Selina Solutions Concise Mathematics Class 6 Chapter 8: H.C.F and L.C.M Exercise 8(C) give accurate answers in finding L.C.M by using various methods mentioned in this exercise. These solutions are designed by a set of subject matter experts to help students improve their problem-solving skills. Each and every step of Selina Solutions helps students understand in-depth and solve the textbook questions effortlessly. Students, who go through these solutions, will surely score good marks in exams. Selina Solutions Concise Mathematics Class 6 Chapter 8 H.C.F and L.C.M Exercise 8(C) PDF can be downloaded by students, from the links available below

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### Access Selina Solutions Concise Mathematics Class 6 Chapter 8: H.C.F and L.C.M Exercise 8(C)

#### Exercise 8(C) page no: 71

**1. Using the common multiple method, find the L.C.M. of the following:**

**(i) 8, 12 and 24**

**(ii) 10, 15 and 20**

**(iii) 3, 6, 9 and 12**

**Solution:**

(i) 8, 12 and 24

We get,

L.C.M = 4 Ã— 3 Ã— 2

= 24

Hence, L.C.M. of 8, 12 and 24 = 24

(ii) 10, 15 and 20

We get,

L.C.M = 2 Ã— 2 Ã— 3 Ã— 5

= 60

Hence, L.C.M. of 10, 15 and 20 = 60

(iii) 3, 6, 9 and 12

We get,

L.C.M. = 2 Ã— 2 Ã— 3 Ã— 3

= 36

Hence, L.C.M. of 3, 6, 9 and 12 = 36

**2. Find the L.C.M. of each of the following groups of numbers, using (i) the prime factor method and (ii) the common division method:**

**(i) 18, 24 and 96**

**(ii) 100, 150 and 200**

**(iii) 14, 21 and 98**

**(iv) 22, 121 and 33**

**(v) 34, 85 and 51**

**Solution:**

(i) 18, 24 and 96

By using prime factor method, L.C.M. of 18, 24 and 96 are given below

Prime factors of 18 = 2 Ã— 3 Ã— 3

Prime factors of 24 = 2 Ã— 2 Ã— 2 Ã— 3

Prime factors of 96 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3

âˆ´ L.C.M. = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3

= 288

By using common division method, L.C.M. of 18, 24 and 96 are given below

âˆ´ L.C.M. = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3

= 288

(ii) 100, 150 and 200

By using prime factor method, L.C.M. of 100, 150 and 200 are given below

Prime factor of 100 = 2 Ã— 2 Ã— 5 Ã— 5

Prime factor of 150 = 2 Ã— 3 Ã— 5 Ã— 5

Prime factor of 200 = 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5

âˆ´ L.C.M. = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 5 Ã— 5

= 600

By using common division method, L.C.M. of 100, 150 and 200 are given below

âˆ´L.C.M. = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 5 Ã— 5

= 600

(iii) 14, 21 and 98

By using prime factor method, L.C.M. of 14, 21 and 98 are given below

Prime factor of 14 = 2 Ã— 7

Prime factor of 21 = 3 Ã— 7

Prime factor of 98 = 2 Ã— 7 Ã— 7

âˆ´ L.C.M. = 2 Ã— 3 Ã— 7 Ã— 7

= 294

By using common division method, L.C.M. of 14, 21 and 98 are given below

âˆ´ L.C.M. = 2 Ã— 3 Ã— 7 Ã— 7

= 294

(iv) 22, 121 and 33

By using prime factor method, L.C.M. of 22, 121 and 33 are given below

Prime factor of 22 = 2 Ã— 11

Prime factor of 121 = 11 Ã— 11

Prime factor of 33 = 3 Ã— 11

âˆ´ L.C.M. = 2 Ã— 3 Ã— 11 Ã— 11

= 726

By using common division method, L.C.M. of 22, 121 and 33 are given below

âˆ´ L.C.M. = 2 Ã— 3 Ã— 11 Ã— 11

= 726

(v) 34, 85 and 51

By using prime factor method, L.C.M. of 34, 85 and 51 are given below

Prime factor of 34 = 2 Ã— 17

Prime factor of 85 = 5 Ã— 17

Prime factor of 51 = 3 Ã— 17

âˆ´ L.C.M. = 2 Ã— 3 Ã— 5 Ã— 17

= 510

By using common division method, L.C.M. of 34, 85 and 51 are given below

âˆ´ L.C.M. = 2 Ã— 3 Ã— 5 Ã— 17

= 510

**3. The H.C.F. and the L.C.M. of two numbers are 50 and 300 respectively. If one of the numbers is 150, find the other one.**

**Solution:**

Given

H.C.F. = 50

L.C.M. = 300

One number = 150

We know that,

Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers

For other number,

50 Ã— 300 = 150 Ã— other number

15000 / 150 = other number

100 = other number

Hence, the other number is 100

**4. The product of two numbers is 432 and their L.C.M. is 72. Find their H.C.F.**

**Solution:**

Given

Product of two numbers = 432 and L.C.M.= 72

We know that,

Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers.

Now, to find H.C.F

H.C.F. Ã— 72 = 432

H.C.F.= 432 / 72

H.C.F. = 6

Hence, H.C.F. = 6

**5. The product of two numbers is 19,200 and their H.C.F. is 40. Find their L.C.M.**

**Solution:**

Given

Product of two numbers = 19200 and H.C.F. = 40

We know that,

Product of H.C.F. and L.C.M. of two numbers is equal to product of those two numbers

Now, to find L.C.M.

40 Ã— L.C.M. = 19200

L.C.M. = 19200 / 40

L.C.M. = 480

Hence, L.C.M. = 480

**6. Find the smallest number which, when divided by 12, 15, 18, 24 and 36 leaves no remainder.**

**Solution:**

The given numbers L.C.M. will be the least number which is exactly divisible 12, 15, 18, 24 and 36 and leaves no remainder

L.C.M. = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5

= 360

Hence, smallest required number = 360

**7. Find the smallest number which, when increased by one is exactly divisible by 12, 18, 24, 32 and 40.**

**Solution:**

First, let us find out the L.C.M. of 12, 18, 24, 32 and 40

L.C.M. = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5

= 1440

This can be written as

= 1439 + 1

Hence, 1439 is the smallest number which, when increased by one is exactly divisible by the given numbers

**8. Find the smallest number which, on being decreased by 3, is completely divisible by 18, 36, 32 and 27.**

**Solution:**

First, let us solve for L.C.M. of 18, 36, 32 and 27

L.C.M. = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3

= 864

This can be written as

= 867 â€“ 3

Hence, 867 is the smallest number which, when decreased by 3 is exactly divisible by the given numbers