Topic outline

  • Playing With Numbers (RD Sharma –Chapter 2-Solution of Exercise 2.10)

    • RD Sharma - Exercise 2.10 - Solutions Page

      Playing With Numbers

      Exercise 2.10 (Some Application of L.C.M)

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

      Solution:

      We know that the smallest number divisible by 24, 36 and 54 is their L.C.M. Therefore, the required number must be 5 more than their L.C.M.

       

      2

      24, 36, 54

      2

      12, 18, 27

      3

       6,   9,  27

      3

       2,   3,  9

      3

       2,   1,  3

      2

       2,   1,  1

       

       1,   1,  1

       

      L.C.M = 2x2x2x3x3x3 = 216

      Hence, the required number = 216+5 = 221

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

      Solution:

      We know that the smallest number divisible by 33 and 39 is their L.C.M. Therefore, the required number must be 5 more than their L.C.M.

       

      3

      33, 39

      11

      11, 13

      13

       1, 13

       

       1,  1

       

      L.C.M = 3x11x13 = 429

      Hence, the required number = 429+5 = 434

       

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

      Solution:

      We know that the least number divisible by 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 is their L.C.M.

       

      1

      1, 2, 3, 4, 5, 6, 7, 8, 9, 10

      2

      1, 2, 3, 4, 5, 6, 7, 8, 9, 10

      2

      1, 2, 3, 2, 5, 3, 7, 4, 9, 5

      2

      1, 1, 3, 1, 5, 3, 7, 2, 9, 5

      3

      1, 1, 3, 1, 5, 3, 7, 1, 9, 5

      3

      1, 1, 1, 1, 5, 1, 7, 1, 3, 5

      5

      1, 1, 1, 1, 5, 1, 7, 1, 1, 5

      7

      1, 1, 1, 1, 1, 1, 7, 1, 1, 1

       

      1, 1, 1, 1, 1, 1, 1, 1, 1, 1

       

      L.C.M = 1x2x2x2x3x3x5x7 = 2520

       


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

      Solution:

      We know that the smallest number divisible by 35, 56 and 91 is their L.C.M. Therefore, the required number must be 7 more than their L.C.M.

       

      7

      35, 56, 91

      2

       5,   8,  13

      2

       5,   4,  13

      2

       5,   2,  13

      5

       5,   1,  13

      13

       1,   1,  13

       

       1,   1,  1

       

      L.C.M = 2x2x2x5x7x13 = 3640

      Hence, the required number = 3640+7 = 3647

        

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

      Solution:

      Section A – 32 Students

      Section B – 36 Students

      We know that the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B is L.C.M of 32 and 36.

       

      2

      32, 36

      2

      16,  18

      2

       8,    9

      2

       4,    9

      2

       2,    9

      3

       1,    9

      3

       1,    3

       

       1,    1

       

      L.C.M = 2x2x2x2x2x3x3 = 288

      Hence, the required number = 288 books

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

      Solution:

      We know that the minimum distance each should walk so that he can cover the distance in complete steps is the L.C.M of 80cm, 85cm and 90cm.

      5

      80, 85, 90

      2

      16, 17, 18

      2

       8,  17,  9

      2

       4,  17,  9

      2

       2,  17,  9

      3

       1,  17,  9

      3

       1,  17,  3

      17

       1,  17,  1

       

       1,  17,  1

       

      L.C.M = 2x2x2x2x3x3x5x17 = 12240 cm or 122m 40 cm

        

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

      Solution:

      We first compute L.C.M of 8, 15 and 21.

       

      2

      8, 15, 21

      2

      4, 15, 21

      2

      2, 15, 21

      3

      1, 15, 21

      5

      1,  5,  7

      7

      1,  1,  7

       

      1,  1,  1

       

      L.C.M = 2x2x2x3x5x7 = 840

      840 x 119 = 99960

      840 x 120 = 100800 (nearest to 100000 but greater than 100000)

      840 x 121 = 101640

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

      Solution:

      We first compute L.C.M of 6 and 8.

       

      2

      6, 8

      2

      3, 4

      2

      3, 2

      3

      3, 1

       

      1, 1

       

      L.C.M = 2x2x2x3 = 24

      Therefore, the bus stop at the 24th block of flats.

        

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

      Solution:

      We first compute L.C.M 220m and 300m

       

      2

      220, 300

      2

      110, 150

      5

       55,   75

      5

       11,   15

      3

       11,    3

      11

       11,    1

       

       1,      1

       

      L.C.M = 2x2x3x5x5x11 = 3300m

      Next heap which lies at the foot of a pole will at the distance of 3300m.

       

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

      Solution:

      L.C.M of 28 and 32

       

      2

      28, 32

      2

      14, 16

      2

       7,   8

      2

       7,   4

      2

       7,   2

      7

       7,   1

       

       1,   1

       

      L.C.M = 2x2x2x2x2x7 = 224

      Smallest number which leaves remainder 8 and 12 when divided by 28 and 32 = LCM –(8+12) = 224 - 20 = 204