## Topic outline

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

• 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