## NCERT Class 6 Mathematics Third Chapter Playing with Numbers Exercise 3.7 Solutions

## 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. **

Solution: The maximum value of weight will be the HCF of 75 and 69.

It is found as follows:

Hence, 75 = 3 × 5 × 5

And 69 = 3 × 23

The common factor of 75 and 69 is 3.

Thus, the HCF of 75 and 69 is 3.

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

**(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 distance covered by each one of them is required to be the same as well as minimum. The required minimum distance each should walk would be the lowest common multiple of the measures of their steps. Thus, we find the LCM of 63, 70 and 77.

The LCM of 63, 70 and 77 is 6930.

The required minimum distance is 6930 cm.

**(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: We need HCF of 825, 675 and 450.

It is found as follows:

Hence, 825 = 3 × 5 × 5 × 11

675 = 3 × 3 × 3 × 5 × 5

450 = 2 × 3 × 3 × 5 × 5

The common factors of 825, 675 and 450 are 3 × 5 × 5.

Thus, the HCF of 825, 675 and 450 is 3 × 5 × 5 = 75.

Therefore, the longest tape is 75 cm.

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

Ans: The required number is 120.

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

Ans: The number 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? **

Solution: Therefore, LCM of 48, 72, 108= 2 x 2 x 3 x 3 x 2 x 2 x 3= 432 sec

So, they will change again simultaneously after 432 sec i.e. 7 minutes 12 sec.

**In case you are missed :- Previous Chapter Exercise Solution**

**(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 such a container will be the HCF of 403, 434 and 465.

Hence, 403 = 13 × 31

434 = 2 × 7 × 31

465 = 3 × 5 × 31

The common factor of 403, 434 and 465 is 31.

Thus, the HCF of 403, 434 and 465 is 31.

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

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

Solution: Therefore, LCM of 6, 15 and 18 = 2 × 3 × 3 × 5 = 90.

So, the least number is (90 + 5) = 95.

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

Solution: 1152.

**(10) Find the LCM of the following numbers: **

(a) 9 and 4 = 9 × 4 = 36

(b) 12 and 5 = 12 × 5 = 60

(c) 6 and 5 = 6 × 5 = 30

(d) 15 and 4 = 15 × 4 = 60

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

Ans: Here, in each case LCM is a multiple of 3.

Yes, in each case LCM = the product of two numbers

**(11) Find the LCM of the following numbers in which one number is the factor of the other. **

(a) 5, 20

Ans: LCM of 5 and 20 is 5 × 4 = 20

(b) 6, 18

Ans: LCM of 6 and 18 is 2 × 3 × 3 = 18

(c) 12, 48

Ans: LCM of 12 and 48 is 2 × 2 × 3 × 4 = 48

(d) 9, 45

Ans: LCM of 9 and 45 is 3 × 3 × 5 = 45

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

Ans: The LCM of the given numbers in each case is the larger of the two numbers.

**In case you are missed :- Next Chapter Exercise Solution**