WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple

Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple

Question 1. Find the H.C.F. of the following numbers by factorization 24, 36, 54

Solution:

Given:

24, 36, 54

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 1.1

∴ 24 = 2 x 2 x 2 x 3

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 1.2

∴ 36 = 2 x 2 x 3 x 3

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WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 1.3

∴ 54 = 2 x 3 x 3 x 3

∴ The common prime factors of given 3 numbers are 2, 3.

∴ The required H.C.F. = 2 x 3

= 6.

 

Question 2. Find the H.C.F. of the following numbers by division method: 160, 165, 305

Solution:

Given: 160, 165, 305

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 2.1

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 2.2

 

The required H.C.F. = 5.


Question 3. Find the H.C.F. of the following numbers by short division method 165, 264, 286

Solution :

Given: 165, 264, 286

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 3

 

∴ The required H.C.F = 11.

 

Question 4. Find the L.C.M. of the following numbers by factorization: 36, 60, 72

Solution:

Given: 36, 60, 72

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 4.1

 

∴ 36 = 2 x 2 x 3 x 3.

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 4.2

 

∴ 72 = 2 x 2 x 2 x 3 x 3

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 4.3

 

∴ 60 = 2 x 2 x 3 x 5

∴ The required L.C.M = 2 x 2 x 3 x 3 x 2 x 5

= 360

 

Question 5. Find the L.C.M. of the following numbers by the short division method :

1. 24, 36, 45, 60

Solution:

Given: 24, 36, 45, 60

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 5 Q 1

 

2. 105, 119, 289

Solution:

Given: 105, 119, 289

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 5 Q 2

 

Question 6. Find the H.C.F. and L.C.M. of the following quantities :

1. 2 m 28 cm; 3 m 42 cm; 4 m 56 cm

Solution:

Given: 2 m 28 cm; 3 m 42 cm; 4 m 56 cm

2 m 28 cm = (2 x 100 + 28) cm = 228 cm

3 m 42 cm = (3 x 100 + 42) cm =. 342 cm

4 m 56 cm = (4 x 100 + 56) cm = 456 cm

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 6 Q 1

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 6 Q 1.1

 

∴  H.C.F. of 228, 342, 456 = 114.

∴ The required H.C.F. = 114 cm = 1 m 14 cm

Again,

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 6 Q 1.2

 

 

L.C.M. of 228, 342, 456 = 2x2x3x 19 x3x2 = 1368.

The required L.C.M. = 1368 cm = 13 m 68 cm.


2. 6
paisa 50; Rs. 5 Paisa 20; Rs. 7 Paisa 80.

Solution:

Given: 6 paisa 50; Rs. 5 Paisa 20; Rs. 7 Paisa 80.

Rs. 6 paisa 50 = 650 paisa;

Rs. 5 paisa 20 = 520 paisa;

Rs. 7 paisa 80 = 780 paisa.

Now,

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 6 Q 2

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 6 Q 2.1

∴ The H.C.F of 650, 520, and 780 = 130.

∴ The required H.C.F. = 130 paise = Re 1 Paisa 30.

Again,

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 6 Q 2.2

 

∴ L.C.M. of 650, 520, 780 = 10 x 13 x 2 x 5 x 2 x 3 = 7800

∴ The required L.C.M = 7800 Paisa = Rs. 78.

 

Question 7. Prove of the following numbers that the product of the numbers is equal to the product of their H.C.F. and L.C.M. 87, 145

Solution:

Given: 87, 145

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 7

 

∴ The H.C.F of 87 and 145 = 29.

Again,

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 7 Q 1

 

∴ L.C.M. of 87 and 145 = 29 x 3 x 5 = 435.

∴ Product of the H.C.F. and L.C.M. of the numbers 87, 145 = 29 x 435 = 12615. Product of the numbers = 87 x 145 = 12615.

The product of the numbers = Product of their H.C.F. and L.C.M. (Proved).

 

Question 8. Which least number is exactly divisible by 15, 20, 24, and 32?

Solution:

Given: 15, 20, 24, And 32

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 8

 

∴ L.C.M. of 15, 20, 24, 32 = 2 x 2 x 2 x 3 x 5 x 4 = 480.

The required least number will be the L.C.M. of 15, 20, 24, and 32, and this last number will be exactly divisible by the given numbers.

∴ The required number = 480.

 

Question 9. By which greatest number 306, 810, and 2214 will be exactly divisible?

Solution:

Given: 306, 810, And 2214

The required greatest number will be the H.C.F. of 306, 810, and 2214 because H.C.F. will be the greatest number by which 306, 810, and 2214 will be divisible.

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 9

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 9.1

 

∴ The H.C.F. of 306, 810, and 2214 = 18.

∴ The required greatest number = is 18.

 

Question 10.

1. Find the L.C.M. of 145 and 232 with the help of their H.C.F.

Solution:

Given: 145 And 232

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 10. Q 1

 

∴ H.C.F. of 145 and 232 = 29.

We know that, the product of two numbers = Their H.C.F x L.C.M.

∴ L.C.M of two given numbers = Product of the numbers / H.C.F

= 145 x 232 / 29

= 1160.

∴ The required L.C.M. = 1160.

 

2. Find the H.C.F. of 144 and 384 with the help of their L.C.M.

Given: 144 and 384

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 10. Q 2

 

∴ L.C.M. of 144 and 384 = 2 x 2 x 2 x 2 x 3 x 3 x 8 = 1152.

We know that the product of two given numbers = Their H.C.F x L.C.M.

∴ H.C.F of two numbers = Product of two given numbers

= 144 x 384 / 1152

= 48.

∴ The required H.C.F. = 48.

 

Question 11. Find two pairs of numbers whose H.C.F. is 16 between 50 and 100.

Solution:

Given:

50 and 100

Since the H.C.F. of two numbers is 16, let the numbers be 16m and 16n where m and n are two co-prime natural numbers.

Now, 2 and 3 are co-prime natural numbers, and 2 x 16 = 32, and 3 x 16 = 48. But they do not lie between 50 and 100.

3 and 4 are co-prime natural numbers and taking m = 3, n = 4, we get 16 x 3 = 48, 16 x 4 = 64.

But 48 does not lie between 50 and 100.

Now, if we take m = 4 and n = 5 as 4, 5 are two co-prime natural numbers, and 16 x 4 = 64, 16 x 5 = 80.

Here both 64 and 80 lie between 50 and 100.

Again if we take m = 5 and n = 6 as 5,-6 are also two co-prime natural numbers, and 5 x 16 = 80, 6 x 16 = 96.

Here both 80 and 96 lie between 50 and 100.

∴ The required two pairs of numbers are 64, 80, and 80, 96.

 

Question 12. The H.C.F. and L.C.M. of the two numbers are 145 and 2175 respectively. If one number is 725, then what is the other number?

Solution:

We know that, Product of two numbers = Their H.C.F. x L.C.M.

Here H.C.F. = 145 and L.C.M. = 2175

∴ Product of two numbers = H.C.F. x L.C.M. = 145 x 2175

Since one number = 725 (given),

∴ The other number = 145 x 2175 / 725

= 435.

 

Question 13. Which least number is subtracted from 5834 so that the result of subtraction is divisible by 20, 28, 32, and 35?

Solution:

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 13

 

∴ L.C.M. of 20, 28, 32, 35 = 2 x 2 x 5 x 7 x 8 = 1120. So 1120 is the least number that is divisible by each of the numbers 20, 28, 32, and 35.

∴ Any multiple of 1120 is also divisible by each of the given numbers.

Now the multiple of 1120 are

1120 x 1 = 1120,

1120 x 2 = 2240,

1120 x 3 = 3360,

1120 x 4 = 4480,

1120 x 5 = 5600,

1120 x 6 = 6720

But the number 5834 lies between 5600 and 6720 and is very near to 5600 and 5834 > 5600.

Now, if we subtract (5834 – 5600) = 234 from 5834 the remainder is 5600, which is a multiple of 1120 and so 5600 is divisible by each of the numbers 20, 28, 32, and 35.

∴ The required least number = is 234.

 

Question 14. Find the greatest number that will divide 2300 and 3500 leaving the remainder 32 and 56 respectively.

Solution:

We have to find the greatest number that will divide 2300 and 3500 leaving the remainder 32 and 56 respectively.

2300 – 32 = 2268;

3500 – 56 = 3444.

Therefore, the numbers 2268 and 3444 must be divisible by the required greatest number and so the H.C.F. of 2268 and 3444 will be the required greatest number.

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 14

 

∴ The H.C.F. of 2268 and 3444 = 84.

∴ The required number = is 84.

 

Question 15. Find the greatest number which will divide 650, 775, and 1250 so as to leave the same remainder in each case.

Solution:

Since the required number when divides 650, 775, and 1250 leaves the

same remainder in each case, therefore (775 – 650) or 125 and (1250 – 775) or 475 must be divisible by the required greatest number. So the required greatest number will be the H.C.F. of 125 and 475.

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 15

 

∴ The H.C.F. of 125 and 475 = 25.

∴ The required greatest number is 25.

 

Question 16. The sum of two numbers is 384 and the H.C.F. of two numbers is 48. What are the two possible numbers?

Solution:

Given:

The sum of two numbers is 384 and the H.C.F. of two numbers is 48

The H.C.F. of two numbers is 48.

Let the numbers be 48x and 48y, where x and y are two co-prime natural numbers.

Again, by the questions,

48x + 48y = 384

or, 48 (x + y) = 384,

or, x + y = 384/48

= 8

or, x + y = 8

or, y = 8 – x ………………….(1)

8 = 1+7 = 2 + 6 = 3 + 5= 4 + 4

Among them, (1, 7), (3, 5), (5, 3), and (7, 1) are co-prime natural numbers

and they are values of (x, y)

∴ When x = 1, y = 7, the numbers are 48 x 1 = 48, 48 x 7 = 336.

When x = 3, y = 5, the numbers are 48 x 3 = 144, 48 x 5 = 240.

When x = 5, y = 3, the numbers are 48 x 5 = 240, 48 x 3 = 144.

When x = 7, y = 1, the numbers are 48 x 7 = 336, 48 x 1 = 48.

∴ The two possible numbers are (48, 336)

or, (336, 48) and (144, 240)

or, (240, 144).

The two possible numbers (48, 336)

 

Question 17. From which least number 4000 is subtracted so that the remainder is divisible by 7, 11, and 13?

Solution:

Here 7, 11, and 13 are prime to each other.

∴ L.C.M. of 7, 11, and 13 = 7 x 11 x 13 = 1001.

So the least number which is divisible by each of 7, 11, and 13 is 1001.

The required number is (1001 + 4000) = 5001.

 

Example 18. The H.C.F. and L.C.M. of two numbers are 12 and 720 respectively. What are the possible numbers?

Solution:

Since the HCF of two numbers is 12, let the numbers be 12jc and 12y where x and y are two co-prime natural numbers.

Now the L.C.M. of 12jc and 12y = 12xy (x, y are co-prime to each other.)

By the given question, 12xy = 720

or, xy = .60

∴ 60

= 1 x 60

= 2 x 30

= 3 x 20

= 4 x 15

= 5 x 12

= 6 x 10

∴ Among these the pairs (1, 60), (3, 20), (4, 15), and (5, 12) are co-prime and they are the values of (;c, y). _

When x = 1, y = 60;

when x = 3, y = 20;

when x = 4, y = 15;

when x = 5, y = 12.

The numbers are

12 x 1 = 12, 12 x 60 = 720;

12 x 3 = 36; 12 x 20 = 240;

12 x 4 = 48; 12 x 15 = 180;

12 x 5 = 60; 12 x 12 = 144.

∴ The possible required numbers are 12, 720; 36, and 240; 48, 180; 60, and 144.

 

Question 19. Find the number which is divisible by 28, 33, 42, and 77 and which is nearer to 98765.

Solution:

Given:

28, 33, 42, and 77 and which is nearer to 98765

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 19

 

∴ 924 is the least number that is divisible by the given numbers.

Now,

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 19 Q 1

 

98765 – 821 = 97944 is divisible by 924 and so 97944 is divisible by each of the given numbers 28, 33, 42, 77.

Again, 924 x 107 = 98868 is also divisible by 924 and so 98868 is divisible by each of the numbers 28, 33, 42, 77.

Since 98765 – 97944 = 821 and 98868 – 98765 = 103, the number 98868 is very near to 98765.

Hence the required number is 98868.

 

Question 20. Find the least number which is divisible by 13 and which when divided by 8, 12, 16, and 20 will leave a remainder of 1 in each case.

Solution:

Given:

8, 12, 16, And 20

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 20

 

The L.C.M. of 8, 12, 16, 20 = 2 x 2 x 2 x 3 x 2 x 5 = 240.

This implies that 240 is the least number that is divisible by each of the numbers 8, 12, 16, and 20.

But we have found the least number which is divisible by 13 and the number which when divided by each of the numbers 8, 12, 16, 20 will leave the remainder of 1 in each case.

Let the number (240 k + 1) be divisible by 13 where k is a positive integer.

Putting k = 1, 2, 3, 4, etc, we get,

240 k + 1 = 240 x 1 + 1 = 241 which is not divisible by 13.

240 k + 1 = 240 x 2 + 1 = 481 which is divisible by 13.

The required least number is 481.

 

Question 21. The circumference of the front wheel of a carriage is 1 m 4 dm and the circumference of the rear wheel is 2 1/2 times that of the front wheel. Find the least distance to be covered so that both wheels complete their full rotation simultaneously.

Solution :

Given:

The circumference of the front wheel of a carriage is 1 m 4 dm and the circumference of the rear wheel is 2 1/2 times that of the front wheel.

1 m 4 dm = (1 x 10 + 4) dm = 14 dm.

The circumference of the front wheel = 14 dm.

So the circumference of the rear wheel = (14 x 2 1/2)dm

= 14 x 5/2 dm

= 35 dm.

Here the least required distance will be the L.C.M. of 14 dm and 35 dm.

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 21

 

L.C.M. of 14 dm and 35 dm = (7x2x5) dm

= 70 dm

= 7 m.

The least distance that the carriage will go so that both the front wheel and the rear wheel will rotate complete full rotation is 7 meters.

 

Question 22. There are 3 traffic signals at the crossing of 3 different ways. The lights of 3 traffic signals are changed at an interval every 16 sec, 28 sec, and 40 sec respectively. If all the lights in the traffic signals changed their lights simultaneously at 8 a.m. in the morning, when will they again change their lights simultaneously?

Solution:

Given:

There are 3 traffic signals at the crossing of 3 different ways. The lights of 3 traffic signals are changed at an interval every 16 sec, 28 sec, and 40 sec respectively. If all the lights in the traffic signals changed their lights simultaneously at 8 a.m. in the morning

The required time will be the L.C.M. of 16 sec, 28 sec, and 40 sec.

 

WBBSE Solutions For Class 6 Maths Arithmetic Chapter 10 Highest Common Factor And Least Common Multiple Question 22

 

L.C.M. of 16 sec, 28 sec, 40 sec = (2 x 2 x 2 x 2 x 7 x 5) sec = 560 sec.

= 9 m sec.

Since the first time, the traffic signals simultaneously change their light again they will change their lights at (8 am + 9 m 20 sec) or 8 hr 9 m 20 sec in the morning.

 

 

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