- Staff Selection Commission Mathematics 1999 to 2017 - LCM and HCF Part 3

Staff Selection Commission Mathematics - LCM and HCF



Ans .

2


  1. Explanation :

    The LCM of 5, 6, 7 and 8 = 840
    Required number = 840 k + 3 which is exactly divisible by 9 for some value of k. Now, 840 k + 3 = 93 × 9 k + (3k + 3) When k = 2, 3k + 3 = 9, which is divisible by 9.
    Required number = 840 × 2 + 3 = 1683





Ans .

1


  1. Explanation :

    Using Rule 5, Here,12 – 2 = 10; 16 – 6 = 10; 24 – 14 = 10
    Now, LCM of 12, 16 and 24 = 48
    The greatest 4–digit number exactly divisible by 48 = 9984
    Required number = 9984 – 10 = 9974





Ans .

1


  1. Explanation :

    Using Rule 5, LCM of 15, 20 and 35 = 420
    Required least number = 420 + 8 = 428





Ans .

3


  1. Explanation :

    Using Rule 5, The smallest number divisible by 12 or 10 or 8 = LCM of 12, 10 and 8 = 120
    Required number =120 + 6 = 126





Ans .

2


  1. Explanation :

    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.





Ans .

3


  1. Explanation :

    A makes one complete round of the circular track in 5= 2 hours,
                                                                                     5
                                                                                     2
    B in 5
           3
    hours and C in 5 hours.
                             2
    That is after 2 hours A is at the starting point, B after 5 hours
                                                                                    3
    and C after 5 hours.
                        2
    Hence the required time = LCM of 2, 5
    3 and 5
    2 hours = LCM of 2,5,5
    HCFof3,2
    =10
    1 = 10 hours.





Ans .

1


  1. Explanation :

    Required time = LCM of 200, 300, 360 and 450 seconds = 1800 seconds





Ans .

1


  1. Explanation :

    LCM of 4, 6, 8, 14 = 168 seconds = 2 minutes 48 seconds They ring again at 12 + 2 min. 48 sec. = 12 hrs. 2 min. 48 sec





Ans .

4


  1. Explanation :

    1 1 hours = 90 minutes
       2
    1 hour and 45 minutes = 105 minutes
    1 hour = 60 minutes
    LCM of 30 minutes, 60 minutes, 90 minutes and 105 minutes.
    3   30, 60, 90, 105
    5   10, 20, 30, 35
    2   2, 4, 6, 7
         1, 2, 3, 7
    LCM = 3 × 5 × 2 × 2 × 3 × 7 = 1260 minutes 1260 minutes= 1260 = 21 hours
                                                     60
    The bell will again ring simultaneously after 21 hours.
    Time will be = 12 noon + 21 hours = 9 a.m.





Ans .

1


  1. Explanation :

    The LCM of 5, 6, 8 and 9 = 360 seconds = 6 minutes





Ans .

2


  1. Explanation :

    LCM of 20, 30 and 40 minutes = 120 minutes Hence, the bells will toll together again after 2 hours i.e. at 1 p.m





Ans .

1


  1. Explanation :

    The difference between divisor and the corresponding remainder is equal. LCM of 3, 5, 7 and 9 = 315 Largest 4-digit number = 9999
    315)9999(31
          945
          549
          315
           234
    Number divisible by 315 = 9999 – 234 = 9765 Required number = 9765 – 2 = 9763





Ans .

3


  1. Explanation :

    Required time = LCM of 6, 7, 8, 9 and 12 seconds = 504 seconds





Ans .

2


  1. Explanation :

    Using Rule 2, LCM = LCM of 2, 4, 5
                                                     HCF of 3, 9, 6
    = 20
        3





Ans .

2


  1. Explanation :

    LCM of 3, 4, 5, 6, 7, 8 = 840
    840 ) 10000 ( 11
             840
             1600
             840
             760
    Since, the remainder 760 is more than half of the divisor 840. The nearest number = 10000 + (840 – 760) = 10080





Ans .

2


  1. Explanation :

    Using Rule 8, The largest number of 4-digits is 9999. L.C.M. of divisors
    2   12, 15, 18, 27
    3   6, 15, 9, 27
    3   2, 5, 3, 9
         2, 5, 1, 3
    LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540
    Divide 9999 by 540, now we get 279 as remainder. 9999 – 279 = 9720 Hence, 9720 is the largest 4-digit number exactly divisible by each of 12, 15, 18 and 27.





Ans .

2


  1. Explanation :

    The smallest number divisible by 16, 20 and 24 = LCM of 16, 20 and 24
    2   16, 20, 24
    2   8, 10, 12
    2   4, 5, 6
         2, 5, 3
    LCM = 2×2×2×2×5×3 = 22 × 22 × 5 × 3
    Required complete square number = 22 × 22 ×52 × 32 = 3600





Ans .

2


  1. Explanation :

    LCM of 25, 50 and 75 = 150
    On dividing 43582 by 150, remainder = 82
    150 ) 43582 ( 290
             300
              1358
              1350
              82
    Required number = 43582 + (150 – 82) = 43650





Ans .

2


  1. Explanation :

    Required number = (LCM of 24, 32, 36 and 54) – 5
    Now,
    2   24, 32, 36, 54
    2   12, 16, 18, 27
    2   6, 8, 9, 27
    3   3, 4, 9, 27
    3   1, 4, 3, 9
         1, 4, 1, 3
    LCM = 2 × 2 × 2 × 3 × 3 × 3 × 4 = 864
    Required number = 864 – 5 = 859





Ans .

2


  1. Explanation :

    2   20, 28, 32, 35
    2   10, 14, 16, 35
    5   5, 7, 8, 35
    7   1, 7, 82, 7
         1, 1, 8, 1

    LCM = 2 × 2 × 5 × 7 × 8 = 1120
    Required number = 5834 – 1120 = 4714





Ans .

4


  1. Explanation :

    The LCM of 6, 12 and 18 = 36 = 62





Ans .

2


  1. Explanation :

    Using Rule 8, LCM of 10, 15 and 20 = 60
    Largest 4-digit number = 9999
    60 ) 9999 ( 166
    60
    399
    360
    399
    360
    39
    Required number = 9999 – 39 = 996





Ans .

4


  1. Explanation :

    Using Rule 4,
    Required number = (LCM of 15, 20, 36 and 48) + 3
    2   15, 20, 36, 48
    2   15, 10, 18, 24
    3   15, 5, 9, 12
    5   5, 5, 3, 4
         1, 1, 3, 4
    LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720
    Required number = 720 + 3 = 723





Ans .

3


  1. Explanation :

    Required distance = LCM of 63, 70 and 77 cm. = 6930 cm.
    7   63, 70, 77
         9, 10, 11
    LCM = 7 × 9 × 10 × 11 = 6930





Ans .

2


  1. Explanation :

    Required answer = LCM of 36, 40 and 48 seconds = 720 seconds
    = 720 minutes = 12 minutes
          60
    2   36, 40, 48
    2   18, 20, 24
    2   9, 10, 12
    3   9, 5, 6
         3, 5, 2
    LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720





Ans .

2


  1. Explanation :

    2   60
    2   30
    3   15
         5
    60 = 2 × 2 × 3 × 5 i.e.,
    Numbers = 2, 3, 4 and 5
    Required sum = 2 + 3 + 4 + 5 = 14





Ans .

1


  1. Explanation :

    LCM of x and y = 161
    xy = 23 × 7
    x = 23; y = 7
    3y – x = 3 × 7 – 23 = 21 – 23 = – 2





Ans .

1


  1. Explanation :

    Required time = LCM of 48, 72 and 108 seconds
    2   48, 72, 108
    2   24, 36, 54
    2   12, 18, 54
    3   6, 9, 27
    3   2, 3, 9
         2, 1, 3
    LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds = 7 minutes 12 second
    Required time = 10 : 07 : 12 hours





Ans .

1


  1. Explanation :

    Maximum number of students = The greatest common divisor = HCF of 1001 and 910 = 91





Ans .

3


  1. Explanation :

    Using Rule 7, Required number = HCF of (989 – 5) and (1327 – 7) = HCF of 984 and 1320 = 24 \ HCF = 24





Ans .

2


  1. Explanation :

    Using Rule 3,
    HCF of 2 4 , and 6
                3   5      7
    = HCF of 2, 4 and 6
    LCM of 3, 5 and 7\
    = 2
    105





Ans .

1


  1. Explanation :

    Using Rule 7, The greatest number N = HCF of (1305 – x ), (4665 – x) and (6905 – x), where x is the remainder = HCF of (4665 – 1305), (6905– 4665) and (6905 – 1305) = HCF of 3360, 2240 and 5600
    2240 ) 3360 ( 1
    2240
    1120 ) 2240 ( 2
    2240
    x
    N = 1120 Sum of digits = 1 + 1 + 2 + 0 = 4





Ans .

1


  1. Explanation :

    Using Rule 7,
    The number will be HCF of 307 – 3 = 304 and 330 – 7 = 323.
    304 ) 323 ( 1
    304
    19 )304( 16
    19
    114
    114
    x
    Required number = 19





Ans .

3


  1. Explanation :

    Using Rule 7, 3026 –11 = 3015 and 5053 –13 = 5040 Required number = HCF of 3015 and 5040
    3015 ) 5040 ( 1
    3015
    2025 ) 3015 ( 1
    2025
    990 )2025 ( 2
    1980
    45 )990( 22
    90
    90
    90
    x
    Required Number = 45





Ans .

1


  1. Explanation :

    Using Rule 7, We have to find HCF of (1657 – 6 = 1651) and (2037 – 5 = 2032) 1651 = 13 × 127 2032 = 16 × 127
    HCF = 127 So, required number will be 127.





Ans .

1


  1. Explanation :

    Using Rule 7, Let x be the remainder. Then, (25 – x), (73 – x ), and (97 – x ) Will be exactly divisible by the required number.
    Required number = HCF of (73 –x ) – (25 –x ), (97 –x ) – (73 –x ) and (97 –x ) – (25 –x ) = HCF of (73 –25), (97 –73), and (97 –25) = HCF of 48, 24 and 72 = 24





Ans .

2


  1. Explanation :

    Using Rule 7, Required number = HCF of (110 – 2) and (128 – 2) = HCF of 108 and 126 = 18





Ans .

3


  1. Explanation :

    Required maximum capacity of container = HCF of 75 l and 45 l Now, 75 = 5 × 5 × 3 45 = 5 × 3 × 3
    HCF = 15 litres





Ans .

4


  1. Explanation :

    Length of the floor = 15 m 17 cm = 1517 cm Breadth of the floor = 9m 2 cm = 902 cm. Area of the floor = 1517 × 902 cm2 The number of square tiles will be least, when the size of each tile is maximum.
    Size of each tile = HCF of 1517 and 902 = 41
    Required number of tiles = 1517 x 902 = 814
    41 x 41





Ans .

1


  1. Explanation :

    Number of books in each stack = HCF of 336, 240, 96 = 48
    240 ) 336 ( 1
          240
          96 ) 240 ( 2
          192
                48 ) 96 ( 2
                96
                x
    Total number of stacks =
    360 - 240 + 96
    48  48  48 =7 + 5 + 2 = 14





Ans .

3


  1. Explanation :

    First of all we find the HCF of 945 and 2475. HCF = 45 Illustration :
    945 )2475( 2
          1890
                585 ) 945 ( 1
                585
    360 ) 585 ( 1
                      360
                      225 ) 360 ( 1
                            225
                            135) 225 ( 1
                                  135
                                  90 ) 135 ( 1
                                              90
                                        45 ) 90 ( 2
                                        90
                                        x
    Maximum number of animals in each flock = 45
    Required total number of flocks
    945 + 2475 = 21 + 55 = 76
    45 45





Ans .

2


  1. Explanation :

    Maximum quantity in each can = HCF of 21, 42 and 63 litres = 21 litres Required least number of cans
    21 + 42 + 63
    21 21 21
    = 1 + 2 + 3 = 6





Ans .

3


  1. Explanation :

    Using Rule 7, Required number = HCF of 411 – 3 = 408; 684 – 4 = 680 and 821 – 5 = 816 HCF of 408 and 816 = 408 HCF of 408 and 680
          408 ) 680 ( 1
             408
                272 ) 408 ( 1
    272
                136 ) 272 ( 2
                272
                x
    Required number = 136





Ans .

4


  1. Explanation :

    Required number = HCF of 200 and 320 = 40
    200 ) 320 ( 1
          200
         120 ) 200 ( 1
              120
             80 ) 120 ( 1
                  80
                 40 ) 80 ( 2
                      80
                     x





Ans .

3


  1. Explanation :

    As the height of each stack is same, the required number of
    books in each stack
    = HCF of 84, 90 and 120
    84 = 2 × 2 × 3 × 7
    90 = 2 × 3 × 3 × 5
    120 = 2 × 2 × 2 × 3 × 5
    HCF = 2 × 3 = 6





Ans .

2


  1. Explanation :

    Using Rule 7,
    Required number
    = HCF of (729 – 9)
    = 720 and (901 – 5)
    = 896
    720 ) 896 ( 1
          720
         176 ) 720 ( 4
              704
             16 ) 176 ( 11
                  16
                 16
                      16
                     x





Ans .

1


  1. Explanation :

    Greatest capacity of measuring vessel
    = HCF of 403 litres, 434 litres
    and 465 litres
    = 31 litres
    HCF of 403 and 434
         403 ) 434 ( 1
              403
             31 ) 403 ( 13
                  31
                 93
                      93
                     x
    HCF of 31 and 465
    31) 465 ( 15
        31
        155
        155
        x





Ans .

4


  1. Explanation :

    Minimum number of rows = Maximum number of fruits in each row

    HCF of 24, 36 and 60 = 12
    Minimum number of rows 24 + 36 + 60
    12  12  12
    2 + 3 + 5 = 10





Ans .

4


  1. Explanation :

    Using Rule 7, Required number = HCF of 2300 – 32 = 2268 and 3500 – 56 = 3444
         2268 ) 3444 ( 1
              2268
             1176 ) 2268 ( 1
                  1176
                 1092 ) 1176 ( 1
                      1092
                     84 ) 1092 ( 13
                      84
                     252
                      252
                     x
    HCF = 84





Ans .

3


  1. Explanation :

    HCF of numbers = 12 Let the numbers be 12x and 12y
    where x and y are co–prime.
    According to the question,
    12x × 12y = 2160
    xy =
    Required numbers = 12 × 3 = 36 and 12 × 5 = 60





Ans .

2


  1. Explanation :

    Required number = HCF of 390, 495 and 300 = 15
         390 ) 495 ( 1
              390
             105 ) 390 ( 3
                  315
                 75 ) 105 ( 1
                      75
                     30 ) 75 ( 1
                      60
                     15 ) 30 ( 2
                      30
                     x
    HCF of 15 and 300 = 15





Ans .

4


  1. Explanation :

    First of all we find HCF of 391 and 323.
         323 ) 391 ( 1
              323
             68 ) 323 ( 4
                  272
                 51 ) 68 ( 1
                      51
                     17 ) 51 ( 3
                      51
                     x
    Number of classes = 17





Ans .

3


  1. Explanation :

    Maximum length of each piece = HCF of 1.5 metre and 1.2 metre = 0.3 metre
                 12 ) 15 ( 1
                      12
                     3 ) 12 ( 4
                      12
                     x
    HCF of 1.5 and 1.2 metre = 0.3 metre