Problems on LCM and HCF

☛ Read Basics of LCM and HCF Here, Before starting Practice 


1. Find the least number which when increased by 4 is exactly divisible by 8, 16, 24, 30 and 32 ?
a)   480
b)   484
c)   476
d)   472
e)   None of these

2. What is the greatest number of five digits which when 3769 is added to it will be exactly divisible by 5, 6 , 10, 12, 15 and 18 ?
a)   4309
b)   99459
c)   100539
d)   99911
e)   None of These

3. Find the minimum number of square tiles required to pave the floor of a room of 2m 50cm long and 1m 50cm broad ?
a)    50
b)   750
c)   45
d)   15
e)   None of these

4. Five bells toll together at the intervals of 5, 6, 8, 12 and 20 seconds respectively. Find the number of times they toll together in one hour's time (Inclusive of the toll at the beginning)
a)   120
b)   31
c)   30
d)   5
e)   None of These

5. A milk man has three different kinds of milk 493liters, 551 liters and 435 liters. Find the minimum number of equal size containers required to store all the milk without mixing.
a)   29
b)   51
c)   58
d)   49
e)    None of these

6. The circumference of the front and back wheels of a vehicle are 6 3/14 m and 8 1/18 m respectively. At any given moment, a chalk mark is put on the point of contact of each wheel with the ground. Find the distance traveled by the vehicle so that both the chalk marks are again on the ground at the same time
a)   217.5 m
b)   435 m
c)    412m
d)   419m
e)   None of these

7. The LCM of two numbers is 28 times of their HCF. The sum of their LCM and HCF is 1740. If one of the numbers is 420, the other number is
a)   150
b)   225
c)   180
d)   240
e)   None of these

8. Two persons A and B walk around a circular track whose radius is 1.4 km. A walks at a speed of 176 meters per minute while B walks at a speed of 110 meters per minute. if they both start at the same time, from the same point and walk in the same direction, at what interval of time would they both be at the same starting point again? (in Hours)
a)   6 2/3
b)   2 1/3
c)   5 1/4
d)   3 2/3
e) None of these

9. Find the least number which when divided by 8, 9, 15, 24, 32 and 36 leaves remainders 3, 4, 10, 19, 27 and 31 respectively?
a)   2880
b)   2885
c)   2974
d)   2875
e)   None of these

10. Find the greatest number which when divide 357, 192 and 252 leaves same remainder in each case
a)   45
b)   1
c)   15
d)   Cant be determinde
e)    None of these




Solutions :


1.  LCM of 8, 16, 24, 30 and 32 is 480
     So, Required number is 480 - 4  =  476

2.  LCM of 5, 6, 10, 12 and 18 is 540
      On dividing  (99999 + 3769)  by 540, the remainder is  88
      So, the required number is 99999 - 88 = 99911

3.  HCF of 250 cm and 150 cm is 50 cm, which is the side of the tile

     So, the required number of tiles  =  (250 X 150) / (50X50)  =  15

4.  Time after which all the bells tol together is the LCM of 5, 6, 8, 12 and 20.
       i.e., 120 seconds = 20   minutes
      The number of times they toll together in one hour =  60/2  = 30 + 1 (beginning tone)
         So, the answer is 31

5.  As minimum number of containers are required, the size of the container should be maximum and the size is also equal. so size of the container will be HCF of 493, 551 and 435 i.e., 29
     So, required number of containers is =  (493+551+435) / 29  =  51

6.  The required distance is the LCM of 6 3/14  and 8 1/18
      LCM of 6 3/14 and 8 1/18 =  LCM (87/14, 145/18)
      =>  LCM(87,145) / HCF (14,18)  =  435/2  =  217.5m
7.  LCM =  28 HCF
      LCM + HCF = 1740
       =>  28  HCF + HCF = 1740
        HCF =  1740/29  =  60 and LCM = 28X60
      if a and b are two numbers, then
       LCM of (a & b) X HCF of (a & b) = aXb
      28X60X60 = x X 420  = >  x = 240
         So, the other number is 240


8.   Circumference of the track is 2πr
        = 2 X (22/7) X 1400m = 8800 m


         Time taken by A to complete one round
           =  (8800m) / (176m/min)  =  (8800X60)/(176) =  3000 sec

        Time taken by B to complete one round

          = (8800) / (110m/min)  =  (8800X60) / (110)  =  4800 sec

         Time they meet together at the starting point is LCM of 3000 and 4800 sec

           i.e., 24000 sec   =  6 2/3 hours

            So, they meet at the starting point after 6 2/3 hours

9.   We can observe that the difference between the numbers and their remainders is same 

         i.e., 8-3  =  9-4  =  15-10  =  24-19  =  32-27  =  36-31  =  5

        So, required answer is 

                                                LCM (8, 9, 15, 24, 32, 36) - 5

                    = > 2880 - 5 = 2875

10.   Required answer is 

               HCF (357-192, 192-252, 357-252)

                 = > HCF (165, 60, 105) = 15