SSC Mathematics Topic Wise Solved Papers – Number System and HCF & LCM

SSC Mathematics Topic Wise Solved Papers – Number System and HCF & LCM

SSC Mathematics Previous Year Question Papers English Reasoning General Awareness

1. The H.C.F. and L.C.M. of two numebrs are 8 and 48 respectively. If one of the numbers is 24, then the other number is (SSC CGL 1^st Sit. 2010)
(a) 48
(b) 36
(c) 24
(d) 16

2. The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20,28,32 and 35, is (SSC CGL 1^st Sit. 2010)
(a) 1120
(b)  4714
(c) 5200
(d)  5600

3. The ninth term of the sequence 0, 3,8,15,24, 35,…. is (SSC CGL 1^st Sit. 2010)
(a) 63
(b) 70
(c)  80
(d)  99

4. A number, when divided by 114, leaves remainder 21. If the same number is divided by 19, then the remainder will be (SSC CGL 1^st Sit. 2010)
(a) 1
(b) 2
(c)  7
(d) 17

5. Two numbers are in the ratio 3 : 4. Their L.C.M. is 84. The greater number is (SSC CGL 1^st Sit. 2010)
(a) 21
(b) 24
(c) 28
(d) 84

6. The sixth term of the sequence 2.6. 11. 17 is (SSC CGL 1^st Sit. 2010)
(a) 24
(b) 30
(c) 32
(d) 36

7. A number, when divided by 136, leaves remainder 36. Ifthe same number is divided by 17, the remainder will be (SSC CGL 2^nd Sit. 2010)
(a) 9
(b) 7
(c) 3
(d) 2

8. A 4-digit number is formed by repeating a 2-digit number such as 1515,3737, etc. Any number of this form is exactly divisible by (SSC CGL 2^nd Sit. 2010)
(a) 7
(b) 11
(c) 13
(d) 101

9. The H.C.F. and L.C.M. of two numbers are 12 and 336 respectively. If one of the numbers is 84, the other is (SSC CGL 2^nd Sit. 2010)
(a) 36
(b) 48
(c) 72
(d) 96

10. The sum of two numbers is 36 and their H.C.F and L.C.M. are 3 and 105 respectively. The sum of the reciprocals of two numbers is (SSC CGL 2^nd Sit. 2010)
(a) 
(b) 
(c) 
(d) 

11. If ‘n’ be any natural number, then by which largest number (n³ – n) is always divisible ?  (SSC CGL 2^nd Sit. 2010)
(a) 3
(b) 6
(c) 12
(d) 18

12. How many perfect squares lie between 120 and 300 ? (SSC CGL 2^nd Sit. 2010)
(a)  5
(b)  6
(c)  7
(d)  8

13. The remainder when 3²¹ is divided by 5 is (SSC CGL I^s‘Sit. 2011)
(a) 1
(b) 2
(c)  3
(d) 4

14. The last digit of (1001 )²⁰⁰⁸ + 1002 is (SSC CGL 1^st Sit. 2011)

(a) 0
(b)  3
(c) 4
(d)  6

15. If x * y = (x + 3)² (y-l), then the value of 5 *4 is (SSC CGL 1^st Sit. 2011)

(a)  192
(b)  182
(c) 
(d)  356

16. The L.C.M. of three different numbers is 120. Which of the following cannot be their H.C.F.? (SSC CGL 1^st Sit. 2011)
(a)  8
(b) 12
(c)24
(d) 35

17. A number when divided by 49 leaves 32 as remainder. This number when divided by 7 will have the remainder as (SSC CGL 1^st Sit. 2011)
(a) 4
(b) 3
(c) 2
(d) 5

18. The traffic lights at three different road crossings change after 24 seconds, 36 seconds and 54 seconds respectively. If they all change simultaneously at 10 : 15 :00 AM, then at what time will they again.change simultaneously? (SSC CGL 1^st Sit. 2011)
(a) 10:16:54 AM
(b) 10:18:36 AM
(c) 10:17:02 AM
(d) 10:22:12AM

19. The least number, which is to be added to the greatest number of 4 digits so that the sum may be divisible by 345, is (SSC CGL 2^nd Sit. 2011)
(a) 50
(b) 6
(c) 60
(d) 5

20. If 17200 is divided by 18, the remainder is (SSC CGL 2^nd Sit. 2011)
(a) 1
(b) 2
(c) 16
(d) 17

21. The unit digit in the sum of (124)³⁷² + (124)³⁷³ is (SSC CGL 2^nd Sit. 2011)
(a) 5
(b) 4
(c) 2
(d) 0

22. If a * b = a^b, then the value of 5 * 3 is (SSC CGL 2^nd Sit. 2011)
(a) 125
(b) 243
(c) 53
(d) 15

23. Which one of the following will completely divide 5⁷¹ + 5⁷²+ 5⁷³? (SSC CGL 2^nd Sit. 2011)
(a) 150
(b) 160
(c) 155
(d) 30

24. L.C.M. oftwo numbers is 120 and their H.C.F. is 10. Which of the following can be the sum of those two numbers? (SSCCGL 2^nd Sit. 2011)
(a) 140
(b) 80
(c) 60
(d) 70

25. When ‘n’ is divisible by 5 the remainder is 2. What is the remainder when n² is divided by 5? (SSCCGL 2^nd Sit. 2011)
(a) 2
(b) 3
(c) 1
(d) 4

26. Four runners started running simultaneously from a point on a circular track. They took 200 seconds, 300 seconds, 360 seconds and 450 seconds to complete one round. After how much time they meet at the starting point for the first time? (SSC CGL 2^nd Sit. 2011)
(a) 1800 seconds
(b) 3600 seconds
(c) 2400 seconds
(d) 4800 seconds

27. The greatest number that can divide 140, 176, 264 leaving remainders of 4,6, and 9 respectively is (SSC Sub. Ins. 2012)
(a) 85
(b) 34
(c) 17
(d) 2

28. There are 4 terms in an A.P. such that the sum of two means is 110 and product of their extremes is 2125. The 3^rd term is (SSC Sub. Ins. 2012)
(a) 65
(b) 75
(c) 55
(d) 45

29. The number nearest to 75070 which is divisible by 65, is (SSC CGL 1^st Sit. 2012)
(a) 75070
(b) 75075
(c) 75010
(d) 75065

30. The least number which when divided by 35, 45, 55 leaves the remainder 18,28,38 respectively is (SSC CGL 1^st Sit. 2012)
(a) 3448
(b) 3482
(c) 2468
(d) 3265

31. A three-digit number 4a3 is added to another three-digit number 984 to give the four digit number 13b7 which is divisible by 11. Then the value of (a + b) is:  (SSC CGL 1^st Sit. 2012)
(a) 11
(b) 12
(c) 9
(d) 10

32. The greatest number that will divide 19,35 and 59 to leave the same remainder in each case is: (SSC CGL 1^st Sit. 2012)
(a) 9
(b) 6
(c) 7
(d) 8

33. The next term of the series -1,6,25,62,123,214,______s: (SSC CGL 1^st Sit. 2012)
(a) 345
(b) 143
(c) 341
(d) 343

34. The next term of the series 1, 5 12,24,43 is (SSC CGL 1^st Sit. 2012)
(a) 51
(b) 62
(c) 71
(d) 78

35. The least multiple of 13 which when divided by 4, 5, 6, 7 leaves remainder 3 in each case is (SSC CGL 2^nd Sit. 2012)
(a) 3780
(b) 3783
(c) 2520
(d) 2522

36. What would be the sum of 1+3 + 5 + 7 + 9+11 + 13 + 15 +……………up to 15th term? (SSC CGL 2^nd Sit. 2012)
(a) 250
(b) 240
(c) 225
(d) 265

37. The least number which when divided by 48,64,90,120will leave the remainders 38,54,80,110 respectively, is (SSC CGL 2^nd Sit. 2012)
(a) 2870
(b) 2860
(c) 2890
(d) 2880

38. If 13 + 23 +…………….+ 93 = 2025, then the approx, value of (0.11)³ + (0.22)³ +………………+ (0.99)³ is (SSC CGL 2^nd Sit. 2012)
(a) 0.2695
(b) 0.3695
(c) 2.695
(d) 3.695

39. With a two digit prime number, if 18 is added, we get another prime number with digits reversed. How many such numbers are possible? (SSC CGL 2^nd Sit 2012)
(a) 2
(b) 3
(c) 0
(d) 1

40. If x=  and y =  than the value of x³ + y³  is:  (SSC Sub. Ins. 2013)
(a) 950
(b) 730
(c) 650
(d) 970

41. The ratio oftwo numbers is 3 : 4 and their HCF is 5. Their LCM is: (SSC Sub. Ins. 2013)
(a) 10
(b) 60
(c) 15
(d) 12

42. L.C.M. of  is
(a) 
(b) 
(c) 
(d) 

43. ‘a’ divides 228 leaving a remainder 18. The biggest two-digit value of ‘a’ is (SSC CHSL 2013)
(a) 30
(b) 70
(c) 21
(d) 35

44. If the sum of the digits of any integer lying between 100 and 1000 is subtracted from thenumber, the result always is (SSC CHSL 2013)
(a) divisible by 5
(b) divisible by 6
(c) divisible by 2
(d) divisible by 9

45. The fifth term ofthe sequence for which t₁= 1, t₂ = 2 and t_n + 2 = t_n + 1, is is (SSC CGL 1^st Sit. 2013)
(a) 5
(b) 10
(c) 6
(d) 8

46. Product of two co-prime numbers is 117. Then their L.C.M. is (SSC CGL 2013)
(a) 13
(b) 39
(c) 117
(d) 9

47. A number x when divided by 289 leaves 18 as the remainder. The same number when divided by 17 leaves y as a remainder. The value ofy is (SSC CGL 2nd Sit. 2013)
(a) 3
(b) 1
(c) 5
(d) 2

48. The sum of the squares of the digits of the largest prime number in two digits is (SSC Multitasking 2014)
(a) 148
(b) 130
(c) 97
(d) 118

49. Find the number lying between 900 and 1000 which when divided by 38 and 57 leaves in each case a remainder 23. (SSC Multitasking 2014)
(a) 912
(b) 926
(c) 935
(d) 962

50. The next term of the sequence,

51. Three tankers contain 403 litres, 434 litres, 465 litres of diesel respectively. Then the maximum capacity of a container that can measure the diesel of the three container exact number oftimesis (SSC Sub. Ins. 2014)
(a) 31 litres
(b) 62 litres
(c) 41 litres
(d) 84 litres

52. The H.C.F. and L.C.M. of two numbers are 44 and 264 respectively. If the first number is divided by 2, the quotient is 44. The other number is (SSC CHSL 2014)
(a) 147
(b) 528
(c) 132
(d) 264

53. A teacher wants to arrange his students in an equal number of rows and columns. Ifthere are 1369 students, the number of students in the last row are (SSC CHSL 2014)
(a) 37
(b) 33
(c) 63
(d) 47

54. The first term of an Arithmetic Progression is 22 and the last term is -11. If the sum is 66, the number of terms in the sequence are: (SSC CHSL 2014)
(a) 10
(b) 12
(c) 9
(d) 8

55. If the product of first fifty positive consecutive integers be divisible by 7ⁿ, where n is an integer, then the largest possible value of n is (SSC CGL 1^st Sit. 2014)
(a) 7
(b) 8
(c) 10
(d) 5

56. The smallest five digit number which is divisible by 12,18 and 21 is: (SSC CHSL 2015)
(a) 50321
(b) 10224
(c) 30256
(d) 10080

57. If 1³ + 2³ …………+10³-3025, then the value of 2³ + 4³ +……….+ 20³ is: (SSC CHSL 2015)
(a) 5060
(b) 12100
(c) 24200
(d) 7590

58. The least number that should be added to 2055 so that the sum is exactly divisible by 27: (SSC CGL 1^st Sit. 2015)
(a) 24
(b) 27
(c) 31
(d) 28

59. The least number which when divided by 6, 9, 12, 15, 18 leaves the same remainder 2 in each case is: (SSC CGL 2^nd Sit. 2015)
(a) 178
(b) 182
(c) 176
(d) 180

60. What least value must be assigned to ‘*’ so that the numbers 451*603 is exactly divisible by 9? (SSC CGL 1^st Sit. 2016)
(a) 7
(b) 8
(c) 5
(d) 9

61. If X and Y are the two digits of the number 347 XY such that the number is completely divisible by 80, then what is the value of X + Y? (SSC CGL 2017)
(a) 2
(b) 4
(c) 6
(d) 8

62. How many numbers are there from 300 to 650 which are completely divisible by both 5 and 7? (SSC CGL 2017)
(a) 8
(b) 9
(c) 10
(d) 12

63. Which value among is the largest? (SSC CGL 2017)
(a) 
(b)
(c)
(d)

64. By which least number should 5000 be divided so that it becomes a perfect square? (SSC CGL 2017)
(a) 2
(b) 5
(c) 10
(d) 25

65. What is the LCM (least common multiple) of 57 and 93? (SSC CHSL 2017)
(a) 1767
(b) 1567
(c) 1576
(d) 1919

66. Product of digits of a 2-digit number is 27. If we add 54 to the number, the new number obtained is a number formed by interchange of the digits. Find the number. (SSC CHSL 2017)
(a) 39
(b) 93
(c) 63
(d) 36

67. The least number of five digits exactly divisible by 88 is: (SSC MTS 2017)
(a) 10088
(b) 10023
(c) 10132
(d) 10032

68. Of the three numbers, the first is twice the second, and the second is twice the third. The average of the reciprocal of the numbers is 7/12. The numbers are: (SSC MTS 2017)
(a) 20,10,5
(b) 4,2,1
(c) 36,18,9
(d) 16,8,4

69. What is the smallest value that must be added to 709, so that the resultant is a perfect square? (SSC Sub. Ins. 2017)
(a) 8
(b) 12
(c) 20
(d) 32

70. Which one among is the smallest number? (SSC Sub. Ins. 2017)
(a)
(b)
(c)
(d) All are equal

71. If 34N is divisible by 11, then what is the value of N (SSC Sub. Ins. 2017)
(a) 1
(b) 3
(c) 4
(d) 9

Hints & Solution

1. (d)

2. (b)

3. (c)

4. (b)
If the first divisor is a multiple of second divisor. Then, remainder by the second divisor.
∴ Remainder = 21 ÷5-19 = 2

5. (c)

6. (c)

7. (d) 
If the first divisor be a multiple of the second divisor, then required remainder = remainder obtained by dividing the first remainder (36)
by the second divisor (17) = 2
∵ 17 is a factor of 136
∴ Remainder when 36 is divided by 17 = 2

8. (d)
xyxy=xy x 100+xy
= xy(100+ 1) = 101 xxy
Hence, the number is exactly divisible by 101.

9. (b)

10. (c)

11. (b)

12. (c)

13. (c)

14. (b)
Last digit of (1001 )²⁰⁰⁸ + 1002 = 1 +2 = 3

15. (c)

16. (d)
HCF must be a factor of LCM from option 35 is not factor of 120.

Alternate Method: If two number are in the form of ax and bx then x is H.C.F and a x b x x is their L.C.M Hence L.C.M is always divisible by H.C.F.

17. (a)
Here, the first divisor i.e. 49 is multiple of second divisor i.e. 7.
∴ Required remainder = Remainder obtained on
dividing 32 by    7 =4

18. (b)
LCM of 24,36 and 54 seconds
= 216 seconds = 3 minutes 36 seconds
∴ Required time = 10 :15 :00+
3 minutes 36 seconds = 10 :18 : 36 a.m.

19. (b)

20. (a)

21. (d)

Unit’s digit of the sum = 6+4=0

22. (a)

23. (c)

24. (d)

25. (d)
Required remainder = Remainder obtained by dividing 2² by 5.
Remainder =4

26. (a)
Required time = LCM of 200,300,360 and 450 seconds
= 1800 seconds.

27. (c)

28.(a)

29. (b)

30. (a)

31. (d)

32. (d)

33. (c)

34. (c)

35. (b)

36. (c)

37. (a)

38. (c)

39. (a)

40. (d)

41. (b)

42. (c)

43.(b)
228 – 18 = 210 is exactly divisible biggest two digit no. i.e. 70

44. (d)
(100 x+ 10 y + z)-(x+y + z) = 99 x + 9 y
= 9(11 x + y)

45. (d)
t_{n + 2} = t_n+t_n+1
t₂ – t₂ +t₂ = 3
t₄ = t₂ +t₂ = 3 + 2 = 5
t₅ = ^t4 ^{+ t}3 ^{= 3 + 5 = 8}

46. (c)
HCF of two-prime numbers = 1
∴ Product of numbers = their LCM =117

47. (b)
Here, the first divisor (289) is a multiple of second
divisor (17).
∴ Required remainder = Remainder obtained on dividing 18 by 17 = 1

48. (b)
Largest two digit prime number is 97
9²+ ₇²= 81 +49 = 130

49. (c)
L.C.M of (38,57)= 114
Multiple of 114 between 900 and 1000 = 912
number which leaves 23 = 912 + 23 = 935

50. (a)
H.C.F. of 403,434 and 465 is 31.

52. (c)

53. (a)
If they are equal number of rows and columns then,

54. (b)

55. (b)
Product of first fifty positive consecutive integers = 1 x 2 x

56. (d)

57. (c)

58. (a)

59. (b)

60. (b)

61. (a)
347XY as 347X0. Since 8 is a factor of 80.
347X0 is divisible by 8. It means last three digits 7X0 is divisible by 8.
Hence, X is 2 or 6
if X = 6, number is 34760. But this is not divisible by 80.
ifX = 2, number is 34720, which is divisible by 80.
Therefore, number is 34720 with X = 2 and Y = 0.
∴ x + y = 2 + 0 = 2.

62. (c)
LCM of 5 and 7 = 35
So, the numbers divisible by both 5 and 7 are multilpe of 35. Between 300 and 650. We have 10 multiple of 35. They are: 315,350,385,420,455,490,525,560,595, 630.

63. (a)

64. (a)
According to option,
5000 ÷ 2 = 2500
Hence, 2500 is a perfect square of 50.

65. (a)
LCM of 57 and 93,

66. (a)
Let digit at ten’s place be x and digit at unit’s place be y.
∴The number = 10x + y
When digit are interchanged, the new number = 10y + x

67. (d)
The smallest number of 5 digits = 10000

68. (c)
Let third number = x then,
second number = 2x first number = 4x
According to question

69. (c)
According to question

70. (b)
Here,

71. (a)
A number is divisible by 11, if difference of the sum of the digits at the even places and sum of digits at odd places is either 0 (zero) or a multiple of 11.