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3 RMAT—Success Master 1 Number System Introduction In the decimal number system, numbers are expressed by means of symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, called digits. Here, 0 is called an insignificant digit whereas 1, 2, 3, 4, 5, 6, 7, 8, 9 are called significant digits. We can express a number in two ways : Notation Representing a number in figures is known as notation as 350. Numeration Representing a number in words is known as numeration as ‘Five hundred and forty five’. Indian System of Numeration Ten Arb Arb Ten Crore Crore Ten Lac 1010 109 108 107 106 Lac Ten Thousand Thousand Place Place 105 104 103 Hundred Place 102 Ten Place Unit Place 101 100 Face Value and Place Value of a Digit Face Value It is the value of the digit itself eg, in 3452, face value of 4 is ‘four’, face value of 2 is ‘two’. Place Value It is the face value of the digit multiplied by the place value at which it is situated eg, in 2586, place value of 5 is 5 × 10 2 = 500. Categories of Numbers Natural Numbers (N ) Numbers that are used in counting are called natural numbers. If N is the set of natural numbers, then we write N = {1, 2, 3, 4, 5, 6, ......... } The smallest natural number is 1. Whole Numbers (W ) When zero is included in the set of natural numbers, the numbers are known as whole numbers. If W is the set of whole numbers, then we write W = { 0, 1, 2, 3, 4, 5, .......... } The smallest whole number is 0. Integers (I ) Integers are whole numbers, negative of whole numbers including zero. If I is the set of integers, then we write I = { ........... – 3, – 2, – 1, 0, 1, 2, 3, .......... } Rational Numbers Numbers which can be p expressed in the form of , where p and q are both q integers and q ≠ 0 are called rational numbers. Basic Rules on Natural Numbers 1. One digit numbers are from 1 to 9. There are 9 one digit numbers. ie, 9 × 10 0 . 2. Two digit numbers are from 10 to 99. There are 90 two digit numbers. ie, 9 × 10 1. 3. Three digit numbers are from100 to 999. There are 900 three digit numbers ie, 9 × 10 2. In general the number of n digit numbers are 9 × 10 ( n − 1). 4. Sum of the first n natural numbers ie, n (n + 1) . 1+ 2 + 3 + 4 + K + n = 2 5. Sum of the squares of the first n natural numbers ie, n (n + 1) (2n + 1) . 12 + 2 2 + 3 2 + 4 2 + K + n 2 = 6 6. Sum of the cubes of the first n natural numbers ie, 2 n (n + 1) 13 + 23 + 33 + K + n3 = 2 4 Mathematical Ability : Number System 3 −7 , , 5, – 2, ........... . 2 9 There exists infinite number of rational numbers between any two rational numbers. Irrational Numbers Non-recurring and non-terminating decimals are irrational numbers. These numbers p cannot be expressed in the form of . q eg, eg, Real Numbers 3, 5, 29, ............ . Real number includes both rational and irrational numbers. Different Types of Numbers Even Numbers Numbers which are exactly divisible by 2 Extra Punch ! are called even numbers. eg, 2, 4, 6, ..... . Sum of first n even l 2 is the only even prime number. numbers = n ( n + 1) l 1 is not a prime number because it has two equal factors. Odd Numbers Numbers which are not exactly divisible l Every prime number greater than 3 can be written in by 2 are called odd numbers. eg, 1, 3, 5, ..... . Sum of first n odd the form of (6K + 1) or (6K − 1) where K is an integer. numbers = n 2 l There are 15 prime numbers between 1 and 50 and Prime Numbers Prime numbers are divisible by one 10 prime numbers between 50 and 100. and ‘itself’ only. eg, 2, 3, 5, 7, 11, ...... . Relative Prime Numbers Two numbers are said to be relatively prime if they do not have any common factor other than 1. eg, (3, 5), (4, 7), (11, 15), (15, 4), ...... Twin Primes Two prime numbers which differ by 2 are called twin primes. eg, (3, 5), (5, 7), (11, 13), .... . Note 1 is neither prime nor composite. Composite Numbers Numbers which are not prime are called composite numbers. eg, 4, 6, 9, 15, ...... . Perfect Numbers A number is said to be a perfect number, if the sum of all its factors excluding itself is equal to the number itself. eg, Factors of 6 are 1, 2, 3 and 6. Sum of factors excluding 6 = 1 + 2 + 3 = 6. 6 is a perfect number. ∴ Other examples of perfect numbers are 28, 496, 8128 etc. Tests for Divisibility Divisibility By 2 A number is divisible by 2 when the digit at ones place is 0, 2, 4, 6 or 8. eg, 3582, 460, 28, 352, .... . Divisibility By 3 A number is divisible by 3 when sum of all digits of a number is a multiple of 3. eg, 453 = 4 + 5 + 3 = 12. 12 is divisible by 3 so, 453 is also divisible by 3. Divisibility By 4 A number is divisible by 4, if the number formed with its last two digits is divisible by 4. eg, If we take the number 45024, the last two digits form the number 24. Since, the number 24 is divisible by 4, the number 45024 is also divisible by 4. Divisibility By 5 A number is divisible by 5, if its last digit is 0 or 5. eg, 10, 25, 60, ..... . Divisibility By 6 A number is divisible by 6, if it is divisible by both 2 and 3. eg, 48, 24, 108, ..... . Divisibility By 7 A number is divisible by 7 when the difference between twice the digit at ones place and the number formed by other digits is either zero or a multiple of 7. eg, 658 65 − 2 × 8 = 65 − 16 = 49 As 49 is divisible by 7 the number 658 is also divisible by 7. 5 RMAT—Success Master Divisibility By 8 A number is divisible by 8, if the number formed by the last 3 digits of the number is divisible by 8. eg, If we take the number 57832, the last three digits form 832. Since, the number 832 is divisible by 8, the number 57832 is also divisible by 8. Divisibility By 9 A number is divisible by 9, if the sum of all the digits of a number is a multiple of 9. eg, 684 = 6 + 8 + 4 = 18. 18 is divisible by 9 so, 684 is also divisible by 9. Divisibility By 10 A number is divisible by 10, if its last digit is 0. eg, 20, 180, 350, ............. . Divisibility By 11 When the difference between the sum of its digits in odd places and in even places is either 0 or a multiple of 11. Extra Punch ! eg, 30426 l If p and q are coprimes and both are factors of a 3 + 4 + 6 = 13 number r, then their product p × q will also be a 0+ 2 = 2 factor of r. eg, Factors of 24 are 1, 2, 3, 4, 6, 8, 12 13 − 2 = 11 and 24 prime factors of 24 are 2 and 3, which are As the difference is a multiple of 11 the number 30426 is also coprime. Product of 2 × 3 = 6, 6 is also a factor also divisible by 11. of 24. l If ‘p’ divides ‘q’ and ‘r’, then ‘p’ also divides their In a sum of division, we have four quantities. sum or difference. eg, 4 divides 12 and 20. They are (i) Dividend (ii) Divisor (iii) Quotient and (iv) Sum of 12 and 20 is 32 which is divisible by 4. Remainder. Difference of 20 and 12 is 8 which is divisible by 4. These quantities are connected by a relation. l If a number is divisible by another number, then it (a) Dividend = Divisor × Quotient + Remainder. must be divisible by each of the factors of that (b) Divisor = (Dividend – Remainder) ÷ Quotient. number. 48 is divisible by 12. Factors of 12 are 1, 2, (c) Quotient = (Dividend – Remainder) – Divisor. 3, 4, 6, 12. So, 48 is divisible by 2, 3, 4 and 6 also. Illustration 1 In a sum of division the dividend is 258, the quotient is 17 and the remainder is 3. Find the divisor. Solution. Divisor = (Dividend – Remainder) ÷ Quotient = (258 − 3) ÷ 17 = 15 Hence, the divisor is 15. Illustration 2 In a sum of division, the quotient is 110, the remainder is 250, the divisor is equal to the sum of the quotient and remainder. What is the dividend ? Solution. Divisor = (110 + 250) = 360 Dividend = (360 × 110) + 250 = 39865 Hence, the dividend is 39865. Illustration 3 Find the total number of numbers upto 600 which are divisible by 14. Solution. Divide 600 by 13, the quotient obtained is 46. Thus, there are 46 numbers less than 600 which are divisible by 14. Factors and Multiples Factor A number which exactly divides a given number is called a factor of the given number. eg, 24 = 1 × 24, 2 × 12, 3 × 8, 4 × 6 Thus, 1, 2, 3, 4, 6, 8, 12 and 24 are factors of 24. Number of factors of a number If N is a composite number such that N = a m b n c o K where a, b, c, ... are prime factors of N and m, n, o, ... are positive integers, then the number of factors of N is given by the expression ( m + 1) ( n + 1) ( o + 1) ........ . Extra Punch ! l l l l l l 1 is a factor of every number. A number is a factor of itself. The smallest factor of a given number is 1 and the greatest factor is the number itself. If a number is divided by any of its factors, the remainder is always zero. Every factor of a number is either less than or at the most equal to the given number. Number of factors of a number are finite. 6 Mathematical Ability : Number System Illustration 4 Find the number of factors that 224 has Solution. 224 = 25 × 71 Hence, 224 has (5 + 1) (1 + 1) = 6 × 2 = 12 factors. Multiple Extra Punch ! A multiple of a number is a number obtained by multiplying it by a natural number eg, Multiples of 5 are 5, 10, 15, 20, ....... Multiples of 12 are 12, 24, 36, 48, ..... l l l l Every number is a multiple of 1. The smallest multiple of a number is the number itself. We cannot find the greatest multiple of a number. Number of multiples of a number are infinite. LCM & HCF of Numbers Lowest Common Multiple (LCM) The lowest common multiple of two or more given numbers is the least of their common multiples. Multiples of 25 are 25, 50, 75, 100, 125, 150, .... Multiples of 30 are 30, 60, 90, 120, 150, 180, .... Lowest common multiple is 150. Highest Common Factor (HCF) The highest common factor of two or more given numbers is the largest of their common factors. Factors of 20 are 1, 2, 4, 5, 10, 20. Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36. Common factors are 1, 2 and 4. Highest common factor is 4. LCM and HCF by Division Method Illustration 5 Find the LCM and HCF of 12 and 52. Solution. To find LCM 2 2 3 13 12 52 6 26 3 13 1 13 1 1 Thus, LCM of 12 and 52 is 2 × 2 × 3 × 13 = 156 Thus, LCM of 12 and 52 is 2 × 2 × 3 × 13 = 156 To find HCF 12)52(4 48 4) 12 (3 12 × So, 4 is the HCF of 12 and 52. Illustration 6 Find the LCM and HCF of 15, 20 and 30. Solution. LCM 5 2 3 2 LCM = 2 × 2 × 3 × 5 = 60. 15 3 3 1 1 20 4 2 2 1 30 6 3 1 1 7 RMAT—Success Master HCF Let us first find HCF of any two numbers (15 and 20). 15)20(1 15 5)15(3 15 × So, 5 is the HCF of 15 and 20. Now, we find HCF of 5 and 30. 5)30(6 30 × Thus, HCF of 15, 20 and 30 is 5. Illustration 7 Find the HCF of 321 and 428. Solution. 321)428(1 321 107)321(3 321 × HCF of 321 and 428 is 107. LCM of Numerators HCF of Denominators HCF of Numerators To find HCF of fractions = LCM of Denominators To find LCM of fractions = Illustration 8 Find the LCM and HCF of 3 4 9 , , . 5 15 8 LCM of 3, 4, 9 36 − 36 = HCF of 5, 15, 8 1 HCF of 3, 4, 9 1 HCF of fractions = = LCM of 5, 15, 8 120 Solution. LCM of fractions = Illustration 9 Arrange the following in ascending order 3 4 7 , , . 5 7 10 Solution. LCM of 5, 7 and 10 = 70 3 3 × 14 42 4 4 × 10 40 7 7 × 7 49 ; = ; = = = = = 5 5 × 14 70 7 7 × 10 70 10 10 × 7 70 4 3 7 ∴ Ascending order is , , . 7 5 10 Extra Punch ! l l l l l The least number which is exactly divisible by a, b and c is the LCM of a, b, c. The greatest number that will divide a, b, c is the HCF of a, b, c. If x is a factor of a and b, then x is also a factor of a + b, a − b and ab. HCF of given numbers must be a factor of their LCM. The product of the LCM and HCF of two numbers is equal to the product of the two numbers. Illustration 10 LCM of two numbers is 56 times of their HCF. Sum of LCM and HCF is 456. If one of them is 56. Find the other. Solution. Let the HCF be x; then LCM = 56x x + 56x = 456 Now, ⇒ x=8 ∴ HCF = 8 and LCM = 448 8 × 448 So, other number = = 64 8 × 448 = 56 × other number 56 Practice Exercises A. Model Questions 1. Find the sum of the smallest four-digit number and the largest five-digit number. (a) 199900 (b) 100999 (c) 211000 (d) 211999 2. A positive integer which when added to 999 gives a sum which is greater than when it is multiplied by 999. Which of the following could be the value of the positive integer ? (a) 3 (b) 4 (c) 1 (d) 2 3. What is the total number of prime numbers less than 100 ? (a) 25 (b) 21 (c) 29 (d) 27 4. If 43p is divisible by 3, then what is the largest value p can take ? (a) 8 (b) 2 (c) 5 (d) 0 5. If the number 6pq 5 is divisible by both 3 and 5, which of the following digits can replace p and q? (a) 5,9 (b) 9,5 (c) 5,7 (d) 9,7 6. There are four prime numbers written in ascending order. The product of the first three is 1001 and that of the last three is 2431. Find the first prime number. (a) 7 (b) 11 (c) 3 (d) 5 7. What is the remainder when 152 × 698 is divided by 5. (a) 1 (b) 2 (c) 4 (d) 3 8. What is the remainder when 75 3 is divided by 4 ? (a) 1 (b) 1 (c) 3 (d) 2 9. What is the remainder when 412 × 363 × 312 is divided by 15 ? (a) 2 (b) 6 (c) 4 (d) 3 10. Find the largest four digit number exactly divisible by 55. (a) 9900 (b) 9955 (c) 9945 (d) 9960 11. What must be added to 581173 to make it exactly divisible by 11 ? (a) 1 (b) 2 (c) 0 (d) 3 12. An even number, total of whose digits is 27, is always divisible by (a) 18 (b) 9 (c) 2 (d) 6 13. By what least number 250 must be multiplied to get a multiple of 15. (a) 3 (b) 5 (c) 2 (d) 15 14. What is the sum of all numbers between 250 and 550, which are divisible by 19. (a) 6635 (b) 6640 (c) 6000 (d) 6695 15. A man is typing numbers 5 to 700 on his computer. How many times does he press the key of his computer ? (a) 988 (b) 1988 (c) 1888 (d) 1990 16. Find the smallest number to be added to the least three digit number so that it is exactly divisible by 45. (a) 35 (b) 25 (c) 45 (d) 15 17. What least number must be added to 15763 so that it is exactly divisible by 18 ? (a) 5 (b) 18 (c) 12 (d) 13 18. What least number must be subtracted from 178669 so that it is exactly divisible by 36 ? (a) 18 (b) 1 (c) 36 (d) 5 19. How many numbers between 1000 and 3050 are divisible by 24 ? (a) 86 (b) 82 (c) 78 (d) 92 20. Find the numbers of prime factors contained in the product of 2311 × 74 × 35 . (a) 25 (b) 17 (c) 30 (d) 20 21. A number when divided by 225 gives a remainder of 32 . What will be the remainder when the same number is divided by 15 ? (a) 4 (b) 2 (c) 3 (d) 1 22. Two numbers when divided by a certain divisor give remainders 35 and 20 respectively and when their sum is divided by the same divisor, the remainders is 15. Find the divisor. (a) 30 (b) 40 (c) 35 (d) 45 23. What is the smallest number by which 1250 be divided in order to make it a perfect square ? (a) 5 (b) 25 (c) 2 (d) 4 24. What is the smallest number by which 3000 be divided in order to make it a perfect cube ? (a) 27 (b) 9 (c) 3 (d) 6 25. A certain number when successively divided by 5 and 7 leaves the remainders 3 and 4 respectively. What is the remainder, if the same number is divided by 35. (a) 21 (b) 23 (c) 24 (d) 29 26. What is the difference between the greatest and the smallest five digit numbers formed using the digits 7, 0, 3, 4 and 2 without repeating the digits ? (a) 65672 (b) 53978 (c) 54672 (d) 65978 27. A boy had to do a multiplication. Instead of taking 35 as one of the multipliers, he took 53. As a result the product went up by 540. What is the new product? (a) 1590 (b) 1450 (c) 1550 (d) 1420 9 RMAT—Success Master 28. If both 52 and 73 are factors of the number p × 43 × 72 × 135 , then what is the least possible value of p ? (a) 175 (b) 145 (c) 190 (d) 156 29. A worker earns ` 40 the first day and spends ` 25 on the second day. He earns ` 40 on the third day and spends ` 25 on the fourth day and so on. On which day would he have ` 100. (a) 10 (b) 9 (c) 8 (d) 11 30. The sum of two numbers is 85 and their difference is 9. What is the difference of their squares ? (a) 765 (b) 845 (c) 565 (d) 645 31. Find the least number, which is exactly divisible by 20, 36 and 128. (a) 5766 (b) 5760 (c) 5780 (d) 5790 32. Find the greatest four digit number which when divided by 8, 9 and 11 gives a remainder 7 in each case. (a) 9504 (b) 9541 (c) 9511 (d) 9527 33. Find the least four digit number which when divided by 5, 6 and 7 gives a remainder 3 in each case. (a) 1053 (b) 1063 (c) 1073 (d) 1083 34. A shepherd has 324 sheeps, 396 goats and 428 cows. He wants to arrange them into groups of equal size without mixing. Find the least number of groups required to arrange them. (a) 285 (b) 287 (c) 284 (d) 289 35. The LCM of 69, 115 and a third number is 2760. Their HCF is 23. Find the third number. (a) 170 (b) 186 (c) 172 (d) 184 36. A number 475 is divided into two parts in such a way that the LCM and HCF of the two parts is 2250 and 25 respectively. Find the two numbers. (a) 250, 225 (b) 200, 275 (c) 175, 300 (d) 240, 235 37. The LCM of two numbers 45 times their HCF one number is 250 and the sum of their HCF and LCM is 2300. Find the other number. (a) 400 (b) 450 (c) 460 (d) 420 38. The sum of the first 100 numbers, 1 to 100 is always divisible by (a) 2 (b) 2 and 4 (c) 2, 4 and 8 (d) None of these 39. Find the HCF of 36a 2b3 c2,18a3 b2c4 and 12ab4c3 . (a) 6ab2c2 (b) 12a 2bc (c) 6a 2bc2 (d) 12ab2c2 40. The length, breadth and height of a room are 6 m 30 cm, 5 m 85 cm and 3 m 60 cm respectively. What will be the greatest length of a tape which can measure the dimensions of room exact number of times ? (a) 45 cm (b) 50 cm (c) 60 cm (d) 40 cm 41. The traffic lights at three different road crossings change after every 50 s, 75 s and 100 s respectively.If they start changing simultaneously at 10 am, after how much time will they change again simultaneously ? (a) 6 min (b) 5 min (c) 3 min (d) 8 min 42. The number of students in three sections of a medical college are 40, 45 and 60. Find the minimum number of books required for the class library for equal distribution in the three sections. (a) 340 (b) 360 (c) 320 (d) 380 43. Three boys steps off together from the same spot. Their steps measures 30 cm, 35 cm and 40 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps ? (a) 860 cm (b) 840 cm (c) 820 cm (d) 875 cm 44. Find the unit digit in the expansion of 4223 . (a) 4 (b) 8 (c) 6 (d) 2 45. Find the unit digit in the expansion of 32545 is (a) 0 (b) 5 (c) 1 (d) 2 46. Find the unit digit in the expansion of 31448. (a) 6 (b)4 (c) 8 (d) 2 47. A worker was engaged for a certain number of days and was promised to be paid ` 1755. He remained absent for some days and was paid ` 1365 only. What were his daily wages ? (a) ` 182 (b) ` 195 (c) ` 185 (d) ` 192 48. A heap of stones can be made up into groups of 7. When made up into groups of 16, 18, 20 and 25 there are 4 stones left in each case.How many stones at least can there be in the heap ? (a) 18000 (b) 18004 (c) 18002 (d) 18006 49. If 2/7 of a number increased by 25 gives 45. Find the number (a) 81 (b) 63 (c) 72 (d) 70 50. Out of four consecutive prime numbers, the product of first three is 385 and the product of the last three is 1001. Find the last number. (a) 7 (b) 11 (c) 13 (d) 17 51. Three numbers are in the ratio 3 : 5 : 6. Sum of the greatest and the smallest is equal to the sum of the middle and 16. Find the smallest number. (a) 12 (b) 20 (c) 24 (d) 16 52. Sum of squares of two numbers is 60 and difference of the squares is 12. Find the sum of two numbers. (a) 4 (b) 10 (c) 6 (d) 8 53. The product of two consecutive numbers is 600. Then, the sum of their squares is (a) 1203 (b) 1206 (c) 1201 (d) 1204 54. Thrice a number diminished by 2 is 19. Find the number. (a) 5 (b) 7 (c) 4 (d) 6 55. A sum of ` 640 is distributed among three persons P, Q and R. Q gets ` 70 more than P. R gets `. 80 more than Q. Find the share of Q. (a) ` 210 (b) ` 215 (c) ` 200 (d) ` 220 10 Mathematical Ability : Number System 56. If the numerator of a fraction is increased by 2 and the 2 denominator is decreased by 1, then it becomes . If the 3 numerator is increased by 1 and the denominator is 1 increased by 2, then it becomes . Find the fraction. 3 2 2 1 1 (b) (c) (d) (a) 9 7 6 5 57. A and B had some candies each. If A gave 5 candies to B, then B will have twice as many as A. Instead of that, if B were to give 5 candies to A, then they will both have the same number of candies. How many did A have ? (a) 28 (b) 21 (c) 26 (d) 25 58. There were 240 multiple choice questions in an entrance examination of a medical college. A candidate was given 1 2 marks for every correct answer and penalised mark 2 for every wrong answer. The candidate scored 160 marks in the paper having answered all the questions. Find the number of correct answers given by the candidate. (a) 108 (b) 112 (c) 116 (d) 104 59. 126 is divided into two parts such that 8 times the first part added to 5 times the second part is 810. Find the first part. (a) 60 (b) 66 (c) 70 (d) 56 60. A worker was engaged for a certain number of days and was promised to be paid ` 480. He was absent for six days and was paid ` 432 only. What was his daily wages ? (a) 432 (b) 60 (c) 65 (d) 70 61. Find the LCM of 8, 10, 24. (a) 100 (b) 120 (c) 240 (d) 48 62. Find the LCM of 3, 9, 12, 15, 6 (a) 240 (b) 90 (c) 180 (d) 120 63. Find the LCM of 23 × 34 × 52, 22 × 33 × 7, 52 × 72 (a) 23 × 34 × 52 × 72 (b) 23 × 52 × 72 4 2 2 (c) 3 × 5 × 7 (d) 23 × 72 × 5 4 8 3 64. Find the LCM of , , 3 9 5 (a) 20 (b) 24 (c) 1/24 (d) 1/20 65. Find the HCF of 40 and 64. (a) 4 (b) 8 (c) 2 (d) 10 66. Find the HCF of 120, 150, 180. (a) 30 (b) 60 (c) 50 (d) 10 67. Find the HCF of 2378 and 3280. (a) 86 (b) 84 (c) 80 (d) 82 68. Find the HCF of 22 × 33 × 52,23 × 32 × 5, 52 × 7 (a) 22 × 32 × 5 × 7 (b) 23 × 33 × 5 3 2 (c) 2 × 3 × 7 (d) 32 × 52 × 7 2 12 1 69. Find the HCF of , , 5 11 3 1 5 2 12 (a) (b) (c) (d) 165 165 165 165 70. Find the LCM and HCF of 0.25, 0.5, 0.75 (a) 7.5, 0.05 (b) 0.75, 0.5 (c) 0.75, 0.05 (d) 7.5, 0.5 71. The product of 2 numbers is 3024 and their LCM is 36. Find their HCF. (a) 88 (b) 82 (c) 84 (d) 86 72. LCM of two numbers is 12 times of their HCF. Sum of LCM and HCF is 195. If one of them is 60. Find the other. (a) 48 (b) 45 (c) 52 (d) 36 73. The sum of two numbers is 135 and their HCF is 9. How many such pairs of numbers can be formed ? (a) 6 (b) 2 (c) 5 (d) 4 74. HCF of two numbers is 12 and their product is 3600. How many such pairs of numbers can be formed ? (a) 0 (b) 1 (c) 2 (d) 4 75. The LCM of two numbers is 36 times that of their HCF. The product of numbers is 3600. Find their HCF. (a) 10 (b) 15 (c) 8 (d) 20 76. Two numbers are in the ratio 3 : 5. Their, LCM is 600. Find the sum of the numbers. (a) 320 (b) 160 (c) 340 (d) 280 77. Find the least number which is exactly divisible by 12, 15, 20 and 27. (a) 650 (b) 520 (c) 600 (d) 540 78. Find the least number which when increased by 3 is exactly by 10, 12, 14 and 16. (a) 1680 (b) 1677 (c) 1697 (d) 1670 79. Find the smallest number which when decreased by 4 is exactly divisible by 9, 12, 15 and 18. (a) 188 (b) 182 (c) 186 (d) 184 80. Find the least number which when divided by 12, 15, 20 and 24 leaves the same remainder 2 in each case. (a) 122 (b) 120 (c) 124 (d) 126 81. Find the least number of girls so that they can be arranged in the groups of 15 or 20 or 30. (a) 60 (b) 90 (c) 45 (d) 80 82. Find the least perfect square number which is exactly divisible by 3, 4, 5, 6 and 8. (a) 3600 (b) 3900 (c) 3500 (d) 3200 83. Find the greatest number less than 800, which is divisible by 8, 12, 28. (a) 624 (b) 672 (c) 636 (d) 684 84. Find the least number which when divided by 12, 15 and 20 leaves remainders 4, 7 and 12 respectively. (a) 48 (b) 60 (c) 52 (d) 64 85. Find the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3 but when divided by 9 leaves no remainder. (a) 1683 (b) 1725 (c) 1685 (d) 1723 86. Find the least number which when divided by 15, 20 and 24 leaves 4 as remainder but when divided by 13 leaves no remainder. (a) 364 (b) 370 (c) 484 (d) 244 87. Five bells begin to toll together and toll at intervals of 24, 40, 64, 72 and 120 s. After what interval of time will they toll again together ? (a) 42 min (b) 36 min (c) 48 min (d) 54 min 11 RMAT—Success Master 88. Four wheels moving 12, 25, 20 and 30 revolutions in a minute starting at a certain point on the circumference downwards. After what interval of time will they come together again in the same position ? (a) 30 s (b) 20 s (c) 15 s (d) 45 s 93. Find the greatest number which will divide 321,428 and 535 exactly. (a) 105 (b) 107 (c) 109 (d) 102 89. Three athletes start together to travel the same way around on a circular path of 12 km with speeds of 3 km/ph, 4 km/ph and 6 km/ph respectively. They meet at the same point again after what time ? (a) 12 h (b) 8 h (c) 16 h (d) 10 h 95. Find the least number of square tiles required to pave the floor of a room of 4 m 50 cm long and 5 m 25 cm broad. (a) 40 (b) 42 (c) 36 (d) 38 94. Find the greatest possible length which can be used to measure 4 m 3 cm, 4 m 34 cm and 4 m 65 cm exactly. (a) 31 cm (b) 29 cm (c) 28 cm (d) 32 cm 96. Find the greatest number which will divide 676, 481 and 767 exactly. (a) 11 (b) 13 (c) 17 (d) 19 90. The traffic lights at three different road crossings change after every 48,72 and 108 s respectively .If they change simultaneously at 9 am,at what time will they change again simultaneously ? (a) 10 min 10 s (b) 7 min 12 s (c) 7 min 10 s (d) 12 min 10 s 97. Find the greatest number which divides 48, 63 and 78 leaving 3 as the remainder in each case. (a) 11 (b) 18 (c) 12 (d) 15 91. Find the smallest 4 digit number that is exactly divisible by 8,10 and 12. (a) 1080 (b) 1100 (c) 1050 (d) 1120 99. Find the greatest number that will divide 1196, 209 and 1449 to leave remainders 6,5 and 4 respectively. (a) 15 (b) 17 (c) 21 (d) 14 92. Find the greatest five digit number exactly divisible by 6, 8, and 12. (a) 99924 (b) 99988 (c) 99984 (d) 99972 100. Find the greatest number that will divide 228, 275 and 321 to leave remainder 3, 5, and 6 respectively. (a) 45 (b) 36 (c) 42 (d) 39 98. Find the greatest number which divides 121, 134 and 147 leaving 4 as the remainder in each case. (a) 11 (b) 19 (c) 15 (d) 13 B. MBA Entrances’ Gallery 1. The number of boys in a class is three times the number of girls. Which one of the following numbers cannot represent the total number of children in the class ? (a) 48 (b) 44 [MAT Feb 2010] (c) 42 (d) 40 2. If 8 + 12 = 2, 7 + 14 = 3, then 10 + 18 = ? (a) 10 (b) 4 (c) 6 [IP 2009] (d) 18 3. Which among the following is the least number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, when divided by 8 leaves a remainder of 7, when divided by 7 leaves a remainder of 6, when divided by 6 leaves a remainder of 5, when divided by 5 leaves a remainder of 4, when divided by 4 leaves a remainder of 3, when divided by 3 leaves a remainder of 2, when divided by 2 leaves a remainder of 1 ? [IP 2009] (a) 7559 (b) 839 (c) 5039 (d) None of these 4. What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24 ? [IP 2009] (a) 264 (b) 259 (c) 269 (d) None of these 5. What will be the value of the remainder when, [IP 2009] 1693 × 797 × 3395 is divided by 17 ? (a) 16 (b) 5 (c) 15 (d) 1 6. 25 times a number is 775. The number is [IP 2008] (a) 31 (b) 30 (c) 32 (d) None of these 7. Which of the following cannot be a digit in the unit’s place of a perfect square ? [IP 2008] (a) 7 (b) 1 (c) 5 (d) 0 8. The smallest number, which when added to the sum of square of 9 to 10 gives a perfect square is [IP 2008] (a) 0 (b) 3 (c) 8 (d) 15 9. (9992 − 9982) is equal to (a) 1 (c) 1997 [IP 2008] (b) 999 (d) 998 10. A number when divided by 119 leaves the remainder 19. If the same number is divided by 17, the remainder will be [IP 2008] (a) 19 (b) 10 (c) 7 (d) 2 11. The LCM of two numbers is 63 and their HCF is 9. If one of the numbers is 27, the other number will be [IP 2008] (a) 9 (b) 21 (c) 17 (d) 189 12. If n is any positive integer, then 34n − 43 n is always divisible by [IP 2008] (a) 7 (b) 17 (c) 112 (d) 145 13. The number zero (0) is surrounded by the same 2 digit number on both (left and right) sides: for example 25025, 67067 etc. The largest number that always divides such a number is [IP 2008] (a) 7 (b) 11 (c) 13 (d) 1001 14. A number when divided by 45 leaves the remainder 30. If the number is divided by 5, the remainder must be (a) 3 (b) 2 [IP 2008] (c) 1 (d) 0 12 Mathematical Ability : Number System 15. A boy was asked to write 25 ⋅ 92, but he wrote 2592. The numerical difference between the values of 25 ⋅ 92 and 2592 is [IP 2008] (a) 0 (b) 1 (c) 2 (d) 3 16. The HCF of 168, 189 and 231 reduced by 8, gives (a) 13 (b) 14 [IP 2008] (c) 17 (d) 21 17. Which of the following fractions is less than 7 8 and [RMAT 2008] greater than 1 3 ? (a) 1 4 (b) 23 24 (d) 17 24 (c) 11 12 18. Ashok had to do a multiplication. Instead of taking 35 as one of the multipliers, he took 53. As a result, the product went up by 540. What is the new product ? (a) 1050 (b) 1590 [RMAT 2008] (c) 1440 (d) None of these 19. If a, b, c are real numbers such that a − b + c < 1, a + b + c > −1 and 9a + 3b + c < −4. What is the sign of a ? (a) ‘+’ only (b) ‘−’ only [MPMET 2008] (c) ‘+’ or ‘−’ (d) cannot say 20. Which of the following is a multiple of 88 ? (a) 1392578 (b) 138204 [MPMET 2008] (c) 1436280 (d) 143616 21. How many of the natural numbers from 1 to 1000 have none of their digits repeated ? [MPMET 2008] (a) 589 (b) 585 (c) 738 (d) 739 22. PQ ⋅ RS × D ⋅ AB [MPMET 2008] In the multiplication problem shown above, if each of these letters represents a single digit number from 1 to 9, both inclusive and ‘.’ denotes decimal point, then the product of the two numbers must be (a) between 10 and 1000 (b) between 1 to 1000 (c) at most 100 (d) greater than 100 23. Find the values of x and y, if the number 5372 xy is divisible by 45. [MPMET 2008] (a) {(1, 5);(5, 5)} (b) {(1, 0);(5, 5)} (c) {(5, 5):(0, 5)} (d) {(3, 0);(5, 0)} a3 + b3 + c3 . abc [MPMET 2008] (b) 0 (d) 2 24. If a + b + c = 0, find the value of (a) 3 (c) 1 25. If a and b are distinct primes, the LCM of a x and by were x and y are positive integers is [MPMET 2007] (a) (ab) LCM of x and y (b) (ab)xy (c) a x / by HCF of (x, y) (d) a x by 26. How many numbers lying between 3000 and 4000 can be formed with the digits 1, 2, 3, 4, 5, and 6 ? The digits are [MPMET 2007] not repeated in any one number. (a) 40 (b) 104 (c) 60 (d) 82 27. The LCM of two numbers is 280 and the ratio of the numbers of 7 : 8. Find the numbers. [MPMET 2006] (a) 70 and 80 (b) 42 and 48 (c) 35 and 40 (d) 28 and 32 28. When a number is divided by 7, the remainder is 4. When the same number is divided by 6, the remainder [OJEE 2009] is 3. What is the number ? (a) 39 (c) 49 (b) 37 (d) 52 29. What is the least number to which when 5 is added, the sum is divisible by 12, 15 and 18 respectively ? (a) 165 (b) 175 [OJEE 2008] (c) 185 (d) 195 30. The largest four digit number which is a multiple of [OJEE 2008] 8, 10, 12 and 15 is (a) 0120 (b) 9600 (c) 9840 (d) 9960 31. Which of the following pairs is equal in value, given that ‘a’ is a real number ? [OJEE 2008] (a) 812 a and 243a 10 (b) 125a 12 and 15625a 16 (c) 4a 4 and 8a 6 (d) 494a 3 and 3433 a 4 32. What is the number which when successively divided by 4, 5 and 6 leaves respective remainders of 3, 4 and 5 with last quotient as 2 ? [OJEE 2008] (a) 247 (b) 231 (c) 351 (d) 359 Answers with Solutions A. Model Questions 1. (b) The smallest four-digit number = 1000 The largest five-digit number = 99999 Required sum = 100999 ∴ 2. (c) Let the positive integer = x Then, 999 + x > 999x By Trial and error when x = 1 1000 > 999 So, the required positive integer is 1. 3. (a) The prime numbers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. These add upto 25. 4. (a) A number is divisible by 3 when sum of the digits is divisible by 3. 4+3+ P = 7+ P So, P can be 2, 5, 8. So the largest value is 8. 5. (d) If p is 9 and q is 7. The number is 6975. 6975 is divisible by both 3 and 5. 6. (a) Let the four prime numbers be a, b, c, d. abc = 1001 and bcd = 2431 abc 1001 7 = = bcd 2431 17 a 7 = ⇒a = 7 ∴ d 17 7. (a) The remainder when 152 is divided by 5 = 2 The remainder when 698 is divided by 5 = 3 Their product = 2 × 3 = 6 The remainder when 6 is divided by 5 = 1. 8. (c) The remainder when 75 is divided by 4 is 3. ∴The remainder when 3 × 3 × 3 is divided by 4 is 3. So, the required remainder is 3. 9. (b) 41, 36 and 31 when divided by 15, leave remainders 11, 6 and 1 respectively. ∴412 × 363 × 312 when divided by 15, leaves remainder 112 × 63 × 12 = 26136. As remainder cannot exceed the divisor. ∴ Remainder is 6 ie, remainder obtained on dividing 26136 by 15. 10. (b) The largest four digit number = 9999. 55)9999(181 55 449 440 99 55 44 The required number = 9999 − 44 = 9955. 11. (a) Sum of digits at add places (from left side) = 5 + 1 + 7 = 13 Sum of digits at even places (from left side) = 8 + 1 + 3 = 12 Whatever we add to the given number will increase the sum of digits which includes the units digit ie, to 12. If we add 1 to given number difference between the sum of digits becomes 13 − 13 = 0 ∴1 is to be added to the given number. 12. (a) An even number is always divisible by 2. Further any number sum of whose digits is 27 ie; (2 + 7 = 9) is always divisible by 9. ∴ The number is divisible by both 2 and 9. ∴ The number is divisble by 2 × 9 = 18. 13. (d) 15 = 3 × 5 ∴Any multiple of 15 must be divisible by 3 and 5. 250 = 2 × 5 × 5 × 5. ∴250 is divisible by 5, but not by 3. So, 250 must be multiplied by 3 to make it a multiple of 15. 14. (c) Since, 250 = 13 × 19 + 3 Terms divisible by 19 upto 250 = 13. And 550 = 28 × 19 + 18 ∴Term divisible by 19 upto 550 = 28. ∴Total terms divisible by 19 between 250 and 550 = 28 − 13 = 15. Midterm (Average) between 250 to 550 = 400. ∴Sum of 15 terms = 400 × 15 = 6000 15. (b) One digit numbers from 5 to 9 = 5 Two digit numbers from 10 to 99 = 90 Three digit numbers from 100 to 700 = 601 Total number of digits that are to be used = (5 × 1) + (90 × 2) + (601 × 3) = 5 + 180 + 1803 = 1988 16. (a) The least three digit numbers = 100 45)100(2 90 10 So, the smallest number to be added = 45 − 10 = 35. 17. (a) 18)15763(875 144 136 126 103 90 13 ∴The least number that should be added = (18 − 13) = 5. 14 Mathematical Ability : Number System 18. (b) 36)178669(4963 144 346 324 226 216 109 108 1 ∴The least number that should be subtracted = 1. 19. (a) Quotient when 1000 and 3050 are divided by 24 are = 41 and 127. ∴Between 1000 and 3050 there are 127 − 41 = 86 numbers. 20. (d) Number of prime factors = 11 + 4 + 5 = 20. 21. (b) Let the number be N. N = 225Q + 32 where Q the quotient can have the values 1, 2, 3, etc. N = 15 × 15Q + (15 × 2) + 2 Divide N by 15, we see that the remainder is 2. 22. (b) In such questions the divisor is r1 + r2 − r3 . Divisor = 35 + 20 − 15 = 40 ∴ 23. (c) 1250 = 5 × 5 × 5 × 5 × 2 So, when 1250 is divided by 2, we get a perfect square. 24. (c) 3000 = 2 × 2 × 2 × 5 × 5 × 5 × 3 So, when 3000 is divided by 3, we get a perfect cube. 25. (b) 5 N 7 X −3 should include 52. , 73 is a factor of the given number. The number has 72. Hence, P should include 71 to make it 73 . ∴P should be atleast 52 × 7 = 175 to have 52 and 73 as its factors. 29. (b) His savings for every 2 days = ` 40 − ` 25 = ` 15 His savings in 8 days = ` 60 On the 9th day he earns ` 40. On the 9th day he has ` 60 + ` 40 = `100. 30. (a) Let the two numbers be x and y. x + y = 85 x− y =9 Solving x = 47 and y = 38. The difference of their squares is 472 − 382 = 765. 31. (b) Required number = LCM of 20, 36, 128 = 5760. 32. (c) The LCM of 8, 9 and 11 = 792. The greatest four-digit number = 9999. 792)9999(122 792 2079 1584 495 The greatest four-digit number divisible by 792 is 9999 − 495 = 9504. Now, the greatest four-digit number divisible by 792 and leaving a remainder 7 is 9504 + 7 = 9511. 33. (a) The LCM of 5, 6 and 7 = 210. Y −4 X = 7Y + 4 N = 5 (7Y + 4) + 3 = 35Y + 23 If Y = 1, N = 58 ∴When 58 is divided by 35 the remainder is 23. 26. (b) The greatest five-digit number that can be formed using the given digits = 74320 The smallest five-digit number that can be formed using the given digits = 20342. ∴ The required difference = 74320 − 20342 = 53978 27. (a) Let the number that the boy wanted to multiply be x. He was expected to find the value of 35 x. Instead, he found the value of 53x. The difference between the value that he got and what he was expected to get is 540. ie, 53x − 35x = 450 x = 30 ∴The new product = 30 × 53 = 1590. 28. (a) If 52 is a factor of the given number. The number does not have a power or multiple of 11 as its factor. Hence, P The least four- digit number = 1000. 210)1000(4 840 160 The least four-digit number divisible by 210 is 1000 + (210 − 160) = 1050. Now, the least four-digit number divisible by 210 and leaving a remainder 3 is 1050 + 3 = 1053. 34. (b) HCF of (324, 396, 428) = 4 So, the number of groups required 324 396 428 = + + 4 4 4 = 81 + 99 + 107 = 287 35. (d) Let the third number be 23x LCM of 69, 115 and 23x is 23 69, 115, 23x 3, 5, x LCM = 23 × 3 × 5 × x = 345x Now, 345x = 2760 x=8 ∴The third number = 23x = 23 × 8 = 184. 15 RMAT—Success Master 36. (a) Let the two numbers be 25x and 25 y. 25x + 25 y = 475 x + y = 19 Also, the LCM of 25x and 25 y = 25xy. 25xy = 2250 xy = 90 (x − y)2 = (x + y)2 − 4xy 192 = (x + y)2 − 4 (90) Solving x− y=1 Using Eqs. (i) and (iii) x = 10 and y = 9. ∴The two numbers are 250 and 225. ...(i) ...(ii) …(iii) 37. (b) Let the HCF of the two numbers be x. So, the LCM = 45x Now, x + 45x = 2300 x = 50 HCF = 50, LCM = 45x = 45 × 50 = 2250. Product of the two numbers = Product of the LCM and HCF of the two numbers. Let the other number be N. N × 250 = 50 × 2250 N = 450. 38. (a) The sum of the first n natural numbers n (n + 1) = 2 = The sum of the first 100 natural numbers. 100 × 101 = = 50(101) 2 As 101 is an odd number and 50 is divisible by 2. So, the sum is always divisible by 2. 39. (a) HCF of 36, 18 and 12 is 6. The common factors are a, b, c and their highest powers common to all are 1, 2, 2 respectively ∴ Required HCF = 6ab2c2 40. (a) Greatest length of tape = HCF of 630, 585 and 360 = 45 cm 41. (b) LCM of 50, 75 and 100 = 300 s. ∴Traffic lights will change simultaneously after 300 s. = 5 min. 42. (b) LCM of 40, 45 and 60 = 360. ∴ 360 is the minimum number of books required for equal distribution in each section. 43. (b) LCM of 30, 35 and 40 = 840 cm. 840 cm is the minimum distance they should cover in complete steps. 44. (b) The unit digit comes from the unit place. So, we have to find the unit digit is the expansion of 223. Now, 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64 ........ We see that the unit digit is 21 is the same as 25 . ∴The unit digit in 223 is the same as 23 = 8. 45. (b) The unit digit in 32545 is the same as that of 545 . Now, 51 = 5, 52 = 25, 53 = 125, 54 = 625, ........ We see that the unit digit is always 5. ∴The unit digit in 32545 is 5. 46. (a) The unit digits in the expansion of 31448 is the same as that of 448. Now, 41 = 4, 42 = 16 , 43 = 64,44 = 256. We see that the unit digit in 41 is the same as that in 43 . So, the unit digit is 448 is the same as is 448 = 6. 47. (b) The daily wages of the worker is the HCF of 1755 and 1365. HCF of 1755 and 1365 is 195. ∴The daily wages of the worker is Rs 195. 48. (b) LCM of 16, 18, 20 and 25 = 3600. 1st number = 3600 × 1 + 4 = 3604 which is not divisible by 7. 2nd number = 3600 × 2 + 4 = 7204 which is not divisible by 7. 3rd number = 3600 × 3 + 4 = 10804 which is not divisible by 7. 4th number = 3600 × 4 + 4 = 14404 which is not divisible by 7. 5th number = 3600 × 5 + 4 = 18004 which is divisible by 7. ∴There can be 18004 stones. 49. (d) Let the number be x. 2 x + 25 = 45. On solving x = 70 7 50. (c) Let the four numbers be p, q, r and s. pqr = 385, qrs = 1001 qrs 1001 = pqr 385 13 = 4 s 13 = p 5 So, p = 5 and s = 13. ∴The required number is 13. 51. (a) Let the numbers be 3x, 5x and 6x. Now, 3x + 6x = 5x + 16 x=4 So, the smallest number is 3x = 12. 52. (b) Let the number be x and y. x2 + y2 = 60, x2 − y2 = 12 Adding 2x2 = 72 x = 6 and y = 4 ∴Required sum is 10. 53. (c) Let the two numbers be x and x + 1 x (x + 1) = 600 On solving x = 24, so the numbers are 24 and 25. The sum of their squares is = 242 + 252 = 1201 16 Mathematical Ability : Number System 54. (b) Let the number be x. 62. (c) 2 3, 9, 12, 15, 6 2 3, 9, 6, 15, 3 On solving x = 17 3 3, 9, 3, 15, 3 ∴The required number is 7. 3 1, 3, 1, 5, 1 3x − 2 = 19 55. (a) P + Q + R = 640 ...(i) Q = 70 + P 5 R = 80 + (70 + P ) On solving, we get Q = Rs 210. x 56. (b) Let the fraction be y x+2 2 = y −1 3 x+1 1 = y+2 3 x 2 On solving, we get = . y 7 ...(iii) 1, 1, 1, 1, 1 LCM = 2 × 2 × 3 × 3 × 5 = 180 63. (a) LCM of given numbers = 23 × 34 × 52 × 72 ...(i) ...(ii) (Take the greatest power of each term) LCM of numerators 64. (b) LCM of fraction = HCF of denomenators LCM of 4, 8, 3 24 = = = 24 HCF of 3, 9, 5 1 65. (b) b + 5 = 2 (a − 5) b −5 = a + 5 On solving, we get a = 25. Then, ...(i) ...(ii) 58. (b) Let the number of wrong answers marked = w Let the number of right answers marked = r ...(i) r + w = 240 w ...(ii) 2r − = 160 2 On solving, we get r = 112. 59. (a) Let the two parts be x and 126 − x. 8x + 5 (126 − x) = 810 8x + 630 − 5x = 810 3x = 810 − 630 = 180 x = 60 So, the first part is 60. 60. (b) Let the number of days for which the worker works = x For x days he is paid Rs 480 For (x − 6) days he is paid Rs 432. 480 His daily wages = x 432 = x −6 On solving, x = Rs 60 2 2 2 3 5 8, 10, 24 4, 5, 12 2, 5, 6 1, 5, 3 1, 5, 1 1, 1, 1 LCM = 2 × 2 × 2 × 3 × 5 = 120 40)64(1 40 24)40(1 24 16)24(1 16 8)16(2 16 0 57. (d) Let A have a candies and B have b candies 61. (b) 1, 1, 1, 5, 1 ...(ii) ∴ HCF = 8 66. (a) Step 1 120)150(1 120 30)120(4 120 0 Step 2 30) 180 (6 180 0 HCF of 30 and 180 is 30. HCF of 120 and 150 is 30. ∴The required HCF is 30. 67. (d) 2378)3280(1 2378 902)2378(2 1804 574)902(1 574 328)574(1 328 246)328(1 246 82) 246(3 246 0 ∴The required HCF is 82. 17 RMAT—Success Master 68. (a) HCF = 22 × 32 × 5 × 7 (Take the least powers of common terms). HCF of numerator 69. (a) HCF of fraction = HCF of denominator HCF of 2, 12, 1 1 = = LCM of 5, 11, 3 165 25 5 75 70. (a) LCM = LCM of , , 100 10 100 LCM of 25, 5, 75 75 = = = 7.5 HCF of 100, 10, 100 10 25 5 75 HCF = HCF of , , 100 10 100 HCF of 25, 5, 75 5 = = = 0.05 LCM of 100, 10, 100 100 71. (c) Product of (LCM × HCF) = Product of 2 numbers 36 × HCF = 3024 3024 HCF = = 84 36 72. (b) Let the HCF = x 77. (d) Required number = LCM of 12, 15, 20 and 27 2 12, 15, 20, 27 2 6, 15, 10, 27 3 3, 15, 5, 27 3 1, 5, 5, 9 3 1, 5, 5, 3 5 1, 5, 5, 1 1, 1, 1, 1 ∴ LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540. 78. (b) Required number = (LCM of 10, 12, 14, 16) −3 2 10, 12, 14, 16 2 5, 6, 7, 8 2 5, 3, 7, 4 2 5, 3, 7, 2 3 5, 3, 7, 1 5 5, 1, 7, 1 7 1, 1, 7, 1 LCM = 12x x + 12x = 195 13x = 195 x = 15 ∴HCF = 15 and LCM = (12) (15) = 180. 180 × 15 Other number = = 45. 60 73. (d) The two numbers are always the multiples of the HCF ∴Let the two numbers are 9x and 9y. 9x + 9 y = 135 ⇒ x + y = 15 Now, the possible values of x and y are (1, 14), (2, 13), (3, 12), (4, 11), (5, 10), (6, 9), (7, 8). Now, consider only the coprime pairs. These are (1, 14), (2, 13), (4, 11), (7, 8). ∴4 pairs of numbers can be formed whose sum is 135 and HCF is 9. 74. (b) Let the two numbers be 12x and 12 y. 12x × 12 y = 3600 xy = 25 Possible values of x and y are (1, 25) (5, 5). But (5, 5) are not coprimes. ∴Only 1 pair of number can be formed. 75. (a) Let HCF = x LCM = 36x 36x × x = 3600 x2 = 100 ⇒ x = 10 ∴ HCF is 10. 76. (a) Let the numbers be 3x and 5x. LCM is 15x = 600 x = 40 The two numbers are 120 and 200. Sum = 320. 1, 1, 1, 1 LCM = 2 × 2 × 2 × 2 × 3 × 5 × 7 = 168 ∴Required number = 1680 − 3 = 1677. 79. (d) Required number = (LCM of 9, 12, 15, 18) + 4 2 9, 12, 15, 18 2 9, 6, 15, 9 3 9, 3, 15, 9 3 3, 1, 5, 3 5 1, 1, 5, 1 1, 1, 1, 1 LCM = 2 × 2 × 3 × 3 × 5 = 180 ∴Required number = 180 + 4 = 184 80. (a) Required number = (LCM of 12, 15, 20, 24) + 2 2 12, 15, 20, 24 2 6, 15, 10, 12 3 3, 15, 5, 6 2 1, 5, 5, 2 5 1, 5, 5, 1 1, 1, 1, 1 LCM = 2 × 2 × 2 × 3 × 5 = 120 Required number = 120 + 2 = 122 81. (a) Required number = (LCM of 15, 20, 30) 2 15, 20, 30 2 15, 10, 15 5 15, 5, 15 3 3, 1, 3 1, 1, 1 LCM = 2 × 2 × 3 × 5 = 60 18 Mathematical Ability : Number System 82. (a) LCM of 3, 4, 5, 6 and 8 is 120 2 2 2 3 5 3, 4, 5, 6, 8 3, 2, 5, 3, 4 3, 1, 5, 3, 2 3, 1, 5, 3, 1 1, 1, 5, 1, 1 1, 1, 1, 1, 1 LCM = 2 × 2 × 2 × 3 × 5 = 120 120 = 23 × 3 × 5 ∴The required perfect square = (23 × 3 × 5) × 2 × 3 × 5 = 3600 83. (b) LCM of 8, 12 and 28 = 168. 2 2 2 3 7 8, 12, 28 4, 6, 14 2, 3, 7 1, 3, 7 1, 1, 7 1, 1, 1 LCM = 2 × 2 × 2 × 3 × 7 = 168 168)800(4 672 128 ∴The required number = 800 − 128 = 672 84. (c) LCM of 12, 15 and 20 = 60 2 2 3 5 12, 15, 20 6, 15, 10 3, 15, 5 1, 5, 5 1, 1, 1 LCM = 2 × 2 × 3 × 5 = 60 Now, 12 − 4 = 8, 15 − 7 = 8 and 20 − 12 = 8 So, required number = 60 − 8 = 52 85. (a) LCM of 5, 6, 7 and 8 = 840 2 2 2 3 5 7 5, 6, 7, 8 5, 3, 7, 4 5, 3, 7, 2 5, 3, 7, 1 5, 1, 7, 1 1, 1, 7, 1 1, 1, 1, 1 LCM = 2 × 2 × 2 × 3 × 5 × 7 = 840 Multiples of 840 = 840, 1680, ........... ∴Required number = 1680 + 3 = 1683 Which is exactly divisible by 9. 86. (a) LCM of 15, 20 and 24 = 120 2 2 2 3 5 15, 20, 24 15, 10, 12 15, 5, 6 15, 5, 3 5, 5, 1 1, 1, 1 LCM = 2 × 2 × 2 × 3 × 5 = 120 Multiples of 120 = 120, 240, 360, 480, ..... ∴Required number = 360 + 4 = 364 Which is exactly divisible by 13. 87. (c) LCM of 24, 40, 64, 72 are 124 = 2880 s 2 2 2 2 2 2 3 3 5 24, 40, 64, 72, 120 12, 20, 32, 36, 60 6, 10, 16, 18, 30 3, 5, 8, 9, 15 3, 5, 4, 9, 15 3, 5, 2, 9, 15 3, 5, 1, 9, 15 1, 5, 1, 3, 5 1, 5, 1, 1, 5 1, 1, 1, 1, 1 LCM = 26 × 32 × 5 = 2880 2880 ∴Required time = = 48 min 60 88. (c) Time taken for 1 revolution 1 1 1 1 min = , , , 12 25 20 30 LCM of Numerators 1 ∴ Required time = = min HCF of Denominators 4 1 = × 60 = 15 s 4 89. (a) Time taken to complete 1 revolution 12 12 12 h = 4, 3, 2 h = , , 3 4 6 ∴Required time = LCM of 4, 3 and 2 = 12 h. 90. (b) LCM of 48, 72 and 180 = 432 s. 2 2 2 3 3 2 3 48, 72, 108 24, 36, 54 12, 18, 27 6, 9, 27 2, 3, 9 2, 1, 3 1, 1, 3 1, 1, 1 LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 Traffic lights will change simultaneously after 432 s = 7 min 12 s 91. (a) The smallest four–digit number exactly divisible by 8, 10 and 12 should also be divisible by the LCM of 8, 10 and 12. LCM of 8, 10 and 12 = 120. Smallest four-digit number = 1000 120)1000(8 960 40 So, the required number = 1000 + (120 − 40) = 1000 + 80 = 1080 92. (c) The greatest five-digit number divisible by 6, 8 and 12 should also be divisible by the LCM of 6, 8 and 12. LCM of 6, 8 and 12 = 24. 19 RMAT—Success Master The greatest five-digit number = 99999 24)99999(416 96 39 24 159 144 15 So, the required number = 99999 − 15 = 99984 93. (b) Required number is the HCF of 321, 428 and 535. Step 1 Find HCF of 321 and 428 321)428(1 321 107)321(3 321 0 HCF of 321 and 428 = 107 Step 2 Find HCF of 107 and 535. 107)535(5 535 0 So, HCF of 321, 428 and 535 = 107. 94. (a) Required measure = HCF of 403, 434 and 465. Step 1 Find HCF of 403 and 434. 403)434(1 403 31)403(13 31 93 93 0 HCF of 403 and 434 = 31. Step 2 Find HCF of 31 and 465 31)465(15 31 155 155 0 HCF of 31 and 465 = 31. ∴Required measure = 31 cm. 95. (b) Side of square tile = HCF of 450 and 525. = 75 cm 450)525(1 450 75)450(6 450 0 ∴Required number of tiles = 450 × 525 = 42. 75 × 75 96. (b) Required number is the HCF of 676, 481 and 767 = 13. 97. (d) Required number = HCF of (48 − 3), (63 − 3) and (78 − 3) = HCF of 45, 60, 75 = 15 98. (d) Required number = HCF of (121 − 4), (134 − 4) and (147 − 4). = HCF of 117, 130, 143 = 13 99. (b) Required number = HCF of (1196 − 6) , (209 − 5) and (1449 − 4) = HCF of 1190, 204, 1445 = 17 100. (a) Required number = HCF of (228 − 3), (275 − 5) and (321 − 6) = HCF of 225, 270, 315 = 45 B. MBA Entrances’ Gallery 1. (c) The ratio of Boys : Girls = 3 : 1 So, 42 is not possible as it is not divisible by (3 + 1) = 4 6. (a) Let the number be n ⇒ n × 25 = 775 ⇒ n = 31 2. (a) 8 + 12 = 20 ⇒ 2 + 0 = 2 7 + 14 = 21 ⇒ 2 + 1 = 3 ∴ 10 + 18 = 28 ⇒ 2 + 8 = 10 7. (a) Perfect square are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ..... Clearly 7 cannot be at unit’s digit. 3. (d) (10 − 9) = (9 − 8) = (8 − 7) = (7 − 6) = (6 − 5) = (5 − 4) = (4 − 3) = (3 − 2) = 1 LCM of 10, 9, 8, 7, 6, 5, 4, 3, 2, = 2520 ∴Required number = 2520 − 1 = 2519 4. (b) LCM of 8, 11, and 24 = 264 Required number = 264 − 5 = 259 ∴ 5. (c) Remainder when 1693 divided by 17 = 10 Remainder when 797 divided by 17 = 15 Remainder when 3395 divided by 17 = 12 ∴Required number = Remainder when 10 × 15 × 12 divided by 17 = 1800 divided by 17 = 15 8. (d) 92 + 102 = 81 + 100 = 181, 196 is the nearest perfect square. Hence, 15 should be added. 9. (c) 9992 − 9982 = (999 + 998) (999 − 998) = (1997) (1) = 1997 10. (d) The number is = 119n + 19 = 138, 257 When these numbers are divided by 17, then remainder is 2. 11. (b) Product of numbers = HCF × LCM ⇒ 27 × n = 63 × 9 63 × 9 ⇒ n= = 21 27 20 Mathematical Ability : Number System 12. (b) 34(1) − 43(1) = 81 − 64 = 17, 34( 2) − 43( 2) = 812 − 642 = 2465 It is always divisible by 17. 13. (d) The largest such number is 1001. 14. (d) In this case the remainder must be 0. 15. (a) The difference between 25 ⋅ 92 and 2592 ⇒ 32 × 81 ⇒ 2592 − 2592 = 0 16. (a) HCF of 168, 189, 231 is 21. Then, difference between 21 − 8 = 13. 7 1 17. (d) = 0. 875, = 0. 333, 8 3 1 23 = 0. 25, = 0. 958 4 24 11 17 = 0. 916, = 0. 708 12 24 Clearly, 0.708 lies between 0.333 and 0.875. 17 . ∴ Required fraction is 24 18. (b) Suppose one of the multiplier is x then, 53x − 35x = 540 18x = 540 x = 30 ∴ New multiplication = 30 × 53 = 1590 . 19. (b) Let us rearrange the equations : a −b+ c=1−k a + b + c = −1 + l 9 a + 3 b + c = −4 − m Where k, l, m are positive. Suppose λ and µ are two numbers such that in the expression, 9a + 3b + c + λ (a − b + c) + µ (a + b + c), the terms b and c vanish. ∴ − λ + µ = −3 λ + µ = −1 −4 µ= = −2 and λ = 1 ∴ 2 Hence, 9a + 3b + c + 1 (a − b + c) − 2 (a + b + c) = 9a + 3b + c + a − b + c − 2a − 2b − 2c = 8 a = −4 − m + 1 − k + 2 − 2 l = −1 − k − m − 2 l which is less than 0, since k, m and l are all + ve. ∴ a <0 It’s sign is ‘−’ ve only. 20. (d) Only 143616 is a multiple of both 11 and 8 so it is a multiple of 88. 21. (c) We have to consider all the one-digit, two-digit and three-digit numbers and see how many of these have distinct digits. (i) One-digit numbers, 1 to 9 are all distinct. (ii) Two-digit numbers, ten’s place cannot be filled by 0. Hence, 9 possibilities. Having filled that, 8 possibilities for the unit’s place and also zero. Hence, 9 × 9 = 81 numbers are possible. (iii) In a similar way, the three-digit numbers can be determined Hundred’s place = 9 Ten’s place = 9 Unit’s place = 8 ∴ Number of three-digit numbers = 9 × 9 × 8 = 648 ∴ Total numbers = 9 + 81 + 648 = 738. 22. (a) PQ ⋅ RS is less than 100 D ⋅ AB is less than 10 ∴ PQ ⋅ RS × D ⋅ AB must be less than 1000 and definitely greater than 10. 23. (b) If 5372 xy is to be divisible by 45, it must be divisible by 9 and also by 5. y must be 0 or 5 for divisibility by 5. For divisibility by 9. 5 + 3 + 7 + 2 + x + y must be a multiple of 9. ie, 17 + x + y must be a multiple of 9. ∴ If y = 5, x must be 5 If y = 0, x must be 1 ie, x = 1 or 5 and y = 0 or 5 ∴ {(1, 0);(5, 5)} is the correct answer. 24. (a) Since, a + b + c = 0, a + b = − c (a + b)3 = (− c)3 3 3 a + b + 3ab(a + b) = − c3 a3 + b3 + 3ab(− c) = − c3 a3 + b3 − 3abc = − c3 a3 + b3 + c3 = 3abc a3 + b3 + c3 ∴ =3. abc 25. (d) LCM of a x and by Since, a and b are distinct primes there is no common factor. Hence, LCM = a x by . 26. (c) The thousands place can be filled by 3 only. Then, the hundreds place can be filled in 5 ways , the tens place in 4 ways and the units place in 3 ways. Total number of ways is 1 × 5 × 4 × 3 or 60. 27. (c) Let the two numbers are 7x and 8x and LCM is 56x. It is given that LCM = 280 ie, 56x = 280 and x = 5 ie, Numbers are 35 and 40 . 28. (a) From the given options Required number = 39 29. (b) LCM of 12, 15, 18 = 180 So, required number = 180 − 5 = 175 30. (d) LCM of 8, 10, 12 and 15 = 20 We know that 83 × 120 = 9960 and 84 × 120 = 10080 So, 9960 is the required number. 31. (c) 4a 4 = 2a 2 8a 6 = 2a 2 32. (d) 359 is the required number. As 359 ÷ 4 (remainder is 3) 89 ÷ 5 (remainder is 4) 17 ÷ 6 (remainder is 5)