NCERT Solutions for Class 6 Maths Chapter 3 - Playing with Numbers Exercise 3.7

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.

Ans: It is given renu purchases two bags of fertilizer of weights 75kg and 69kg.

To find the maximum weight which can measure the weight of the fertilizer an exact number of times, we need to find the Highest Common Factor of 75 and 69.

Now list the factors of 75 and 69.

Factors of \[75 = 5 \times 3 \times 5\]

Factors of \[69 = 23 \times 3\]

We can see that the Highest Common Factor of 75 and 69 is 3.

Therefore the Maximum required weight is 3kg.

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

Ans: The step measure of the three boys is given as 63cm, 70cm, and 77cm.

To find the maximum distance each should cover so that all the three boys can cover the distance in complete steps, we need to find the Least Common Multiple of 63, 70 and 77.

Now let us find the LCM(63,70,77).

$ 2\left| \!{\underline {\,  {63,70,77} \,}} \right.   \\  3\left| \!{\underline {\,  {63,35,77} \,}} \right.   \\ 3\left| \!{\underline {\,  {21,35,77} \,}} \right.   \\  5\left| \!{\underline {\,  {7,35,77} \,}} \right.   \\  7\left| \!{\underline {\,  {7,7,77} \,}} \right.   \\   11\left| \!{\underline {\,  {1,1,11} \,}} \right.   \\   1\left| \!{\underline {\,  {1,1,1} \,}} \right.   $

LCM (63,70, 77) = \[7 \times 9 \times 10 \times 11 = 6930cm\]

Therefore, the maximum distance each of the boys should cover is 6930cm.

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.

Ans: It is given that, 

Length of the room = 825cm.

Breadth of the room = 675cm.

Height of the room = 450cm.

To find the measurement of the tape we need to find the Highest Common Factor of 825, 675, 450.

Factors of \[825 = 5 \times 5 \times 3 \times 11\]

Factors of \[675 = 5 \times 5 \times 3 \times 3 \times 3\]

Factors of \[450 = 2 \times 3 \times 3 \times 5 \times 5\]

Therefore the HCF (825, 675, 450) =\[3 \times 5 \times 5 = 75\]

So, the length of the tape that can measure the three dimensions of the room exactly is 75cm.

4. Determine the smallest 3-digit number which is exactly divisible by 6, 8 and12.

Ans: To find the smallest 3-digit number that is exactly divisible by 6, 8 and 12 , first we need to find the Least Common Multiple of 6, 8 and 12.

$  2\left| \!{\underline {\,  {6,8,12} \,}} \right.   \\  2\left| \!{\underline {\,  {3,4,6} \,}} \right.   \\  2\left| \!{\underline {\,  {3,2,3} \,}} \right.   \\  3\left| \!{\underline {\,  {3,1,3} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1,1} \,}} \right.    $

Therefore the LCM (6,8,12) = \[2 \times 2 \times 2 \times 3 = 24\]

The smallest three digit number divisible by 24 is the number that is exactly divisible by 6, 8 and 12.

120 is the smallest 3-digit number divisible by 24.

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

5. Determine the largest 3-digit number which is exactly divisible by 8, 10 and12.

Ans: To find the largest 3-digit number that is exactly divisible by 8, 10 and 12, we need to find the Least Common Multiple of 8, 10 and 12.

$  2\left| \!{\underline {\,  {8,10,12} \,}} \right.   \\  2\left| \!{\underline {\,  {4,5,6} \,}} \right.   \\  2\left| \!{\underline {\,  {2,5,3} \,}} \right.   \\  3\left| \!{\underline {\,  {1,5,3} \,}} \right.   \\  5\left| \!{\underline {\,  {1,5,1} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1,1} \,}} \right.    $

Therefore the LCM (8,10,12) = \[2 \times 2 \times 2 \times 3 \times 5 = 120\]

The largest 3-digit number divisible by 120 is the number that is exactly divisible by 8, 10 and 12.

960 is the largest 3-digit number divisible by 120

So, the largest 3-digit number that is exactly divisible by 8, 10 and 12 is 960.

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?

Ans: It is given that 3 different traffic lights change after every 48 seconds, 72 seconds and 108 seconds.

In order to find after how many seconds the traffic light changes simultaneously we need to find the Least Common Multiple of 48, 72 and 108.

$  2\left| \!{\underline {\,  {48,72,108} \,}} \right.   \\  2\left| \!{\underline {\,  {24,36,54} \,}} \right.   \\  2\left| \!{\underline {\,  {12,18,27} \,}} \right.   \\  3\left| \!{\underline {\,  {6,9,27} \,}} \right.   \\  3\left| \!{\underline {\,  {2,3,9} \,}} \right.   \\  2\left| \!{\underline {\,  {2,1,3} \,}} \right.   \\  3\left| \!{\underline {\,  {1,1,3} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1,1} \,}} \right.    $

Therefore the LCM (48,72,108) =\[2 \times 2 \times 2 \times 3 \times 3 \times 2 \times 3 = 432\]sec.

After 432sec the 3 different traffic signals will change simultaneously. 

i.e. if the change simultaneously at 7.am then,

432 sec = 7 minutes 12 seconds.

Therefore the time they will change simultaneously again = 7.am+7minutes 12 seconds

= 7:07:12am

The time the 3 different traffic signal changes simultaneously is 7:07:12am

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 the exact number of times. 

Ans: It is given that 3 tankers can contain a diesel of 403 litres, 434 litres and 465 litres. In order to find the maximum capacity of the container we need to find the Highest Common Factor of 403, 434 and 465 litres.

Factors of \[403 = 31 \times 13\]

Factors of \[434 = 7 \times 2 \times 31\]

Factors of \[465 = 31 \times 3 \times 5\]

Therefore the HCF (403, 434, 465) = 31.

Thus 31 litres is the maximum capacity of a container that can measure the diesel of three containers an exact number of times.

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

Ans: To find the least number divided by 6, 15 and 18, we need to find the Least Common Multiple of 6, 15 and 18.

$  2\left| \!{\underline {\,  {6,15,18} \,}} \right.   \\  3\left| \!{\underline {\,  {3,15,9} \,}} \right.   \\  3\left| \!{\underline {\,  {1,5,3} \,}} \right.   \\  5\left| \!{\underline {\,  {1,5,1} \,}} \right.   \\  1\left| \!{\underline {\, {1,1,1} \,}} \right.    $

The LCM (6, 15, 18) = \[2 \times 3 \times 3 \times 5 = 90\]

90 is the least number that is divided by 6, 15 and 18. 

Therefore, the least number which when divided by 6, 15, and 18 leaves remainder 5 is 90+5 = 95.

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

Ans: To find the smallest 4-digit number divisible by 18, 24 and 32 we need to find the Least Common Multiple of 18, 24 and 32.

$  2\left| \!{\underline {\,  {18,24,32} \,}} \right.   \\  2\left| \!{\underline {\,  {9,12,16} \,}} \right.   \\  2\left| \!{\underline {\,  {9,6,8} \,}} \right.   \\  2\left| \!{\underline {\,  {9,3,4} \,}} \right.   \\  2\left| \!{\underline {\,  {9,3,2} \,}} \right.   \\  3\left| \!{\underline {\,  {9,3,1} \,}} \right.   \\  3\left| \!{\underline {\,  {3,1,1} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1,1} \,}} \right.   $

The LCM (18, 24, 32)=\[2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 288\]

The smallest 4-digit number that is divisible by 288 is the number that is exactly divisible by 18, 24, and 32.

1152 is the smallest 4-digit number divisible by 288.

So the smallest 4-digit number that is divisible by 18, 24 and 32 is 1152.

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

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

1. 9 and 4

Ans: LCM of 9 and 4:

$  2\left| \!{\underline {\,  {9,4} \,}} \right.   \\  2\left| \!{\underline {\,  {9,2} \,}} \right.   \\  3\left| \!{\underline {\,  {9,1} \,}} \right.   \\  3\left| \!{\underline {\,  {3,1} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1} \,}} \right.   $

   LCM (9, 4) = \[2 \times 2 \times 3 \times 3 = 36\]

   Therefore LCM (9, 4) = 36

2. 12 and 5

Ans:  LCM of 12 and 5:

$  2\left| \!{\underline {\,  {12,5} \,}} \right.   \\  2\left| \!{\underline {\,  {6,5} \,}} \right.   \\  3\left| \!{\underline {\,  {3,5} \,}} \right.   \\  5\left| \!{\underline {\,  {1,5} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1} \,}} \right.    $

   LCM (12, 5) = \[2 \times 2 \times 3 \times 5 = 60\]

   Therefore the LCM (12, 5) = 60

3. 6 and 5

Ans:  LCM of 6 and 5

$  2\left| \!{\underline {\,  {6,5} \,}} \right.   \\  3\left| \!{\underline {\,  {3,5} \,}} \right.   \\  5\left| \!{\underline {\,  {1,5} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1} \,}} \right.    $

   The LCM (6, 5) = \[2 \times 3 \times 5 = 30\]

   Therefore the LCM (6, 5) = 30

4. 15 and 4

Ans:  The LCM of 15 and 4:

$  2\left| \!{\underline {\,  {15,4} \,}} \right.   \\  2\left| \!{\underline {\,  {15,2} \,}} \right.   \\  3\left| \!{\underline {\,  {15,1} \,}} \right.   \\  5\left| \!{\underline {\,  {5,1} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1} \,}} \right.    $

   LCM (15, 4) = \[2 \times 2 \times 3 \times 5 = 60\]

LCM(15, 4) = 60

The common property obtained in all the LCMs is that all the LCM is the multiple of 3.

Yes, the LCMs are the product of two numbers in each case.

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

What do you observe in the result obtained?

1. 5, 20

Ans: LCM of 5 and 20:

$  2\left| \!{\underline {\,  {5,20} \,}} \right.   \\  2\left| \!{\underline {\,  {5,10} \,}} \right.   \\  5\left| \!{\underline {\,  {5,5} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1} \,}} \right.    $

The LCM (5, 20) =\[2 \times 2 \times 5 = 20\]

LCM (5, 20) = 20

2. 6, 18

Ans:    LCM of 6 and 18:

$  2\left| \!{\underline {\,  {6,18} \,}} \right.   \\  3\left| \!{\underline {\,  {3,9} \,}} \right.   \\  3\left| \!{\underline {\,  {1,3} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1} \,}} \right.    $

   The LCM (6, 18) = \[2 \times 3 \times 3 = 18\]

    Therefore LCM (6, 18) = 18

3. 12, 48

Ans:   LCM of 12 and 48:

$  2\left| \!{\underline {\,  {12,48} \,}} \right.   \\  2\left| \!{\underline {\,  {6,24} \,}} \right.   \\  2\left| \!{\underline {\,  {3,12} \,}} \right.   \\  2\left| \!{\underline {\,  {3,6} \,}} \right.   \\  3\left| \!{\underline {\,  {3,3} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1} \,}} \right.   $

   The LCM (12, 48) = \[2 \times 2 \times 2 \times 2 \times 3 = 48\]

   Therefore the LCM (12, 48) = 48

4. 9, 45

Ans: The LCM of 9 and 25:

$ 3\left| \!{\underline {\,  {9,45} \,}} \right.   \\  3\left| \!{\underline {\,  {3,15} \,}} \right.   \\  5\left| \!{\underline {\,  {1,5} \,}} \right.   \\  1\left| \!{\underline {\,  {1,1} \,}} \right.    $

   The LCM (9, 45) = \[3 \times 3 \times 5 = 45\]

   Therefore LCM (9, 45) = 45

From the answers of the LCM we can clearly observe that the LCM of the two given numbers is equal to that of the larger number given.

Conclusion

Class 6 Maths 3.7 Exercise in Chapter 3, "Playing with Numbers," marks an important step in your mathematical journey by solidifying students understanding of various number concepts. This exercise focuses on reinforcing the properties of numbers, factors, multiples, and the identification of prime and composite numbers. By diligently working through Class 6 Ex 3.7, students have honed their skills in recognizing and applying these properties, which are fundamental to more advanced mathematical operations. Mastering these concepts will greatly benefit you in future chapters and other areas of mathematics. The ability to identify and work with factors, multiples, primes, and composites is crucial for solving more complex problems efficiently.

Class 6 Maths Chapter 3: Exercises Breakdown

CBSE Class 6 Maths Chapter 3 Other Study Materials

Chapter-Specific NCERT Solutions for Class 6 Maths

Given below are the chapter-wise NCERT Solutions for Class 6 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.