Topic outline

  • Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.11)

    • Page icon
      RD Sharma - Exercise 2.11 - Solutions Page

      Playing With Numbers

      Exercise 2.11 (Some Properties of H.C.F and L.C.M of Given Numbers)

      Question 1 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      i. 25, 65

      We compute the H.C.F and L.C.M of 25 and 65.

       

      5

      25, 65

       

       5,  13

       

      H.C.F = 5  and  L.C.M = 5x5x13 = 325

      Now, H.C.F x L.C.M = 5x325 = 1625

      And, Product of the numbers = 25x65 = 1625

      Therefore, Product of the numbers = Product of their H.C.F and L.C.M



      ii. 117, 221

       

      13

      117, 221

       

       9,    17

       

      H.C.F = 13  and  L.C.M = 9x13x17 = 1989

      Now, H.C.F x L.C.M = 13x1989 = 25857

      And, Product of the numbers = 117x221 = 25857

      Therefore, Product of the numbers = Product of their H.C.F and L.C.M

       

      iii. 35, 40

       

      5

      35, 40

       

       7,  8

       

      H.C.F = 5  and  L.C.M = 5x7x8 = 280

      Now, H.C.F x L.C.M = 5x280 = 1400

      And, Product of the numbers = 35x40 = 1400

      Therefore, Product of the numbers = Product of their H.C.F and L.C.M

        

      iv. 87, 145

       

      29

      87, 145

       

       3,   5

       

      H.C.F = 29  and  L.C.M = 3x5x29 = 435

      Now, H.C.F x L.C.M = 29x435 = 12615

      And, Product of the numbers = 87x145 = 12615

      Therefore, Product of the numbers = Product of their H.C.F and L.C.M 

       

      v. 490, 1155

       

      5

      490, 1155

       

       98,   231

       

      H.C.F = 5  and  L.C.M = 5x98x231 = 113190

      Now, H.C.F x L.C.M = 5x113190 = 565950

      And, Product of the numbers = 490x1155 = 565950

      Therefore, Product of the numbers = Product of their H.C.F and L.C.M

       


      Question 2 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      i. 117, 221

       

      13

      117, 221

       

       9,    17

       

      H.C.F = 13  and  L.C.M = 9x13x17 = 1989

       

      ii. 234, 572

       

      26

      234, 572

       

       9,    22

       

      H.C.F = 26  and  L.C.M = 9x22x26 = 5148

        

      iii. 145, 232

       

      29

      145, 232

       

        5,    8

       

      H.C.F = 29  and  L.C.M = 5x8x29 = 1160

       

      iv. 861, 1353

       

      123

      861, 1353

       

       7,      11

       

      H.C.F = 123  and  L.C.M = 7x11x123 = 9471

       

      Question 3 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      H.C.F of two numbers = 6

      L.C.M of two numbers = 180

      One of the number = 30

      Other number = H.C.F x L.C.M / One of the number

                           = 6x180 / 30

                           = 1080/30

                           = 36

       

      Question 4 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      H.C.F of two numbers = 16

      Product of the two numbers = 3072

      L.C.M = Product of two numbers/ H.C.F

               = 3072/16

               = 192

       


      Question 5 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      H.C.F of two numbers = 145

      L.C.M of two numbers = 2175

      One of the number = 725

      Other number = H.C.F x L.C.M / One of the number

                           = 145x2175 / 725

                           = 435

                     

      Question 6 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      We know that the H.C.F of given numbers must divide their L.C.M completely. But, 16 do not divide 380 completely.

      So, there can be no two numbers with 16 as their H.C.F and 380 as their L.C.M