ML Aggarwal Playing With Numbers Exe-4.4 Class 6 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-4.4 Questions for Playing With Numbers as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-6.

## ML Aggarwal Playing With Numbers Exe-4.4 Class 6 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 6th Chapter-4 Playing With Numbers Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-4.4 Questions Edition 2023-2024

### Playing With Numbers Exe-4.4

ML Aggarwal Class 6 ICSE Maths Solutions

Page-81

#### Question 1. Find the H.C.F. of the given numbers by prime factorization method :

(i) 28, 36
(ii) 54, 72, 90
(iii) 105, 140, 175

#### Answer:

(i) Prime factorization of given number are

28 = 2 x 2 x 7
36 = 2 x 2 x 3 x 3

Notice that 2 occurs as a common prime factor at least 2 times in both given
numbers.
so, H.C.F =  2 x 2-4

(ii) Prime factorization of the given numbers are
54 = 2 x 3 x 3 x 3
72 = 2 x 2 x 2 x 3 x 3
90 = 2 x 3 x 3 x 5

Notice that 2 occurs as a common prime factor at least one times and 3 at least two times in given numbers.

so, H.C.F = 2 x 3 x 3 = 18

(iii) Prime factorization of given numbers are:

105 = 3 x 5 x 7
140 = 2 x 2 x 5 x
175 = 5 x 5 x 7

Notice that 5 occurs as a common prime factor at least one time and 7 one time
given numbers.

so, H.C.F = 5 x 7 = 35

#### Question 2. Find the H.C.F. of the given numbers by division method:

(i) 198, 429
(ii) 20, 64, 104
(iii) 120, 144, 204

#### Answer:

(i) 198, 429

Last reminder = 0

So, H.C.F. = 33 (it is last divisor)

(ii) 20, 64, 104

Find the HCF of 20 and 64

HCF of 20 and 64 = 4

Now, find the HCF of 4 and 104

So , HCF of 20, 64, 104 = 4

(iii) 120, 144, 204

Find the HCF of 120 and 144

HCF of 120 and 204 is 12

Now, find the HCF of 12 and 144

So , HCF of 120, 144, 204 = 12

#### Question 3. Fill in the blanks:

(i) HCF of two consecutive natural numbers is ……
(ii) HCF of two consecutive odd numbers is …..
(iii) HCF of two consecutive even numbers is ……

#### Answer:

(i) HCF of two consecutive natural numbers is 1.

(ii) HCF of two consecutive odd numbers is 1.
(iii) HCF of two consecutive even numbers is 2.

#### Answer:

When 257 is divided by the required number, 5 is left as a remainder.

So, 257-5 = 252 (ex. 252 is exactly divisible)
by that number.

Similarly 329-5 = 324 (is exactly divisible by that number).

So, 252 and 324 are both divisible by that number.
Thus, the required number is the H.C.E. of 252 and 324.

Hence, the required number 36

#### Answer:

Numbers are 623, 729 and 841 and remainders are 3, 9. 1 respectively
So, Numbers will be 623-3 = 620

729 – 9 = 720
841 – 1 = 840

Now, let us find the H.C.E. of 620, 720 and 840

The HCF of 620, 720, and 840 is 20

So, The largest number = 20

#### Answer:

Weight of two bags = 75 kg and 69 kg

Factor of 75 is = 3 x 5 x 5

and Factor of 69 is  = 3 x 23

So, HCF of 75 and 69 is = 3

The maximum value of weight which can measure the weight of the rich exact number of times is 2 kg

#### Answer:

Three tankers capacity is 403, 434, 465

So, maximum capacity of a container, that can measure the diesel of three tankers

= HCF of 403, 434, 465 = 31

So, Reburied measure = 31 ltr

—  : End of ML Aggarwal Playing With Numbers Exe-4.4 Class 6 ICSE Maths Solutions :–

Thanks

Share with your friends

You might also like
Leave a comment