Topic outline

  • Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.8)

    • RD Sharma - Exercise 2.8 - Solutions Page

      Playing With Numbers

      Exercise 2.8 (Some Application of H.C.F)

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

      Solution:

      Clearly, the required number is the H.C.F of the numbers.

      615-6 = 609, 963-6 = 957

      Now, first we find the H.C.F of 609 and 957.

       

      The H.C.F of 609 and 957 is 87. Therefore, required number is 87.

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

      Solution:

      Clearly, the required number is the H.C.F of the numbers.

      285-9 = 276, 1249-7 = 1242

      Now, first we find the H.C.F of 285 and 1242.

              

      The H.C.F of 276 and 1242 is 138. Therefore, required number is 138.

       

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

      Solution:

      Clearly, the required number is the H.C.F of the numbers.

      626-1 = 625, 3127-2 = 3125, 15628-3 = 15625

      Now, first we find the H.C.F of 625, 3125 and 15628.

      Let’s take 625 and 3125 for the H.C.F

            

      The H.C.F of 625, 3125 and 15628 is 625. Therefore, required number is 625.

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

      Solution:

      Length = 8m 25cm = 825 cm

      Breadth = 6m 75cm = 675 cm

      Height = 4m 50 cm = 450 cm

      Now we find the H.C.F of 825, 675 and 450

                         

      5

      825, 675, 450

      5

      165, 135,  90

      3

        33,   27, 18

       

        11,    9,   6

       

      Thus, H.C.F. of 825, 675 and 450 = 5x5x3 = 75

      Therefore, longest rod which can measure the three dimensions of the room exactly= 75 cm

              

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

      Solution:

      Length = 20m 16cm = 2016cm

      Breadth = 15m 60cm = 1560cm

      Now we find the H.C.F of 2016 and 1560

       

      2

      2016, 1560

      2

      1008,  780

      2

       504,   390

      3

       252,   195

       

        84,     65

       

      H.C.F 2x2x2x3 = 24 cm

       

      Therefore, least number of stones = Area of courtyard/Area of square stone

                                                        = 2016 x 1560 / 24 x 24

                                                        = 5460

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

      Solution:

      Length = 7m = 700 cm

      Breadth = 3m 85cm = 385 cm

      Height = 12m 95 cm = 1295 cm

      Now we find the H.C.F of 700, 385 and 1295

       

      5

      700, 385, 1295

      7

      140,  77,  259

       

       20,   11,  37

       

      H.C.F = 5X7 = 35cm

      Longest tape which can be used to measure exactly the lengths 7m, 3m 85cm and 12m 95cm is 35cm.

       

      Question 7 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      No. of goats = 105

      No. of donkeys = 140

      No. of cows = 175

      Now we find the H.C.F of 105, 140 and 175.

       

      5

      105, 140, 175

      7

        21,  28,  35

       

         3,    4,    5

       

      H.C.F = 5X7 = 35

      Therefore, 35 animals went in each trip.

      Question 8 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      No. of chocolate of brand 1=24

      No. of chocolate of brand 2 = 15

      The required number is the L.C.M of the numbers.

       

      3

      24, 15

      5

       8,  5

      8

       8,  1

       

       1,  1

       

      L.C.M = 3X5X8 = 120

      No. of boxes of kind I = 120/24 = 5

      No. of boxes of kind II = 120/15 = 8

       

      Question 9 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      No. of pencils in each packet = 24

      No. of crayons in each packet = 32

      The required number is the L.C.M of the numbers.

       

      2

      24, 32

      2

      12, 16

      2

       6,   8

      2

       3,   4

      2

       3,   2

      3

       3,   1

       

       1,   1

       

      L.C.M = 2x2x2x2x2x3 = 96 

      No. of pencils packets = 96/24 = 4

      No. of crayons packets = 96/32 = 3

      Question 10 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      i. 161/207

       23x7/23x9 = 7/9

       

      ii. 296/481

       37x8/37x9 = 8/9

       

      Question 11 (Refer Book - Mathematics Class VI R.D. Sharma)

      Solution:

      I kind of oil =120 l

      II kind of oil = 180 l

      III kind of oil = 240 l

      The required number is the H.C.F of the numbers.

       

      2

      120, 180, 240

      2

       60,   90,  120

      3

       30,   45,   60

      5

       10,   15,   20

       

        2,     3,    4

       

      H.C.F = 2X2X3X5 = 60 l

      60 l is the capacity of each tin.