Youth Competition Times RRB MATHEMATICS Chapterwise Solved Papers (Computer Based Test) CHIEF EDITOR A.K. Mahajan EDITOR Advocate Abhishek Singh WRITERS Ambuj Kumar, Anand Soni, Kamlesh Shrivastava, Vinit Shrivastava, Hemant Sharma, Rahul Varma COMPUTER GRAPHICS BY Balkrishna, Charan Singh, Vinay Sahu EDITORIAL OFFICE Youth Competition Times 12, Church Lane Prayagraj-211002 Mob. : 9415650134 Email : [email protected] website : www.yctbooks.com PUBLISHER DECLARATION Edited and Published by A.K. Mahajan printed by Printex India, Prayagraj. Youth Competition Times 12, Church Lane, Prayagraj In order to publish the book, full care has been taken by the editor and the publisher, Still your suggestions and queries are welcomed. Rs. : 495/- In case of any dispute, the judicial area will be Prayagraj.
CONTENTS Analysis chart of RRB Paramedical, JE, ALP, NTPC, Group-D, RPF SI & Constable Pre Exam Questions Papers.................................................................................................................7 RRB Paramedical, JE, ALP, NTPC, Group-D, RPF SI & Constable Papers Through Pie Chart and Bar Graph............................................................................................ 8 Section-1 : Number System ...................................................................................................... 9-47 Type 1 Problems based on Divisibility ..............................................................................................................9 Type 2 Problems based on specificity of digits ...............................................................................................15 Type 3 Problems based on composite and prime numbers ...........................................................................17 Type 4 Problems based on basic notion of numbers ......................................................................................20 Type 5 Problems based on Rational and Irrational numbers.......................................................................32 Type 6 Problems based on powers and exponents of numbers.....................................................................37 Type 7 Problems based on unit digit and factorization of numbers.............................................................40 Type 8 Problems based on place value and numerical value ........................................................................43 Type 9 Miscellaneous........................................................................................................................................44 Section-2 : Decimal Fractions ................................................................................................ 48-80 Type 1 Problems based on finding the smallest and largest fractions..........................................................48 Type 2 Problems based on ascending and descending order of fractions....................................................51 Type 3 Problems based on Terminating and Recurring Decimal Values of Fractions...............................54 Type 4 Problems based on Bar of Decimal numbers .....................................................................................57 Type 5 Problems based on finding the simplest value of the fraction ..........................................................60 Type 6 Problems based on the addition of a fraction and its reciprocal .....................................................63 Type 7 Problems based on the sum and difference of fractions....................................................................64 Type 8 Problems based on the value of decimal and fractional expressions ...............................................70 Type 9 Problems based on options ..................................................................................................................73 Type 10 Problems based on finding the fraction ..............................................................................................73 Type 11 Miscellaneous ........................................................................................................................................78 Section-3 : Indices and Surds ................................................................................................... 81-96 Type 1 Problems based on finding square root of numbers..........................................................................81 Type 2 Problems based on finding the square root of decimal numbers .....................................................86 Type 3 Problems based on sum and product of square root of numbers.....................................................87 Type 4 Problems based on finding the square root of fractional numbers..................................................91 Type 5 Miscellaneous........................................................................................................................................94 Section-4 : Simplification ..................................................................................................... 97-125 Type 1 Simple problems related to addition, subtraction, multiplication and division..............................97 Type 2 Problems based on the rule of BODMAS.........................................................................................100 Type 3 Problems based on formulas and on finding the value of a term of an expression.......................105 Type 4 Problems based on simplification of decimal and fractional expressions......................................113 Type 5 Simplification of numbers based on surds .......................................................................................120 Type 6 Miscellaneous......................................................................................................................................123 2
Section-5 : Lowest Common Multiple & Highest Common Factor ...............................126-152 Type 1 Common problems related to LCM..................................................................................................126 Type 2 Remainder problems related to LCM ..............................................................................................128 Type 3 Common problems related to HCF ..................................................................................................131 Type 4 Remainder problems related to HCF ...............................................................................................135 Type 5 Decimal and Fractional Problems related to LCM and HCF ........................................................137 Type 6 Combined problems of LCM and HCF............................................................................................138 Type 7 Problems based on Ratio, Sum, Difference and Factor of LCM and HCF ...................................141 Type 8 Problems related to square tiles ........................................................................................................145 Type 9 Problems based on Alarm/Bell/Light etc..........................................................................................146 Type 10 Divisibility problems based on LCM and HCF ...............................................................................148 Section-6 : Percentage ........................................................................................................ 153-185 Type 1 Problems based on Population ..........................................................................................................153 Type 2 Problems based on Exam and Students ...........................................................................................155 Type 3 Problems based on Income, Expenditure and Savings ...................................................................161 Type 4 Problems based on Percentage of Numbers .....................................................................................164 Type 5 Problems based on Percentage change.............................................................................................170 Type 6 Problems based on investment and business....................................................................................174 Type 7 Problems related to finding percentage quantity ............................................................................177 Type 8 Problems based on finding percentage.............................................................................................178 Type 9 Problems based on percentage change in area ................................................................................184 Type 10 Problems based on voting ..................................................................................................................183 Section-7 : Profit & Loss .................................................................................................... 186-208 Type 1 Problems based on finding the percentage of profit and loss .........................................................186 Type 2 Problems based on finding the cost price .........................................................................................190 Type 3 Problems based on finding the selling price.....................................................................................195 Type 4 Problems based on finding the amount of profit and loss...............................................................198 Type 5 Problems based on finding the ratio of values .................................................................................200 Type 6 Problems based on profit or loss of cost/sale of two articles...........................................................201 Type 7 Problems based on goods bought and sold at a particular rate .....................................................202 Type 8 Miscellaneous......................................................................................................................................206 Section-8 : Discount ............................................................................................................ 209-218 44444444 44 Section-9 : Ratio & Proportion.......................................................................................... 219-239 Type 1 Problems based on ratio of two parts ...............................................................................................219 Type 2 Problems based on ratio of three parts.............................................................................................225 Type 3 Problems based on new ratio due to increase or decrease in original ratio ..................................227 Type 4 Problems based on consecutive ratio and proportion ....................................................................233 Type 5 Problems based on Coins/balls etc ....................................................................................................235 Type 6 Miscellaneous......................................................................................................................................237 Section-10 : Partnership .................................................................................................... 240-248 Type 1 Problems based on finding capital in partnership .............................................................................240 Type 2 Problems based on sharing of profits in partnership ........................................................................244 Type 3 Miscellaneous ........................................................................................................................................247 3
Section-11 : Work & Time ................................................................................................. 249-282 Type 1 Problems based on the involvement of two persons in the work....................................................249 Type 2 Problems based on the involvement of three persons in the work ................................................252 Type 3 Problems based on the involvement of a group of persons in the work ........................................260 Type 4 Problems based on part of work and remaining work etc..............................................................264 Type 5 Problems based on leaving and joining in the middle of the work.................................................267 Type 6 Problems based on work efficiency and wages etc ..........................................................................275 Type 7 Miscellaneous......................................................................................................................................279 Section-12 : Alligation ........................................................................................................ 283-291 Type 1 Problems based on the value and quantity of substances in a mixture .........................................283 Type 2 Problems based on finding ratio of substances in a mixture .........................................................286 Type 3 Miscellaneous .....................................................................................................................................290 Section-13 : Pipe & Cistern ................................................................................................ 292-320 Type 1 General Problems based on Pipe and Cistern .................................................................................292 Type 2 When one Pipe is fills and the other Pipe is empties .......................................................................296 Type 3 When more than two Pipes work together ......................................................................................300 Type 4 When a Pipe is turned on or off in the middle .................................................................................306 Type 5 When the Pipe is opened alternately.................................................................................................312 Type 6 Problems based on capacity ..............................................................................................................315 Type 7 Miscellaneous .....................................................................................................................................316 Yeeie-14 : Simple Interest).................................................................................................. 321-346 Type 1 Problems based on finding simple interest.......................................................................................321 Type 2 Problems based on finding the Principal .........................................................................................324 Type 3 Problems based on finding the amount ............................................................................................332 Type 4 Problems based on finding the rate ..................................................................................................334 Type 5 Problems based on finding the time..................................................................................................341 Type 6 Miscellaneous......................................................................................................................................345 Section-15 : Compound Interest........................................................................................ 347-383 Type 1 Problems based on finding the Principal .........................................................................................347 Type 2 Problems based on finding the Amount ...........................................................................................351 Type 3 Problems based on finding the Compound Interest ........................................................................356 Type 4 Problems based on Simple and Compound Interest .......................................................................361 Type 5 Problems based on the difference between simple and compound interest ..................................369 Type 6 Problems based on finding the rate ..................................................................................................374 Type 7 Problems based on finding the time..................................................................................................378 Type 8 Miscellaneous......................................................................................................................................381 Section-16 : Problems Based on Age ................................................................................. 384-418 Type 1 Problems based on finding the present age of a person ..................................................................384 Type 2 Problems based on finding the age of two persons ..........................................................................400 Type 3 Problems based on finding the sum and difference of ages ............................................................404 Type 4 Problems based on ratio of ages........................................................................................................410 Type 5 Problems based on finding the age at a particular time .................................................................414 Type 6 Miscellaneous......................................................................................................................................415 4
Section-17 : Average........................................................................................................... 419-434 Type 1 Problems based on the average of consecutive numbers ................................................................419 Type 2 Problems based on examination, students and marks ....................................................................421 Type 3 Problems based on average age and average weight.......................................................................425 Type 4 Problems based on finding the value of any one result ...................................................................428 Type 5 Average problems based on table .....................................................................................................429 Type 6 Miscellaneous......................................................................................................................................431 Section-18 : Speed, Time & Distance ................................................................................ 435-458 Type 1 Problems based on finding speed ......................................................................................................435 Type 2 Problems based on finding time ........................................................................................................440 Type 3 Problems based on finding distance..................................................................................................445 Type 4 Problems based on Average speed ....................................................................................................454 Type 5 Proportional questions of speed, time and distance ........................................................................458 Section-19 : Train ............................................................................................................... 459-480 Type 1 Simple problems related to train ......................................................................................................459 Type 2 When the train crosses a person or a pole........................................................................................461 Type 3 When the train crosses another moving person...............................................................................463 Type 4 When the train crosses a platform or bridge ...................................................................................465 Type 5 When the train crosses a platform and a person or a pole, etc. .....................................................470 Type 6 Problems based on two trains having same direction .....................................................................471 Type 7 When two trains start in opposite directions from two places .......................................................473 Type 8 Problems based on average speed of trains......................................................................................477 Type 9 Miscellaneous......................................................................................................................................478 Section-20 : Boat & Stream ............................................................................................... 481-489 Type 1 Problems based on finding the speed of stream...............................................................................481 Type 2 Problems based on finding the speed of boat/person etc ................................................................485 Type 3 Problems based on finding the average speed .................................................................................486 Type 4 Problems based on finding the ratio of the speeds ..........................................................................487 Type 5 Problems based on finding the distance and time ...........................................................................488 Section-21 : Mensuration ................................................................................................... 490-554 Type 1 Problems based on Triangles ............................................................................................................490 Type 2 Problems based on Quadrilateral .....................................................................................................498 Type 3 Problems based on Circle ..................................................................................................................500 Type 4 Problems based on Square ................................................................................................................506 Type 5 Problems based on Rectangle............................................................................................................514 Type 6 Problems based on Cube....................................................................................................................524 Type 7 Problems based on Cuboid ................................................................................................................527 Type 8 Problems based on Cylinder..............................................................................................................530 Type 9 Problems based on Cone....................................................................................................................534 Type 10 Problems based on Sphere/Hemisphere ...........................................................................................538 Type 11 Problems based on Prism/Pyramid...................................................................................................541 Type 12 Miscellaneous ......................................................................................................................................542 Section-22 : Algebra............................................................................................................ 555-596 Type 1 Problems based on Arithmetic and Geometric Progression...........................................................555 Type 2 Problems based on LCM and HCF of Algebraic Expressions .......................................................560 Type 3 Problems based on Linear Equations ...............................................................................................562 Type 4 Problems based on Algebraic formulas............................................................................................563 5
Type 5 Problems based on Divisibility of Polynomials ................................................................................574 Type 6 Problems based on Factors of Polynomials......................................................................................578 Type 7 Problems based on Quadratic Equation and its Discriminant.......................................................580 Type 8 Problems based on Algebraic Expressions.......................................................................................589 Type 9 Problems based on Sets......................................................................................................................591 Type 10 Miscellaneous......................................................................................................................................596 Section-23 : Trigonometry ................................................................................................. 597-635 Type 1 Problems based on Trigonometric Functions ..................................................................................597 Type 2 Problems based on exponents of trigonometric functions ..............................................................610 Type 3 Problems based on Angular value ....................................................................................................614 Type 4 Problems based on angular values of consecutive trigonometric functions ..................................619 Type 5 Problems based on height and Distance ...........................................................................................622 Type 6 Miscellaneous......................................................................................................................................633 Section-24 : Co-ordinate Geometry................................................................................... 636-645 Type 1 Problems based on finding the coordinate point .............................................................................636 Type 2 Problems based on part made up of Dots.........................................................................................639 Type 3 Problems based on finding the Equation .........................................................................................642 Type 4 Miscellaneous......................................................................................................................................643 Section-25 : Geometry ........................................................................................................ 646-693 Type 1 Problems based on Triangles ............................................................................................................646 Type 2 Problems based on Quardrilateral ...................................................................................................658 Type 3 Problems based on Rhombus ............................................................................................................660 Type 4 Problems based on Parallelogram ....................................................................................................663 Type 5 Problems based on Trapezium..........................................................................................................666 Type 6 Problems based on Circle ..................................................................................................................667 Type 7 Problems based on tangent to circle .................................................................................................674 Type 8 Problems based on Polygons .............................................................................................................680 Type 9 Problems based on Complementary/Supplementary Angle)..........................................................687 Type 10 Miscellaneous ......................................................................................................................................688 Section-26 : Elementary Statistics/Probability................................................................. 694-731 Type 1 Problems based on Mean of Data .....................................................................................................694 Type 2 Problems based on Median of Data ..................................................................................................702 Type 3 Problems based on Mode of Data .....................................................................................................710 Type 4 Problems based on Standard Deviation ...........................................................................................712 Type 5 Problems based on Range..................................................................................................................715 Type 6 Miscellaneous......................................................................................................................................717 Probability ..........................................................................................................................................728 Section-27 : Data Interpretation........................................................................................ 732-772 Type 1 Problems based on Pie-Chart............................................................................................................732 Type 2 Problems based on Chart ..................................................................................................................741 Type 3 Problems based on Bar Graph ..........................................................................................................754 Type 4 Problems based on Line Graph.........................................................................................................767 Section-28 : Miscellaneous ................................................................................................. 773-784 6
Analysis chart of Question Papers of Various Previous Exams of RRB S.N. Exams Exam year Total Total Maths questions 2020-21 question 30 × 133 = 3990 1. RRB NTPC 2019 Stage-1 paper 133 2. RPF Constable 2018 2019 17 35 × 17 = 595 3. RPF SI 2018 2019 23 35 × 23 = 805 4. RRB JE 2018 2019 38 38 × 30 = 1140 5. RRB ALP 2018 2019 18 18 × 40 = 720 Stage-2 2019 6 18 × 7 = 126 2018 30 25 × 30 = 750 6. RRB Paramedical 2019 2018 135 25 × 135 = 3375 7. RRB ALP/Tech. 2018 Stage-1 8. RRB Group D 2018 9. RRB NTPC 2015 2017 9 35 × 9 = 315 Stage-2 10. RRB NTPC 2015 2016 63 30 × 63 = 1890 Stage-1 Total 472 13,706 Note– In this book, out of total 472 papers of JE, ALP, NTPC, RPF Constable, RPF SI and Group D exams conducted by RRB, out of total 13706 questions asked from General Mathematics, total 9236 questions of General behavior have been removed and chapterwise compilation of 4470 questions of different types has been presented. In this book, every effort has been made by the Examination Special Committee to accommodate maximum variety of questions, so that the examinees can be made aware of the variety of questions asked by RRB. 7
Trend Analysis of Previous Year RRB Paramedical, JE, ALP, NTPC, Group-D, RPF SI & Constable, Papers Through Pie Chart and Bar Graph 8
01. Number System Type - 1 697 = 17+ (n–1)×17 680 = (n–1)× 17 1. When a number n is divided by 5, the 40 = n –1 remainder is 2. When n2 is divided by 5, the n = 41 remainder will be: Hence, required number (n) = 41 (a) 3 (b) 1 4. When 19300 is divided by 20, find the (c) 4 (d) 0 remainder. RRB NTPC 07.01.2021 (Shift-I) Stage Ist (a) 2 (b) 1 Ans. (c) : Number = Divisor × Quotient + Remainder (c) 3 (d) 4 According to question, RRB NTPC 29.01.2021 (Shift-II) Stage Ist n=5×q+2 Ans. (b) : From question, On squaring both the sides, 19 300 ⇒ (20 − )1 300 ⇒ 0 + (–1)300 = 1(Remainder) n2 = 25q2 + 4 + 20q 20 20 On dividing by 5 – 5. Which of the following is the greatest three n2 = 5q2 + 4 + 4q or n2 = 5(5q2 + 4q) + 4 55 digit number that is divisible by 13? (a) 990 (b) 575 Hence, required remainder will be 4. (c) 988 (d) 908 2. How many numbers of the first 100 positive RRB NTPC 18.01.2021 (Shift-II) Stage Ist integers are divisible by 3 or 4 without a Ans. (c) : Greatest three digit number = 999 remainder? On dividing by 13 = 999 = 76 11 (a) 50 (b) 5 (c) 58 (d) 85 13 13 RRB NTPC 08.02.2021 (Shift-II) Stage Ist Q 999 divided by 13 leaves remainder 11. Ans. (a) : Total number of positive integers which is ∴ The greatest three digit number divisible by 13 = 999 divisible by 3 = 100 = 33 – 11 = 988 3 6. The number 93248x6 are divisible by 11. Then Total number of positive integers which is divisible by 4 = 100 = 25 digit x is equal to. (b) 2 (a) 5 4 (c) 8 (d) 7 RRB NTPC 03.02.2021 (Shift-I) Stage Ist Total number of positive integers which is divisible by Ans. (d) : Divisibility rule of 11–In a given number if 12 = 100 = 8 the difference of sum of all digit even place and placed 12 at odd place is zero or multiple of 11, then that number Hence, the total number of positive integers which is will also be divisible by 11. divisible by 3 or 4. (9+2+8+6)–(3+4+x) = (33 + 25 – 8) 25 – (7+x) = 11 = 50 18–x = 11 x = 18 – 11 3. How many numbers between 1 and 700 are Hence, x = 7 completely divisible by 17? 7. (4143 + 4343) is divisible by: (a) 42 (b) 41 (a) 86 (b) 74 (c) 45 (d) 46 (c) 12 (d) 84 RRB NTPC 29.01.2021 (Shift-II) Stage Ist RRB NTPC 25.01.2021 (Shift-II) Stage Ist Ans. (b) : Numbers between 1 and 700 which are Ans. (d) : (xn + an) is divisible by (x + a), if the value exactly divisible by 17. of n is odd 17, 34 ...........697. Q 43 is a odd number, therefore (4143 + 4343) will be l = a + (n–1) × d divisible by 41 + 43 = 84 Number System 9 YCT
8. If pq is a two-digit number, then pq – qp will Ans. (b) : The greatest number of 4 digits = 9999 307)9999(32 be completely divisible by: − 921 (a) 9 (b) 7 789 (c) 6 (d) 5 614 RRB NTPC 07.04.2021 (Shift-II) Stage Ist Ans. (a) : Let the two digit number (pq) = 10x + y Then, qp = 10y + x 175 According to the question, Hence, the smallest number to be added = 307–175 pq – qp = 132 = 10x + y – (10y + x) 12. How many numbers from 3 to 60 are odd = 10x + y – 10y – x = 9x – 9y numbers that are exactly divisible by 5? = 9 (x – y) (a) 7 (b) 5 Hence pq – qp will be completely divisible by 9. (c) 8 (d) 6 RRB NTPC 09.01.2021 (Shift-I) Stage Ist 9. If n is a natural number then n3–n is always Ans. (d) : Odd numbers between 3 to 60 which divisible by…………… divisible by 5. (a) 8 (b) 6 (c) 5 (d) 4 5, 15, 25, 35, 45, 55 So total number of odd numbers from 3 to 60 which are RRB NTPC 05.04.2021 (Shift-II) Stage Ist exactly divisible by 5 = 6. Ans. (b) : Q n is a natural number. ∴ n3 − n = n(n2 −1) = n(n +1)(n −1) 13. How many numbers between 300 and 1000 are divisible by 7? n(n+1) (n–1) {Multiplication of three consecutive (a) 994 (b) 301 natural numbers} (c) 101 (d) 100 On putting the value of n = 2 RRB NTPC 09.01.2021 (Shift-I) Stage Ist n3 – n = n (n + 1) (n –1) = 2×3×1 = 6 Hence, it will always divisible by 6. Ans. (d) : Total number of numbers between 1 and Note- The multiplication of three consecutive natural 1000 which are divisible by 7 numbers will be always divisible by 6. = 1000 = 142 10. A number when divided by 7 leaves a 7 remainder 4. What will be the remainder when the square of the same number is divided by 7? Total number of numbers between 1 and 300 which are divisible by 7 (a) 2 (b) 4 = 300 = 42 (c) 1 (d) 3 7 RRB NTPC 29.01.2021 (Shift-I) Stage Ist Hence, Total number of numbers between 1 and 300 which are divisible by 7 between 300 and 1000 Ans. (a) : Let, Quotient = n = 142 − 42 = 100 Number = Divisor × Quotient + Remainder Number = 7× n + 4 (Given, Remainder = 4) 14. Find the greatest number of five digits, which is On putting n = 1, Number = 7×1+ 4 = 11 exactly divisible by 468. On dividing the number by 7, (a) 99684 (b) 99486 (c) 99864 (d) 99468 RRB NTPC 04.01.2021 (Shift-II) Stage Ist Remainder = 4 Ans. (a) : The greatest number of five digits = 99999 Hence, on dividing the square of 11 by 7 468) 99999 ( 213 Remainder = (11)2 = 121 = 2 936 77 639 11. The smallest positive number which must be 468 added to the greatest number of 4 digits in order that the sum may be exactly divisible by 1719 307 is: (b) 132 1404 (a) 307 (d) 176 315 (c) 306 RRB NTPC 17.01.2021 (Shift-II) Stage Ist Required number = 99999 – 315 = 99684 Number System 10 YCT
15. In between 250–1000, how many numbers are Ans. (d) : When the sum of all the digits of a number is divisible by 9, then number will be divisible by 9. completely divisible by 5, 6 & 7. Given number– (a) 5 (b) 7 • 111.............1 (n digits) • When n = 1, number is 1, which is not divisible by 9. (c) 6 (d) 3 • When n = 2, number is 11, which is a prime number RRB NTPC 29.12.2020 (Shift-II) Stage Ist and thus not divisible by 9. • When n = 3, number is 111 and 1+1+1=3, which is Ans. (d) : LCM of 5, 6, 7 – not divisible by 9. 2 5,6, 7 3 5,3, 7 5 5,1, 7 7 1,1, 7 1,1,1 2 × 3 × 5 × 7 = 210 • When n = 9, number is 111111111 and 1+1+1+1+1+ Q Numbers from 250 to 1000 which are divisible by 5, 6, 7 will be always divisible by 210 or in multiples of 1+1+1+1= 9, which is divisible by 9 210. Hence, the least possible value of n is 9. Therefore, the numbers are 210 × 1, 210 × 2, 210 × 3, 19. A number when divided by 280 leaves 73 as the 210 × 4, 210 × 5 ....... 210, 420, 630, 840, .......... remainder. When the same number is divided Hence, the required numbers = 3 by 35, the remainder will be: 16. The largest four-digit number that is exactly (a) 4 (b) 2 divisible by 83 is: (c) 3 (d) 7 (a) 9936 (b) 9954 RRB NTPC 16.01.2021 (Shift-I) Stage Ist (c) 9960 (d) 9966 Ans. (c) : Let number = N RRB NTPC 20.01.2021 (Shift-I) Stage Ist N = 280K + 73 = (35×8) K + 70 + 3 Ans. (c) : The largest four-digit number = 9999 = 35 (8K + 2) + 3 83)9999(12 N = 35m + 3.....(i) (where, m = 8 K + 2) 83 or N = 35q + r ......(ii) On comparing both equation, 169 r = 3 166 Hence, on dividing the same numbers by 35 the 39 remainder will be 3. Therefore required number · 9999 – 39 = 9960 20. The least number that is divisible by all the Hence, 9960 is the largest four-digit number which is numbers from 2 to 10 is– exactly divisible by 83. (a) 2520 (b) 100 17. (47)25 – 1 is exactly divisible by: (c) 504 (d) 9 (a) 21 (b) 24 RRB NTPC 10.01.2021 (Shift-I) Stage Ist (c) 23 (d) 19 Ans. (a) : Required number = LCM of 2, 3, 4, 5, 6, 7, RRB NTPC 18.01.2021 (Shift-I) Stage Ist 8, 9, 10 = 2, 3, (2×2), 5, (2×3), 7, (2×2×2), (3×3)×(2×5) Ans. (c) : (47)25 −1 = 2×2×2×3×3×5×7 = 2520 an – bn is completely divisible by (a – b) 21. How many numbers greater than 2 and less When n = odd numbers, As per the question than 30 are divisible by 1 and themselves n = 25 .....(Odd number) (a) 9 (b) 29 a = 47, b = 1 Then, (c) 27 (d) 11 RRB NTPC 10.01.2021 (Shift-I) Stage Ist a – b = 47 – 1 Ans. (a) : Prime number-The numbers which is only = 46 divisible by 1 and itself are known as prime number. The prime numbers greater than 2 and less than 30 are– = 2 × 23 Hence, 4725 – 1 is divisible by 23. = 3, 5, 7, 11, 13, 17, 19, 23, 29 = Total 9 numbers 18. If 111 ..... 1 (n digits) is divisible by 9, then the Hence, the required number = 9 least value of n is: 22. 371+372+373+374+375 is divisible by: (a) 18 (b) 12 (a) 8 (b) 5 (c) 3 (d) 9 (c) 11 (d) 7 RRB NTPC 18.01.2021 (Shift-I) Stage Ist RRB NTPC 08.01.2021 (Shift-II) Stage Ist Number System 11 YCT
Ans. (c) : 371 + 372 + 373 + 374 + 375 27. Find the smallest four digit number that is divisible by 47. ( )= 371 30 + 31 + 32 + 33 + 34 (a) 1200 (b) 1025 = 371 (1+ 3 + 9 + 27 + 81) (c) 1034 (d) 1360 = 371 ×121 RRB Group-D – 22/09/2018 (Shift-III) = 371 ×112 Ans. (c) : The smallest four digit number = 1000 Hence, given series will be divisible by 11. 21 47 1000 23. The smallest 5 digit number that leaves a remainder of 6 when divided by 7 is : 94 (a) 10009 (b) 10002 60 (c) 10003 (d) 10007 47 RRB NTPC 28.12.2020 (Shift-I) Stage Ist 13 Ans. (b) : Smallest number of 5 digits =10000 Hence, the smallest four digit number divisible by 47, 10000 7 Remainder=4 = 1000 + (47–13) 1000+34 = 1034 Required number = 10000 + (6 – 4) = 10002 28. Find the least 6 digit number that is a multiple 24. N is a whole number which when, divided by 6 of 18. leaves the remainder 4. Find the remainder (a) 100000 (b) 999900 (c) 100008 (d) 100006 when 2N is divided by 6. RRB NTPC 29.04.2016 Shift : 1 (a) 4 (b) 8 Ans : (c) The smallest 6 digit number = 100000 (c) 2 (d) Zero RRB NTPC 28.04.2016 Shift : 1 5555 Ans : (c) Let the quotient be “a” when N is divided by 6. 18 100000 ∴ N = 6a + 4……….(i) 90 By equation (i) ×2, 100 2N = 2 × 6a + 8 90 2N = 12a + 6 + 2 2N = 6 (2a +1) + 2 100 Hence, the required remainder will be 2. 90 25. Find the largest number of four digit that is 100 90 completely divisible by 49. (a) 9998 (b) 9994 (c) 9992 (d) 9996 10 RRB RPF-SI -10/01/2019 (Shift-II) The remainder is 10, hence 18 – 10 = 8 is added to the RRB Group-D – 18/09/2018 (Shift-II) number will make it completely divisible. Ans : (d) The largest 4-digit number is 9999. Hence, the required number = 100000 + 8 = 100008 49 ) 9999 (204 29. A student divided a number by 12 instead of 21 98 and received 35. Find the correct answer. 199 (a) 20 (b) 15 (c) 26 (d) 25 196 RRB JE - 26/05/2019 (Shift-II) 3 Ans : (a) Let the number be x. Hence, the required number = 9999-3 = 9996, which is According to the question, exactly divisible by 49. On dividing by 12, 26. What should be added to 135642 to get the x = 35 12 largest six digit number? x = 35 × 12 x = 420 (a) 864350 (b) 863357 The number is 420 Dividing 420 by 21- (c) 864357 (d) 864347 420 = 20 RRB Group-D – 29/10/2018 (Shift-III) 21 Ans : (c) Let the required number is x. Adding x to 135642 to get a largest six digit number ∴ 135642 + x = 999999 x = 999999 – 135642 x = 864357 Hence, the correct answer = 20 Number System 12 YCT
30. Find the least number to be added to 231228 to ⇒ 83x = 14528 – 3 ⇒ 83x = 14525 make it exactly divisible by 33. (a) 3 (b) 4 ⇒ x = 14525 ⇒ x = 175 83 (c) 2 (d) 1 RRB JE - 27/05/2019 (Shift-III) 34. If the number x4461 is divisible by 11, find the Ans : (a) From question, value of x. 7006 (a) 2 (b) 4 33 231228 (c) 3 (d) 5 231 RRB Group-D – 17/09/2018 (Shift-I) × × ×228 Ans : (d) Rule of divisibility by 11- If the difference between sum of digits at even places 198 and the sum of digits at odd places of a number is 0 or 30 is divisible by 11, then that number will also be divisible by 11. Hence, the required number = 33 – 30 = 3 The number – x 4 4 6 1 31. Which of the following numbers is not divisible x + 4 + 1 – ( 4 + 6) = 0 by 8? x + 5 – 10 = 0 (a) 12676 (b) 11504 x=5 (c) 12832 (d) 12360 35. Which number is divisible by 9? RRB RPF Constable -24/01/2019 (Shift-II) (a) 56112 (b) 89445 Ans : (a) From options, (c) 49653 (d) 58556 (a) 12676 =1584.5 RRB Group-D – 03/10/2018 (Shift-II) 8 Ans : (c) If the sum of all the digits of a number is (b) 11504 = 1438 divisible by 9, the number will also be completely 8 divisible by 9. (c) 12832 = 1604 8 Hence, from options- (a) The sum of the digits of 56112 = 15 ( × ) (b) The sum of the digits of 89445 = 30 ( × ) (d) 12360 = 1545 (c) The sum of the digits of 49653 = 27 (√) 8 (d) The sum of the digits of 58556 = 29 ( × ) Hence, the number 12676 is not divisible by 8. Hence, the number divisible by 9 = 49653 32. 276x1, is divisible by 3. What is the sum of the 36. Which of the following numbers is divisible by 6? possible values of x? (a) 12378 (b) 12363 (a) 18 (b) 21 (c) 12370 (d) 12388 (c) 12 (d) 15 RRB Group-D – 05/12/2018 (Shift-I) RRB RPF SI-12/01/2019 (Shift-I) Ans. (d) If the sum of all digits of a number is divisible Ans : (a) Divisibility rule by 6 – If a given number is divisible by both 2 and 3 then the number will also be by 3, then the number will be divisible by 3. divisible by 6. 276x1, is divisible by 3. Divisibility rule by 2 – If the unit digit of a given 2 + 7 + 6 + x + 1 = 16 + x number is divisible by 2, then the number will also be The number will be completely divisible by 3, by divisible by 2. putting the possible values of x as 2, 8, and 5. Divisibility rule by 3 – If the sum of all the digits of the Hence, the sum of the possible values of x = 2+8+5 =15 number is divisible by 3, then the number will also be divisible by 3. 33. By dividing 14528 by a certain number, Suresh From option (a)- 1 + 2 + 3 + 7 + 8 = 21 gets 83 as quotient and 3 as remainder. What is = 21 = 7 3 the divisor? (b) 185 (a) 165 (d) 175 (c) 195 RRB RPF SI -06/01/2019 (Shift-III) Hence, the number 12378 is divisible by 6. Ans : (d) Let the divisor is ‘x’. 37. Choose the missing digit 'x' from the options Given- Dividend = 14528 given for the number 987x54, so that the Quotient = 83 number is completely divisible by 6. Remainder = 3 (a) 2 (b) 5 Dividend = (Divisor × Quotient + Remainder) (c) 3 (d) 1 RRB Group-D – 18/09/2018 (Shift-I) ⇒ 14528 · (x × 83) ± 3 Number System 13 YCT
Ans. (c) : The given number will be divisible by 6 if it Ans : (a) 310 = 3×3×3×3×3×3×3×3×3×3 is divisible by 2 and 3. = 59049 Divisibility rule by 2 – If the unit digit of a number is ∴ 59049 divisible by 2, then the number will also be divisible by 7 2. = 4 remainder The unit digit of given number is 4, which is divisible by 2. 41. Which of the following numbers is divisible by 12? Divisibility rule by 3 – If the sum of all the digits of the (a) 93412 (b) 63412 given number is divisible by 3, then the number will also be divisible by 3. (c) 73412 (d) 83412 ⇒ 9 + 8 + 7 + x + 5 + 4 = 33 + x RRB ALP & Tec. (31-08-18 Shift-II) 33 Ans. (d) : The number which is divisible by 12, should From option (c) on putting x = 3 be divisible by 3 and 4 also. If the sum of all the digits of a number is divisible by 3, the number will also be divisible by 3. 36 = 12 If the last two digit of a number are divisible by 4, the 3 number will also be divisible by 4. Hence, the value of x will be 3. From option (d), 38. What number should be deducted from 1265 to Then 8+3+4+1+2 = 18, Which is divisible by 3. The last 2-digit of the number are 12, Which is also make it divisible by 29 exactly? divisible by 4. (a) 15 (b) 16 Hence, the number 83412 is divisible by 12. (c) 18 (d) 17 RRB NTPC 05.04.2016 Shift : 3 42. Which of the following numbers is divisible by 9? Ans : (c) (a) 56765 (b) 47862 43 (c) 54321 (d) 87654 29 1265 RRB ALP & Tec. (30-08-18 Shift-III) 116 Ans : (b) If the sum of all the digits of a number is ×105 divisible by 9, the number will be divisible by 9. Hence, from option (b), 87 47862 ⇒ 4+7+8+6+2 = 27, which is divisible by 9. 18 Hence option (b) will be true. Hence, 18 should be subtracted from 1265 to make it 43. If 3x2 + ax + 4 is perfectly divisible by x − 5, completely divisible by 29. then the value of a is: (b) − 5 39. Find the least number to be added to 1739 so (a) − 12 (d) − 15.6 (c) − 15.8 that it is exactly divisible by 11. RRB ALP & Tec. (09-08-18 Shift-II) (a) 11 (b) 2 (c) 1 (d) 10 Ans : (c) According to the question, RRB NTPC 30.03.2016 Shift : 1 3x2 + ax + 4 = 0 --------- (i) Ans : (d) To get the required number divide 1739 by 11 Q equation (i), is divisible by ( x −5 ) then subtract the remainder from the divisor. Hence, 158 ⇒ x–5=0 11 1739 ⇒ x=5 11 Putting the value of x in equation(i), 3(5)2 + a × 5 + 4 = 0 ×63 55 75 + 5a + 4 = 0 ×89 5a = – 79 88 a = – 15.8 ×1 44. The product of 4 consecutive numbers is Hence, the required number will be 11 – 1 = 10. always divisible by which of the following 40. Find the remainder, when 310 is divided by 7. numbers? (a) 4 (b) 3 (a) 10 (b) 22 (c) 5 (d) 6 (c) 24 (d) 48 RRB NTPC 18.04.2016 Shift : 3 RRB RPF SI -05/01/2019 (Shift-I) Number System 14 YCT
Ans : (c) Let 4 consecutive numbers are n, (n+1), (n+2) 47. What is the smallest four digit number formed and (n+3) respectively. by using the digits 3, 5, 0, 6? According to the question, (a) 3056 (b) 0356 The Product of four consecutive numbers (c) 0536 (d) 3506 = n(n+1)(n+2)(n+3) RRB NTPC 08.02.2021 (Shift-I) Stage Ist Where n = 1,2,3,……. Ans. (a) : The smallest four-digit number formed by Putting n = 1, 3,5,0,6 = 3056 Product, 48. What is the smallest five-digit number formed by using the digits 2, 3, 4, 0, 5? · 1 (1+ 1) (1+ 2) (1 + 3) = 1 × 2 × 3 × 4 = 24 (a) 23045 (b) 20435 Putting n = 2, (c) 02345 (d) 20345 Product of numbers, RRB NTPC 04.02.2021 (Shift-I) Stage Ist = 2×3×4×5 = 24 × 5 = 120 Ans. (d) : Largest 5 digit number = 99999 Hence, the product of 4 consecutive numbers is always Smallest 5 digit number = 10000 divisible by 24. The smallest five digit number that can be formed from 45. When the number (5)501 is divided by 126 then the digits 2, 3, 4, 0, 5 is = 20345 the remainder will be? 49. Find sum of the smallest and the largest (a) 117 (b) 121 positive numbers of 6 digits which contains only digits 0, 4, 6 and each of these digits (c) 89 (d) 125 appears at least once. RRB ALP CBT-2 Mec. & Diesel 21-01-2019 (Shift-I) Ans. (d) : (a) 666444 (b) 604604 ( )( )= 5 501 = 53 167 (c) 666666 (d) 1066646 126 126 RRB NTPC 09.02.2021 (Shift-II) Stage Ist ( )−1 167 Ans. (d) : According to the question- Q Smallest 6 digit no = 400006 = 126 Greatest 6 digit no = 666640 = −1 ∴ Required sum = 400006 + 666640 = 1066646 126 50. How many times is digit 3 comes in counting Remainder = 125 from 301 to 399? (a) 119 (b) 11 (c) 121 (d) 21 Type - 2 RRB NTPC 10.01.2021 (Shift-II) Stage Ist Ans. (a) : In Counting from 301 to 399, the digit 3 46. In a five digit number, the digit in the hundred's comes a total of 119 times. place is 2 and the digit in the unit's place is twice 51. Find the two-digit number such that the sum of its digits is 8 and the digits of the number get the digit in the hundred's place. The digit at reversed when 36 is added to it. thousands place is zero. The digit in the ten thousand's place is the sum of the digit in the (a) 71 (b) 35 hundred's place and the digit in the unit's place. (c) 62 (d) 26 The digit in the ten's place is the digit in the ten RRB NTPC 15.02.2021 (Shift-II) Stage Ist thousand's place minus 1. The number is: Ans. (d) : Let number = 10x+y (a) 60234 (b) 60224 Accoding to the question, (c) 60254 (d) 60264 x+y=8 …(i) RRB NTPC 09.02.2021 (Shift-I) Stage Ist (10x+y) + 36= 10y+ x 9y – 9x = 36 Ans. (c) : Let us assume the number = abcde y – x = 4 …(ii) On solving equation (i) and equation (ii) As per question, c=2 x=2 e=2×c y=6 e=2×2 Hence, required number = 10x + y = 10 × 2 + 6 = 26 e=4 52. If the number 2893#$ is divisible by 8 and 5, b=0 a=2+4 then one possible choice of the digits that come a=6 in the place of # and $ can be: d=6–1 (a) 0, 2 (b) 2, 2 d=5 (c) 0, 0 (d) 2, 0 Putting all values, then the required number = 60254 RRB NTPC 13.03.2021 (Shift-II) Stage Ist Number System 15 YCT
Ans. (d) : Divisibility rule of '5' ⇒ if a number has '0' 57. The sum of the greatest and smallest numbers or '5' in its unit digit then it is completely divisible by 5. Divisiblity rule of '8' ⇒ if the last three digits of a given of six digits is: number are divisible by '8' then number will be always divisible by 8. (a) 100000 (b) 199999 from option 'd' (c) 999999 (d) 1099999 On putting the value of # = 2 and $ = 0 RRB NTPC 08.02.2021 (Shift-I) Stage Ist 289320 ⇒ 57864 5 Ans. (d) : According to question, 289320 ⇒ 36165 Greatest number of six-digit · 999999 8 Smallest number of six-digit · 100000 Hence, option (d) will be correct. Hence required sum = 999999 + 100000 = 1099999 58. The least number consisting of five - digit which is divisible by 97 is x. What is the sum of 53. If the largest 4–digit number is subtracted the digits of x? (b) 15 from the smallest 6-digit number, then the (a) 13 remainder will be: (c) 17 (d) 16 (a) 90000 (b) 99991 RRB ALP CBT-2 Physics & Maths 21-01-2019 (Shift-III) (c) 80001 (d) 90001 Ans. (c) : Minimum five - digit number = 10000 RRB NTPC 04.02.2021 (Shift-II) Stage Ist 97 10000(103 Ans. (d) : The smallest number of 6 – digit = 100000 −97 300 The largest number of 4 – digit = – 9999 −291 ×9 Required number = 90001 Hence, five - digit number that is divisible by 97 54. How many significant digits are there to the x = 10000 + (97 –9) x = 10000 + 88 right of the decimal point in the product of x = 10088 Required sum = 1 + 0 + 0 + 8 + 8 95.75 and 0.02554? = 17 (a) 5 (b) 3 (c) 4 (d) 6 RRB NTPC 11.02.2021 (Shift-I) Stage Ist Ans. (d) : 95.75 × 0.02554 = 2.445455 So the number obtained as the product of 95.75 and 59. How many total tens digit in the calculation 0.0254 will have 6 significant digits to the right of the from series 1 to 99? decimal point. (a) 98 (b) 90 55. What is the value of the digits A and B? (c) 99 (d) 100 BA × B3 = 57 A RRB RPF Constable -17/01/2019 (Shift-II) (a) A = 2, B = 4 (b) A = 3, B = 5 Ans : (b) The number of tens digit from 1 to 10 = 1 (c) A = 5, B = 2 (d) A = 5, B = 3 The number of tens digit from 11 to 90 = 80 RRB NTPC 09.02.2021 (Shift-I) Stage Ist The number of tens digit from 91 to 99 = 9 Ans. (c) : From option (c) Putting the values of A and B Hence, the total number of tens from series 1 to 99 in the equation. = 1 + 80 + 9 = 90 A = 5, B = 2 60. Find two consecutive numbers where thrice the BA × B3 = 57A first number is more than twice the second 25 × 23 = 575 number by 5. 575 = 575 (a) 5 and 6 (b) 6 and 7 Hence, option (c) will be correct. (c) 7 and 8 (d) 9 and 10 56. The difference between the greatest and the RRB NTPC 28.03.2016 Shift : 1 smallest six-digit numbers is: Ans : (c) Let the two consecutive numbers be x and (a) 988888 (b) 999999 x+1. (c) 888888 (d) 899999 According to the question- RRB NTPC 04.02.2021 (Shift-I) Stage Ist 3x = 2(x +1) + 5 Ans. (d) : The largest six digit number is 999999 ⇒ 3x = 2x + 7 The smallest six digit number is 100000 ⇒x=7 ∴ Required difference = 999999 – 100000 = 899999 Hence, the required consecutive numbers will be 7 and 8. Number System 16 YCT
61. How many times does the digit 2 come in place 67. One-third of the sum of all the prime numbers of tens in counting from 1 to 100? greater than 5 but less than 18 is the square of: (a) 20 (b) 11 (a) 3 (b) 5 (c) 10 (d) 19 (c) 6 (d) 4 RRB NTPC 31.03.2016 Shift : 1 RRB NTPC 08.04.2021 (Shift-I) Stage Ist Ans : (c) From the digit come in place of tens in Ans. (d) : counting, 1 to 10 = 0 time From 11 to 20 = 1 time Prime numbers greater than 5 but smaller than 18 = 7, From 21 to 30 = 9 times 11,13, 17 From 31 to 100 = 0 times According to the question- = 7 +11+13 +17 ∴ Total required number = 1 + 9 = 10 3 62. How many times does the digit 5 come in the counting from 1 to 100? (a) 21 (b) 22 = 48 = 16 = (4)2 (c) 20 (d) 19 3 RRB RPF SI-16/01/2019 (Shift-I) Hence, required number = 4 Ans : (c) The total numbers in which 5 comes from 1 68. Which of the following is a prime number? to 49 = 5 (a) 143 (b) 173 From 50 to 60, such number = 11 And from 61 to 100, such number = 4 (c) 123 (d) 213 Hence, total required number = 5 + 11 + 4 = 20 RRB NTPC 15.03.2021 (Shift-I) Stage Ist Type - 3 Ans. (b) : Prime number are the numbers, which are only divisible by 1 and itself. From the given options- 63. The greatest prime number less than 200 is: (a) 143 is divisible by 11, so it is not a prime number. (a) 199 (b) 193 (b) 173 is divisible by 1 and itself, so it is a prime (c) 197 (d) 191 number. RRB NTPC 21.01.2021 (Shift-II) Stage Ist (c) 123 is divisible by 3, so it is not a prime number. Ans. (a) : The greatest prime number less than 200 is (d) 213 is divisible by 3, so it is not a prime number. 199. 69. Find the sum of prime no. between 50 and 60. 64. Which of the following numbers is prime? (a) 118 (b) 114 (a) 323 (b) 571 (c) 110 (d) 112 (c) 513 (d) 715 RRB NTPC 31.01.2021 (Shift-I) Stage Ist RRB NTPC 02.03.2021 (Shift-II) Stage Ist Ans. (d) : The prime number between 50 and 60– Ans. (b) : According to option, 53 and 59 571 is a prime number. Whereas 323 is divisible by 17, Required Sum = 53 + 59 = 112 513 is divisible by 3 and 715 is divisible by 5. 65. Find the smallest three digit prime number? 70. Find the number of all prime numbers less (a) 107 (b) 109 than 55. (c) 103 (d) 101 (a) 18 (b) 17 RRB NTPC 23.07.2021 (Shift-II) Stage Ist (c) 16 (d) 15 Ans. (d) : The smallest three-digit prime number = 101 RRB NTPC 30.12.2020 (Shift-I) Stage Ist 66. Which of the following pairs of numbers are Ans. (c) : The number of all prime numbers less than 55 co-prime? is 16 (a) 28, 81 (b) 12, 27 i.e. ⇒ (2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, (c) 21, 56 (d) 36, 20 47, 53) RRB NTPC 23.07.2021 (Shift-II) Stage Ist 71. The number of pairs of twin primes between 1 Ans. (a) : Co-prime numbers are the numbers whose common factor is only 1. and 100 are: Hence, in the given option (28, 81) are co-prime (a) 7 (b) 8 numbers. (c) 10 (d) 9 RRB NTPC 26.07.2021 (Shift-I) Stage Ist Number System 17 YCT
Ans. (b) : The number of pairs of twin primes between 162 = 256 1 and 100 are 8. The numbers are - 172 = 289 Then, divide the given number by all the prime numbers {(3,5),(5,7),(11,13),(17,19),(29,31),(41,43),(59,61),(71,73)} Note- Twins prime numbers are that numbers whose below 16 and 17. If the number is not divisible by any number then it is a prime number. difference is 2. ⇒ 263 (is less than the square of 17) 72. The number that has factors other than 1 and Which is not divisible by 2,3,5,7, 11 and 13. Hence, it is a prime number. itself is called a ............. number. (a) Prime Number (b) Composite Number 77. Find the largest two-digit prime number. (c) Even Number (d) Odd Number (a) 93 (b) 89 RRB NTPC 26.07.2021 (Shift-I) Stage Ist (c) 91 (d) 97 Ans. (b) : Composite Number:- Numbers which have RRB JE - 23/05/2019 (Shift-II) more than two factors. Ans : (d) The number which is divisible by only 1 and Ex- 4, 6, 8 ------- Prime Number:- Numbers which have only two factor itself is called prime number. Hence, It is clear that the largest two digit prime 1 and itself is called prime number. 73. Find the number of prime number less than 20. number = 97 (a) 9 (b) 7 78. What will be the product of the smallest prime (c) 10 (d) 8 number (except 0) and any whole number? RRB NTPC 06.04.2021 (Shift-II) Stage Ist (a) Always 0 Ans. (d) : Prime mumber less than 20. (b) Always 1 2, 3, 5, 7, 11,13, 17 and 19 (c) Always even number (d) Always odd number Hence the number of prime number less than 20 = 8 74. Three prime number are arranged in RRB RPF Constable -20/01/2019 (Shift-II) descending order. If the product of the first two Ans : (c) The smallest prime number = 2, is 323 and that of the last two is 221, then what The result of the product of any whole number(except is the value of the biggest prime number? 0) and 2 is always an even number. (a) 17 (b) 19 79. Find the sum of the prime numbers between 50 (c) 13 (d) 23 and 80. RRB NTPC 04.03.2021 (Shift-I) Stage Ist (a) 392 (b) 390 Ans. (b) : Let the consecutive prime numbers are x, y (c) 463 (d) 396 and z in which x is biggest prime number. According to the question, RRB RPF Constable -18/01/2019 (Shift-I) x × y = 323 Ans : (c) Sum of prime numbers between 50 and 80 = Taking 53 + 59 + 61 + 67 + 71 + 73 + 79 = 463 x = 19 y = 17 80. The sum of which four odd prime numbers is 19 × 17 = 323 34? (b) 3, 5, 7, 9 Taking y = 17 and z = 13 (a) 1, 3, 5, 7 And 17 × 13 = 221 (c) 3, 5, 11, 13 (d) 3, 7, 11, 13 So, the biggest prime number is = 19 RRB NTPC 04.04.2016 Shift : 2 75. How many of the integers between 109 and 121, Ans : (d) From option- (d) 3 + 7 +11+13 = 34 both inclusive, are prime numbers? (a) 1 (b) 0 81. In a prime number……... (c) 2 (d) 3 (a) There are more than two divisors. RRB NTPC 08.02.2021 (Shift-I) Stage Ist (b) The number divided by itself and 1. Ans. (c) : Two integers (109, 113) between 109 and 121 (c) It has no divisor. both inclusive are prime numbers. (d) Is not a positive integer. 76. Which of the following numbers is prime? RRB NTPC 30.03.2016 Shift : 2 (a) 263 (b) 243 Ans : (b) A prime number is divided by only itself and 1. (c) 253 (d) 273 RRB RPF Constable -17/01/2019 (Shift-III) 82. How many total prime numbers are in first 200 Ans : (a) To identify a prime number, compare the odd natural numbers? given number with its nearest square. For example (a) 45 (b) 49 option (a), 263 (Which is between the squares of 16 and (c) 50 (d) 46 17) RRB Group 'D' 07/12/2018 (Shift-I) Number System 18 YCT
Ans : (a) Total prime numbers in first 200 odd natural Ans : (d) The prime numbers between 60 and 100 = 61, numbers = 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 67, 71, 73, 79, 83, 89, 97 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, Hence, Total 8 prime numbers will be between 60 and 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 100. 173, 179, 181, 191, 193, 197, 199 = 45 90. Which of the following numbers is a prime 83. Which of the following pairs are co-primes? number? (a) 121 (a) 348, 296 (b) 114, 213 (b) 141 (c) 59, 97 (d) 3025, 4920 (c) 181 (d) 161 RRB Group-D – 20/09/2018 (Shift-II) RRB ALP & Tec. (21-08-18 Shift-II) Ans : (c) Such two numbers are called co-prime whose Ans : (c) Prime numbers are divisible by 1 and itself HCF is 1. only. In option (c) 59, 97 is the appropriate co-prime pair in Example:- 5,11,13,19 the alternatives. From options, 84. Which of the following numbers is divisible? Factors of the numbers, (a) 719 (b) 709 181 = 1, 181 121 = 1, 11, 121 (c) 729 (d) 739 141 = 1, 3, 47, 141 161 = 1, 7, 23, 161 RRB Group-D – 20/09/2018 (Shift-I) Hence from the above it is clear that '181' is a prime Ans. (c) The number- 729 is divisible by 3, 9 and 81. number. 85. How many prime numbers are in first 100 natural numbers? (a) 25 (b) 27 91. Which of the following pairs is NOT a pair of (c) 24 (d) 26 RRB Group-D – 26/11/2018 (Shift-III) twin primes? Ans : (a) Prime numbers in first 100 natural numbers = (a) 11, 13 (b) 71, 73 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, (c) 131, 133 (d) 191, 193 59, 61, 67, 71, 73, 79, 83, 89, 97 RRB ALP & Tec. (21-08-18 Shift-II) Therefore, total such numbers are 25. Ans : (c) From option (c), 86. Find out which of the following sets form co- Factor of 131 = 131, 1 and prime numbers. Factors of 133 = 1, 7, 19, 133. (a) (12, 7) (b) (21, 42) Hence, this pair is not a pair of twin primes. (c) (3, 9) (d) (43, 129) 92. Which of the following is the prime number series from 1 to 20? RRB NTPC 18.01.2017 Shift : 1 (a) 3, 5, 7, 11, 13, 17, 19 (b) 2, 5, 7, 9, 11, 13, 17, 19 Ans : (a) Co-prime numbers- The set of two such (c) 2, 3, 5, 7, 11, 13, 17, 19 numbers whose HCF is 1, is called co-prime numbers. (d) 1, 2, 3, 5, 7, 11, 13, 17, 19 ∴ In option (a), HCF of the numbers (12, 7) = 1 RRB ALP & Tec. (20-08-18 Shift-I) 87. Which of the following is an odd composite number? (b) 17 Ans : (c) Prime numbers:- The number which is (a) 13 (c) 12 (d) 15 RRB NTPC 18.01.2017 Shift : 2 divisible by 1 and itself only. Ans : (d) In the given options odd composite number Hence, from options, the required series will be 2,3,5,7,11,13,17,19. will be 15. 88. Find the sum of first 8 odd prime numbers. 93. Calculate the difference between the largest (a) 77 (b) 98 and the smallest two-digit prime number. (c) 75 (d) 100 (a) 82 (b) 83 RRB NTPC 19.04.2016 Shift : 2 (c) 84 (d) 86 Ans : (b) First 8 odd prime numbers = 3, 5, 7, 11, 13, RRB RPF Constable -17/01/2019 (Shift-I) 17, 19, 23 Ans : (d) The largest two digit prime number = 97 Sum of the numbers = 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 The smallest two digit prime number = 11 = 98 Hence, the required number = 97 – 11 = 86 89. How many prime numbers are between 94. Which of the following number is not positive integers 60 and 100? composite? (a) 9 (b) 6 (a) 209 (b) 203 (c) 7 (d) 8 (c) 161 (d) 109 RRB NTPC 06.04.2016 Shift : 1 RRB ALP & Tec. (14-08-18 Shift-I) Number System 19 YCT
Ans : (d) The number is called composite number. = x+y xy which is formed by multiplying whole numbers. Hence, 209 = 11 × 19 From equation (i) and (ii) = 40 ⇒ 2 203 = 7 × 29 60 3 161 = 7 × 23 But, 109 cannot be expressed in the form of factors (except 1). So it is not composite. 98. What is the sum of the cubes of the natural numbers from 5 to 14? Type - 4 (a) 10930 (b) 10925 (c) 10935 (d) 10920 95. Pragya invited male and females to her RRB NTPC 04.02.2021 (Shift-I) Stage Ist birthday party in the ratio of 7 : 6. If the Ans. (b): The sum of the cubes of natural number number of males in the party were 56, then the n (n +1) 2 total number of guests attending the party were? = 2 (a) 48 (b) 104 (c) 108 (d) 112 Sum of cubes of all natural numbers from 5 to 14 RRB NTPC 07.01.2021 (Shift-II) Stage Ist · [Sum of cubes of number 1 to 14] – [Sum of cubes Ans. (b) : Let number of males = 7x of numbers 1 to 4 ] and, number of female = 6x According to the question- 14(14 + 1) 2 4(4 + 1) 2 2 2 7x = 56 = − x=8 = (105)2 – (10)2 ∴ Total number of guests = 7x + 6x = 11025 – 100 = 10925 = 13x = 13 × 8 99. If the difference between squares of two consecutive positive odd integers is 56, then the = 104 two consecutive odd integers are. 96. What is the sum of the cube of the natural (a) 17,19 (b) 13,15 numbers from 1 to 10, both inclusive? (c) 11,13 (d) 15,17 (a) 3023 (b) 3025 RRB NTPC 07.01.2021 (Shift-I) Stage Ist (c) 3024 (d) 3022 Ans. (b) : Suppose first odd number = a RRB NTPC 08.02.2021 (Shift-I) Stage Ist and, second consecutive odd number = a+2 Ans. (b) : The sum of the cube of the natural numbers According to the question, from 1 to 10– (a+2)2 – (a)2 = 56 a2 + 4 + 4a − a2 = 56 = 13 + 23 + 33 + 43 + 53 + 63 + 73 + 83 + 93 + 103 a = 52 = 13 = 10 ×11 2 Q Σn3 = n(n + 1) 2 4 2 2 = 100 × 121 = 3025 First Number (a) = 13 4 Second Number (a + 2) = 13+2 = 15 97. The sum of two numbers is 40 and their 100. An orchard has 5776 trees and the product is 60. The sum of their reciprocals is: arrangement of trees is such that there are as many rows as there are trees in a row. Then the (a) 3 (b) 3 number of rows is: 42 (a) 48 (b) 76 (c) 2 (d) 1 (c) 65 (d) 56 32 RRB NTPC 13.03.2021 (Shift-II) Stage Ist RRB NTPC 04.02.2021 (Shift-I) Stage Ist Ans. (b) : Let the number of trees be X and the number Ans. (c) : Let the two numbers are x and y of rows also X. According to the question, According to the question, x + y = 40 .............(i) X × X = 5776 and x × y = 60 ..........(ii) X2 = 5776 Sum of reciprocal of numbers = 1 + 1 X = 76 xy Hence, the number of rows (X) = 76 Number System 20 YCT
101. What is the sum of the squares of the numbers 104. The sum of two numbers is 25 and their from 3 to 18? difference is 15. The ratio of the numbers is? (a) 2103 (b) 2102 (a) 3:2 (b) 5:3 (c) 2101 (d) 2104 (c) 4:1 (d) 2:3 RRB NTPC 09.02.2021 (Shift-II) Stage Ist RRB NTPC 04.01.2021 (Shift-I) Stage Ist Ans. (d) Ans. (c) : Let the numbers be a and b. Sum of squares of the first 'n' terms = n (2n +1)(n +1) According to the question, a + b = 25 _____ (i) 6 a - b = 15 _____ (ii) Sum of squares of numbers form 3 to 18 By equation (i) and (ii) ( ) ( )= 12 + 22 + 32 + 42 + ....... +182 – 12 + 22 ⇒ a = 25 +15 = 20 = 18(18× 2 +1)(18 +1) – 5 2 b = 25 −15 = 5 6 = 18× 37 ×19 – 5 2 Hence, the ratio of the numbers a : b = 20 : 5 = 4 : 1 6 105. The sum of two number is 16 and their product = 2109 – 5 is 63. The sum of their reciprocal is equal to: = 2104 (a) 16 (b) 63 102. The sum of two numbers is 20 and their 63 16 product is 96. What is the difference between (c) 8 (d) 60 63 63 the two numbers? (a) 4 (b) 5 RRB NTPC 04.01.2021 (Shift-I) Stage Ist (c) 6 (d) 8 Ans. (a) : Let the numbers be x and y RRB NTPC 08.02.2021 (Shift-II) Stage I According to the question, Ans. (a) : Let the two numbers are x and y. x + y = 16 ________ (i) According to the question, and x × y = 63 ________ (ii) x + y = 20 then, 1 + 1 = ? xy = 96 xy From, x – y = (x + y)2 − 4xy x + y = 16 xy 63 = (20)2 − 4× 96 = 400 − 384 106. The difference between two numbers which are = 16 in the ratio 5 : 3 is 50. What is the product of the numbers? =4 (a) 1035 (b) 9375 103. If the sum of two numbers is 30 and the (c) 8575 (d) 9975 product is 50, then the sum of their reciprocals RRB NTPC 08.04.2021 (Shift-II) Stage Ist is: Ans. (b) : Let the numbers are 5x, 3x As per question, (a) 3 (b) 5 5x–3x = 50 5 3 2x = 50 (c) 2 (d) 5 x = 25 5 2 Hence, the product of two numbers = 5x×3x = 15x2 RRB NTPC 29.01.2021 (Shift-II) Stage I = 15×252 = 9375 Ans. (a) : Let the numbers be x and y – 107. 3 of a number is 10 more than half of the Given, 5 second number. If 8 is subtracted from 3 of the x + y = 30 .........(i) 7 xy = 50 .........(ii) first number, then it becomes 4 less than half of The sum of reciprocals of numbers the second number. What is the sum of the two =1+1 =x+y numbers? x y xy = 30 = 3 (a) 56 (b) 57 50 5 (c) 54 (d) 55 RRB NTPC 08.04.2021 (Shift-II) Stage Ist Number System 21 YCT
Ans. (b) : Let the no. be x and y x2 – y2 = 243 According to the question, (x–y) (x+y) = 243______(ii) 3 x – y = 10 Putting value of (x + y) from eqn (i) in eq (ii), 52 (x–y) × 27 = 243 6x – 5y = 100 ...(i) (x–y) = 243 = 9 27 and 3 x – 8 = y – 4 72 So, difference between the numbers = x–y = 9 6x – 7y = 56 ...(ii) 111. What is the sum of the squares of the numbers On subtracting equation (ii) from equation(i) from 1 to 12? 2y = 44 y = 22 (a) 655 (b) 660 x = 100 + 5× 22 = 35 {from equation (i)} (c) 650 (d) 665 6 RRB NTPC 04.02.2021 (Shift-II) Stage Ist Hence, sum of two numbers = x+y = 35+22 = 57 Ans. (c) : 12 + 22 + 32 + ........... +122 108. The ratio of five numbers are 1:2:3:4:5 and From, Sum of the square of the first n natural numbers their sum is 30. Find the sum of second and n (n +1)(2n +1) fifth number? = 6 (a) 15 (b) 14 = 12×13× 25 = 650 (c) 13 (d) 12 RRB NTPC 05.04.2021 (Shift-II) Stage Ist 6 Ans. (b) : Let the number are x, 2x, 3x, 4x, 5x. 112. Find the least number which must be added to According to the question, the number 6412 to get a perfect square. x+2x+3x+4x+5x=30 (a) 149 (b) 129 15x=30 ⇒ x = 2 (c) 181 (d) 150 Then the sum of (second+fifth) number = 2x+5x=7x RRB NTPC 12.01.2021 (Shift-II) Stage Ist = 7×2=14 Ans. (a) : (80)2 = 6400 109. There are 2401 students in a school. The PT (81)2 = 6561 teacher wants all of them to stand in rows and Hence on adding 6561 – 6412 = 149, 6412 will be the columns. Find the number of rows, if the perfect square. number of rows is equal to the number of 113. Out of four consecutive numbers, the sum of the first two numbers is equal to the fourth columns. number. What is half of the sum of the four (a) 29 (b) 39 numbers. (c) 49 (d) 19 (a) 14 (b) 7 RRB NTPC 10.02.2021 (Shift-II) Stage Ist (c) 9 (d) 2 Ans. (c) : Let number of Rows = x RRB NTPC 12.01.2021 (Shift-II) Stage Ist then number of columns · x Ans. (b) : Let four consecutive numbers be x, (x + 1), (x +2) and (x + 3) Number of students in school = 2401.....(given) According to question, Q Number of rows × Number of columns = 2401 x + (x +1) = x +3 ∴ x × x = 2401 x=2 x2 = 2401 Half of the sum of four number = 4x + 6 = 2x + 3 x = 49 2 Hence, the number of rows (x) = 49 =2×2+3 =7 110. The sum of two numbers is 27 and the 114. 24 mango trees, 56 apple trees and 72 orange difference of their squares is 243. What is the trees have to be planted in rows such that each difference between the numbers? row contains the same number of trees of one (a) 42 (b) 9 variety only. Find the minimum number of (c) 72 (d) 3 rows in which the above mentioned trees may RRB NTPC 05.02.2021 (Shift-I) Stage Ist be planted. Ans. (b) : let us the numbers be x and y respectively. (a) 15 (b) 18 Given, (c) 17 (d) 19 x + y = 27––––––(i) RRB NTPC 04.01.2021 (Shift-II) Stage Ist Number System 22 YCT
Ans. (d) : 117. One-fourth of one-eight of a number is 300. (Number of total columns × Number of total rows) What is one fifth of the same number? 8×3 8×7 (a) 1900 (b) 1910 8×9 8 (3 + 7 + 9) = Total number of trees (c) 1920 (d) 1890 Total number of rows = 3 + 7 + 9 = 19 RRB NTPC 03.03.2021 (Shift-I) Stage Ist Ans. (c) : Let the required number = x According to the question, 115. What is the sum of the cubes of the first four x× 1 × 1 = 300 8 4 natural numbers? x = 300× 32 ⇒ x = 9600 (a) 96 (b) 84 (c) 100 (d) 1000 Then, 9600 × 1 = 1920 5 RRB NTPC 23.07.2021 (Shift-I) Stage Ist Ans. (c) : First four natural numbers– 118. Two-fifth of one-fourth of three-seventh of a number is 15. What is the half of that number? ⇒ 1, 2, 3, 4 Cube, (a) 375 (b) 175 (1)3 = 1 (c) 300 (d) 170 (2)3 = 8 (3)3 = 27 RRB NTPC 09.02.2021 (Shift-I) Stage Ist (4)3 = 64 Ans. (b) : Let the number is x Sum of cubes of the first four natural numbers According to the question, = 1 + 8 + 27 + 64 x × 3 × 1 × 2 = 15 = 9 + 27 + 64 745 = 36 + 64 x = 350 = 100 then, half of that number = 350 116. 6 of the people present in a hall are sitting in 2 11 = 175 9 of the chairs available, and the rest are 119. Instead of multiplying a number by 2, Rahul 14 divided it by 2 and got the answer as 2. What standing. If there are 30 empty chairs, how should be the actual answer? many people in the hall are standing? (a) 4 (b) 8 (a) 40 (b) 35 (c) 6 (d) 2 (c) 30 (d) 45 RRB NTPC 25.01.2021 (Shift-I) Stage Ist RRB NTPC 23.07.2021 (Shift-I) Stage Ist Ans. (b) : Let the no. = x Ans. (d) : If number of total chair · x According to question, Then empty chair = x − 9x = 5x actual answer = 2x 14 14 and from the question Whereas, x × 5 = 30 14 x =2 2 x = 84 (Number of total chair) x=4 Hence, number of people sitting on the chair. ⇒ 84 – 30 = 54 Actual answer = 2x If total people are y then, = 2×4 =8 y × 6 = 54 120. In a reunion of class XII, out of 45 students, 30 11 students participated in the function. If all or y = 99 people present in the function shake hands with one Q Number of standing people y 1 − 6 = y × 5 other, find the total number of handshakes. 11 11 (a) 870 (b) 435 Hence, Number of standing people = 99 × 5 = 45 people (c) 841 (d) 900 11 RRB NTPC 12.01.2021 (Shift-I) Stage Ist Number System 23 YCT
Ans. (b) : Total number of handshakes Ans : (d) Let the number be x. = n (n −1) x − x = x − 30 2 7 = 30 (30 −1) 7x − x = x − 30 2 7 = 15 × 29 6x = 7x – 210 = 435 x = 210 121. The difference of two numbers is 5. If their 125. If the product of two numbers is 24, and their product is 336, find the sum of the numbers. square’s sum is 52, then find their sum. (a) 21 (b) 37 (a) 5 (b) 10 (c) 28 (d) 51 (c) 15 (d) 20 RRB JE - 26/06/2019 (Shift-I) RRB RPF Constable -24/01/2019 (Shift-I) Ans. (b) Let the numbers be x and y respectively. Ans :(b) Let the numbers be x and y. x–y=5 (i) According to the question, xy = 336 (ii) (x + y)2 = (x–y)2 + 4xy x. y = 24………..(1) x2 + y2 = 52……(2) From equation (i) and (ii), ∵ (x + y)2 = x2 + y2 + 2xy (x + y)2 = (5)2 + 4 × 336 (x + y)2 = 25 + 1344 (x + y)2 = 1369 = 52 + 2×24 = 52 + 48 = 100 (x + y) = 1369 x + y = 100 = 10 x + y = 37 126. If 10 is subtracted from the 5 times of a Hence, the required sum of the numbers =37 number, then that number will be equal to the 122. If x + y = 11, then (-1)x+(–1)y is equal to ....... number found when adding 8 to 4 times of that (where x and y are whole numbers). number, what is that number? (a) -1 (b) 1 (a) 15 (b) 18 (c) 2 (d) 0 (c) 22 (d) 21 RRB JE - 23/05/2019 (Shift-I) RRB RPF Constable -25/01/2019 (Shift-III) Ans : (d) Given – Ans. (b) : Let the number be x. According to the question, x + y = 11 (–1)x + (–1)y = ? 5x – 10 = 4x + 8 x = 18 Note- When the sum of two whole numbers is an odd Hence, the required number will be 18. number then one will be even and second will be odd. 127. When 8 times of a number is added to 4, the Hence, + (–1)odd/even = 0 result obtained is the smallest 3-digit number. (–1)even/odd 123. From a cloth of 30 m long, 12 pieces each What is that number? measuring 225 cm are cut and sold. How much (a) 12 (b) 10 is left of the original length? (c) 15 (d) 8 (a) 1/3 (b) 1/9 (c) 1/10 (d) 3/10 RRB RPF Constable -22/01/2019 (Shift-II) RRB JE - 23/05/2019 (Shift-III) Ans : (a) Let the number be x, Ans : (c) The total length of the cloth = 30 m [1m = 100 Q The smallest 3-digit number = 100 cm] = 3000 cm Total length of the cloth that is sold = 225 × 12 = 2700 According to the question, cm The length of the remaining cloth = 3000 – 2700 = 300 8x + 4 = 100 cm 8x = 96 The remaining part = 300 = 1 x = 96 = 12 3000 10 8 Hence, the required number will be 12. 124. If 1/7 of a number is subtracted from the 128. The sum of two numbers is 22. Five times of number, the result is 30 less than the number. one number is equal to 6 times the other. Find Find the number. the larger of the two numbers. (a) 105 (b) 140 (a) 12 (b) 15 (c) 120 (d) 210 (c) 10 (d) 16 RRB JE - 24/05/2019 (Shift-III) RRB JE - 25/05/2019 (Shift-I) Number System 24 YCT
Ans : (a) Let the numbers are x and y, Ans : (d) Let the number be x, According to the question, According to question, x + y = 22 .....(i) x × 1 × 2 = 32 and 5x = 6y .......(ii) 43 x = 6y x = 32 × 6 = 192 5 132. If the sum of two numbers is 13 and the sum of Putting the value of x in equation (i) - their squares is 97, what is their product? 6 y + y = 22 5 (a) 72 (b) 36 11 y = 22 (c) 110 (d) 84 5 y = 22 × 5 = 10 RRB JE - 28/06/2019 (Shift-III) 11 Ans. (b) Let both the numbers are X and Y. y = 10 Given, xy = ? ∴ x = 6 ×10 = 12 x + y = 13, and x2 + y2 = 97, 5 Q (x + y)2 = x2 + y2 + 2xy ...... (i) On putting the values in equation (i), (13)2 = 97 + 2xy 169 = 97 + 2xy 2xy = 169 – 97 Hence, the larger number is 12. xy = 72 129. If doubling a number and adding 20 to the 2 result gives the same answer as multiplying the xy = 36 number by 8 and subtracting 4 from the 133. Which of the fraction given below, when added product, find the number. to 13 , gives 1? 5 (a) 3 (b) 4 (c) 6 (d) 2 (a) − 48 (b) − 7 30 5 RRB JE - 25/05/2019 (Shift-II) Ans : (b) Let the number be = x (c) − 28 (d) − 8 10 15 According to the question, 2x + 20 = x × 8 – 4 RRB Group-D – 19/09/2018 (Shift-II) 2x + 20 = 8x – 4 Ans. (a) : Let the fraction be x. 24 = 6x According to the question, x=4 x + 13 = 1 130. The product of two numbers is 9375. The 5 quotient, when the largest number is divided x = 1− 13 by the smallest number is 15. Find the sum of 5 these numbers. (b) 380 x = −8 (a) 400 5 (c) 425 (d) 395 or , x = −8× 6 = −48 RRB JE - 30/05/2019 (Shift-II) 5×6 30 Ans : (a) Let the smaller number be = x 134. Shalini, Tanvir and Rashid shared a cake. ∴ Larger number = 15x Shalini had 1 part of it, Tanvir had 1 part of 64 According to the question, x × 15x = 9375 it and Rashid had the remaining part. What 15x2 = 9375 was fraction of Rashid’s cake? x2 = 625 (a) 5 x = 25 first number 6 ∴ 15x = 15 × 25 = 375 second number (b) 3 Hence, the sum of the numbers = 375 + 25 = 400 5 131. If 2/3rd of 1/4th of a number is 32. Find the (c) 13 number. 15 (a) 202 (b) 198 (d) 7 (c) 196 (d) 192 12 RRB RPF-SI -13/01/2019 (Shift-III) RRB Group-D – 31/10/2018 (Shift-II) Number System 25 YCT
Ans : (d) Shalini’s share of the cake = 1 part 137. If the product of two numbers is thrice of their 6 sum, if 1st number is 12 find the 2nd number. Tanvir’s share of the cake = 1 part (a) 2 (b) 3 4 (c) 4 (d) 5 Total share of Shalini and Tanvir’s cake =1+ 1 = 2+3 = 5 RRB NTPC 04.04.2016 Shift : 1 6 4 12 12 Ans : (c) Let the 2nd number be x. x × 12 = (x + 12) × 3 12x = 3x + 36 Hence, Rashid’s share of the cake = 1− 5 = 7 part 9x = 36 12 12 Hence, x = 4 135. The sum of two numbers is 9. The sum of their 138. Two partners M and N buy a car. M pays his share of 3 th of the total cost of the car. M pays reciprocals is 1/2. One of the number is. (a) 2 (b) 4 7 (c) 5 (d) 6 `31,540 less as compared to N. What is the cost RRB Group-D – 17/09/2018 (Shift-III) of the car? Ans. (d) : Let the first number be x and the second (a) `2,32,680 (b) `2,03,175 number be y. (c) `2,20,780 (d) `1,85,780 According to the question, RRB ALP & Tec. (31-08-18 Shift-III) x + y = 9 .........(i) 1 + 1 = 1 ....... (ii) Ans : (c) Let the cost of the car is ` x xy2 According to the question, From equation (i), x+y=9 M’s share = 3x y=9–x 7 N’s share = 3x + 31540 7 From equation (ii) Then, 3x + 31540 + 3x = x 1+1 =1 77 xy2 x+y = 1 x = 31540×7 xy 2 x = `2,20,780 9×2 =1 xy 139. If 2 part of a pizza costs ` 300, then 3 part of a 35 pizza will cost: 2 × 9 = xy (a) `180 (b) `250 On putting the value of y, (c) `225 (d) `270 18 = x (9 – x) RRB ALP & Tec. (30-08-18 Shift-I) 18 = 9x – x2 Ans : (d) The cost of 2/3 part of the pizza = `300 x2 – 9x + 18 = 0 Then, the cost of 1 share of the pizza = 300 × 3 = `450 x2 – 6x – 3x + 18 = 0 2 x(x – 6) – 3(x – 6) = 0 The cost of 3/5th share of the pizza = 450 × 3 (x – 3) (x – 6) = 0 5 (x – 3) = 0 or x = 3 = 90×3 = `270 (x – 6) = 0 or x = 6 140. When 472 pieces of plywood, each 0.23 cm 136. If the sum of two numbers is 26 and their thick, are placed on top of each other, what difference is 12. Find the difference of their would be the height of the pillar in metre? squares. (b) 312 (a) 10.856 (b) 1.0856 (a) 296 (d) 336 (c) 324 RRB NTPC 05.04.2016 Shift : 2 (c) 108.56 (d) 1.856 RRB ALP & Tec. (29-08-18 Shift-III) Ans : (b) Let the numbers be x and y. Ans : (b) The required height of the pillar, x + y = 26 = 0.23× 472 = 1.0856 metre 100 x – y = 12 ∴ The difference of the squares, 141. 15 small rods, each of length 23 2 m are joined = x2 – y2 7 = (x + y) (x – y) to make a big rod. What is the length of the big = 26 × 12 = 312 rod? Number System 26 YCT
(a) 349 3 m (b) 349 1 m 145. If 1 of a number multiplied by 2 of the same 7 7 53 (c) 349 2 m (d) 349 5 m number gives 480, then the number is? 7 7 (a) 60 (b) 70 RRB ALP & Tec. (21-08-18 Shift-I) (c) 80 (d) 100 Ans : (c) The length of each rod = 23 2 = 163 m RRB NTPC 10.01.2021 (Shift-II) Stage Ist 77 Ans. (a) : Let, number = x In this way, the length of big rod = 15× 163 According to the question- 7 x × 1 × x × 2 = 480 53 = 2445 = 349 2 m 77 2x2 = 480 15 142. Find the smallest four digit number which is a x2 = 240 ×15 perfect square. (a) 1000 (b) 1024 (c) 1081 (d) 1064 x2 = 3600 x = 60 RRB NTPC 04.04.2016 Shift : 1 Ans : (b) The smallest 4 digit number = 1000 32 146. One-fourth of a number is equal to three- eighth of another number. If 30 is added to the 3 1000 first number, then it becomes six times that of +3 9 the second number. The first number is: 62 100 (a) 12 (b) 20 2 124 (c) 10 (d) 15 −24 RRB NTPC 13.01.2021 (Shift-II) Stage Ist Hence, the smallest 4 digit perfect square number = Ans. (c) : Let the first number is x and the second 1000 + 24 = 1024 number is y then, 143. A number when multiplied by 6 gives 108 . According to the question, 5 125 x =3y The number is: 48 (a) 625 (b) 648 x = 3y ... (i) 648 625 2 (c) 18 (d) 25 And x + 30 = 6y ... (ii) 25 18 Substituting the value of x from equation (i) in equation RRB NTPC 15.03.2021 (Shift-II) Stage I (ii)- Ans. (c) : Let the number = x 3 y + 30 = 6y 2 As per question x× 6 = 108 or x = 108× 5 3 y − 6y = −30 5 125 6 ×125 2 ⇒ x = 18 −9y = −30 2 25 144. Four fifths of a number is 12 more than three y = 20 3 fourths of the number. Find the number. (a) 120 (b) 160 From equation (i)- (c) 200 (d) 240 RRB NTPC 30.12.2020 (Shift-II) Stage Ist x = 3 × 20 23 Ans. (d) : Let the number = x According to the question, x = 10 4 x − 3 x = 12 147. Calculate the positive number which when 54 16x −15x = 12 added by 15 is equal to 100 times the reciprocal 20 of the number. (a) 10 (b) 20 x = 240 (c) 5 (d) 15 Hence the number is 240. RRB NTPC 15.02.2021 (Shift-II) Stage Ist Number System 27 YCT
Ans. (c) : Let the positive number is x x2 – x = 812 According to the question, x2 – x – 812 = 0 x +15 = 1 ×100 x x2 – 29x + 28x – 812 = 0 x2 +15x = 100 x (x – 29) + 28(x – 29) = 0 x2 +15x −100 = 0 (x – 29)(x + 28) = 0 x2 + 20x − 5x −100 = 0 x – 29 = 0 x = 29 x(x + 20) − 5(x + 20) = 0 (x + 20) (x − 5) = 0 150. The sum of 4 consecutive odd numbers is 160. Find the smallest number. x=5 (a) 27 (b) 37 Hence the number is 5. (c) 35 (d) 25 RRB NTPC 01.02.2021 (Shift-I) Stage Ist 148. A number consists of 3 digits whose sum is 18 and Ans. (b) : Let the 4 consecutive odd numbers is the middle digit is equal to the sum of other two. x, x +2, x + 4, x + 6 If the number increased by 297 when its digits are According to the question, reversed, then what is the number? (x) + (x + 2) + (x + 4) + (x + 6) = 160 (a) 585 (b) 495 4x + 12 = 160 (c) 396 (d) 486 4x = 148 RRB NTPC 01.02.2021 (Shift-I) Stage Ist x = 148 Ans. (c) : Let the digits of number are x, y and z 4 respectively. x = 37 Given, Hence, the smallest number (x) = 37 x + y + z = 18 ____ (i) 151. There are two numbers with the difference of And, y = x + z 14 between them and the difference of their On putting the value of y in equation (i), squares is 56. What are those numbers? x + x + z + z = 18 (a) 9, –5 (b) 2, 16 2x + 2z = 18 (c) 3, 17 (d) 23, –9 x + z = 9 ____ (ii) RRB NTPC 22.02.2021 (Shift-I) Stage Ist According to the question, Ans. (a) : Let the two numbers be x and y respectively. 100x + 10y + z + 297 = 100z + 10y + x According to the question, 99x + 297 = 99z x – y = 14 ………….(i) x + 3 = z ____(iii) On putting the value of z in equation (ii), And x2 – y2 = 56 ………….(ii) (x + y)(x – y) = 56 .......(From, x2 – y2 = (x + y) (x – y) x+x+3=9 From equation (i) 2x = 6 x + y = 4 ………(iii) x=3 From equation (i) and equation (iii), On putting the value of x in equation (ii), x = 9, y = – 5 x+z=9 3+ z = 9 152. The sum of half, one-third and one-fifth of a z=6 number exceeds the number by 12. What is the From equation (i), y=x+z number? y = 3+6 y=9 (a) 144 (b) 360 Hence, the number will be 396. (c) 444 (d) 122 RRB NTPC 10.02.2021 (Shift-II) Stage Ist Ans. (b) : Let the number = x According to the question, 149. If a positive number is subtracted from its x 1 + 1 + 1 − x = 12 2 3 5 square, we get 812. Find the number. (a) 25 (b) 23 31x − x = 12 30 (c) 27 (d) 29 RRB NTPC 01.02.2021 (Shift-I) Stage Ist x = 12 30 Ans. (d) : Let the number = x and square of number = x2 x = 360 According to the question, Number System 28 YCT
153. A number when reduced by 22 1 % becomes Ans. (d) : Let the number is x 2 According to the question, 217, find the number. x – 40 = x × 60 100 (a) 315 (b) 212 (c) 280 (d) 420 x − 60x = 40 100 RRB NTPC 29.12.2020 (Shift-II) Stage Ist Ans. (c) : Let the number is x 40x = 40 According to the question, 100 x 100% − 22 1 % = 217 x = 100 2 157. The 5th part of a number when divided by 3 x × 77 1 % = 217 yields three times half of tenth part of half of 2 80. What is the number? x = 217 ×100 × 2 155 (a) 60 (b) 90 (c) 45 (d) 44 RRB NTPC 20.01.2021 (Shift-I) Stage Ist x = 280 Ans. (b) : Let the number is x 154. When 38 is added to 30% of a number. The According to the question, 1 1 ×1 result is 50. What is the number? x× 5 80 × 2 10 (a) 20 (b) 80 = × 3 (c) 60 (d) 40 3 2 RRB NTPC 23.02.2021 (Shift-I) Stage Ist Ans. (d) : Let the number = x x = 40× 1 × 1 × 3 According to the question, 15 10 2 x × 30 + 38 = 50 x = 90 100 158. If three-fourth of a number is 50 more than its x × 30 = 50 − 38 = 12 one-third, then find the number. 100 (a) 140 (b) 130 x × 30 = 100 × 12 (c) 120 (d) 100 x = 1200 = 40 30 RRB NTPC 16.01.2021 (Shift-I) Stage Ist Ans. (c) : Let the number is x According to the question, Hence, number (x) = 40 3 x = 1 x + 50 43 155. The sum of two numbers is 20 and the 3 x – 1 x = 50 43 difference of their squares is 80. Select both the 9x – 4x = 50 numbers from the given alternatives. 12 5x = 600 (a) 15, 5 (b) 13, 7 (c) 11, 9 (d) 12, 8 RRB NTPC 15.02.2021 (Shift-I) Stage Ist Ans. (d) : Let the numbers are x and y x = 120 According to the question, 159. The sum of three consecutive odd numbers is x + y = 20 … (i) more than first number of it by 20. Find the x2 – y2 = 80 largest number among them. (x –y) (x+y) = 80 (a) 13 (b) 9 From equation (i), (c) 11 (d) 7 x–y=4 … (ii) RRB JE - 28/06/2019 (Shift-III) From equation (i) and (ii), Ans. (c) Let the three consecutive odd numbers are x, x x = 12, y = 8 +2, x +4. According to the question, 156. When 40 is subtracted from a number, it x + x+2 + x+4 = x + 20 reduces to its 60%. What is the number? 3x + 6 = x + 20 (a) 130 (b) 160 2x = 14 (c) 200 (d) 100 x=7 RRB NTPC 09.02.2021 (Shift-I) Stage Ist Hence, the required number = x + 4 = 7 + 4 = 11 Number System 29 YCT
160. Three times the first of three consecutive odd x – y = 4 ………(ii) integers is 3 more than two times the third. By adding equation (i) and (ii), Find the third integer. 2x = 14 x=7 (a) 15 (b) 13 (c) 11 (d) 9 And y = 3 RRB JE - 26/06/2019 (Shift-III) Hence, the changed number (10y + x) = 10 × 3 +7 = 37 Ans : (a) Let three consecutive odd integers = x, x+2, 163. The sum of a two digit number and the number x+4 made by interchanging its digits is 132. If the According to the question, difference of the digits is 4, find the number. 3x = 2(x + 4) +3 (a) 37 (b) 84 3x = 2x + 8 + 3 (c) 73 (d) 62 x = 11 Hence, the third integer = x + 4 = 11 + 4 = 15 RRB RPF-SI -16/01/2019 (Shift-III) Ans : (b) Let the tens digit of the number is x and the 161. On adding 18 to a two digit number, the digits unit digit is y. of the number are interchanged. The product So, the number = 10x + y of the digits is ‘8’. Find the number. According to the question, (a) 42 (b) 18 x − y = 4.................(i) And, 10x + y + 10y + x = 132 (c) 32 (d) 24 11x + 11y = 132 RRB JE - 27/06/2019 (Shift-I) Ans : (d) Let the unit digit = x. tens digit = y x + y = 12..............(ii) The number = 10y + x From equation (i) and (ii), x–y=4 Given, xy = 8 ---(i) x + y = 12 2x = 16 According to the question, x = 8, y = 4 Hence, the required number = 10x + y = 10 × 8 + 4 = 84 10y + x + 18 = 10x + y 164. The sum of the digits of a two digit number is 9x – 9y = 18 12. The new number formed when the digits are interchanged is 18 more than the original x–y=2 number. What is the original number? On putting the value of x = 8 y ∴ 8−y=2 y 8 – y2 = 2y (a) 39 (b) 48 y2 + 2y – 8 = 0 y2 + 4y – 2y – 8 = 0 (c) 75 (d) 57 RRB Group-D – 26/11/2018 (Shift-III) y (y + 4) –2 (y + 4) = 0 Ans : (d) Let the tens digit of the number is x and the (y –2) (y + 4) = 0 unit digit is y. y=2 Given, x + y = 12 .......(i) Hence, the two digit number = 10x + y On putting the value of y in equation (i)- The number obtained by interchanging the place of the digits = 10y + x x×2=8 According to the question, x=4 10 y + x = 10x + y + 18 Hence required number = 10y + x = 10 × 2 + 4 = 24 9y – 9x = 18 9x – 9y = –18 162. The sum of the digits of a two digit number is x − y = −2 ............(ii) 10. When the digits are interchanged is reduced the number to 36. Find the changed number. (a) 82 (b) 73 (c) 37 (d) 28 By adding equation (i) and (ii), x + y = 12 RRB RPF Constable -17/01/2019 (Shift-III) x – y = –2 2x = 10 Ans : (c) Let the number = 10 x + y x=5 According to the question, x + y = 10 …….(i) The number obtained by interchanging digits = (10y + x) According to the question y=7 Hence, the required number = 10x+y = 10×5+7 = 57 (10 x+y) – (10y + x) = 36 ⇒ 9x – 9y = 36 Number System 30 YCT
165. The sum of the digits of a two digit number is Ans : (b) Let the tens digit be x. 9. Also nine times of this number is twice the And the unit digit be y. number obtained by reversing the order of the The number = 10x + y digits. Find the number. According to the question, (a) 19 (b) 18 (10x + y) – (10y + x) = 45 9x – 9y = 45 (c) 28 (d) 30 RRB Group-D – 05/11/2018 (Shift-III) Hence, the required difference will be x – y = 5 Ans. (b) : Let the tens digit is x and the unit digit is y. 168. The sum of the digits of a two digit number is ∴ The number = 10x + y 11. If the digits are interchanged, the number According to the first condition, decreases to 63. Find the number. x + y = 9 .....(i) (a) 83 (b) 92 According to the second condition, (c) 29 (d) 38 (10x + y) × 9 = (10y + x) × 2 RRB NTPC 04.04.2016 Shift : 3 90x + 9y = 20y + 2x Ans : (b) Let the tens digit be x and the unit digit be y 88x = 11y y = 8x of the number. Putting the value of y in equation (i), ∴ The number = 10x + y x + 8x = 9 x=1 According to the question- Putting the value of x in equation (i), x + y = 11 ..... (i) And 10y + x = 10x + y − 63 1+y=9 9x − 9y = 63 y=8 Hence, the number = 10x + y x−y=7 ..... (ii) = 10 × 1 + 8 = 18 By adding equation (i) and (ii) 166. The sum of the digits of a two digit number is 2x = 18 ⇒ x = 9, y = 2 11. The new number formed when the digits Hence, the number = 10x + y = 10 × 9 + 2 = 92 interchanged is 45 less than the original 169. The sum of the digits of a two digit number is number. Find the original number. 9. When 27 is added to the number, the place of (a) 92 (b) 56 the digits are interchanged. Find the number. (c) 65 (d) 83 (a) 45 (b) 36 RRB Group-D – 15/10/2018 (Shift-III) (c) 18 (d) 27 Ans. (d) : Let the tens digit of the number is a and the RRB NTPC 03.04.2016 Shift : 1 unit digit is b. So, the number = 10a + b Ans : (b) Let the unit digit be x in the two digit number. According to the question, Then, a + b = 11 .....(i) According to the question, 10b + a = 10 a + b – 45 9a – 9b = 45 The tens digit = 9 – x And the number = 10 (9 – x) + x 10 (9 – x) + x + 27 = 10x + 9 - x ⇒ 90 – 10x + x + 27 = 9x + 9 ⇒ 90 + 27 – 9 = 18x a – b = 5 .....(ii) ⇒ 18x = 108 On adding equation (i) and (ii), x=6 a + b = 11 Then, the number =10 (9-x) + x a −b =5 =10 (9-6) + 6 =36 2a = 16 a=8 170. The sum of the digits of a two digit number is Putting the value of a in equation (i), 13. If those digits are interchanged, the number gets decreased by 27. Find the changed number. 8 + b = 11 (a) 85 (b) 76 b=3 (c) 67 (d) 58 Hence, the required number = 10 × 8 + 3 = 83 RRB NTPC 02.04.2016 Shift : 1 167. The difference between a number of two digits Ans : (d) Let the tens digit is x, and the new number formed when the digits The unit digit = 13 – x are interchanged is 45. Find the difference ∴ The number = 10 × x + (13 –x) between the two digits. According to the question, (a) 4 (b) 5 10 × (13 –x) + x = 10 × x + (13–x) – 27 (c) 6 (d) 7 130 – 10x + x = 10x + 13 – x – 27 RRB NTPC 05.04.2016 Shift : 2 18 x = 144 Number System 31 YCT
x = 8 x − y = −2 --------------------- (ii) ∴ The changed number, By adding equation (i) and (ii) - 2x = 8 ⇒ x = 4 , y = 6 = 10 × (13–x) + x = 10 × (13 – 8) + 8 Hence, The required number = 10 × 4 + 6 = 46 = 10 × 5 + 8 = 58 The sum of a two digit number is 9. The 174. The sum of a two digit number and the number 171. formed by interchanging its digits, is 99. Find number is reduces from 45, when the digits are the number if the difference of the digits is 3. interchanged, find the changed number. (a) 27 (b) 63 (a) 45 (b) 72 (c) 45 (d) 54 (c) 63 (d) 27 RRB NTPC 10.04.2016 Shift : 3 RRB NTPC 02.04.2016 Shift : 2 Ans : (d) Let the tens digit be = x Ans : (b) Let the unit digit be y and the tens digit be x. ∴ The number = 10x + y And the unit digit be = y Number = 10 x + y According to the question, Given, x + y = 9 ……..(1) According to the question, (10x + y) + (10y + x) = 99 (10 x + y) – (10 y + x) = 45 11x + 11y = 99 9x – 9y = 45 x + y = 9 ………(i) x – y = 5 ...... (2) x – y = 3 ……..(ii) By adding equation (i) and (ii), Equation (1) ± (2) 2x = 12 x=6 2x = 14 ⇒ x = 7 From equation (i), y = 3 From, equation (1), ∴ The required number = 10x+y = 10×6+3 = 60+3= 63 y = 9–7 = 2 175. The sum of the digits of a two digit number is Hence, The required number = 10y+x = 10×2+7 = 27 5. When the digits are reversed the number 172. The sum of digits of a two-digit number is 10. decreases by 9. Find the changed number. When the digits are reversed, the number (a) 32 (b) 23 decreases by 54. Find the new number. (c) 41 (d) 14 (a) 73 (b) 28 RRB NTPC 28.04.2016 Shift : 3 (c) 82 (d) 37 Ans : (b) Let the tens digit of the number be x and the RRB NTPC 02.04.2016 Shift : 3 unit digit be y. Ans : (b) Let the tens digit of the number is x and the ∴ the number = 10x + y unit digit is y. According to first condition, ∴ The number = 10x + y x + y = 5 ……(i) According to the question, The obtained number after reversing the digits = 10y+x x + y = 10 − − − − − − − − −(i) According to the question, And 10x + y = 10y + x + 54 (10x+y) − (10y+x) = 9 ⇒ 9x – 9y = 9 ⇒ 9x − 9y = 54 ⇒ x − y = 6 − − − − −(ii) ⇒ x – y = 1 .........(ii) By adding equation (i) and (ii), By adding equation (i) and (ii), 2x = 16 ⇒ x = 8, y = 2 2x = 6 Hence, the new number = 10y + x = 10 × 2 + 8 = 28 x=3 From equation (ii) 173. The sum of digits of a two-digit number is 10. 3–y=1 When the digits are interchanged, the number y = 3–1 = 2 increases by 18. Find the number. Hence, the changed number = 10y + x (a) 46 (b) 64 = 10 × 2 + 3 = 23 (c) 19 (d) 28 RRB NTPC 29.03.2016 Shift : 1 Type - 5 Ans : (a) Let the tens digit of the number is x and the unit digit is y. 176. 0.23 is ∴ The number = 10x + y (a) An irrational number (b) A rational number According to the question- (c) A prime number (d) A composite number x + y = 10 --------------- (i) RRB NTPC 20.01.2021 (Shift-I) Stage Ist And 10x + y = 10y + x − 18 9x − 9y = −18 Number System 32 YCT
Ans. (b) : Let us assume 181. Which of the following rational number lies y = 0.23.....(i) between 1 and 1 . Multiplying by 100 in equation (i)- 100y = 23⋅ 23 ........ (ii) 42 (a) 1 (b) 1 6 8 Subtracting eqn (i) from eqn (ii) (c) 3 (d) 3 99y = 23 5 8 y = 23 (Rational number ) RRB NTPC 31.01.2021 (Shift-II) Stage Ist 99 Ans. (d) : 177. ( 3 + 11)2 is a/an 1+1 1+ 2 =3 =4 2 =4 (a) Natural number (b) Whole number 2 28 (c) Irrational number (d) Rational number Therefore, rational number 3 will lie between 1 and RRB NTPC 20.01.2021 (Shift-I) Stage Ist 84 Ans. (c) : 1. 2 ( 3 + 11)2 = 3 +11+ 2× 3 × 11 -40 ( 3 + 11)2 = 14 + 2 33 182. Express as a rational number whose 56 Therefore ( 3 + 11)2 is an irrational number numerator is –5. (b) – 5 178. The product of 2 and 3 is: (a) – 5 8 6 (a) Sometimes a rational number and sometimes (c) – 5 (d) – 5 an irrational number 7 18 (b) Equal to 4 RRB NTPC 23.07.2021 (Shift-II) Stage Ist (c) A rational number Ans. (c) : − 40 = − 8× 5 = − 5 (d) An irrational number 56 8× 7 7 RRB NTPC 20.01.2021 (Shift-I) Stage Ist It is clear that option (c) is the required rational number. Ans. (d) : From above question, (3 5 + 125 ) 2 × 3 = 6 (irrational number) ( )183. ...............is 80 + 6 5 An irrational number is a real number that can't be (a) A rational number expressed in the form p/q , q≠0 (b) A natural number for example - 2, 5, 7, etc. (c) An integer 179. The number of rational number between 5 and (d) An irrational number 7 is: RRB NTPC 13.01.2021 (Shift-I) Stage Ist (a) 2 (b) 0 Ans. (a) : Given, (c) Infinite (d) 1 3 5 + 125 RRB NTPC 19.01.2021 (Shift-II) Stage Ist 80 + 6 5 Ans. (c) : Note:- There are infinite number of rational 5+5 5+6 numbers between any two integers. Hence, there are =3 5 4 5 infinite number of rational numbers that occurs between 5 and 7. = 8 5 = 8 = 4 ( rational number) 180. 3 + 2 5 is : 10 5 10 5 (a) Rational number (b) Irrational number (c) Composite number (d) Natural number Therefore 3 5 + 125 is a rational number RRB NTPC 07.01.2021 (Shift-II) Stage Ist 80 + 6 5 Ans. (b) : Irrational number: The set of real numbers 184. Number 0.232323 can be written in rational that cannot be represented in form of p q is called form as: irrational number that means the number which is not (a) 23 (b) 23 rational is called irrational number. 999 99 Example- 2, 3 ..... (c) 23 (d) 23 9 990 ∴ 3 + 2 5 is an irrational number. RRB NTPC 30.12.2020 (Shift-I) Stage Ist Number System 33 YCT
Ans. (b) : 0.232323.... Ans : (b) Irrational numbers is a real number which = 0.23 cannot be expressed as p/q. (where p and q are integers = 23 and q is not 0). It means, irrational number cannot be expressed as 99 fractions. for example 2 and π are irrational number. 185. Which of the following rational number lies 189. Which of the following is not an irrational? between 9.2 and 10.5? (a) 5428 (b) 6084 (a) 9.15 (b) 9.55 (c) π (d) 7652 (c) 10.67 (d) 9.08 RRB RPF Constable -18/01/2019 (Shift-III) RRB NTPC 03.03.2021 (Shift-I) Stage Ist Ans. (b) : The real numbers which cannot be expressed as p/q, where p and q are integers and q is not 0, are Ans. (b) Q 9.55 is the rational number lies between 9.2 called irrational numbers. These numbers are and 10.5. 186. Which of the following is a rational number represented by QC or QI . between 5 and 7 ? For example- 2,1+ 3, π (a) 4 1 (b) 1 1 6084 = 78 × 78 = 78 (Rational number) 5 5 190. Denote 0.125 as a rational number. (c) 2 2 (d) 3 1 5 5 (a) 119/993 (b) 113/990 (c) 125/999 (d) 100/999 RRB NTPC 20.01.2021 (Shift-I) Stage Ist RRB JE - 25/05/2019 (Shift-I) Ans. (c) : 5 = 2.23 and 7 = 2.64 Ans : (c) Let x = 0.125 From the given options, x = 0.125125 ......... (i) (a) 4 1 = 21 = 4.2 (b) 11 = 6 = 1.2 1000x = 125.125125 ....... (ii) 55 55 From equation (ii)- equation (i) – 999x = (125.125125.......) – (0.125125.....) (c) 2 2 = 12 = 2.4 (d) 3 1 = 16 = 3.2 999x = 125.0 55 55 Hence 2 2 ,is a rational number between 5 and 7 . x = 125 999 5 187. Which of the following is not a rational 191. Find the value of the denominator of 1 ( )number? 5+ 3 32 + 42 , 12.96, 125 and 900 in rational number. (a) 12.96 (b) 900 (5− 3) (b) 5 + 3 22 (c) 125 (d) 32 + 42 (a) 22 (5− 3) RRB NTPC 05.01.2021 (Shift-I) Stage Ist (c) 5 − 3 (d) Ans. (c) : 32 + 42 = 9 +16 = 25 = 5 → Rational 20 20 number RRB Group-D – 29/10/2018 (Shift-III) 12.96 = 1296 ×10−2 = 36 = 18 → Rational number 10 5 Ans : (a) 125 = 5× 5× 5 = 5 5 → Irrational Number According to the question- 900 = 30 × 30 = 30 → Rational Number 1 (5− 3) 3) 3) = (5+ 3)(5− (5 + Hence, 125 is not a rational number. (5− 3) 188. Which of the following is not a rational ( )= 2 (5)2 − number? 3 (a) 3 1728 (b) π (5− 3) (5− 3) (c) 2.487627287 (d) 8.36712846781 = 25 − 3 = 22 RRB RPF-SI -05/01/2019 (Shift-I) Number System 34 YCT
192. Which of the following square roots is 195. From the given options, find the rational irrational? number between the range 2/4 and 0.6. (a) 21025 (b) 18025 (a) 11 (b) 21 25 40 (c) 13225 (d) 15625 (c) 3 (d) 11 RRB Paramedical Exam – 21/07/2018 (Shift-I) 4 4 Ans. (b) : 21025 = 5× 5× 29× 29 RRB NTPC 19.01.2017 Shift : 2 = 145 (Rational number) Ans : (b) From option (b) 18025 = 5× 5× 7 ×103 The rational number between 2 = 0.5 and 0.6 = 135.257 (Irrational number) 4 = 21 = 0.525 13225 = 5× 5× 23× 23 40 = 5×23=115 (Rational number) Hence, 0.5< 0.525< 0.6 15625 = 5× 5× 5× 5× 5× 5 196. Which of the following numbers is irrational? = 5×5×5=125(Rational number) (a) 3 64 (b) 64 Hence, it is clear that the square root of 18025 is (c) 6 64 (d) 4 64 irrational number. RRB ALP & Tec. (30-08-18 Shift-I) 193. Find the rational value of the denominator of Ans : (d) From options, ( )1 1 1/(2+ 3 ). 3 64 = (64)3 = 43 3 = 4 (Rational number) (a) 2 + 3 (b) 2 − 3 ( )1 1 (c) 1 (d) 4 + 3 64 = (64)2 = 82 2 = 8 ( Rational number) RRB Group-D – 22/10/2018 (Shift-III) ( )1 1 Ans : (b) Rationalizing the denominator of 1 , 6 64 = (64)6 = 26 6 = 2 ( Rational number) 2+ 3 4 64 = 4 16 × 4 4 = 2 × 4 4 = (Irrational number) 1 ×2− 3 2+ 3 2− 3 197. Among the following which is a rational number? ( )= 2 − 3 (a) 3 2 (b) 3 8 2 (c) 3 4 (d) 3 12 22 − 3 RRB ALP & Tec. (13-08-18 Shift-III) = 2− 3 = 2− 3 Ans : (b) Rational number can be written as p/q ;(q ≠ 0). 4−3 From option (b), 194. Find the rational value of the denominator of 3 8 = 2 is rational number 1 198. Which of the numbers given below is NOT (5 + 2 3) rational number? (a) (5 − 2 3) (b) (5 − 2 3) (a) 64 (b) 3 64 12 13 (c) 5 − 2 3 (c) 3 8 (d) 8 (d) 5 + 2 3 13 13 RRB ALP & Tec. (09-08-18 Shift-II) Ans : (d) 64 = 8 (Rational number) RRB Group-D – 25/10/2018 (Shift-II) 3 64 = 4 (Rational number) Ans : (b) Rationalizing the denominator of the given 3 8 = 2 (Rational number) 8 = 2 2 (Irrational number) fraction, Hence 2 2 is not a rational number. = 1 × (5 − 2 3) (5 + 2 3) (5 − 2 3) = (5 − 2 3) [(a+b)(a-b) = a2-b2] (A number which we can write as p/q where p and q (5)2 − (2 3)2 both are integers but q ≠ 0 is called rational numbers.) 199. All irrational numbers are---------numbers. = 5−2 3 = 5−2 3 (a) Integers (b) Imaginary 25 −12 13 (c) Whole (d) Real RRB NTPC 19.01.2017 Shift : 3 Number System 35 YCT
Ans : (d) All irrational numbers are real numbers. Ans. (a) From options, As- 2 (a) 4 1024 = 4 4 4 (Irrational number) (b) 101024 = 2 (Rational number) 200. Which of the following is an irrational? (a) 1000000 (b) 3 1000000 (c) 1024 = 32 (Rational number) (c) 6 1000000 (d) 4 1000000 (d) 5 1024 = 4 (Rational number) RRB Group-D – 08/10/2018 (Shift-II) 204. Which of the following is not a rational Ans : (d) From options- number? (a) 1000000 (a) 5 32 (b) 3 64 100 ×100 ×100 = 10 ×10 ×10 = 1000 (Rational) (c) 4 32 (d) 3 27 ( )(b) 1 = 100 RRB Group-D – 28/09/2018 (Shift-I) 3 1000000 = 1003 3 (Rational) Ans : (c) From options, ( )(c) 106 1 = 10 5 32 = 2(Rational) 6 1000000 = 6 (Rational) (d) 4 1000000 = 10 4 100 (Irrational) 3 64 = 4 (Rational) 3 27 = 3 (Rational) 201. Which of the following is an irrational number? 4 32 = 24 2 (Irrational) (a) 4 4 (b) 3 8 205. Which from the following is a rational (c) 16 (d) 6 1 number? RRB Group-D – 22/09/2018 (Shift-III) (a) 5 1551 (b) 3 1331 Ans. (a) : Irrational number – The number which (c) 1221 (d) 4 1441 cannot be expressed as p/q. Example- 3, 6.......4 4 RRB Group-D – 11/10/2018 (Shift-III) Ans : (b) A rational number can be written as p/q ;(q ≠ 0). From options– Hence, From option (b) 3 1331 = 3 11×11×11 = 11 Therefore, 11/1 is a rational number. (a) 4 4 = (22 )14 = 21 = 2 (Irrational number ) 2 (b) 3 8 = 3 2× 2× 2 = 2 (Rational number) 206. Whose square root from the following numbers (c) 16 = 2 × 2 × 2 × 2 = 2 × 2 = 4 (Rational number) is a rational number? (d) 6 1 = 1 (Rational number) (a) 576 (b) 512 (c) 480 (d) 544 Note– π is an irrational number while 22/7 is a rational RRB Group-D – 07/12/2018 (Shift-III) number. Ans : (a) From the options, 202. Which of the following is a rational number? (a) 576 = 24 (a) 3 2 − 2 (b) 3 8 − 2 (b) 512 = 16 2 (c) 3 4 + 4 (d) 3 12 +1 (c) 480 = 4 30 RRB Group-D – 25/09/2018 (Shift-I) (d) 544 = 4 34 Ans : (b) Hence, square root of 576 is 24, which is a rational Rational number – Such numbers which can be number. expressed as p/q ;(q ≠ 0), are called rational numbers. 207. Whose square root from the following numbers Example:- 3 8, 4 etc; will be rational? Irrational number – Such numbers which cannot be (a) 46232 (b) 46233 expressed as p/q. Example:- π ,3 2, 2 etc; (c) 14448 (d) 34225 From options, RRB Group-D – 06/12/2018 (Shift-II) 3 8 − 2 = 0 is a rational number while others 3 2 − 2 , 3 4 + 4 and 3 12 +1 are irrational numbers. Ans. (d) : From options– 203. Which of the following is an irrational (a)46232 = 46232 = 215.016 number? (b)46233 = 46233 = 215.0186 (c)14448 = 14448 = 120.199 (a) 4 1024 (b) 10 1024 (d)34225 = 34225 = 185 (c) 1024 (d) 5 1024 Hence, The square root of 34225 is 185, which is a RRB Group-D – 27/09/2018 (Shift-I) rational number. Number System 36 YCT
208. Whose square root from the following numbers Ans. (b) : is irrational? ( )On rationalizing the given expression 1 , 2+ 3 (a) 5184 (b) 4465 (1× 2 − 3) (c) 3025 (d) 8836 = (2+ 3)(2− 3) (2− 3) (2− 3) RRB Group-D – 05/12/2018 (Shift-II) = (4 − 3) = 1 Ans. (b) From options, (a) 5184 = 72 (b) 4465 = 5 ×19 × 47 (Irrational) (c) 3025 = 55 (d) 8836 = 94 Hence, the square root of 4465 is irrational. Type - 6 209. Whose square root from the following numbers is rational? (b) 344 213. Find the least number which when added to (a) 336 1780 makes the sum a perfect square. (c) 320 (d) 324 (a) 46 (b) 49 RRB Group-D – 04/12/2018 (Shift-III) (c) 69 (d) 72 Ans. (d) From option (d), RRB JE - 27/05/2019 (Shift-II) 324 = 18×18 = 18 Ans : (c) On adding 69 to the number 1780 it will be Hence, the square root of 324 will be 18, which is a 1849, which is a perfect square number. rational number. Thus- 210. Whose square root from the following numbers 1780 + 69 = 1849 will be irrational? 1849 = 43 × 43 (43)2 = 1849 (a) 6441 (b) 9604 (c) 7921 (d) 5776 214. Find the smallest integer whose cube is equal to RRB Group-D – 03/12/2018 (Shift-II) itself. (b) 2 Ans : (a) From options, (a) -1 (d) 0 (c) 1 RRB JE - 22/05/2019 (Shift-I) 6441 = 80.25 is irrational 9604 = 98× 98 = 98 Ans : (a) -1 and 1 are such integers whose cube is equal 7921 = 89 ×89 = 89 to itself. 5776 = 76 × 76 = 76 Hence, the smallest integer = -1 Q (-1)3 = -1 Hence, the square root of 6441 will be irrational. 215. If the cube of a number is subtracted from 211. Whose square root from the following numbers (153)2 the result gives 1457. Find the number. is a rational number? (a) 18 (b) 16 (a) 144 (b) 136 (c) 28 (d) 24 (c) 128 (d) 120 RRB JE - 24/05/2019 (Shift-I) RRB Group-D – 01/12/2018 (Shift-II) Ans : (c) Let the number be x. Ans : (a) From options– According to the question, 144 = 12 (153)2 – x3 = 1457 136 = 2 34 x3 = (153)2 – 1457 128 = 8 2 x3 = 23409 – 1457 x3 = 21952 120 = 2 30 ∴ x = 3 21952 = 3 28× 28× 28 = 28 Hence, the square root of 144 is 12, that is a rational 216. Five times of a positive integer is 3 less than number. twice of its square. Find the integer. ( )212. Express 1 as a rational number. (a) 3 (b) 8 2+ 3 (c) 2 (d) 5 (a) 5 − 2 3 /12 ( )(b) 2 − 3 /1 RRB RPF Constable -19/01/2019 (Shift-I) Ans : (a) Let the positive integer is x. ( )(c) 5 − 2 3 /13 (d) 5 + 2 3 /13 According to the question- 5x = 2x2 –3 RRB Group-D – 08/10/2018 (Shift-I) 2x2 – 5x – 3 = 0 Number System 37 YCT
2x2 – 6x + x – 3 = 0 221. Which of these is a perfect square? 2x (x–3) + 1 (x –3) = 0 (x – 3) (2x + 1) = 0 (a) 9801 (b) 9887 (c) 9013 (d) 9016 x–3=0 RRB JE - 01/06/2019 (Shift-III) 2x + 1 = 0 Ans. (a) From option (a), x = 3 or x = –½ (Invalid) 217. Which of these square numbers cannot be expressed as the sum of two prime numbers? (a) 81 (b) 49 (c) 121 (d) 144 RRB JE - 30/05/2019 (Shift-II) Ans : (c) Hence, 9801 is a perfect square of 99. 81 → 2 + 79 (both of which are prime number) 49 → 2 + 47 (both of which are prime number) 222. If the last digit of the square of a number is 1. 144 → 3 + 141(both of which are prime number) 121 → 2 + 119 (but 119 is not prime number) Find the last digit of its cube. (a) Only 9 (b) 1 or 9 (c) Any odd number (d) Only 1 Hence, option (c) cannot be expressed as the sum of two RRB JE - 27/06/2019 (Shift-I) prime numbers. Ans : (b) Let the number be 9. The last digit of whose 218. Three times the square of a number square is 1. Which is as follows- subtracting by 4 times the number is equal to 92 · 81 50 more than the number. Find the number. Last digit of 729 which is cube of 9 = 9 Let the number be 11. The last digit of whose square is 1. (a) 5 (b) 4 Which is as follows- (c) 6 (d) 10 112 · 121 RRB JE - 28/05/2019 (Shift-II) The last digit of the cube of 11- Ans : (a) Let the number be = x 113 · 1331 According to the question, Hence the last digit = 1 Hence the number will be 1 or 9. 3x2 – x × 4 = x + 50 3x2 – 4x – x – 50 = 0 3x2 – 5x – 50 = 0 3x2 – 15x + 10x – 50 = 0 223. The sum and the difference of two numbers are 3x (x – 5) + 10 (x – 5) = 0 25 and 3 respectively. Find the difference of (x – 5) (3x + 10) = 0 their squares. x–5=0 (a) 165 (b) 75 x=5 (c) 154 (d) 140 219. Which of the following is not a perfect square? RRB JE - 27/06/2019 (Shift-III) (a) 2025 (b) 16641 Ans : (b) Let the two numbers are x and y (c) 1250 (d) 9801 According to the question RRB RPF Constable -20/01/2019 (Shift-I) x + y = 25 ----(i) Ans : (c) From options- x–y=3 ----(ii) 1250 = (35.36)2 is not a perfect square x2 – y2 = (x + y) (x–y) = 25 × 3 = 75 2025 = (45)2 Hence, the difference of their squares = 75 16641= (129)2 224. How many perfect squares are there between 9801 = (99)2 100 and 200? Hence 1250 is not a perfect square, while others are (a) 7 (b) 4 perfect squares. (c) 6 (d) 5 220. Which of these numbers is not a sum of two RRB JE - 27/06/2019 (Shift-III) squares? Ans : (b) Perfect square numbers greater than 100 or nearest to 100 = 121 = (11)2 (a) 41 (b) 13 Perfect square numbers smaller than 200 or nearest to (c) 23 (d) 37 200 = 196 = (14)2 The numbers from (11)2 to (14)2 are = 121, 144, 169, RRB JE - 26/06/2019 (Shift-I) 196 Therefore, there will be 4 such perfect square numbers Ans : (c) From options- (b) 13 = 22 + 32 between 100 and 200. (a) 41 = 52 + 42 (d) 37 = 62 + 12 (c) 23 Hence the number 23 is not the sum of two squares. Number System 38 YCT
225. Find the least number that should be added to 228. What smallest number should be added to the 4042 to make it a perfect square. sum of squares of 15 and 14, so that the (a) 41 (b) 54 resulting number is a perfect square? (c) 64 (d) 58 RRB Group-D – 22/09/2018 (Shift-III) (a) 17 (b) 20 (c) 11 (d) 9 RRB NTPC 29.03.2016 Shift : 1 Ans. (b) : Square root of 4042, Ans : (b) 152 + 142 = 225 + 196 = 421 63 Let the number to added be x, 421+ x = 441 6 4042 ⇒ x = 441− 421 = 20 +6 36 Hence, the required number = 20 123 442 229. Calculate the sum of squares of numbers from 369 1 to 9. (b) 285 (a) 284 (d) 380 73 (c) 385 Square of 64 = 64 × 64 = 4096 RRB NTPC 27.04.2016 Shift : 1 Hence, the required number = 4096–4042 = 54 Ans : (b) The sum of squares of first n numbers By adding 54, the number 4042 will become a perfect = n (n +1)(2n +1) square. 6 226. Divide the number 137592 by the smallest ∴ The sum of squares from 1 to 9 will be- number that leaves no remainder and quotient is a perfect cube. Find the cube root of the = 9(9 +1)(18 +1) = 9×10×19 = 285 quotient. 66 (a) 8 (b) 2 230. Calculate the sum of squares of number from 1 (c) 4 (d) 6 to 10. RRB Group-D – 05/12/2018 (Shift-II) (a) 384 (b) 285 Ans. (d) 137592 = 2×2×2×3×3×3×7×7×13 (c) 385 (d) 380 Hence, it is clear that, dividing 137592 by 7 × 7 × 13 = RRB NTPC 30.04.2016 Shift : 2 637 will leave no remainder And quotient 216 will be a perfect cube. Ans : (c) The sum of squares of first n numbers = n(n +1) (2n +1) 216 = 2×2×2×3×3×3 6 Hence, the required cube root = 2×3 = 6 The sum of squares of the numbers from 1 to 10 will 227. A positive number exceed its square root by 30. be- Find the number. = 10(10 +1) (20 +1) = 10 ×11× 21 = 385 (a) 16 (b) 36 66 (c) 25 (d) 49 231. Find the least number which should be added RRB NTPC 02.04.2016 Shift : 3 to 7864, to make it a perfect square. Ans : (b) Let the number be x, then- (a) 61 (b) 57 x = x + 30 (c) 71 (d) 79 x − 30 = x RRB Group-D – 11/12/2018 (Shift-III) On squaring in both side- Ans : (b) Adding 57 to 7864, gives 7921 which is a perfect square of 89. ( )(x − 30)2 = 2 Hence, it is clear that adding 57 to 7864 will make the x x2 + 900 − 60x = x number a perfect square. x2 − 60x − x + 900 = 0 232. The number 4050 becomes a perfect square x2 − 61x + 900 = 0 when multiplying by a positive integer. Find x2 − 36x − 25x + 900 = 0 the square root of the number. x (x − 36) − 25(x − 36) = 0 (a) 95 (b) 80 (c) 90 (d) 85 (x − 36)(x − 25) = 0 RRB Group-D – 01/10/2018 (Shift-III) x − 36 = 0 or x − 25 = 0 Ans : (c) Q 4050 = 2 × 3× 3× 3× 3× 5 × 5 x = 36 or x = 25 25 is not more than its square root, which does not Hence, number 4050 becomes a perfect square when follow the condition. multiplied by 2 = 4050×2 = 8100 ∴ The required square root of the number 8100 Hence, the required number will be x = 36. = 2×3×3×5 = 90 Number System 39 YCT
233. Which of the following numbers is a perfect ∴ Number of even factors = 320 – total no. of odd square? (b) 8.1 factors. (a) 0.09 = 320 – {(4 + 1) (3 + 1) (1 + 1)} (c) 0.025 (d) All = 320 – {5 × 4 × 2} RRB NTPC 29.03.2016 Shift : 2 = 320 – 40 Ans : (a) 0.09 = (0.3)2 = 280 Hence, only 0.09 is a perfect square number. 238. Find the digit in the unit's place of 124n + 124(n+1), where n is any whole number. Type - 7 (a) 4 (b) 8 (c) 2 (d) 0 234. For any natural number n, 6n - 5n always ends RRB NTPC 17.02.2021 (Shift-II) Stage Ist with ; Ans. (d) : 124n + 124(n+1) (a) 7 (b) 1 On putting n =1 (c) 5 (d) 3 = 124 + (124)2 RRB NTPC 28.12.2020 (Shift-II) Stage Ist For unit digit 4 + 6 = 10 Ans. (b) : The unit value of 6n–5n for any natural Hence, It is clear that the digit come in the unit place number 'n' will always be 1 because 6 can be any will be '0'. natural number in the power that units number in the 239. What is the unit digit in the following product? power of 5 has its unit digit as 5. 91×92×93×...........×99 235. What is the total number of odd and even (a) 2 (b) 1 divisors of 120, respectively? (c) 4 (d) 0 (a) 12,4 (b) 16,0 RRB NTPC 09.02.2021 (Shift-II) Stage Ist (c) 4,12 (d) 8,8 RRB NTPC 01.02.2021 (Shift-II) Stage I Ans. (d) : Q 91× 92 × 93× 94 × 95× 96 × 97 × 98× 99 Ans. (c) : Divisors of 120– It is clear that multiplying by taking unit digits of all the 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24 numbers will give '0' i.e. where 2×5 comes then its unit 30, 40, 60, 120 digit is always zero. Number of even divisors – 12, 240. Find the number of factors of 4200. Number of odd divisors – 4 (a) 48 (b) 56 236. If the sum of five consecutive multiples of 2 is (c) 64 (d) 46 660, then find the largest number. RRB NTPC 26.07.2021 (Shift-II) Stage Ist (a) 162 (b) 130 Ans. (a) : 4200 = 2×2×2×5×5×3×7 (c) 125 (d) 136 =23×52×31×71 RRB NTPC 15.02.2021 (Shift-II) Stage Ist The number of factors = (3+1) ×(2+1)×(1+1) ×(1+1) Ans. (d) : Let five consecutive multiple of 2 – = 4×3×2×2 2x, 2x+2, 2x+4, 2x+6, 2x+8 = 48 According to the question, 241. How many factors does the number 12288 2x + 2x + 2 + 2x + 4 + 2x + 6 + 2x + 8 = 660 10x+20=660 have? (b) 26 10x=640 (a) 24 x = 64 (c) 28 (d) 22 Hence, largest number = 2x + 8 = 2×64+8 RRB NTPC 23.07.2021 (Shift-I) Stage Ist =128+8 Ans. (b) : 12288 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 136 × 2 × 2 × 3 = 212 × 31 237. How many factors of 27 × 34 × 53 × 7 are even ? Hence numbers of factors = (12 + 1) × (1 + 1) (a) 40 (b) 280 = 13 × 2 (c) 320 (d) 84 = 26 RRB NTPC 31.01.2021 (Shift-I) Stage Ist 242. If a positive number N, when divided by 5 RRB NTPC 14.03.2021 (Shift-I) Stage Ist leaves a remainder 3, then the unit's place digit Ans. (b) : 27 × 34 × 53 × 7 Number of factors. = (7 + 1) (4 + 1) (3 + 1) (1 + 1) of N is? (a) 0 or 5 (b) 0 or 2 =8×5×4×2 (c) 3 or 8 (d) 1 or 5 = 320 RRB NTPC 25.01.2021 (Shift-I) Stage Ist Number System 40 YCT
Ans. (c) : Required positive number 247. Find the unit digit in given factor of (3451)51 × (531)43. = 5K+3 (Q K = 0,1,2…..) = 5×0+3 = 3 (On putting K = 0) (a) 6 (b) 4 = 5×1+3 = 8 (On putting K = 1) (c) 1 (d) 9 RRB RPF-SI -11/01/2019 (Shift-I) Hence, unit digit of N = 3 or 8 Ans : (c) The given expression is (3451)51 × (531)43 243. The unit digit in 4 × 38 × 764 × 1256 is : According to the question it is clear that the unit digit of (a) 6 (b) 8 3451 and 531 is 1, so the unit digit of their product will (c) 4 (d) 5 also be 1. RRB NTPC 28.12.2020 (Shift-I) Stage Ist 248. How many multiples of 28 × 32 × 53 × 75 are even numbers? Ans. (b) : (a) 288 (b) 168 4 × 38 × 764 × 1256 (c) 576 (d) 464 ↓↓ ↓ ↓ RRB Group-D – 06/12/2018 (Shift-II) 6 Ans. (c) : The number of factors of 28 × 32 × 53 × 75 = 4× 8 × 4 × (8 + 1) (2 + 1) (3 + 1) (5 + 1) = 648 = 32 × 24 ∴ The number of even factors (multiples) = 648 – The ↓↓ number of total odd factors = 2× 4 = 648 – {(2 + 1) (3 + 1) (5 + 1)} = 648 – {3 × 4 × 6} Hence unit digit = 8 = 648 – 72 = 576 244. Unit digit of (1373)36 – (1442)20 is - 249. How many factors of 729 are perfect squares? (a) 2 (b) 4 (a) 5 (b) 4 (c) 3 (d) 2 (c) 5 (d) 3 RRB Group-D – 01/10/2018 (Shift-I) RRB ALP CBT-2 Physics & Maths 22-01-2019 (Shift-I) Ans. (c) : The factors of 729, Ans. (c) : (1373)36 − (1442)20 3 729 = ( )3 36 − (2)20 3 243 = ( )3 9×4 − ( 2)5×4 3 81 3 27 = (3)4 − (2)4 39 33 = 81–16 1 = 65 Perfect squares = 3× 3× 3× 3× 3× 3 =5 Hence, total 3 factors of 729 (9,9,9) are perfect squares. 245. How many of the factors of 256 are perfect 250. How many multiples of 29 × 35 ×54 × 76 are odd squares? numbers? (a) 5 (b) 3 (a) 288 (b) 144 (c) 6 (d) 4 (c) 210 (d) 140 RRB ALP & Tec. (20-08-18 Shift-II) RRB Group-D – 06/12/2018 (Shift-III) Ans : (a) Perfect square factors of 256 = 1, 4, 16, 64, 256 Ans. (c) : The required odd multiple number = (5+1) × (4+1) × (6+1) Hence, the required number of perfect square factors = 5 = 6 × 5 × 7 = 210 246. Which of these numbers has the highest 251. Find the last digit of 2136 ? number of divisors? (a) 6 (b) 3 (a) 156 (b) 240 (c) 7 (d) 9 (c) 172 (d) 200 RRB Group-D – 05/12/2018 (Shift-II) RRB JE - 23/05/2019 (Shift-I) Ans. (d) The unit digit of 2136 2136 = (2134 × 2132) Ans : (b) From options– 156 = 22 × 31 × 131 = (2+1) (1+1) (1+1) = 12 (divisor) 1×9=9 240 = 24 × 31 × 51 = (4+1) (1+1) (1+1) = 20 (divisor) 252. The smallest natural number, by which 216 172 = 22 × 431 = (2+1) (1+1) = 6 (divisor) should be multiplied, so that the number of 200 = 23 × 52 = (3+1) (2+1) = 12 (divisor) Hence, It is clear that the number of the divisors of 240 factors of the product is odd? is highest. (a) 4 (b) 6 (c) 12 (d) 8 RRB Group-D – 11/12/2018 (Shift-I) Number System 41 YCT
Ans. (b) 257. Find the unit digit in the following The number of multiples of (216 = 23 × 33) is: (1234)102 + (1234)103 = (3+1) (3+1) = 4 × 4 = 16 (even) (a) 2 (b) 4 The smallest natural number, by which 216 should be multiplied, so that the number of factors of the product (c) 0 (d) 1 is odd = 6 RRB NTPC 28.04.2016 Shift : 2 ∴ Required number of multiples in 216 × 6 = 24 × 34 Ans : (c) Given expression: (1234)102 + (1234)103 The unit digit, = (4)102 + (4)103 = (4 + 1) (4 + 1) = 25 = (42)51 + (42)51 × 41 253. What is the unit digit of [45231632 × 22241632 × = (16)51 + (16)51 × 41 32251632] =6+6×4 (a) 1 (b) 0 = 6 + 2 4 = 30 (c) 4 (d) 5 Hence, the unit digit will be 0. RRB NTPC 18.01.2017 Shift : 3 258. How many factors of 512 are perfect squares? (a) 6 (b) 4 Ans : (b) ( )4523 1632 ×( )2224 1632 ×( )3225 1632 (c) 3 (d) 5 ⇒ (3)4 ×(4)4 × (5)4 RRB Group-D – 28/09/2018 (Shift-I) Ans : (d) The factors of 512 = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 In which = 1, 4, 16, 64, 256 are perfect squares So, the total number of perfect squares factors is 5. 259. Which is the smallest positive integer or 30 ⇒ 0 natural number, when divides 1920 so that the number of factors of quotient is odd? 254. Calculate the total prime factors in the product (a) 40 (b) 10 of {(8)10 × (9)7 × 78 } (c) 20 (d) 30 (a) 45 (b) 54 RRB Group-D – 12/12/2018 (Shift-I) (c) 52 (d) 65 Ans. (d) From options, RRB NTPC 18.04.2016 Shift : 2 Number of factors in 1920 = 48 = 24 ×3 40 Ans : (c) ( )8 10 × (9)7 × 78 is (4+1)(1+1) = 10 (Even) ( ) ( )= (2)3 10 × (3)2 7 × (7)8 Number of factors in 1920 =192 = 26 ×3 10 = 230 × 314 × 78 is (6+1)(1+1) = 14 (Even) Hence, the total prime factors = 30+14+8 = 52 1920 255. Calculate the total prime factors in the product Number of factors in 20 = 96 = 25 ×3 of {(16)7 × (27)6 × 59} is (5+1)(1+1) = 12 (Even) (a) 28 (b) 43 1920 30 (c) 55 (d) 56 Number of factors in = 64 = 26 RRB NTPC 16.04.2016 Shift : 2 is (6+1) = 7 (Odd) Ans : (c) Total prime factors {(16)7 × (27)6 × 59} Hence, option (d) will be required answer. 260. How many factors of the ( ) ( )= 24 7 × 33 6 × 59 number = 228 × 318 × 59 210 × 36 × 53 × 75 are divisible by 2160? = 28 + 18 + 9 = 55 (a) 180 (b) 336 256. Find the unit digit in the product of (4211)102 × (c) 504 (d) 560 (361)52 RRB Group-D – 11/12/2018 (Shift-III) Ans : (c) Factors of 2160 = 24 × 33 × 51 (a) 3 (b) 1 Let the total factors are n. (c) 4 (d) 7 = 210 × 36 × 53 × 75 24 × 33 × 51 RRB NTPC 16.04.2016 Shift : 3 n Ans : (b) n = 26 × 33 × 52 × 75 The required unit digit in (4211)102 × (361)52 So, the total number of factors = (6+1)(3+1)(2+1)(5+1) ⇒ (1)102 × (1)52 = 1×1 = 1 = 7 × 4 × 3 × 6 = 504 Number System 42 YCT
266. The sum of the place values of 3 in 3636 is: Type - 8 (a) 330 (b) 3030 (c) 3 (d) 3003 261. What is the place value of 5 in the number RRB NTPC 25.01.2021 (Shift-II) Stage Ist 56789214? (b) 5×104 Ans. (b) : The place value of 3 in 3636. (a) 5×106 (d) 5×105 (c) 5×107 RRB NTPC 29.01.2021 (Shift-II) Stage I Ans. (c) : The place value of 5 in 56789214 – Sum of place values of 3 = 3000 + 30 = 3030 267. The difference between the place values of 2 262. Find the sum of the place value and the face and 3 in the number 128935 is: value of 7 in the number 53736. (a) 300 (b) 19970 (a) 77 (b) 707 (c) 20000 (d) 30 (c) 770 (d) 777 RRB NTPC 02.03.2021 (Shift-I) Stage Ist RRB NTPC 29.01.2021 (Shift-II) Stage Ist Ans. (b) : Ans. (b) : The place value and the face value of 7 in the number 53736. Place value of 7 = 700 Face value of 7 = 7 Required difference = 20000 – 30 = 19970 Required sum = 700 + 7 268. The sum of the place values of 9 in 96961 is: = 707 (a) 9000 (b) 18 263. In the number 76897, what is the place value of (c) 9090 (d) 90900 8? (b) 8000 RRB NTPC 19.01.2021 (Shift-I) Stage Ist (a) 8 Ans. (d) : Sum of the place value of 9 in number 96961 (c) 800 (d) 80 = 90000+ 900 = 90900 RRB NTPC 09.03.2021 (Shift-II) Stage Ist Hence, option (d) is correct. Ans. (c) : 269. Find the difference between the place values of 8 and 4 in the number 683479. (a) 7 (b) 80000 (c) 79600 (d) 76600 Hence, place value of 8 in 76897 will be 800. RRB NTPC 04.03.2021 (Shift-II) Stage Ist 264. The face value of 8 in 758639 is : Ans. (c) : (a) 8000 (b) 80 (c) 800 (d) 8 RRB NTPC 25.01.2021 (Shift-II) Stage Ist Ans. (d) : In the given number = 758639 The face value of 8 = 8 265. Find the difference of the place and face values Hence, required difference = 80000 – 4000 = 79600 of 6 in 516372 270. Find the sum of the face values of 6 and 5 in (a) 5998 (b) 6698 61827354 (c) 5394 (d) 5994 (a) 60000300 (b) 30 RRB NTPC 25.01.2021 (Shift-II) Stage Ist (c) 40 (d) 11 Ans. (d) : The place values of 6 in 516372– RRB NTPC 12.01.2021 (Shift-II) Stage Ist Ans. (d) : the face values of 6 = 6 Required sum = 6 + 5 = 11 YCT Required difference = 6000 – 6 43 = 5994 Number System
271. The difference between the place values of 2 277. Calculate the sum of the face value and the and 4 in the number 275413 is place value of 7 in the number 3728456. (a) 196600 (b) 2 (a) 700007 (b) 0 (c) 199600 (d) -2 (c) 7 (d) 700000 RRB NTPC 05.03.2021 (Shift-I) Stage Ist RRB Group-D – 01/10/2018 (Shift-I) Ans. (c) : Ans. (a) : The place value of 7 = 7 × 100000 = 700000 The face value of 7 = 7. Hence, the required sum = 700000 + 7 = 700007 278. Find the face value of 4 in 145.390. Difference between the place value of 2 and 4 (a) 40,000 (b) 4 = 200000 – 400 = 199600 (c) 140,000 (d) 45 272. The digit of hundred's place value of 19! is: RRB NTPC 04.04.2016 Shift : 2 (a) 0 (b) 9 Ans : (b) The face value of any digit is the same digit. (c) 4 (d) 1 In the number 145.390 the face value of 4 = 4 RRB NTPC 12.01.2021 (Shift-I) Stage Ist 279. Find the difference between the place value and face value of the digit 9 in the number Ans. (a) : 19! = 19 × 18 × 17 ×16 ×…….. × 1 229301? Number of 5 in 19! = 3 (a) 9292 (b) 8991 So number of zeros = 3 (c) 0 (d) 220 19! · ...........000→100th number RRB NTPC 03.04.2016 Shift : 2 Hence the value of 100th place is 0. Ans : (b) The place value of 9 in the number 229301 = 9 × 1000 = 9000 273. What is the difference between the place and And the face value of 9 = 9 face values of '5' in the number 3675149? Hence, the required difference = 9000- 9 = 8991 (a) 5000 (b) 4995 280. What is the difference between the place value (c) 495 (d) 4990 and face value of 3 in 273965? RRB JE - 23/05/2019 (Shift-I) (a) 2035 (b) 3962 Ans : (b) The place value of 5 in the number 3675149 (c) 2997 (d) 0 = 5×1000 = 5000 RRB ALP & Tec. (31-08-18 Shift-II) And the face value of 5 = 5 Ans. (c) : In the number 273965, Required difference = 5000–5 = 4995 The place value of 3 = 3 x 1000 = 3000 274. What is the place value of 8 in 634785? And the face value = 3 (a) 8 (b) 80 Hence, the required difference = 3000 − 3 = 2997 (c) 800 (d) 80,000 281. The difference between the place values of '4' RRB RPF Constable -20/01/2019 (Shift-I) and '2' in the number 833749502 is: Ans : (b) The place value of 8 in 634785 = 8 ×10 = 80 (a) 49998 (b) 30098 275. Find the sum of the face value and place value (c) 39098 (d) 39998 of 6 in the number 206743? RRB ALP & Tec. (10-08-18 Shift-II) (a) 6749 (b) 12743 Ans : (d) The given number = 833749502 (c) 6006 (d) 12 The place value of 2 = 2 RRB Group-D – 28/11/2018 (Shift-I) The place value of 4 = 4 x 10000 = 40000 Ans : (c) The face value of 6 in the number 206743 = 6 Hence, the required difference = 40000 – 2 = 39998 And the place value of 6 = 6 × 1000 = 6000 The required sum (Face value + Place value) = 6 + 6000 Type - 9 = 6006 276. The difference between two place values of 3 in By how much is 1 th of 432 smaller than 3 th of 64 935071360 is- 282. (a) 29999700 (b) 29999701 216? (c) 2999600 (d) 29999400 RRB Group-D – 23/10/2018 (Shift-I) (a) –90 (b) 72 Ans. (a) : The place value of 3 in 935071360, (c) 90 (d) 162 RRB NTPC 15.03.2021 (Shift-II) Stage Ist Ans. (c) : According to the question- The required difference = 30000000–300 = 29999700 1 part of 432 = 432× 1 = 72 66 Number System 44 YCT
and 3 part of 216 = 216× 3 = 162 (a) `23, `15 (b) `51, `32 44 (c) `15, `23 (d) `32, `51 Required difference = 162 – 72 RRB NTPC 31.01.2021 (Shift-I) Stage Ist = 90 Ans. (c) : Let the fares from city P to Q = `x 283. Terry consumes 1700 mL of milk every day. and the fares from city P to R = `y How many litres of milk will she consume in 5 According to the question, weeks? (a) 59 L (b) 60 L (c) 58.5 L (d) 59.5 L On multiplying by 3 in equation (i) and 2 in equation (ii) RRB NTPC 09.02.2021 (Shift-II) Stage I Ans. (d) : Q Terry consumes in 1 day = 1700 mL ∴ In 5 weeks = 35 days = 1700× 35 From equation (iii) & (iv) we have – 1000 5y = 115 y =`23 = 59500 L 1000 On putting the value of y in equation (i), = 59.5L 284. Mohan earns `60 on first day and spends `50 2x + 3 × 23 = 99 2x + 69 = 99 on the second day. He again earns `60 on the 2x = 99 – 69 x = 30 third day and spends `50 on the fourth day and 2 so on. On which day will he have `200 with him x =`15 before spending? (b) 14th (a) 10th (d) 29th (c) 28th RRB NTPC 24.07.2021 (Shift-II) Stage Ist Ans. (d) : Mohan earns on the first day = `60 Hence the fares from city P to Q and the fares from city and spends on the second day = `50 P to R are `15, `23 respectively. Thus, in 2 days Mohan saves = `10 287. There are 40 persons in a palace. If every Hence, Mohan saves in 28 days= `140 person shakes hands with every other person, Mohan will earn on 29th day = `60 what will be the total number of handshakes? So, On the 29th day Mohan has = 140 + 60 = `200 (a) 750 (b) 780 285. In a farmer's house, there are chickens and (c) 800 (d) 790 goats. The total number of their heads is 42 and RRB NTPC 21.01.2021 (Shift-I) Stage Ist the total number of their legs is 138. Find the Ans. (b) : Total number of handshakes = n (n −1) number of chickens. 2 (a) 15 (b) 18 40(40 −1) (c) 20 (d) 22 RRB NTPC 01.02.2021 (Shift-I) Stage Ist = 2 Ans. (a) : Let the number of chickens = x = 40× 39 2 Number of goats = y = 20 × 39 According to the question, x + y = 42 ____(i) 2x + 4y = 138 ____ (ii) = 780 On solving the equation (i) × 4 and (ii) 288. In a group of 35 persons, 20 are young and 18 4x + 4y = 168 are girls. How many young girls are there in –2x +– 4y =–138 the group ? (b) 3 (a) 1 2x = 30 (c) 18 (d) 2 x = 15 Hence, the number of chickens = x = 15 RRB NTPC 17.01.2021 (Shift-II) Stage Ist Ans. (b) : According to the question, 286. Two bus tickets from city P to Q and three Number of young girls in the group = (20+18) – 35 tickets from city P to R cost `99, but three = 38 – 35 = 3 tickets from city P to Q and two tickets from city P to R cost `91. What are the respective fares from city P to Q and from city P to R. Number System 45 YCT
289. X, Y and Z together earn ` 2,400/- in 15 days, Ans : (a) Let the third number be 100. X and Y together earn ` 1,840/- in 16 days. Y Then according to the question the first number = 125 and Z together earn ` 1,530/- in 18 days. What And the second number = 150 is the daily earning (in `) of Y? Hence, required ratio = First number : Second number = (a) `50 (b) `40 125 : 150 (c) `60 (d) `30 = 5:6 RRB NTPC 05.03.2021 (Shift-I) Stage Ist Ans. (b) : 111 1 Amount earned by X, Y and Z in 1 day = 2400 = 160 294. Solve: 1 + + + + + ...... 15 2 4 8 16 Amount earned by X, Y and Z in 1 day = 1840 = 115 (a) 2 (b) 1/50 16 (c) 3 (d) 1/22 RRB JE - 23/05/2019 (Shift-I) Amount earned by Y and Z in 1 day = 1530 = 85 Ans : (a) 18 1+ 1 + 1 + 1 + 1 + ......... Daily earning of Y · (Daily earning of X and Y 2 4 8 16 This is a geometrical progression- together) ± (Daily earning of Y and Z together) – (Daily earning by X, Y and Z together) a = 1, r = 1 = 115 + 85 – 160 2 = 40 Let the sum be S. 290. The remainder in the expression 27 3 is: S = 1 a r , S∞ = 1 4 − 1− 1 (a) 6 (b) 4 2 (c) 3 (d) 8 = 1 = S∞ = 2 1 RRB NTPC 15.02.2021 (Shift-I) Stage Ist S∞ 2 Ans. (c) : In the given expression 2 Dividend = quotient × divisor + remainder 295. In a school picnic group, 2/9th part were adults = 27 ×4 +3 ∴ Remainder = 3 and the number of children was more than 291. A maximum of how many pieces of exact 17 cm adults by 95. How many children were present length can be cut from a 960 cm long rod? there? (a) 95 (a) 60 (b) 58 (b) 133 (c) 54 (d) 56 (c) 190 (d) 103 RRB NTPC 08.02.2021 (Shift-I) Stage Ist RRB JE - 27/06/2019 (Shift-I) Ans. (d) : According to question Ans : (b) Let the total number of people in the group = x Number of pieces = 960 = 56 + 8 17 17 The number of adults = x × 2 = 2x 99 Hence, number of pieces of exact 17 cm length will be 56. The number of children = x − 2x = 9x − 2x = 7x 9 99 292. If 3/11 < x/3 < 7/11, which of the following can be value of 'x'? (b) 1 7x − 2x = 95 (a) 0.5 (d) 3 99 (c) 2 RRB JE - 23/05/2019 (Shift-I) 7x − 2x = 95 9 Ans : (b) From options, When X = 0.5 then 0.272 < 0.166 < 0.636 (False) 5x = 95 When X = 1 then 0.272 < 0.333 < 0.636 (True) 9 When X = 2 then 0.272 < 0.666 < 0.636 (False) When X = 3 then 0.272 < 1 < 0.636 (False) x = 171 Hence, It is clear that the value of x will be 1. 293. If the first number and the second number is Hence, the number of children = 7x = 7 ×171 = 133 25% and 50% more than the third number 99 respectively, find the ratio between the first 296. Find the value of 52-|8-20|= and second number. (a) 45 (b) 40 (a) 5 : 6 (b) 2 : 1 (c) 65 (d) 64 (c) 6 : 5 (d) 1 : 2 RRB JE - 27/05/2019 (Shift-I) RRB RPF Constable -18/01/2019 (Shift-I) Number System 46 YCT
Ans : (b) The given value = 52− | 8 − 20 | 301. Subtract 64.37 out of 1000.03 and add the = 52− | −12 | resultant obtained from it to the sum of 3.4 and | −A | = A (The value of Mod is always +ve.) 7.56. What will be its value? Hence, the required value = 52 −12 = 40 (a) 948.62 (b) 944.62 (c) 945.62 (d) 946.62 297. If one dozen of apples weigh 1.8 kg, then find RRB Group-D – 08/10/2018 (Shift-III) the number of apples of three boxes whose total Ans : (d) According to the question, weight is 23.25 kg. 1000.03 − 64.37 = 935.66 (a) 280 (b) 155 (c) 465 (d) 215 And RRB RPF-SI -13/01/2019 (Shift-I) 935.66 + (3.4 + 7.56) = 935.66 +10.96 = 946.62 Ans : (b) Total weight = 23.25 kg One apples’s weight = 1.8 kg 302. Seema got ` 50 from her father and purchased 12 toffee for ` 15. Her mother gave her ` 30 but her brother took ` 42 from her. How much The required number of apples, = Total weight = 23.25×12 = 155 money did she have left? 1 apple's weight 1.8 (a) ` 23 (b) ` 24 Number of apples in the box = 155 (c) ` 20 (d) ` 25 298. Pick out the set that forms the factors of 36. RRB Group-D – 23/09/2018 (Shift-II) (a) (2, 3, 4, 6, 9) (b) (2, 3, 4, 6) Ans : (a) Total sum of money that Seema have left (c) (2, 3, 4, 6, 9, 12, 18) (d) (2, 3, 4, 6, 9, 12) = 50-15+30-42 = 80-57 = ` 23 RRB JE - 27/05/2019 (Shift-III) Ans : (c) All the factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36 303. ` 150 of Amit's Pocket money was spent on a pair of shoes and ` 75 on a watch. The total Hence, the required set that is formed by the factors of amount spent was three-fourth of his total 36 will be = (2, 3, 4, 6, 9, 12, 18) 299. The square of a number is 3 more than twice pocket money. What was the amount received the number. What is the possible number. by Amit as pocket money? (a) 1 or 3 (b) 1 or -3 (a) ` 300 (c) -1 or -3 (d) -1 or 3 (b) ` 400 RRB Group-D – 15/10/2018 (Shift-I) (c) ` 375 Ans : (d) Let the number be x (d) ` 250 According to the question, RRB ALP CBT-2 Electrician 22-01-2019 (Shift-I) x2 = 2x + 3 Ans. (a) :Amount spent on shoes = ` 150 x2 - 2x - 3 = 0 Amount spent on watch = ` 75 x2 - 3x + x - 3 = 0 x (x - 3) + 1(x - 3) = 0 Let Amit's pocket money = ` x (x - 3) (x + 1) = 0 According to the question, x-3=0 3x = 150 + 75 x=3 4 x+1=0 x = -1 3x = 4×225 Hence, the possible number is -1 or 3. x = 900 300. 3 8 is directly proportional to- 3 10 + 15 x = ` 300 (a) 11 (b) 11 10 15 So, Amit got the amount for pocket money = ` 300 (c) 6 (d) 3 304. Geeta weighs 11.235 kg. Her sister weighs 1.4 5 15 times her weight. Find the total weight of both. (a) 15.729 kg RRB Group-D – 02/11/2018 (Shift-I) (b) 25.964 kg Ans. (c) 3 + 8 (c) 26.964 kg 10 15 (d) 28.964 kg = 9 +16 = 25 = 5 30 30 6 RRB NTPC 29.03.2016 Shift : 1 Ans :(c) Geeta’s weight = 11.235 kg 5 is directly proportional to 1 = 6 ∴ The weight of Geeta’s sister 6 5 5 = 11.235×1.4 = 15.729 kg 6 The total weight of both of them = 11.235 +15.729 = 26.964 kg Number System 47 YCT
02. Decimal Fractions Type - 1 (c) 5 (d) – 3 2 2 RRB NTPC 23.07.2021 (Shift-II) Stage Ist 1. Which of the following fractions is the Ans. (b) : smallest? – 3 = –1.5 2 (a) 9 (b) 11 11 12 3 = 1.5 2 (c) 8 (d) 10 13 14 RRB NTPC 09.03.2021 (Shift-II) Stage Ist 11 = 2.75 4 Ans. (c) : From option, 5 = 2.5 9 = 0.8181 2 11 It is clear that greatest fraction is 11 11 = 0.916 4 12 4. Which of the following is the smallest fraction? 8 = 0.615 7,7,4,5 13 6957 10 = 0.714 (a) 7 (b) 4 14 6 5 Hence, it is clear that smallest fraction is 8 . (c) 7 (d) 5 9 7 13 2. Which is the smallest fraction among the RRB NTPC 22.02.2021 (Shift-I) Stage Ist following fractions? 8 Ans. (d) : 7 = 1.16, 7 = 0.77, 4 = 0.80, 5 = 0.71 (b) 695 7 3 8 54 , ,, Hence the smallest fraction is 5 7 9 14 8 9 5. Which of the following fractions is the smallest? 4 (a) 9 14 3 5 (a) 7 (b) 7 (c) (d) 8 10 98 (c) 3 (d) 5 RRB NTPC 02.03.2021 (Shift-I) Stage Ist 4 7 Ans. (c) :From question, RRB NTPC 23.02.2021 (Shift-I) Stage Ist 3 = 0.33 8 = 0.57 Ans. (b) : On writing given fraction in descending 9 14 order. 5 = 0.62 4 = 0.44 8 9 735 7 >>> The smallest fraction is 3 9 8 4 7 10 0.875 > 0.75 > 0.714 > 0.70 Hence, 7 will be the smallest fraction. 10 3. Find the greatest fraction out of – 3 , 3 , 11 , 5 : 6. Find the greatest among these fractions. 22 4 2 5/6, 6/11, 2/3, 8/9, 6/7 (a) 3 (b) 11 (a) 2/3 (b) 8/9 24 (c) 5/6 (d) 6/7 RRB JE - 01/06/2019 (Shift-III) Decimal Fractions 48 YCT
Ans. (b) From question :- Ans. (d) : From question, 5 = 0.83 , 6 = 0.54 1 = 0.1 10 6 11 2 = 0.67 , 8 = 0.89 1 = 0.01 100 39 6 = 0.85 9 = 0.009 7 1000 Hence, the greatest fraction is 0.89 = 8 500 = 0.05 9. 9 10000 1000 0.1 > 0.05 > 0.01 > 0.009 7. Find the difference between the greatest and the least fraction among 2/3, 3/4, 4/5, 5/6. Hence, it is clear that the required fraction is (a) 3/5 (b) 1/7 (c) 1/6 (d) 2/5 RRB JE - 22/05/2019 (Shift-I) 11. Which of the following fractions is the greatest? Ans : (c) From question, (b) 9/22 (a) 8/19 2 = 0.66 (Least fraction) (c) 10/23 (d) 11/24 3 RRB RPF-SI -11/01/2019 (Shift-III) 3 = 0.75 Ans : (d) From options 4 8 = 0.421, 9 = 0.409, 10 = 0.43, 11 = 0.458 4 = 0.8 19 22 23 24 5 Hence, the required greatest fraction is 11 . 5 = 0.83 (Greatest fraction) 24 6 12. Arrange the following ratios in decreasing 52 = 5−4 1 order, which number will be the last? − = Hence, the required difference = 11:14, 17:21, 5:7, 2:3 63 6 6 (a) 17:21 (b) 5:7 8. Which of the following is the greatest? (c) 2:3 (d) 11:14 (a) 15/16 (b) 24/25 RRB Group-D – 05/10/2018 (Shift-II) (c) 34/35 (d) 19/20 Ans. (c) The given proportional numbers are- RRB JE - 02/06/2019 (Shift-II) 11 ,17 , 5 , 2 = 33 , 34 , 30 , 28 14 21 7 3 42 42 42 42 Ans. (c) 15/16 = 0.937 The decreasing order of the numbers, 24/25 = 0.96 17 11 5 2 34/35 = 0.97 (the greatest fraction) > >> 19/20 = 0.95 21 14 7 3 Hence, option (c) is the greatest fraction. Hence, the last proportional number will be 2/3. 9. Find the greatest among these fractions. 13. Which of the following fractions is the largest? 1, 2 , 3 , 4 5/11, 3/15, 12/11, 4/7, 9/12 8 12 16 20 (a) 12/11 (b) 3/15 (c) 9/12 (d) 4/7 (a) 3 (b) 4 16 20 RRB RPF-SI -10/01/2019 (Shift-I) Ans : (a) (c) 1 (d) 2 8 12 5 = 0.45, 3 = 0.2, 12 = 1.09, 4 = 0.57, 9 = 0.75 11 15 11 7 12 RRB Group-D – 12/10/2018 (Shift-III) 12 Ans : (b) 1 = 0.125 Hence, the required largest fraction will be = 8 11 10. Find the least among these fractions. 2 = 0.166 1 , 1 , 9 , 500 12 10 100 1000 10000 3 = 0.187 (a) 500 (b) 1 16 10000 100 4 = 0.200 (c) 1 (d) 9 20 10 1000 Hence, it is clear that 4 is the greatest fraction. RRB RPF Constable -17/01/2019 (Shift-I) 20 Decimal Fractions 49 YCT
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