, 9 and 17, respectively? [RBI Clerk 2008] (a) 4 (b) 5 (c) 14 (d) 24 (e) None of the above 38. What will be the greatest number that divides 1023 and 750 leaving remainders 3 and 2, respectively? (a) 68 (b) 65 (C) 78 (d) 19 (e) None of the above 39. In a store, there are 345 L mustard oil, 120 L sunflower oil and 225 L soyabean oil. What will be the capacity of the largest container to measure the above three types of oil? (a) 8 L (b) 20 L (c) 23 L (d) 15 L (e) None of the above 40. The least number which should be added to 2497, so that the sum is exactly divisible by 5, 6, 4 and 3, is (a) 3 (b) 13 (c) 23 (d) 33 (e) None of the above 41. Find the largest number which divides 1305, 4665 and 6905 leaving same remainder in each case. Also, find the common remainder. [CBI Clerk 2009] (a) 1210, 158 (b) 1120, 158 (c) 1120, 185 (d) 1210, 185 (e) None of the above 42. The least number which when divided by 12, 16 and 18 leaves 5 as remainder in each case. Find the number. [RBI Clerk 2009] (a) 139 (b) 144 (c) 149 (d) 154 (e) None of the above 43. What is the greatest number that divides 13850 and 17030 leaves a remainder 17? [CDS 2012] (a) 477 (b) 159 (c) 107 (d) 87 44. The HCF and LCM of two natural numbers are 12 and 72, respectively. What is the difference between the two numbers, if one of the number is 24? [CDS 2012] (a) 12 (b) 18 (c) 21 (d) 24 45. What will be the least number which when divided by 12, 21 and 35 leaves 6 as remainder in each case? [UP Police 2007] (a) 426 (b) 326 (c) 536 (d) 436 46. What is the least number which when diminished by 7, is divisible by each one of 21, 28, 36 and 45? (a) 1255 (b) 1177 (c) 1265 (d) 1267 (e) None of the above What is the least number which when increased by 9, is divisible by each one of 24, 32, 36 and 54? (a) 855 (b) 890 (c) 756 (d) 895 (e) None of the above 48. Find the least number which when divided by 16, 18 and 20 leaves a remainder 4 in each case, but is completely divisible by 7. (a) 2884 (b) 2256 (c) 865 (d) 3332 (e) None of the above 49. What is the greatest four digit number which when divided by 10, 15, 21 and 28 leaves remainders 4, 9, 15 and 22, respectively? [LIC ADO 2008] (a) 9654 (b) 9666 (c) 9664 (d) 9864 (e) None of the above 50. How many numbers are there between 4000 and 6000 which are exactly divisible by 32, 40, 48 and 60? (a) 2 (b) 3 (c) 4 (d) 5 51. Three pieces of timber 84 m, 98 m and 126 m long have to be divided into planks of the same length. What is the greatest possible length of each plank? (a) 14 m (b) 28 m (c) 7 m (d) 21 m (e) None of the above 52. In a school, all the students can stand in a row, so that each row has 5, 9 or 10 students. Find the least number of students in the school. [SSC CGL 2008] (a) 90 (b) 95 (C) 85 (d) 100 Exercise © Higher Skill Level Questions 1. If (x - 6) is the HCF of xL - 2x -24 and x — kx - 6, then what is the value of kl [CDS 2012] (a) 3 (b) 5 (c) 6 (d) 8 2. If a and b are positive integers, then what is the value of 3. What is the HCF of 8 (x5 - x3 + x) and 28(*6+l)? [CDS 2014] {a) 4 (x4 - x2 + 1) (b) x3 - x + 4x2 (c) x3 - x + 3x2 (d) None of these 4. The HCF of {x4 - y4) and (x6 - y6) is [CDS 2013] (a)x2-/2 (b)x-y (c) x3 - y3 (d) x4 - y4 5. The HCF of (x3 - y2 - 2x) and (x3 + x2) is [CDS 2013] (a) x3 - x2 - 2x (b) x2 + x (c) x4-x3-2x2 fc!) x-2 6. What is the HCF of a2b4 + 2a2b2 and (abj7 - Aa bl [CDS 2013] (a) ab (b) a2b3 (c) a2b2 (d) a3b2 7. For any integer n, what is HCF (22ra + 7, 33re + 10) equal to? [CDS 2014] (a) 0 (b) 1 (c) 11 (d) None of these 8. For any integers a and b with HCF (a, 6) = ; what is HCF (o + b, a - b) equal to? [CDS 2013] (a) It is always 1 (b) It is always 2 (c) Either 1 or 2 fc/J None of the above 9. In a fire range, 4 shooters are firing at their respective targets. The first, the second, the third and the fourth shooter hit the target once in every 5 s, 6 s, 7 s and 8 s, respectively. If all of them hit their targets at 9:00 am, when will they hit their targets together again? [CDS 2014] (a) 9 : 04 am (b) 9 : 08 am (c) 9 : 14 am (d) None of the above 10. Three hells chime at intervals of 48, 60 and 90 min, respectively. If all the three bells chime together at 10:00 am, at what time will all the three chime again that day? [IB Grade II 2012] (a) 1:00 pm (b) 2:00 pm (c) 8:00 pm (d) 10:00 pm 11. Five bells begin to toll together at intervals of 9 s, 6 s, 4 s, 10 s and 8 s, respectively. How many times will they toll together in the span of 1 h (excluding the toll at the start)? [Bank Clerks 2007] (a) 5 (b) 8 (c) 10 (d) Couldn't be determined (e) None of the above 12. Monica, Veronica and Rachat begin to jog around a circular stadium. They complete their revolutions in 42 s, 56 s and 63 s, respectively. After how many seconds will they be together at the starting point? [Bank Clerks 2008] (a) 366 (b) 252 (c) 504 (d) Couldn't be determined (e) None of the above 13. A General can draw up his soldiers in the rows of 10, 15 or 18 soldiers and he can also draw them up in the form of a solid square. Find the least number of soldiers with the General. [SSC CGL 2007] (a) 100 (b) 3600 (c) 900 (d) 90 14. 6 bells commence tolling together and toll of intervals are 2, 4, 6, 8, 10 and 12 s, respectively. In 1 h, how many times, do they toll together? (a) 16 (b) 32 (c) 21 (d) 35 (e) None of the above 15. Find the greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm and 12 m 95 cm. (a) 15 cm (b) 25 cm (c) 35 cm (d) 42 cm (e) 45 cm 16. The HCF and LCM of two numbers are 21 and 4641, respectively. If one of the numbers lies between 200 and 300, then find the two numbers. (a) 273, 363 (b) 273, 359 (C) 273, 361 (d) 273, 357 (e) None of the above 17. Find the side of the largest possible square slabs which can be paved on the floor of a room 2 m 50 cm long and 1 m 50 cm broad. Also find the number of such slabs to pave the floor. [LIC AAO 2007] (a) 25, 20 (b) 30, 15 (c) 50, 15 (d) 55, 10 (e) None of the above 18. When in each box 5 or 6 dozens of apples were packed, three dozens were left. Therefore, bigger boxes were taken to pack 8 or 9 dozens of apples. However, still three dozens of apples remained. What was the least number of dozens of apples to be packed? (a) 363 dozens (b) 315 dozens (c) 345 dozens (d) 335 dozens (e) None of the above Answer with Solutions Exercise © Base Level Questions 1. (6) By prime factorisation method, Factors of 8 = 2x2x2 = 23 Factors of 15 = 3 X 5 = 31 X 51 Factors of 24 = 2 X 2 X 2 X 3 = 23 X 31 Factors of 72 = 2x2x2x3x3 = 23 X 32 Here, the prime factors that occur in the given numbers are 2, 3 and 5 and their highest powers are 2 ,3 and 5. ∴ Required LCM = 23 X 32 X 51 = 360 By division method, LCM = 2x2x2x3x5x3 = 360 2. (a) Required LCM = ax2x5x7 = 70a 3. (c) Given prime factors, (23 X 3 X 5Z X 7); (2^ X3Z X 5 X f_ X 11); (2xf X5^].'. Required LCM = Product of prime factors having highest powers = 24 x33 X54 X7Z x 11 4. (a) At the 1st step, we take the two numbers 132 and 204. HCF of 132 and 204 Hence, HCF of 132, 204 and 228 = 12. 5. (c) Given factors, 2X3X7X9; 2X3X9X11; 2X3X4X5.'. Required HCF = Product of common prime factors having least powers = 2x3 6. (c) We know that, 10. (e) Required LCM = (LCM of 250, 100 and 125) X 0.001 LCM of 250, 100 and 125 ∴ LCM = 2X5X5X5X2= 500 ∴ Required LCM = 500 X 0.001 = 0.5 11. (a) Required LCM = (LCM of 25, 12, 200 and 75) X 0.1 LCM of 25, 12, 200, 75 LCM = 2X2X3X5X5X2 = 600 ∴ Required LCM = 600 x 0.1 = 60 12. (d) LCM of two coprimes is equal to their product. 13. (d) HCF of 18 and 15 18 = 2X3 X3 15 = 3X 15 = 3 LCM of 18 and 15 = 2x3x3x5 = 90 /. Product of HCF and LCM of both numbers = 3 X 90 = 270 14. (6) Given, LCM of two numbers = 2376 HCF of two numbers = 33 One of the number = 297. (HCF of two numbers) x (LCM of two numbers) = (First number) X (Second number) 15. (c) Given, LCM = 1989 HCF = 13 1st number =117 and 2nd number = ? According to the formula, Product of LCM and HCF = Product of two numbers 17. (d) We know that, LCM of two numbers must be the multiple of their HCF. In the given options 60 is not a multiple of 8 and hence 60 cannot be the LCM of the numbers. 18. (6) Let the numbers be x and y. According to the question, x + y = 45...(i) ∴ Required LCM = 5x5x4 = 100 19. (d) According to the question, 1 st number = 3 m 2nd number = 4 m where, m = HCF But given, m = 4 We know that, 20. (6) Let numbers are 5x and 6x. Now, HCF of these two numbers is x. We know that, LCM x HCF = Product of two numbers => 480 X x = 5x X 6x => 480x = 30x2 =>x = 16 21. (6) Let the numbers are x, 2x and 3x. Where, x = HCF Given that, x = 23 ∴ The number are 23, 46 and 69. 22. (d) Let the numbers are 3x, 4x and 5x. Where, x=HCF Then, LCM = 60x According to the question, 60x = 1200 x = 20 24. (C) LCM of 11 and 13 will be (11 x 13). Hence, if a number is exactly divisible by 11 and 13, then the same number must be exactly divisible by their LCM or by (11 X 13). 25. (C) Let numbers are 2x and 3x. According to the question, 6x = 48 => x = 8 ( LCM = 6x).". Required sum = (2x + 3x) = 5x = 5 X 8 = 40 26. (6) Let numbers be lOx, 12x, 15x and 18x. Then, LCM = 180x As HCF = x Hence, required LCM = 180 X 3 = 540 27. (6) Given that, product of two numbers = 1500 HCF= 10 According to the formula, Product of two numbers = HCF X LCM => 1500 = 10 X LCM 28. (d) By Hit and Trial From option (d), we can say that the HCF of 36 and 24 is 12 and it is also satisfies the given condition a> b> 12 a = 36 and b = 24 29. (C) Let LCM = m, HCF = n, According to the question, m = 12n,...(i) and m + n = 403...(ii) => 12n + n = 403 [from Eq. (i)] => 13n = 403 30. (a) Given, LCM = 495, HCF = 5 Let 1st number = x, 2nd number = y xy = 495 X 5 [as product of two numbers = HCF x LCM] => xy = 2475 We know that, (x - yf = (x + yf - 4xy = (100)Z -4 X2475 = 10000 -9900 = 100 31. (d) Let HCF = x According to the question, LCM = 20x Given that. HCF + LCM = 2520 Now, LCM = 20x = 20 X 120 = 2400 We know that, 1st number X 2nd number = HCF X LCM ∴ Required answer = 600 X 3 = 1800 32. (c) Let the numbers be 66a and 66jb, where a and b are coprimes. According to the question, 66a + 66£ = 1056 => 66 (a + b) = 1056 /. Possible values of a and b are (a = l,i = 15), (a = 3,6 = 13). (a = 5,i> = ll), (a = 7,£ = 9).". Numbers are (66 X 1,66 X 15), (66 X 3,66 X 13), (66 X 5,66 X 11), (66 X 7,66 X 9). /. Possible number of pairs = 4 33. (6) Required length = LCM of 3 cm, 5 cm and 10 cm = 30 cm 34. (a) Required number = Multiples of LCM Now. LCM of 8. 9. 12. 15 and 18 ∴ LCM =2X2X3X3X2X5 = 360 /. The factors make it clear that to make a perfect square 360 must be multiplied by (2X5). ∴ Required number = 360 X 2 X 5 = 3600 35. (a) Greatest number of 3-digits = 999 LCM of 6, 9, 12 36. (a) Given that, x = 1356, y = 1868, z = 2764 a = b = c = 12 According to the formula, Required number = HCF of [(1356 - 12), (1868 -12), (2764 -12)] = HCF of (1344.1856 and 2752) ∴ HFC of 1344, 1856 and 2752 = 64 Hence, required number = 64 37. (a) Given that, x = 130,7 = 305, z = 245 a = 6, b = 9, c = 17 According to the formula, Required number = HCF of [(130 - 6), (305 - 9), (245 - 17)] = HCF of 124, 296, 228 = 4 38. (a) Given that, x = 1023, y = 750, a = 3, b = 2.'. Required number = HCF of [(1023 - 3), (750 - 2)] = HCF of 1020 and 748.'. Required number = 68 39. (d) Required capacity = HCF of 345 L, 120 L and 225 L /. Required capacity of container to measure the oil = 15 L 40. (c) LCM of 5, 6, 4 and 3 = 60 On dividing 2497 by 60, the remainder is 37. ∴ Number to be added = 60-37 = 23 41. (c) Given that, x = 1305, y = 4665, z = 6905 Then,.. Required number = HCF of 3360, 2240 and 5600 = 1120 On dividing 1305 by 1120, remainder is 185. On dividing 4665 by 1120, re- mainder is 185. On dividing 6905 by 1120, remainder is 185..'. Common remainder =185 42. (c) Given, x = 12, y = 16, z = 18, k = 5 According to the formula, Required number = (LCM of x, y and z) + k = (LCM of 12, 16, 18) + 5 LCM of 12, 16, 18 is ∴ LCM = 2X2X3X4X3 =144.'. Required number = 144 + 5 = 149 43. (b) When divide 13850 and 17030 by the number, the remainder is 17. So, find HCF of(13850-17) and (17030 -17) i.e., 13833 and 17013. Here, 47. (a) LCM of 24, 32, 36, 54 LCM = 2x2x2x3x3x3x4 = 864.'. Required number = LCM — 9 = 864-9 = 855 48. (a) LCM of 16, 18 and 20 = 720.'. Required number = 720k + 4 where, k is a natural number to be divisible by 7, (720 k + 4) will be a multiple of 7. Smallest value of k = 4 ∴ Required number = 720 X 4 + 4 = 2884 49. (a) LCM of 10, 15, 21 and 28 LCM = 2X3X5X7X2 =420 Larqest number of 4-diqits = 9999 /. Remainder = 339.'. Four-digit number divisible by 10, 15, 21 and 28 = 9999 - 339 = 9660 Also, k= 10 -4 = 15-9 = 21-15, 28-22 = 6 /. Required number = (9660 - k) = (9660 - 6) = 9654 50. (C) LCM of 32, 40, 48 and 60 = 480 The number divisible by 480 between 4000 and 6000 are 4320, 4800, 5280 and 5760. Hence, required number of num- bers are 4. 51. (a) Required length = HCF of 84m, 98 m and 126 m HCF of 84, 98 and 126.". Required length = 14 m 52. (Q) Least number of students = LCM of 5, 9, 10 LCM of 5, 9, 10 LCM = 5X9X2 = 90.'. Required number of students = 90 Exercise © Higher SkiU Level Questions 1. (6) Given that, (x-6) is the HCF of xz - 2x - 24 and x2 - kx - 6 i.e.,(x - 6) is a factor of both expressions. Let /(Xj) = x2 - 2xj - 24 and /(x2) = x2 — kxz — 6 Now, f(x{) = /(x2) at (X! = x2 = 6) => (6)2 - 2(6) - 24 = (6)2 - £(6) - 6 [by condition] => 0 =30 -6£:=>6i: = 30 k = 5 3. (a) Let p(x) = 8(x5 - x3 + x) = 4 X 2 X x (x4 - x2 + 1) and q(x) = 28(x6 + 1) = 7X4 [(x2)3 + (l)3] = 4 X 7 X (x2 + 1) (x4 - x2 + 1) ∴HCF of p(x) and q(x) = 4 (x4 - x2 + 1) 4. (£2) Let f(x) = (x4 - y4) = (x2 - y2) (x2 + y2) = (x - y) (x + 7) (x2 + y2) and gr(x) = (x6 - y6) = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x+y)(x2 -xy+ y2) (x - y) (x2 + xy+y2) = (x - y) (x + y) (x2 - xy + y2) (x2 + xy + y2) ∴HCF of [/(x), gr(x)] = (x - y) (x + y) = x2-y2 5. (c) Let /(x) = x3 - xz - 2x = x (x2 - x - 2) = x {x2 - 2x + x - 2) = x {x (x - 2) + 1 (x - 2)} = x (x + 1) (x - 2) and g(x) = x + x = x2 (x+ 1) = xx(x + 1) ∴ LCM of [f(x), gr(x)] = x (x + 1) x (x - 2) = x2 (x + 1) (x - 2) = x2 (x2 - x-2) = x4-x3-2x2 6. (c) a2i>4 + 2a2i2 = a2i>2 (i2 + 2)...(i) and (ab)7 - 4azb9 = a''V - 4a2 jb9 = a2£2 (a5jb5-4jb7)...(ii) From Eqs. (i) and (ii), HCF = aV 7. (6) HCF of (22n + 7,33n + 10) is always 1. Illustration For n = 1, HCF (29,43) => HCF = 1 For n = 2, HCF (51,76) => HCF = 1 For n = 3, HCF (73,109) => HCF = 1 8. (c) Putting arbitrary values of a and £>. Illustration 1 Let a = 9 and b = S HCF (8 + 9,9-8) HCF (17,1) = 1 Illustration 2 Let a = 23 and b = 17 HCF (17+ 23,23-17) HCF (40,6) = 2 Hence HCF(a + b,a- b) can either be 1 or 2. 9. (c) Time after which they will hit the target aaain toaether = LCM (5,6,7 and 8) = 5X3X7X2X4 =840s They will hit together again at 9 : 14 am. 10. (d) We have to find out the LCM of 48, 60 and 90. ∴ LCM = 2x3x2x5x2x2x3 /. Bell chime together again = 10 :00 am+ 12 h= 10 :00 pm 11. (C) The bells will toll together after time in seconds equal to the LCM of 9, 6, 4, 10 and 8. LCM of 9, 6,4, 10 and 8 is ∴ LCM = 2X2X3X3X5X2 = 360 In one hour, the rings will toll together 12. (c) Required time = LCM of 42, 56 and 63 s LCM of 42, 56, 63 is.'. Required time = 2x3x7x4x3= 504 s 13. (c) LCM of 10, 15 and 18 is LCM = 2X3X5X3 = 90 To make it perfect square, we multiply it with 2 x 5 = 10.". Required number of soldiers 90 X10 =900 14. (e) LCM of 2, 4, 6, 8, 10, 12 = (2X2X3X2X5) = 120 /. After every 2 min, they toll together..". Number of times they toll together in one hour 15. (c) Required length = HCF of 7 m, 3 m 85 cm, 12 m 95 cm = HCF of 700 cm, 385 cm, 1295 cm Hence, required length is 35 cm. 16. (d) Let the numbers be 21a and 21Jb, where a and b are coprimes. Then, 21a X 21 b = (21 X 4641) => aft = 221 Two coprimes with product 221 are 13 and 17. ∴ Required number = (21x13,21x17) = (273,357) 17. (c) HCF of 250 cm and 150 cm 18. (a) For such type of questions, remember the following method. Required number of apples = (LCM of 5, 6, 8, 9) + 3 LCM of 5, 6, 8, 9 is ∴ LCM = 2X3X5X4X3= 360.". Required number of apples = 360 + 3 = 363 Chapter 4 Simple and Decimal Fractions A digit which can be represented in p/q form, where q * 0, is called a fraction. Here, p is called the numerator and q is called the denominator. For example 3/5 is a fraction, where 3 is called numerator and 5 is called denominator. or When a unit is divided into any number of equal parts, then these parts are termed as a fraction of the unit. For examnle Tf 1 is to he divided into two ermal narts. then 1 is Simple Fraction Note Simple fraction is also known as vulgar fraction Types of Simple Fractions There are following types of fractions. 1. Proper Fraction When the numerator of a fraction is less than its denominator, then fraction is called proper fraction. 2. Improper Fraction When the numerator of a fraction is greater than its denominator, then fraction is called improper fraction. 3. Compound Fraction A fraction, in which numerator or denominator or both are in fraction, then it is called compound fraction. 4. Inverse Fraction If we inverse the numerator and the denominator of a fraction, then the res- ultant fraction will be the inverse fraction of the original fraction. 5. Mixed Fraction The fraction, which is the combination of integer and fraction, is called mixed fraction. 6. Continuous Fraction It has no certain definition but only say that a fraction contains addi- tional fractions in its denominators, is called continuous fraction. Note lb simplify a continuous fraction, start from bottom and work upwards Decimal Fraction If the fraction has denominator in the powers of 10, then fraction is called decimal fraction. To convert a decimal fraction into a vulgar fraction, place 1 in the denominator under the decim- al point. Then, after removing the decimal point, place as many zeroes after it as the number of digits after the decimal point. Finally, reduce the fraction to its lowest terms. Note Placing zeroes to the right of a decimal fraction, it does not make any change in value Hence, 0.5, 0.50, 0.500 and 0.5000 are equal. If numerator and denominator of a fraction have same number of decimal places, then each of the decimal points be removed Types of Decimal Fractions 1. Recurring Decimal Fraction The decimal fraction, in which one or more decimal digits are repeated again and again, is called recurring decimal fraction. To represent these fractions, a line is drawn on the digits which are repeated. 2. Pure Recurring Decimal Fraction When all the digits in a decimal fraction are repeated after the decimal point, then the decimal fraction is called as pure recurring decimal fraction. To convert pure recurring decimal fractions into simple fractions (vulgar form), write down the repeated digits only once in numerator and place as many nines in the denominator as the num- ber of diaits reneatincr. Since, there is only 1 repeated digit. Therefore, only single 9 is placed in denominator. Since, there are only 2 repeated digits. Therefore, two 9's are placed in denominator. 3. Mixed Recurring Decimal Fraction A decimal fraction in which some digits are repeated and some are not repeated after decimal is called as mixed recurring decimal fraction. To convert mixed recurring decimal fractions into simple fractions, in the numerator, take the difference between the number formed by all the digits after decimal point (repeated digits will be taken only once) and the number formed by non-repeating digits. In the denominator, place as many nines as there are repeating digits and after nine put as many zeroes as the number of non-repeating digits. Important Facts Related to Simple and Decimal Frac- tions ♦ If in a fraction, numerator is equal to denominator, then the value of fraction is equal to 1 ♦ If the numerator of a fraction is always non-zero and denominator is zero, then the value of fraction is infinity (°°). ♦ If the numerator of a fraction is zero and denominator is not equal to zero, then the value of fraction is zero. ♦ If the numerator or denominator of any fraction is either multiplied or divided by same number, then the value of fraction remains unchanged. ♦ If the numerator and denominator have no common factor other than 1, then the fraction is said to be in its lowest form ♦ A fraction is a rational number as it can be expressed in the form of p/q and q * 0 Operations on Simple Fractions Addition of Simple Fractions 1. When Denominators are Same If denominators of fractions are same, then numerators of fractions are added and their addition is divided by denominator. 2. When Denominators are Different If denominators of fractions are not same, then make their denominators equal (by taking their LCM) and then add their numerators. Subtraction of Simple Fractions 1. When Denominators are Same If denominators of fractions are same, then numerators of fractions are subtracted and their subtraction is divided by the denominator. 2. When Denominators are Different If denominators of fractions are not same, then make their denominators equal and then subtract their numerators. Multiplication of Simple Fractions To multiply two or more simple fractions, multiply their numerators and denominators. If fractions are given in mixed form, first convert them into improper fraction and then multiply. Division of Simple Fractions To divide two fractions, first fraction is multiplied by the inverse of second fraction Operations on Decimal Fractions Addition and Subtraction of Decimal Fractions To add or subtract decimal fractions, the given numbers are written under each other such that the decimal points lie in one column and the numbers so arranged can now be added or subtrac- ted as per the conventional method of addition and subtraction. Ex. 1 (i) 353.5 + 2.32 + 43.23 = ? (ii) 1000 - 132.23 = ? Multiplication of Two or More Decimal Fractions Given fractions are multiplied without considering the decimal points and then in the product, decimal point is marked from the right hand side to as many places of decimal as the sum of the numbers of decimal places in the multiplier and the multiplicand together. Ex. 2 (i) 4.3 x 0.13 = ? (ii) 1.12 x 2.3 x 4.325 = ? Sol. (i) 43 x 13 = 559 Sum of the decimal places = (1 + 2) = 3.. Required product = 0.559 (ii) 112 x 23 x 4325 = 11141200 Sum of the decimal places = (2 + 1 + 3) = 6 ∴ Required product = 11.141200 Multiplication of Decimal Fraction by an Integer Given integer is multiplied by the fraction without considering the decimal point and then in the product, decimal is marked as many places before as that in the given decimal fraction. Ex. 3 Find the value of the following. (i) 19.72x4 (ii) 0.0745x10 (iii) 3.52x14 Sol. (i) 19.72 x 4 Multiplying without taking decimal point into consideration 1972 x 4 = 7888 So, 19.72 x 4= 78.88 Since, in the given decimal fraction, decimal point is two places before. So, in the product, decimal point will also be put two places before. Similarly, (ii) 0.0745 x 10 = 0.745 (iii) 3.52 x 14 = 49.28 Dividing a Decimal Fraction by an Integer Do simple division i.e., divide the given decimal number without considering the decimal point and place the decimal point as many places of decimal as in the dividend. Ex. 4 Divide the following. (i) 0.81 * 9 (ii) 1.2875 * 25 (iii) 0.00049 + 7 Division of Decimal Fractions In such divisions, dividend and divisor both are multiplied first by a suitable multiple of 10 to convert di- visor into a whole number and then above mentioned rule of division is followed. Ex. 5 Divide the following. (i) 42 + 0.007 (ii) 0.00048 * 0.8 Comparison of Simple Fractions Following are some techniques to compare fractions. Cross Multiplication Method By Changing Fractions in Decimal Form To compare two or more fractions, first convert fractions into decimal form and then compare. By Equating Denominators of Given Fractions For comparison of fractions, take LCM of the denominators of all fractions, so that the denomin- ators of all fractions are same. Now, the fraction having largest numerator is the largest fraction. By Equating Numerators of Given Fractions For comparison of fractions, take LCM of the numerator of all fractions, so that numerators of all the fractions are same. Now, the fraction having smallest numerator will be largest. Fast Track Formulae to solve the QUESTIONS Formula 1 To represent any fraction in simplified form, divide its numerator and denominator by their HCF. Formula 2 If in the given fractions, the difference between numerator and denominator are same, then frac- tion having larger numerator is the largest and fraction having smaller numerator is the smallest. Ex. 11 Arrange the given fractions in increasing order, Formula 3 if in the given fractions, the numerators are increasing by a definite value and the denominator is also increasing by a definite value but the value of denominator is greater than numerator, then the fraction having smaller numerator will be the smallest fraction and the fraction having larger numerator will be the largest fraction. Sol. Since, in the given fractions, numerator value is increasing by 3 and denominator value is increasing by 6 and 6 > 3. Then, the fraction having larger numerator will be the larger fraction. Formula 4 Ex. 13 Arun was to find 6/7 of a fraction. Instead of multiplying, he divided the fraction by 6/7 and the result obtained was 13/70 greater than original value. Find the fraction given to Arun? Fast Track Practice Exercise© Base Level Questions 12.1088.88 + 1800.08 + 1880.80 = ? [Bank PO 2010] (a) 8790.86 (b) 8890.86 (c) 5588.80 (d) 4769.76 (e) None of the above 13.6435.9 + 7546.4 + 1203.5 = ? [Bank PO 2010] (a) 15188.5 (b) 15185.8 (C) 15155.5 (d) 15815.8 (e) None of the above 14.726.34 + 888.12 - ? = 1001.88 [1DBISO2010] fa; 612.58 (b) 602.64 fc; 654.54 (d) 618.78 fej None of the above (a) X 900 (b) X 940 (c) X 875 (d) X 975 38. A, B, C and D purchase a gift worth X 60. A pays 1/2 of what others are paying, B pays l/3rd of what others are paying and C pays l/4th of what others are paying. What is the amount paid byD? [SSCCGL2013] (a) 14 (b) 15 (c) 16 (d) 13 39. l/4th of number of boys and 3/8th of number of girls participated in annual sports of the school. What fractional part of total number of students participated? [Bank PO 2007] (a) 32% (b) 20% (c) 36% (d) Data inadequate (e) None of the above 40. The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator becomes eight times the numerator, then find the fraction. [SSC CGL 2013] 42. If the fraction a/b is positive, then which of the following must be true? [SSC Multitasking 2014] (aj a > 0 (b) b > 0 (c) ab>0 (d) a + b > 0 Exercise © Higher SkiU Level Questions (a) 0.052 [SSC CPO 2007] (b) 0.698 (c) 0.75 (d) Cannot be determined 4. Find the LCM of the fractions in above question. (a) 13.5 (b) 37.5 (c) 10.5 (d) Cannot be determined (e) None of the above 6. 1/8 part of a pencil is black and 1/2 part of the remaining is white. If the remaining part is blue and length of this blue part is 3% cm, then find the length of the pencil. (a) 6 cm (b) 7 cm (c) 8 cm (d) 9 cm (e) None of the above 7. How many pieces of 13.2 cm can be cut from a 330 cm long rod? [Bank Clerks 2009] (a) 25 (b) 28 (c) 21 (d) 35 (e) None of the above 8. In the year 2011, Shantanu gets ? 3832.5 as his pocket allowance. Find his pocket allowance per day. (a) ? 9.5 (b) ? 10.5 (c) ? 12.5 Ca[)? 11.5 (e) None of the above greatest fraction is divided by the smallest fraction, the result is 7/6, which is greater than the middle frac- tion by 1/3. Find all the three fractions. 16. A man reads 3/8 of a book on a day and 4/5 of the remainder on the second day. If the number of pages still unread are 40, then how many pages did the book contain? (a) 300 (b) 500 (c) 320 (d) 350 17. 4/7 of a pole is in the mud. When 1/3 of it is pulled out, 250 cm of the pole is still in the mud. Find the full length of the pole. (a) 1000 (b) 1100 (c) 950 (d) 1050 (e) None of the above Answer with Solutions Exercise© Base Level Questions Exercise © Higher SkiU Level Questions Chapter 5 Square Root and Cube Root Square If a number is multiplied with itself, then the result of this multiplication is called the square of that number. For example (i) Square of 6 = 6 x 6 = 36 (ii) Square of 12 = 12 x 12 = 144 (iii) Square of 100 = 100 x 100 = 10000 Methods to Find Square Different methods to calculate the square of a number are as follows Multiplication Method In this method, the square of any 2-digit number can be calculated by the following given steps. Step I Square the unit's digit, {If the square has two digits, then write ten's digit as carry.} Step II 2 x Ten's digit X Unit's digit + Carry Step III (Ten's digit)2 + Carry from step II Step IV Now arrange the numbers first write step III number, then step II and at unit's place step I. For example Find the square of 74 Sfepl (4)2 = 16 {Carry = 1} Step II 2 x 7 x 4 + 1 = 57 {Carry = 5} Step III (7)2 + 5 = 49+ 5 = 54 Step IV (74)2 = 5476 Algebraic Method To calculate square by this method, two formulae are used. (i) (a + b)2 = a2 + b2 + 2ab (ii) {a-bf = a2 + b2 -2ab For example The square of 34 is (34)2 = (30 + 4)2 = (30)2 + (4)2 + 2 x 30 x 4 = 900+ 16+240 (34)2 = 1156 Square of Decimal Numbers To find the square of any decimal number, write the square of the number ignoring the decimal and then place the decimal twice the place of the original number starting from unit's place. For example The square of 3.5 is as follows (35)2 = 1225 Here, the decimal is after one-digit in 3.5. Hence, the decimal will be placed twice the place of original number in the result. (3.5)2 = 12.25 Square Root The square root of a number is that number, the square of which is equal to the given number. There are two types of square roots of a number, positive and negative. It is denoted by the sign 'yj~'. For example 49 has two square roots 7 and - 7, because (7)2 = 49 and (- 7)2 = 49. Hence, we can write V49 = ± 7. Methods to Find Square Root Different methods to calculate the square root of a number are as follows Prime Factorisation Method This method has the following steps Step I Express the given number as the product of prime factors. Step II Arrange the factors in pairs of same prime numbers. Step III Take the product of these prime factors taking one out of every pair of the same primes. This product gives us the square root of the given number, Ex. 2 Find the square root of 1024. Sol. Prime factors of 1024 =2x2x2x2x2x2x2x2x2x2 Division Method If it is not easy to evaluate square root using prime factorisation method, then we use division method. The steps of this method can be easily understood with the help of following examples. Ex. 3 Find the square root of 18769. Sol. Step I In the given number, mark off the digits in pairs starting from the unit digit. Each pair and the remaining one-digit (if any) is called a period. Step II Now, 1=1; On subtracting, we get 0 (zero) as remainder. Step III Bring down the next period, i.e. 87. Now, the trial divisor is 1x2 = 2 and trial dividend is 87. So, we take 23 as divisor and put 3 as quotient. The remainder is 18 now. Step IV Bring down the next period, which is 69. Now, trial divisor is 13 x 2 = 26 and trial dividend is 1869. So, we take 267 as dividend and 7 as quotient. The remainder is 0. Step V The process (processes like III and IV) goes on till all the periods (pairs) come to an end and we get remainder as 0 (zero) now. Hence, the required square root = 137 Ex. 4 What is the square root of 151321? Sol. /. Required square root = 389 Properties of Squares and Square Roots The difference of squares of two consecutive numbers will always be equal to the sum of the number i.e., (a2 -b2) = (a + b)(a-b). Here, a>b and (a, b) being consecutive (a — b) =1. If a = 12 and b =11, then (122-112) = (12 + 11)(12-11) = 23 If the square of any number ends with 1, then its square root will end with 1 or 9. If the square of any number ends with 4, then its square root will end with 2 or 8. If the square of any number ends with 5, then its square root will end with 5. If the square of any number ends with 6, then its square root will end with 7 or 6. If the square of any number ends with 9, then its square root will end with 3 or 7. The square of any number always ends with 0,1,4, 5,6 or 9 but will never end with 2,3, 7 or 8, Square root of negative number is imaginary. Important Relations Square Root of Decimal Numbers If in a given decimal number, the number of digits after decimal are not even, then we put a 0 (zero) at the extreme right, So that these are even number of digits after the decimal point. Now, periods are marked as marked in previous explanation starting from right hand side before the decimal point and from the left hand after the decimal digit. For example 156.694 There are odd number of digits after decimal. So, we put a zero after the digit, so that there are even digits after the decimal 156.6940 Now, periods are marked as After the periods are marked, then previous method is used to find the square root Ex.5 Find the square root of 147.1369. Sol. Here, 147.1369 contains even digits after decimal, so there is no need to add zero after the the last digit, now period are marked as 147.1369 Ex.6 Find the square root of 149.597361. Sol. Square Root of a Fraction To find square root of a fraction, we have to find the square roots of numerators and denominators, separ- ately. Note Sometimes, numerator and denominator are not a complete square. In these types of cases, it is better to convert the given fraction into decimal fraction to find the square root. Fast Track Formulae to solve the QUESTIONS Formula 1 If in a given number, the total number of digits are n and if n is even, then square root of that number will have n/2 digits and if n is odd, then Ex. 10 How many digits are there in square root of 1838736. Sol. Since, the total number of digits are 7 and 7 is odd number. Formula 2 If any number has 5 in unit's place, then its square can be calculated as Note Student should understand the dot(-)just as a separation between numbers Ex. 11 Find the square of 125? Sol. (125)2 = 12 (12 + 1) 25= 15625 Fast Track Techniques to solve the QUESTIONS To find smallest and largest n-digit number which is a perfect square. :. 17 Find the 5-digit number which is a perfect square. Sol. The bissest 5-digit number = 99999 Clearly, the biggest 5-digit number which is a perfect square, is less by 143. ∴ Number = 99999 - 143 = 99856 Ex. 18 Find the smallest 4-digit number which is a perfect square. Sol. Smallest 4-digit number = 1000 Clearly, the smallest 4-digit number which is a perfect square, will be greater by 24. ∴ Number = 1000 + 24 = 1024 Cube If a number is multiplied two times with itself, then the result of this multiplication is called the cube of that number. For example (i) Cube of 6=6x6x6=216 (ii) Cube of 8 =8 x8 x8=512 Methods to Find Cube Different methods to calculate the cube of a number are as follows Algebraic Method To calculate cube by this method, two formulae are used. (i) (a + b)3 = a3 + 3ab (a+b) + b3 (ii) (a -bf = a3 -3ab (a-b)- b3 For example The cube of 16 is (16)3 = (10 + 6)3 = (10)3 + 3 x 10 x 6 (10 + 6) + (6)3 = 1000 + 2880 + 216 = 4096 Shortcut Method Step I The answer consists of 4 parts each of which has to be calculated separately, Step II First write down the cube of ten's digit to the extreme left. Write the next two terms to the ricrht of it bv creatine; GP (Geo- metric Proaression) havina Step III Write the double of 2nd and 3rd number below them. Step IV Now, add the number with numbers written below it and write the unit's place digit in a straight line and remaining number is carried forward to the next number. Ex. 19 Find the cube of 35. Sol. Here, unit's digit is 5 and ten's digit is 3. Step I Write the cube of ten's digit at extreme left i.e., (3)3=27 Step II Now, the next two terms on the right will be in a GP of common ratio equals Cube Root Methods to Find Cube Root Method to calculate the Cube root of a number is as follow Prime Factorisation Method This method has following steps. Step I Express the given number as the product of prime factors. Step II Arrange the factors in a group of three of same prime numbers. Step III Take the product of these prime factors picking one out of every group (group of three) of the same primes. This product gives us the cube root of given number. Properties of Cubes and Cube Roots If the cube of a number is of 2 or 3-digits, then its cube root will be of 1-digit If the cube of a number is of 4, 5 or 6 digits, then its cube root will be of 2 digits. If the cube of a number have 0,1, 2,3,4, 5, 6, 7,8,9 in its unit's place, then its cube root will haveO, 1, 8, 7, 4, 5, 6, 3, 2 or 9 in their unit's place, respectively. There are only three numbers whose cube is equal to the number i.e., {of =0 ; (if =1 ; (-If =-1 Multi Concept QUESTIONS Fast Track Practice Exercise O Base Level Questions (a) 0.948 (b) 0.984 (c) 0.988 (d) 938 (e) None of the above 37. Sukhiram plants 15376 orange trees in his garden and arranges them, so that there are as many rows as there are orange trees in each row. Find the number of rows. (a) 125 (b) 124 (c) 128 (d) 135 (e) None of the above 38. What is the least number to be multiplied with 294 to make it a perfect square? [CBI2007] (a) 2 (b) 3 (c) 6 (d) 24 (e) None of the above 39. What least number should be subtracted from 6860, so that 19 be the cube root of the result from this subtraction? [LIC ADO 2009] (a) 1 (b) 2 (c) 3 (d) 5 (e) None of the above 40. What is the least number to be added to 8200 to make it a perfect square? [Bank Clerks 2009] (a) 81 (b) 100 (c) 264 (d) 154 (e) None of the above 41. Find the difference in 777 and its nearest perfect square number. (a) 4 (b) 7 (c) 27 (d) 28 (e) None of the above (a) 14 (b) 2 (c) 15 (d) 8 (e) None of the above 49. Each student of class 10 contributed some money for a picnic. The money contributed by each student was equal to the cube of the total number of students. If the total collected amount was ? 29791, then find the total number of students. [UP Police 2008] (a) 15 (b) 27 (c) 31 (d) 34 50. (Smallest common multiple of 12 and 16) x (Smallest common multiple of 10 and 15) is equal to [CTET 2012] (a) 960 (b) 720 (c) 1440 (d) 480 51. Find the value of 2 x V3 upto three places of decimal. (a) 1.732 (b) 3.464 (c) 4.464 (d) 2.732 (e) None of the above (a) 17 (b) 18 (c) 72 (d) 21 (e) None of the above 54. If (46) is subtracted from the square of a number. The answer so obtained is 485. What is the number? [SBI 2012] (a) 49 (b) 51 (c) 56 (d) 53 (e) None of the above Exercise © Higher Skill Level Ques

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