**NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.1****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.2****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.3****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.4****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.5****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.6****NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7**

**NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Ex 3.7 ****(Rationalised Contents)**

**Ex 3.7 Class 6 Maths Question
1.**

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.**Solution:
**The maximum value of weight which can measure the weight
of the fertiliser exact number of times = HCF of 75 kg and 69 kg

Prime factorisations of 75 and 69 are:

Here, the common factor is 3.

∴ HCF of 75 and 69 = 3

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

**Ex 3.7 Class 6 Maths 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?**Solution:
**The minimum distance that each boy should cover must
be the least common multiple (LCM) of the measure of their steps.

To find the LCM of 63, 70 and 77, we can use division method.

Hence, the required minimum distance is 6930 cm.

**Ex 3.7 Class 6 Maths 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.**Solution:
**The longest tape required to measure the three
dimensions of the room exactly = HCF of 825, 675 and 450

Prime factorisations of 825, 675 and 450 are:

825 = 3 × 5 × 5 × 11

675 = 3 × 3 × 3 × 5 × 5

450 = 2 × 3 × 3 × 5 × 5

Here, common factors are 3, 5, 5.

∴ HCF of 825, 675 and 450 = 3 × 5 × 5 = 75

Hence, the length of the longest tape is 75 cm.

**Ex 3.7 Class 6 Maths Question
4.**

Determine the smallest 3-digit number which is
exactly divisible by 6, 8 and 12.**Solution:
**The smallest 3-digit number is 100.

The smallest number divisible by 6, 8 and 12 = LCM of 6, 8 and 12

Thus, 24 is exactly divisible by 6, 8 and 12. But it
is a 2-digit number.

Since, all the multiples of 24 will also be divisible by 6, 8 and 12.

So, the smallest multiple of 24 in three digits will be just above

100 = (100 – 4) + 24 = 96 + 24 = 120

Hence, the required smallest 3-digit number is 120.

**Ex 3.7 Class 6 Maths Question
5.**

Determine the greatest 3-digit number exactly
divisible by 8, 10 and 12.**Solution:
**To find the LCM of 8, 10 and 12, we have

The greatest 3-digit number is 999.

∴ Multiple of 120 just below 999 = 999 – 39 = 960

Hence, the required greatest 3-digit number is 960.

**Ex 3.7 Class 6 Maths 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?**Solution:**

The** **time at which the traffic lights will
change simultaneously again = LCM of 48, 72 and 108**
**To find the LCM of 48, 72 and 108, we have

So, after 432 seconds, the traffic light will change simultaneously again.

Hence, the required time = 432 seconds = 7 minutes 12 seconds, i.e., 7 minutes 12 seconds past 7 a.m.

**Ex 3.7 Class 6 Maths 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 the three containers exact number of times.**Solution:
**The maximum capacity of the required container = HCF
of 403, 434 and 465

Prime factorisations of 403, 434 and 465 are:

So, the HCF of 403, 434 and 465 = 31

Hence, the maximum capacity of the required container is 31 litres.

**Ex 3.7 Class 6 Maths Question
8.**

Find the least number which when divided by 6, 15
and 18 leave remainder 5 in each case.**Solution:
**The required number is 5 more than the LCM of 6, 15
and 18.

To find the LCM of 6, 15 and 18, we have

Here, 90 is the least number exactly divisible by 6, 15 and 18.

To get a remainder 5, the least number will be 90 + 5 = 95

Hence, the required number is 95.

**Ex 3.7 Class 6 Maths Question
9.**

Find the smallest 4-digit number which is divisible
by 18, 24 and 32.**Solution:
**The smallest 4-digit number is 1000.

To find the LCM of 18, 24 and 32, we have

Since, 288 is the smallest number which is exactly divisible by 18, 24 and 32.

But it is not a 4-digit number.

So, the multiple of 288 just above 1000 is: 1000 – 136 + 288 = 1152

Hence, the required number is 1152.

**Ex 3.7 Class 6 Maths Question
10.**

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?

**Solution:
**(a) To find the LCM of 9 and 4, we have

∴ LCM = 2 × 2 × 3 × 3 = 36

The product 9 and 4 = 9 × 4 = 36

Hence, the LCM of 9 and 4 = Product of 9 and 4

(b) To find LCM of 12 and 5, we have

∴ LCM = 2 × 2 × 3 × 5 = 60The product of 12 and 5 = 12 × 5 = 60

Hence, the LCM of 12 and 5 = Product of 12 and 5

(c) To find the LCM of 6 and 5, we have

∴ LCM = 2 × 3 × 5 = 30

The product of 6 and 5 = 6 × 5 = 30

Hence, the LCM of 6 and 5 = Product of 6 and 5

(d) To find the LCM of 15 and 4, we have

∴ LCM = 2 × 2 × 3 × 5 = 60

The product of 15 and 4 = 15 × 4 = 60

Hence, the LCM of 15 and 4 = Product of 15 and 4

**Ex 3.7 Class 6 Maths Question
11.**

Find the LCM of the following numbers in which one
number is the factor of the other.(a) 5, 20

(b) 6, 18

(c) 12, 48

(d) 9, 45

What do you observe in the results obtained?

**Solution:
**(a) To find the LCM of 5 and 20, we have

∴ LCM = 2 × 2 × 5 = 20

Hence, the LCM of 5 and 20 = 20

(b) To find the LCM of 6 and 18, we have

∴ LCM = 2 × 3 × 3 = 18Hence, the LCM of 6 and 18 = 18

(c) To find the LCM of 12 and 48, we have

LCM = 2 × 2 × 2 × 2 × 3 = 48

Hence, the LCM of 12 and 48 = 48

(d) To find the LCM of 9 and 45, we have

∴ LCM = 3 × 3 × 5 = 45

Hence, the LCM of 9 and 45 = 45

From the above examples, we observe that if one number is a factor of the other
number, then their LCM is the greater number.

**Related Links:**

**NCERT Solutions for Maths Class 7**

**NCERT Solutions for Maths Class 8**

**NCERT Solutions for Maths Class 9**

**NCERT Solutions for Maths Class 10**