SOLUTIONS FOR CBSE CLASS 6 NCERT MATHS CHAPTER 3 HCF & LCM

Find here step-by-step and precisely written Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Playing with Numbers | Exercise 3.6 and Exercise 3.7.

Before moving to Solutions for CBSE Class 6 NCERT Maths HCF & LCM | Ex – 3.6 & Ex – 3.7, it is advised to go through • HCF & LCM

Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.6

Problems in Class 6 Exercise 3.6 require the knowledge of Highest Common Factor (HCF) also known as Greatest Common Divisor (GCD), methods to find HCF and their applications in Word Problems.


EXERCISE 3.6

HCF | GCD

QUESTION 1

Find the H.C.F. of the following numbers:

(i) 18, 48, (ii) 30, 42 , (iii) 18, 60, (iv) 27, 63, (v) 36, 84, (vi) 34, 102, (vii) 70, 105, 175, (viii) 91, 112, 49, (ix) 18, 54, 81, (x) 12, 45, 75

ANSWER

(i) Prime factorisation of 18 and 48 are :
18 = 2 × 3 × 3
48 = 2 × 2 × 2 × 2 × 3

Common factors of 18 and 48 = 2 and 3

Thus, H.C.F. of 18 and 48 = 2 × 3 = 6

(ii) Prime Factorisation of 30 and 42 are:
30 = 2 × 3 × 5
42 = 2 × 3 × 7

H.C.F.(30, 42) = 2 × 3 = 6

(iii) Prime Factorisation of 18 and 60 are
18 = 2 × 3 × 3
60 = 2 × 2 × 3 × 5

Thus, H.C.F.(18, 60) = 2 × 3 = 6

(iv) Prime Factorisation of 27 and 63 are:
27 = 3 × 3 × 3
63 = 3 × 3 × 7

Common factor of 27 and 63 is 3 occuring twice.

Thus, H.C.F.(27, 63) = 3 × 3 = 9

(v) Prime Factorisation of 36 and 84 are:
36 = 2 × 2 × 3 × 3
84 = 2 × 2 × 3 × 7

Thus, H.C.F.(36, 84) = 2 × 2 × 3 = 12

(vi) Prime Factorisation of 34 and 102 are:
34 = 2 × 17
102 = 2 × 3 × 17

Thus, H.C.F.(34, 102) = 2 × 17 = 34

(vii) Prime Factorisation of 70, 105 and 175 are:
70 = 2 × 5 × 7
105 = 3 × 5 × 7
175 = 5 × 5 × 7

Thus, H.C.F.(70, 105, 175) = 5 × 7 = 35

(viii) Prime Factorisation of 91, 112 and 49 are:
91 = 7 × 13
112 = 2 × 2 × 2 × 2 × 7
49 = 7 × 7

Thus, H.C.F.(91, 112, 49) = 7

(ix) Prime Factorisation of 18, 54 and 81 are:
18 = 2 × 3 × 3
54 = 2 × 3 × 3 × 3
81 = 3 × 3 × 3 × 3

Thus, H.C.F.(18, 54, 81) = 3 × 3 = 9

(x) Prime Factorisation of 12, 45 and 75 are:
12 = 2 × 2 × 3
45 = 3 × 3 × 5
75 = 3 × 5 × 5

Thus, H.C.F.(12, 45, 75) = 1 × 3 = 3


ℹ HCF = Highest Common Factor

GCD = Greatest Common Divisor

ℹ HCF of given numbers is the product of common prime factors of the numbers.


❔ How to find HCF | GCD ?


[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.6 ]


QUESTION 2

What is the H.C.F. of two consecutive:

(i) numbers? (ii) even numbers? (ii) odd numbers?

ANSWER

(i) H.C.F. of two consecutive numbers is 1.

(ii) H.C.F. of two consecutive even numbers is 2.

(iii) H.C.F. of two consecutive odd numbers is 1.

QUESTION 3

H.C.F. of co-prime numbers 4 and 15 was found as follows by factorization:

4 = 2 × 2 and 15 = 3 × 5,

since there is no common prime factor, so H.C.F. of 4 and 15 is 0.

Is the answer correct? If not, what is the correct H.C.F.?

ANSWER

0 is not the correct answer.
The correct H.C.F. is 1.


ℹ 1 is the common factor of every number. It is not included in prime factorisation of a number because it is not a prime factor. Also factor 1 is generally omitted because multiplication with 1 doesn’t change the result.

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.6 ]


Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7

Problems in Class 6 Exercise 3.7 require the knowledge of Lowest Common Multiple (LCM), methods to find LCM and their applications in Word Problems.

EXERCISE 3.7

LCM | WORD PROBLEMS ON HCF & LCM

QUESTION 1

Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times.

ANSWER

Maximum value of the weight will be the greatest common diviser (HCF) of 75 and 79.

Prime factorisation of 75 and 69:
75 = 3 × 5 × 5
69 = 3 × 23

Then, H.C.F.(75, 69) = 3

Therefore, maximum value of the weight required is 3 kg.

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7 ]

QUESTION 2

Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

ANSWER

The distance covered by each one of them should be same as well as minimum with complete steps.
The minimum distance with complete steps is the L.C.M of 63, 70 and 77.

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7 ]

L.C.M.(63, 70 and 77) = 7 × 9 × 10 × 11
= 6930 cm = 69 m 30 cm.

Therefore, the minimum distance each should cover is 69 m 30 cm.


ℹ LCM = Least Common Multiple

❔ How to find LCM?

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7 ]


QUESTION 3

The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

ANSWER

The measurement of longest tape = HCF of 450 cm, 675 cm and 825 cm.

Prime factorisation of 450, 675 and 825 are
450 = 2 × 3 × 3 × 5 × 5

675 = 3 × 5 × 5 × 3 × 3

825 = 3 × 3 × 5 × 5 × 11

HCF(450, 675, 825) = 3 × 5 × 5 = 75

Thus, the longest tape which can measure the room exactly is 75 cm.

QUESTION 4

Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

ANSWER

Smallest number divisible by 6, 8 and 12 = L.C.M. of 6, 8 and 12

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7 ]

= 2 × 2 × 2 × 3 = 24

Then, smallest 3-digit number divisible by 6, 8 and 12 must be a multiple of 24.

Multiples of 24 = 24, 48, 72, 96, 120, …

Therefore, smallest 3-digit number divisible by 6, 8 and 12 = 120.

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7 ]

QUESTION 5

Determine the largest 3-digit number which is exactly divisible by 8, 10 and 12.

ANSWER

L.C.M. of 8, 10 and 12

 Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7

= 2 × 2 × 2 × 3 × 5 = 120

Then, largest three digit number divisible by 8, 10 and 12 must be a multiple of 120.

Multiples of 120 = 120, 240, 360, 480, 600, 720, 840, 960, 1080, …

Therefore, the largest three digit number exactly divisible by 8, 10 and 12 = 960.

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7 ]

QUESTION 6

The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m. at what time will they change simultaneously again?

ANSWER

L.C.M. of 48, 72, 108

Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7

= 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432

Then, the lights change simultaneously again after 432 seconds

= 7 minutes 12 seconds

Therefore, the lights change simultaneously again at = 7 : 07 : 12 a.m.

QUESTION 7

Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times.

ANSWER

The maximum capacity of container = H.C.F. of 403, 434 and 465.

Prime factorisation of 403, 434 and 465:

403 = 13 × 31

434 = 2 × 7 × 31

465 = 31 × 35

H.C.F.(403, 434, 465) = 31

Therefore, 31 liters of container can measure the diesel of three containers exact number of times.

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7 ]

QUESTION 8

Find the least number which when divided by 6, 15 and 18, leave remainder 5 in each case.

ANSWER

The number must be 5 more than the least common multiple of the given numbers.

L.C.M. of 6, 15 and 18

Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7

= 2 × 3 × 3 × 5 = 90

Thus, the required number = 90 + 5 = 95.

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7 ]

QUESTION 9

Find the smallest 4-digit number which is divisible by 18, 24 and 32.

ANSWER

L.C.M. of 18, 24 and 32

Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7

= 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

Then, the smallest four-digit number divisible by the given numbers must be a multiples of 288

Multiples of 288 = 288, 576, 864, 1152, …

Thus, the required number = 1152.

QUESTION 10

Find the L.C.M. of the following numbers:

(i) 4 and 9

(ii) 5 and 12

(iii) 5 and 6

(iv) 4 and 15

Observe a common property in the obtained L.C.Ms. Is L.C.M. the product of two numbers in each case?

ANSWER

(i) Prime factorisation of 4 and 9

4 = 2 × 2

9 = 3 × 3

Therefore, L.C.M. of 4 and 9

= 2 × 2 × 3 × 3 = 36 (= 4 × 9)

(ii) Prime factorisation of 5 and 12

5 = 5

12 = 2 × 2 × 3

Therefore, L.C.M. of 5 and 12

= 2 × 2 × 3 × 5 = 60 (= 5 × 12)

(iii) Prime factorisation of 5 and 6

5 = 5

6 = 2 × 3

Therefore, L.C.M. of 5 and 6

= 2 × 3 × 5 = 30 (= 5 × 6)

(iv) Prime factorisation of 4 and 15

4 = 2 × 2

15 = 3 × 5

Therefore, L.C.M. of 4 and 15

= 2 × 2 × 3 × 5 = 60 (= 4 × 15)

We can observe that obtained L.C.M.s are multiple of 3.

Yes, the L.C.M. is the product of two numbers in each case.

[ Solutions for CBSE Class 6 NCERT Maths Chapter 3 HCF & LCM | Exercise 3.7 ]

QUESTION 11

Find the L.C.M. of the following numbers in which one number is the factor of other:

(i) 5, 20

(ii) 6, 18

(iii) 12, 48

(iv) 9, 45

What do you observe in the result obtained?

ANSWER

(i) Prime factorisation of 5 and 20

5 = 5

20 = 2 × 2 × 5

Therefore, L.C.M. of 5 and 20

= 2 × 2 × 5 = 20

(ii) Prime factorisation of 6 and 18

6 = 2 × 3

18 = 2 × 3 × 3

Therefore, L.C.M. of 6 and 18

= 2 × 3 × 3 = 18

(iii) Prime factorisation of 12 and 48

12 = 2 × 2 × 3

48 = 2 × 2 × 2 × 2 × 3

Therefore, L.C.M. of 12 and 48

= 2 × 2 × 2 × 2 × 3 = 48

(iv) Prime factorisation of 9 and 45

9 = 3 × 3

45 = 3 × 3 × 5

Therefore, L.C.M. of 9 and 45

= 3 × 3 × 5 = 45

We can observe that the L.C.M. of two numbers in which one is the factor of the other, is the bigger number.

NCERT MATHS SOLUTION | CLASS 6 | CHAPTER 3 | PLAYING WITH NUMBERS

Exercise 3.1 | Exercise 3.2 | Exercise 3.3

Exercise 3.4 | Exercise 3.5 |

Exercise 3.6 | Exercise 3.7

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