NCERT Class 6 Math Exercise 3.7 Solution of Chapter 3 Playing with Numbers with explanation. Here we provide Class 6 Maths all Chapters in Hindi for cbse, HBSE, Mp Board, UP Board and some other boards.

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NCERT Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.7 Solution in english Medium.

**Class 6 Maths Chapter 3 Exercise 3.7 Solution**

**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.**

**Answer.**

For finding maximum weight, we have to find HCF of 75 and 69.

Factors of 75 = 3x5x5

Factors of 69 = 3×23

HCF = 3

Therefore the required weight is 3 Kg.

**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.**

For finding the minimum distance , we have to find LCM of 63, 70, 77.

LCM of 63, 70, 77 = 7x9x10x11 = 6930 cm.

Therefore, the minimum distance is 6930 cm.

**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 825, 675 and 450.

Factors of 825 = 3x5x5x11

Factors of 675 = 3x3x3x5x5

Factors of 450 = 2x3x3x5x5

HCF = 3x5x5 = 75 cm.

Therefore, the longest tape is 75 cm.

**Question 4.**

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

**Answer.**

LCM of 6, 8 and 12 = 2x2x2x3 = 24.

The smallest 3 digit number = 100

To find the number, we have to divide 100 by 24.

100 = 24×4 +4

Therefore, the required number = 100 + 24 – 4 = 120.

**Question 5.**

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

**Answer.**

LCM of 8, 10 and 12 = 2x2x2x3x5 = 120

The largest three digit number 999.

Now 999/120 = 120×8 + 39.

Therefore, the required number = 999 – 39 = 960.

**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.**

LCM of 48, 72 and 108

= 2x2x2x2x3x3x3 = 432 seconds.

432 seconds = 7 minutes 12 seconds.

Therefore the time = 7 a.m. + 7 minutes 12seconds

= 7:07:12 am.

**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 = HCF (403, 434, 465)

Factors of 403 = 13×31

Factors of 434 = 2x7x31

Factors of 465 = 3x5x31

HCF = 31

Therefore, 31 litres of container is required to measure the quantity.

**Question 8.**

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

**Answer.**

LCM of 6, 15 and 18 = 2x3x3x5 = 90

Therefore, the required number

= 90 + 5 = 95.

**Question 9.**

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

**Answer.**

LCM of 18, 24 and 32 = 2x2x2x2x2x3x3 = 288.

The smallest four digit number = 1000.

Now, 1000/288 = 288×3 + 136

Therefore, the required number is 1000 + 288 – 136 = 1152.

**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?**

**Answer.**

A.

LCM of 9 and 4

= 2x2x3x3

= 36

B.

LCM of 12 and 5

= 2x2x3x5

= 60

C.

LCM of 6 and 5

= 2x3x5

= 30

D.

LCM of 15 and 4

= 2x2x3x5

= 60

Yes, the LCM is equal to the product of two numbers in each case.

And LCM is also the multiple of 3.

**Question 11.**

**Find the LCM of the following numbers in which one number is the factor of other:**

**A. 5, 20**

**B. 6, 18**

**C. 12, 48**

**D. 9, 45**

**What do you observe in the result obtained?**

**Answer.**

A.

LCM of 5 and 20 is

= 2x2x5 = 20

B.

LCM of 6 and 18 is

= 2x3x3 = 18

C.

LCM of 12 and 48 is

= 2x2x2x2x3

= 48

D.

LCM of 9 and 45 is

= 3x3x5

= 45

From these all case, we can conclude that if the smallest number is the factor of largest number, then the LCM of these two numbers is equal to that of large number.