Grade6-LCM-and-HCF.pdf

ID : F-6-LCM-and-HCF [1]

Grade 6 LCM and HCF For more such worksheets visit www.EduGain.com

Answer the questions (1)

3 bells ring at interval of 18, 126, 12 seconds respectively. If the bells ring together at 2 O'clock, when will they ring together again? How many times will they ring together in 126 minutes ?

(2)

(3)

Find the highest common factor (HCF) of following: A)

286, 220

B)

110, 319

C)

242, 220

D)

232, 216

E)

189, 261

F)

119, 98

Along the highway, there are electric poles on one side of the highway and telegraph poles on the other side. The electric poles are all separated by 28 m and the telegraph poles are all separated by 36 m. If an electric and a telegraph pole are both present at the point where the highway begins, after how much distance along the highway will they both be in front of each other again?

(4)

Determine the two numbers nearest to 700000 which are exactly divisible by 2, 4, 3, 6.

(5)

80 cola cans and 115 fruit juice cans need to be stacked in a school canteen. If each stack is to be of the same height and is to contain cans of the same type, what would be the greatest number of cans each stack can have?

(6)

Mark and Kimberly are friends and cricket coach too. Mark goes to Amity school every 4 days and Kimberly goes to Amity every 6 days to deliver coaching classes. If they both delivered the coaching sessions today, how many days in the next 60 days they both would take the session on the same day ?

(7)

A rectangular courtyard with length 7 m 2 cm and breadth 3 m 15 cm is to be paved with square stones of the same size. Find the least number of such stones required.

Choose Choos e correct answer(s) from from given choice cho ice (8)

(9)

The product of any two numbers is equal to product of their a. HCF a. HCF and either of the two numbers

b. HCF b. HCF and difference of the two numbers

c. HCF c. HCF and the sum of the two numbers

d. LCM d. LCM and HCF

If a number is divisible by 12 and 18, it will be necessarily divisible by a. 34 a. 34

b. 39 b. 39

c. 36 c. 36

d. 40 d. 40

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Fill in the blanks (10) The greatest number which divides 157, 258 and 629 leaving remainder 7, 8 and 9 respectively is . (11)

The largest number that divides divides 505 and 253 leaving remainder 1 is

.

(12) Two brands of chocolates are available in packs of 70 and 42 respectively. If Donald wants to buy equal number of chocolates of each kind, then the least number of boxes of each chocolate type that he needs to buy are

and

respectively.

(13) Find the Least Common Multiple (LCM) of following: A)

LCM of 18, 144, 42, 80 is

B)

LCM of 6, 112, 10, 28 is

. D)

.

LCM of 12, 60, 6 is

E)

.

LCM of 63, 210, 42, 30 is .

LCM of 14, 28, 12, 12, 16 is

. (14)

C)

F)

LCM of 8, 42, 10, 35, 20 is .

The two numbers nearest to 400000 which are exactly divisible by 5, 2, 3, 6, 4, 9 are and

.

(15) The lowest natural number which when divided by 24, 240, 12, 30 leaves the remainder of 2 in each case is

.

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Answers (1)

2:04:12 , 30

(2)

A)

22

B)

11

C)

22

D)

8

E)

9

F)

7

(3)

252 m Step 1 Since an electric pole is there at every 28 m, it will be present at distances which are multiples of 28. Step 2 Similarly, a telegraph pole is there at every 36 m. It will thus be present at distances which are multiple of 36. Step 3 Now the distance at which both types of poles will be present again, should be a multiple of both 28 and 36. Therefore, we need to find the Least Common Multiple (LCM) of 28 and 36 in order to find such distance. Step 4 LCM of 28 and 36 = 252 m. Step 5 Therefore, both types of poles will be together again after a distance of 252 m.

(4)

699996, 700008




(5)

5 Step 1 Number of cola cans and fruit juice cans in a school canteen are 80 and 115 respectively. Step 2 Since each stack is to be of the same height and is to contain cans of the same type, so that the greatest number of cans each stack can have the HCF of 80 and 115 Step 3 All prime factors of 80: 2 | 80

2 is a factor of 80

2 | 40

2 is a factor of 40

2 | 20

2 is a factor of 20

2 | 10

2 is a factor of 10

5 |5

5 is a factor of 5

|1 80 = 2 × 2 × 2 × 2 × 5 Step 4 All prime factors of 115: 5

| 1 15

23 | 2 3

5 is a factor of 115 23 is a factor of 23

|1 115 = 5 × 23 Step 5 HCF of 80 and 115 is = 5 =5 Step 6 The greatest number of cans each stack can have 5




(6)

5 Step 1 Mark goes to Amity every 4 th days and Kimberly goes to Amity every 6 th days. So like this they will both will take session on every 12 days. Step 2 In order to find the number of days after which both of them goes together, we need to find a number which is as small as possible and divisible by both 4 and 6. Therefore we need to find LCM of 4 and 6 which is 12 days. Step 3 Now since both of them goes to school every 12 th days. Therefore number of session session in next 60 days they will take = 60 / 12 = 5




(7)

2730 Step 1 The length and breadth of a rectangular courtyard is 7 m 2 cm and 3 m 15 cm respecti r espectively vely Step 2 Convert each dimension into cm. length = 7 m 2 cm Since 1m = 100 cm So that length in cm = 7 × 100 + 2 = 700 + 2 = 702 cm Step 3 breadth = 3 m 15 cm breadth in cm = 3 × 100 + 15 = 300 + 15 = 315 cm Step 4 Therefore the length and breadth of a rectangular courtyard in cm is 702 cm and 315 cm respectively Step 5 If the size of the stones is maximum then least least number of stones are required. Therefore T herefore the maximum size of the stones is the HCF of 702 and 315 Step 6 All prime factors of 702: 2

| 7 02

2 is a factor of 702

3

| 3 51


3

| 1 17


3

| 39

3 is a factor of 39

13 | 1 3


|1 702 = 2 × 3 × 3 × 3 × 13 Step 7 All prime factors of 315: 3 | 31 5


3 | 10 5


5 | 35

5 is a factor of 35

7 |7

7 is a factor of 7

|1 315 = 3 × 3 × 5 × 7 (C) 2015 EduGain (www.EduGain.com)



Step 8 HCF of 702 and 315 is = 3 × 3 =9 Step 9 The least number of such stones required =

length HCF

=

702 9

×

×

breadth HCF

315 9

= 78 × 35 = 2730

(8)

d. LCM d. LCM and HCF

(9)

c. 36 c. 36 If you look at the question carefully, you will notice that a number is divisible by 12 and 18. Now it will be necessarily divisible by the LCM of 12 and 18 i.e 36. Since, we know that if a number is divisible by two numbers, then it must be divisible by the LCM of two numbers.




(10)

10 Step 1 Since remainder of 157, 258 and 629 is 7, 8 and 9 respectively, therefore subtract the remainder 7, 8 and 9 from the number 157, 258 and 629 respectively. respectively. or 157 - 7 = 150 or 258 - 8 = 250 or 629 - 9 = 620 Step 2 The greatest number which divides 157, 258 and 629 leaving leaving remainder 7, 8 and 9 respectively is the HCF of 150, 250 and 620 Step 3 All prime factors of 150: 2 | 15 0


3 | 75

3 is a factor of 75

5 | 25

5 is a factor of 25

5 |5

5 is a factor of 5

|1 150 = 2 × 3 × 5 × 5 Step 4 All prime factors of 250: 2 | 25 0


5 | 12 5


5 | 25

5 is a factor of 25

5 |5

5 is a factor of 5

|1 250 = 2 × 5 × 5 × 5 Step 5 All prime factors of 620: 2

| 6 20


2

| 3 10


5

| 1 55


31 | 3 1


|1 620 = 2 × 2 × 5 × 31 Step 6 HCF of 150, 250 and 620 is = 2 × 5 (C) 2015 EduGain (www.EduGain.com)



= 10 Step 7 Therefore the greatest number which divides 157, 258 and 629 leaving leaving remainder 7, 8 and 9 respectively is = 10




(11)

25 2 Step 1 We have to find the largest number that divides 505 and 253 leaving remainder 1. Step 2 In other words, we have to find the largest number that divides (505-1) and (253-1) leaving leaving no remainder. Such number is the Highest Common Factor (HCF) of: 504 [i.e., 504 [i.e., 505 - 1]and 252 252 [i.e., [i.e., 253 - 1]. Step 3 Let us now find the HCF of 504 and 252. All prime factors of 504: 2 | 50 4


2 | 25 2


2 | 12 6


3 | 63

3 is a factor of 63

3 | 21

3 is a factor of 21

7 |7

7 is a factor of 7

|1 Therefore, 504 = 2 × 2 × 2 × 3 × 3 × 7. Step 4 All prime factors of 252: 2 | 25 2


2 | 12 6


3 | 63

3 is a factor of 63

3 | 21

3 is a factor of 21

7 |7

7 is a factor of 7

|1 Therefore, 252 = 2 × 2 × 3 × 3 × 7. Step 5 The HCF of 504 and 252 is = 2 × 2 × 3 × 3 × 7 = 252. Step 6 Thus, the largest number which divides 505 and 253 leaving remainder 1 is 252.




(12)

3

5 Step 1 Number of chocolates of first type in pack = 70 Number of chocolates of second type in pack = 42 Step 2 If he divides the number of chocolates of first type and second type in packs by the maximum number of chocolates in both packs, then he can find out the least number of boxes of second type and first type of chocolates respectively so that he can buy equal number of chocolates of both types. Step 3 The maximum number of chocolates in both packs is the HCF of 70 and 42. All prime factors of 70: 2 | 70

2 is a factor of 70

5 | 35

5 is a factor of 35

7 |7

7 is a factor of 7

|1 70 = 2 × 5 × 7 Step 4 All prime factors of 42: 2 | 42

2 is a factor of 42

3 | 21

3 is a factor of 21

7 |7

7 is a factor of 7

|1 42 = 2 × 3 × 7 Step 5 HCF of 70 and 42 is = 2 × 7 = 14 Step 6 The least number of boxes of first type he will need to buy = Number of chocolates of second type in pack HCF =

42 14

=3 Step 7 The least number of boxes of second type he will need to buy =




Number of chocolates of first type in pack HCF =

70 14

=5 Step 8 Therefore he will need to buy 3 boxes of first type, 5 boxes of 2nd type.

(13) A) D)

(14) (15)

B)

5 0 40

E)

60

399960

16 8 0 33 6

C) F)

6 30 8 40

400140

24 2



Grade6-LCM-and-HCF.pdf

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