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H.C.F. & L.C.M. (Solved Problems)

Practice material


Q1. Find the L.C.M. of 16, 24, 36 and 54.

  1. 448

  2. 432

  3. 480

  4. 464

Sol : Using Prime Factorization Method 

16 = 2⁴

24 = 2³ x 3¹

36 = 2² x 3²

54 = 2¹ x 3³

L.C.M. of 16, 24, 36 and 54 = 2⁴ x 3³ = 432

                                                             2      8     16             10

Q 2. Find the H.C.F. and L.C.M. of   ── , ── , ── , and  ─── .

                                                            3      9      18            27

                                                      H.C.F. of 2, 8, 16 and 10         2

Sol : H.C.F. of given numbers = ────────────────── = ── .

                                                      L.C.M. of 3, 9, 18 and 27       54

                                                       L.C.M.of 2, 8, 16 and 10          80

        L.C.M.  of given numbers = ────────────────── = ─── .

                                                       H.C.F. of 3, 9, 18 and 27           3


Q3. Find the H.C.F. and L.C.M. of 0.63 ,1.05  and 2.1  .

Sol :  Convert the decimal places into the non-decimal places by multiplying the digits by 100.

Now, H.C.F. of 63 , 105 and 210 = 21

And L.C.M. of 63 , 105 and 210 = 630

But it is required to calculate the H.C.F. and L.C.M. of 0.63 ,1.05  and 2.1.

Hence , Divide the  calculated value H.C.F. and L.C.M. by 100.

H.C.F. of 0.63 ,1.05  and 2.1 =  0.21

L.C.M. of 0.63 ,1.05  and 2.1 = 6.30


Q4. Two numbers are in the ratio of 3 : 4. Their L.C.M. is 144. Find the numbers.

Sol :  a = 3 and b = 4                                                 {Co - Prime Pairs}

         Numbers = ha and hb = 3h and 4h respectively.                   {H.C.F. =  h}

        We clearly know that

                            L.C.M. = h * a * b

                               144   = h * 3 * 4

                                     h = 12

          Hence the numbers are 12 * 3 and 12 * 4 i.e. 36 and 48.

Q5. The H.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the numbers is 77, find the other number.

Sol : We know that,                   

                    H.C.F. x L.C.M. = A x B

                         11 x 693       = 77 x B

                                      B     = 99

       Hence the other number is 99.


Q6. The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is :

Sol : Smaller Number , A = ha = 23 x 13 = 299

         Larger Number ,  B = hb = 23 x 14 = 322


Q 7. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is : 

  1. 9000

  2. 9400

  3. 9600

  4. 9800

Sol : Calculate  L.C.M. of 15, 25, 40 and 75 

Factors of 15 = 3¹ x 5¹

Factors of 25 = 5²

Factors of 40 = 2³ x 5¹

Factors of 75 = 3¹ x 5²

 L.C.M. of 15, 25, 40 and 75  = 2³ x 3¹ x 5² = 600

On dividing the largest four digit number (9999) by 600 , the remainder is 399.

Hence the required number is 9999 - 399 = 9600


Q 8. Three numbers are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is :

  1. 40

  2. 80

  3. 120

  4. 200

Sol : we know that ,

                         L.C.M. = H.C.F. x a x b x c

                         2400   = H.C.F. x 3 x 4 x 5

                         H.C.F. = 40

         

Q 9. The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is :

  1. 74

  2. 94

  3. 184

  4. 364

Sol : Calculate L.C.M. of 6, 9, 15 and 18

Factors of 6 = 2¹ x 3¹

Factors of 9 = 3²

Factors of 15 = 3¹ x 5¹

Factors of 18 = 2¹ x 3²

L.C.M. of 6, 9, 15 and 18 =  2¹ x 3² x 5¹ = 90

Let the required number be (90k + 4) must be completely divisible by 7.

At the least value of k = 4, number is divisible by 7

Required Number =  90 x 4 + 4  = 364 is completely divisible by 7. 


Q 10. The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is

  1. 1677

  2. 1683

  3. 2523

  4. 3363

Sol : Calculate L.C.M. of 5, 6, 7 and 8

Factors of 5 = 5¹

Factors of 6 = 2¹ x 3¹

Factors of 7 = 7¹

Factors of 8 = 2³

L.C.M. of 5, 6, 7 and 8 = 2³ x 3¹ x 5¹ x 7¹ = 840

Let the required number be = 840k + 3 ,must be divisible by 9

At the least value of k = 2, number is divisible by 9

Required number = 840 x 2 + 3 = 1683


Q 11. What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

  1. 196

  2. 630

  3. 1260

  4. 2520

Sol :  Calculate the L.C.M. of 12, 18, 21 and 30

Factors of 12 = 2² x 3¹

Factors of 18 = 2¹ x 3²

Factors of 21 = 3¹ x  7¹

Factors of 30  = 2¹ x 3¹ x 5¹ 

L.C.M. of 12, 18, 21 and 30 = 2² x 3² x 5¹ x 7¹ = 1260

Hence, the required number is 1260 / 2 = 630


Q 12. The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is :

  1. 1008

  2. 1015

  3. 1022

  4. 1035

Sol : Calculate the  L.C.M. of 12, 16, 18, 21 and 28 

 Factors of 12 = 2² x 3¹

Factors of 16 = 2⁴

Factors of 18 = 2¹ x 3²

Factors of 21 = 3¹ x  7¹

Factors of 28 = 2² x  7¹

L.C.M.of 12, 16, 18, 21 and 28 =  2⁴ x 3² x 7¹ = 1008

Hence, Required Number = 1008 + 7 = 1015


Q 13. 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 again at the starting point ?

  1. 26 minutes and 18 seconds

  2. 42 minutes and 36 seconds

  3. 45 minutes

  4. 46 minutes and 12 seconds

Sol : Calculate the  L.C.M. of 252 , 308 and 198 seconds.

       252 = 2² x 3² x 7¹

       308 = 2² x 7¹ x 11¹

       198 = 2¹ x 3² x 11¹

       L.C.M. of 252 , 308 and 198 =   2² x 3² x 7¹ x 11¹  = 2772    

      Hence, A, B and C will meet each other at 2772 seconds = 46 minutes and 12 seconds


Q 14. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

  1. 123

  2. 127

  3. 235

  4. 307

Sol : Calculate the H.C.F. of (1657 - 6)= 1651 and (2037 - 5) = 2032.

        1651 = 13 x 127

         2032 = 16 x 127

        Hence the greatest number which divided 1657 and 2037 leaving  remainders 6 and 5 respectively  is 127.


Q 15. 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 ?

  1. 4

  2. 10

  3. 15

  4. 16ٰ

Sol : Calculate the L.C.M. of 2, 4, 6, 8, 10 and 12 Seconds

       2 = 2¹

       4 = 2² 

       6 = 2¹ x 3¹ 

       8 = 2³

      10 = 2¹ x 5¹

      12 = 2² x 3¹

      L.C.M. of 2, 4, 6, 8, 10 and 12 = 2³ x 3¹ x 5¹ = 120 Seconds = 2 Minutes

      After every 2 Minutes all six bells toll together.

      In 30 Minutes, Bells will toll together 15 Times.

     Total (15 + 1) = 16 Times , Bells will toll together. 

      (1 Time Bell toll at the time of Commence.)


Q 16. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is :

  1. 4

  2. 5

  3. 6

  4. 8

Sol : N = H.C.F. of (6905 - 4665), (4665 -1305) and (6905 - 1305) 

            =  H.C.F. of 3360 , 2240 and 5600

        3360 = 2⁵ x 3¹ x 5¹ x 7¹

        2240 = 2⁶ x 5¹ x 7¹

       

      5600 = 2⁵ x 5² x 7¹

      H.C.F. of 3360, 2240 and 5600 = 2⁵ x 5¹ x 7¹ = 1120

     Sum Of Digits  “N” = 1 + 1 + 2 + 0 = 4


Q 17. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is :

  1. 15 cm

  2. 25 cm

  3. 35 cm

  4. 42 cm

Sol : Calculate the H.C.F. of 700 cm, 385 cm, 1295 cm

      700 = 2² x 5² x 7¹

      385 = 5¹ x 7¹ x 11¹

     1295 =  5¹ x 7¹ x 37¹

     Hence,  H.C.F. of 700 cm, 385 cm, 1295 cm = 5¹ x 7¹ = 35 cm


Q 18. 252 can be expressed as a product of primes as :

  1. 2 x 2 x 3 x 3 x 7

  2. 2 x 2 x 2 x 3 x 7

  3. 3 x 3 x 3 x 3 x 7

  4. 2 x 3 x 3 x 3 x 7 

Sol : Prime Factorization of 252  = 2² x 3² x 7¹

  

Q 19. Which of the following has the highest number of divisors ?

  1. 99

  2. 101

  3. 176

  4. 182

Sol : No. of divisors of 99 = 3² x 11¹ = (2+1) x (1+1) = 3 x 2 = 6

        No. of divisors of 101 = 101¹ = (1 + 1) = 2

        No. of divisors of 176 = 2⁴ x 11¹ = (4+1) x (1 +1) = 5 x 2 =10

        No. of divisors of 182 = 2¹ x 7¹ x 13¹ = (1 + 1) x (1 + 1) x (1 + 1) = 2 x 2 x 2 = 8

        Hence, 176 has the highest number of divisors.


Q 20.  A rectangular floor of dimension 225m x 175m is to be paved by identical square tiles. Find the size of each tile and the number of tiles required.

Sol : Calculate the H.C.F. of 225 and 175

       225 = 3² x 5²

       175 = 5² x 7¹

       H.C.F. of 225 and 175 = 5² = 25

      So,  Size of each tile = 25m x 25m 

       No. of Tiles = Area of the floor / Area of each tile

                            =( 225 x 175) / (25 x 25)

                            = 9 x 7

Hence, No. of tiles required is 63 tiles.



Ask me in the comment section, if you face any problem while solving questions of H.C.F. and L.C.M. . Visit next page for more concepts and problems.

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