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NCERT · Class 6 · Maths

Chapter 3: Playing with Numbers

Practice each question, then open the answer or worked solution.

Exercise 3.7

Textbook page 67 · 11 questions
Q1Written answer

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

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Worked solution

Given weight of two bags of fertiliser = 75 kg and 69 kg

Maximum weight = HCF of two bags weight i.e., (75, 69)

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 1

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 2

75 = 3 × 5 × 5

69 = 3 × 23

HCF = 3

Hence, 3 kg is the maximum value of weight which can measure the weight of the fertiliser exact number of times.

Q2Written answer

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?

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Worked solution

The first boy’s steps measure = 63 cm

The second boy’s steps measure = 70 cm

Third boy’s steps measure = 77 cm

LCM of 63, 70, 77

NCERT solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 3

LCM = 2 × 3 × 3 × 5 × 7 × 11 = 6930

Hence, 6930 cm is the distance each should cover so that all can cover the distance in complete steps.

Q3Written answer

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

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Worked solution

Given the length of a room = 825 cm

The breadth of a room = 675 cm

Height of a room = 450 cm

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 4

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 5

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 6

825 = 3 × 5 × 5 × 11

675 = 3 × 3 × 3 × 5 × 5

450 = 2 × 3 × 3 × 5 × 5

HCF = 3 × 5 × 5 = 75 cm

Hence longest tape is 75 cm which can measure the three dimensions of the room exactly.

Q4Written answer

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

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Worked solution

LCM of 6, 8, 12 = smallest number

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 7

LCM = 2 × 2 × 2 × 3 = 24

Now we need to find the smallest 3-digit multiple of 24

We know that 24 × 4 = 96 and 24 × 5 = 120

Hence, 120 is the smallest 3-digit number which is exactly divisible by 6, 8 and 12

Q5Written answer

Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

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Worked solution

LCM of 8, 10 and 12

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 8

LCM = 2 × 2 × 2 × 3 × 5 = 120

Now we need to find the greatest 3-digit multiple of 120

We may find 120 × 8 = 960 and 120 × 9 = 1080

Hence, 960 is the greatest 3-digit number exactly divisible by 8, 10 and 12

Q6Written answer

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?

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Worked solution

LCM of 48, 72, 108 = time period after which these lights change

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 9

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432

Hence, lights will change together after every 432 seconds

Therefore the lights will change simultaneously at 7 minutes and 12 seconds.

Q7Written answer

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 the three containers the exact number of times.

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Worked solution

HCF of 403, 434, 465 = Maximum capacity of tanker required

403 = 13 × 31

434 = 2 × 7 × 31

465 = 3 × 5 × 31

HCF = 31

Hence, a container of 31 litres can measure the diesel of the three containers the exact number of times.

Q8Written answer

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

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Worked solution

LCM of 6, 15, 18

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 10

LCM = 2 × 3 × 3 × 5 = 90

Required number = 90 + 5

= 95

Hence, 95 is the required number.

 

Q9Written answer

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

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Worked solution

LCM of 18, 24, 32

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 11

LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

Here, we need to find the smallest 4-digit multiple of 288

We find 288 × 3 = 864 and 288 × 4 = 1152

Hence, 1152 is the smallest 4-digit number which is divisible by 18, 24 and 32

 

Q10Written answer

Find the LCM of the following numbers:

(a) 9 and 4 (b) 12 and 5 (c) 6 and 5 (d) 15 and 4

Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

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Worked solution

(a) LCM of 9, 4

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 12

LCM = 2 × 2 × 3 × 3 = 36

(b) LCM of 12, 5

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 13

LCM = 2 × 2 × 3 × 5 = 60

(c) LCM of 6, 5

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 14

LCM = 2 × 3 × 5 = 30

(d) LCM of 15, 4

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7 - 15

LCM = 2 × 2 × 3 × 5 = 60

Yes in each case the LCM of given numbers is the product of these numbers.