• Problems on H.C.F and L.C.M Part-3

    1. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together?
    A.8
    B.11
    C.13
    D.16
    E.None of these

     

    2. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:
    A.68
    B.98
    C.180
    D.364
    E.None of these

     

    3. The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively, is:
    A.11115
    B.15110
    C.15130
    D.15310
    E.None of these

     

    4. The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:
    A.269
    B.275
    C.308
    D.310
    E.None of these

     

    5. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point?
    A.15 minutes 15 seconds
    B.42 minutes 30 seconds
    C.42 minutes
    D.46 minutes 12 seconds
    E.None of these

     

    6. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
    A.30
    B.22
    C.40
    D.60
    E.None of these

     

    7. The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
    A.534
    B.486
    C.544
    D.548
    E.None of these

     

    8. The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
    A.124
    B.100
    C.111
    D.175
    E.None of these

     

    9. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:
    A.1
    B.2
    C.3
    D.5
    E.None of these

     

    10. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
    A.10
    B.14
    C.23
    D.30
    E.None of these

     

    11. If the product of  two numbers is 2496 and HCF is 8,then the ratio of HCF and LCM is
    A)1:32
    B)39:1
    C)1:39
    D)4:63
    E)None of these

     

    12. The greatest possible length  which can be used to measure exactly the lengths 1m 92cm,3m 84cm ,23m 4cm
    A)23
    B)32
    C)36
    D)34
    E)None of these

     

    13. HCF of 4/3, 8/6, 36/63 and 20/42
    A)4/126
    B)4/8
    C)4/36
    D)4/42
    E)None of these

     

     

    14. Find the LCM of 3/8, 9/32, 33/48, 18/72
    A)3/8
    B)8/33
    C)198/8
    D)8/3
    E)None of these

     

    15. The HCF of 2511 and 3402 is
    A)31
    B)42
    C)76
    D)81
    E)None of these

     

    16. If m and n are two whole numbers and if m^n= 49. Find n^m, given that n ≠ 1
    A)118
    B)94
    C)561
    D)128
    E)None of these

     

    17. What will be the least number which when doubled will be exactly divisible by 12,18,21 and 30?
    A)630
    B)360
    C)603
    D)306
    E)None of these

     

    18. HCF and LCM of two numbers are 11 and 385 .If one number lies between 75 and 125 then that number is
    A)123
    B)73
    C)77
    D)154
    E)None of these

     

    19. If the L.C.M of x and y is z, their H.C.F is
    A)xy/z
    B)xyz
    C)(x + y)/z
    D)z/xy
    E)None of these

     

    20. A rectangular courtyard 140cm long 525cm wide is to be paved exactly with square  tiles, all of the same size. what is the largest size of the tile which could be used for the purpose?
    A)64cm
    B)35cm
    C)21cm
    D)28cm
    E)None of these

     

    21. Two containers contain 50 and 125 litres of water respectively. Find the maximum capacity of a container which can measure the water in each container an exact number of times(in litres)
    A)25
    B)11
    C)12
    D)15
    E)None of these

     

    22. Two baskets contain 183 and 242 Apples respectively, which are distributed in equal number among children. Find the largest number of apples that can be given, so that 3 apples are left over from the first basket and 2 from the second.
    A)45
    B)40
    C)60
    D)56
    E)None of these

     

    23. A person has 3 bars whose length are 12,16,24m respectively. He want to cuts the longest possible pieces, all of the same length from each of the 3 bars, what is the length of the each piece, if he is cut without any wastage
    A)12m
    B)20m
    C)6m
    D)4m
    E)None of these

     

    24. Four bells commence tolling together and toll at the intervals of 3,9,12,15 seconds resp. In 60 minutes how many times they will toll together.
    A)20
    B)21
    C)24
    D)30
    E)None of these

     

    25. In a seminar the number of participants in Technology, Economics and Science are 150, 90 and 180 respectively. Find the minimum number of rooms required, where in each room the same number of participants are to be seated and all of them being in the same subject.
    A)27
    B)32
    C)30
    D)25
    E)None of these

     

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  • Answers and Explanations Down

    1. Answer – D (16)
    Explanation – L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
    So, the bells will toll together after every 120 seconds, 
    i.e, 2 minutes.In 30 minutes, 
    they will toll together 30/2 + 1 = 16

     

    2. Answer – D (364)
    Explanation – L.C.M. of 6, 9, 15 and 18 is 90.
    Let required number be 90k + 4, which is multiple of 7.
    Least value of k for which (90k + 4) is divisible by 7 is k = 4.
    Required number = (90 x 4) + 4 = 364.

     

    3. Answer – B (15110)
    Explanation – Here (48 – 38) = 10, (60 – 50) = 10, (72 – 62) = 10,

    (108 – 98) = 10 & (140 – 130) = 10.
    Required number = (L.C.M. of 48, 60, 72, 108, 140) – 10
    = 15120 – 10 = 15110

     

    4. Answer
    Answer – C (308)
    Explanation –
    Other number =[11 x 7700]/275 = 308

     

    5. Answer
    Answer – D (46 minutes 12 seconds)
    Explanation – L.C.M. of 252, 308 and 198 = 2772.
    So, A, B and C will again meet at the starting point in 2772 
    see i.e., 46 min. 12 sec

     

    6. Answer
    Answer – C (40)
    Explanation – Let the numbers be 2x and 3x.
    Then, their L.C.M. = 6x.
    So, 6x = 48 or x = 8.
    The numbers are 16 and 24.
    Hence, required sum = (16 + 24) = 40.

     

    7. Answer
    Answer – D (548)
    Explanation – Required number = (L.C.M. of 12, 15, 20, 54) + 8
    = 540 + 8
    = 548.

     

    8. Answer – C (111)
    Explanation – Let the numbers be 37a and 37b.
    Then, 37a x 37b = 4107
    ab = 3.
    Now, co-primes with product 3 are (1, 3).
    So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).
    Greater number = 111.

     

    9. Answer
    Answer – B (2)
    Explanation – Let the numbers 13a and 13b.
    Then, 13a x 13b = 2028
    ab = 12.
    Now, the co-primes with product 12 are (1, 12) and (3, 4).

     

    10. Answer
    Answer – C (23)
    Explanation – L.C.M. of 5, 6, 4 and 3 = 60. 
    On dividing 2497 by 60, the remainder is 37. 
    Number to be added = (60 – 37) = 23

     

    11. Answer
    Answer – C) 1:39
    Explanation:
    HCF = 8
    LCM = Product/HCF
    Product of the 2 number = 2496
    HCF:LCM = 8 : (2496/8) => 8 : 312 => 1: 39

     

    12. Answer
    Answer -B) 32
    Explanation:
    192 = 4^2×2^2×3
    384 = 4^2×2^2×6
    2304 = 4^2×2×6^2
    HCF = 4^2×2 = 16×2 = 32

    13. Answer
    Answer : A. 4/126
    Explanation:
    HCF of numerator(4,8,36,20) = 4
    LCM of denominator(3,6,63,42) = 126

     

    14. Answer
    Answer -C) 198/8
    Explanation :
    LCM of numerator(3,9,33,18) = 198
    HCF of denominator(8,32,48,72) = 8

     

    15. Answer
    Answer -D) 81
    Explanation :
    2511 = 81×31
    3402 = 81×42
    Hence HCF is 81

     

    16. Answer
    Answer -D) 128
    Explanation :
    49 = 7^2 = m^n
    n^m = 2^7 = 128

     

    17. Answer
    Answer -A) 630
    Explanation :
    LCM of 12,18,21 and 30 = 1260
    Doubled (divide by 2) = 1260/2 = 630

     

    18. Answer
    Answer – C) 77
    Explanation :
    Product of the two number = 11a×11b = 4235
    11a×11ab = 4235/121 = 35
    35 = 7 × 5 (co prime)
    77 × 11 = 77

     

    19. Answer
    Answer – A) xy/z
    Explanation :
    HCF = Product of the 2 number/LCM = XY/Z

     

    20. Answer & Solution
    Answer – B)35cm
    Solution:
    30 cm = 4 × 5 × 7
    525 cm = 5 × 5 × 3 × 7
    Hence common factors are 5 and 7
    Hence LCM = 5 × 7 = 35

     

    21. Answer & Solution
    Answer – A)25
    Solution:
    HCF of 50 and 125 = 25

     

    22. Answer & Solution
    Answer – C)60
    Solution:
    183-3 = 180
    142-2 = 240
    HCF of 240, 180 = 60

     

    23. Answer & Solution
    Answer – D)4m
    Solution:
    HCF of 12, 16, 24 = 4m

     

    24. Answer & Solution
    Answer – B)21
    Solution:
    LCM of 3, 9, 12, 15 = 180s = 3m
    In 60m = 60/3 = 20=> 20+1 = 21

     

    25. Answer & Solution
    Answer – C)30
    Solution:
    HCF of 150, 90 and 180 = 30
    No of participants can be seated in each room = 30