TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4

Students can practice TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Exercise 3.4

Question 1.
Find the HCF of the following numbers by prime factorisation and continued division method.
(i) 18, 27, 36
Answer:
The given numbers are
18, 27, 36

Prime factorisation method:
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 1
The common factor of 18, 27 and 36 is 3 × 3 = 9
Hence, HCF of 18, 27 and 36 is 9.

Continued division method :
The given numbers are 18, 27 and 36.
Let us find the HCF of 18 and 27.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 2
∴ HCF of 18 and 27 is 9.
Let us find the HCF of third number and the HCF of first.two numbers.
Let us find the HCF of 36 and 9.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 3
HCF of 18, 27 and 36 is 9.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4

(ii) 106, 159, 265
Answer:
The given numbers are 106, 159 and 265.

Prime factorisation method:
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 4
∴ The common factor of 106, 159 and 265 is 53.
Hence, HCF of 106,159 and 265 is 53.

Continued division method:
Let us find the HCF of 106 and 159.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 5
The HCF of 106 and 159 is 53.
Let us find the HCF of 265 and 53.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 6
Again the HCF of 265 and 53 is 53.
So, the HCF of 106, 159 and 265 is 53.

(iii) 10, 35, 40
Answer:
The given numbers are 10, 35 and 40.

Prime factorisation method:
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 7
∴ The common factor of 10, 35 and 40 is 5.
Hence, HCF of 10, 35 and 40 is 5.

Continued division method:
The given numbers are 10, 35 and 40.
Let us find the HCF of 10 and 35.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 8
∴HCF of 10 and 35 is 5.
Let us find the HCF of third number and the HCF of first two numbers.
Let us find the HCF of 40 and 5.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 9
∴ HCF of 40 and 5 is 5.
∴ HCF of 10, 35 and 40 is 5.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4

(iv) 32, 64, 96, 128
Answer:
The given numbers are 32, 64, 96 and 128.

Prime factorisation method:
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 10
∴ The common factor of 32, 64, 96 and 128 is 2 × 2 × 2 × 2 × 2 = 32
∴ HCF of 32, 64, 96 and 128 is 32.

Continued division method:
Let us find the HCF of 32 and 64.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 11
∴ HCF of 32 and 64 is 32.
Again find the HCF of 32 and 96.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 12
∴ HCF of 32 and 96 is 32.
Again find the HCF of 32 and 128.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 13
∴ HCF of 32 and 128 is 32.
∴ The HCF of the given numbers is 32.

Question 2.
Find the largest number which is a factor of each of the numbers 504, 792 and 1080.
Answer:
Largest number which is a factor of the given numbers is HCF.
Let us find the HCF of 504 and 792.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 14
∴ HCF of 72 and 1080 is 72.
The HCF of 504, 792 and 1080 is 72.

TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4

Question 3.
The length, breadth and height of a room are 12m, 15m and 18m respectively. Determine the length of longest stick that can measure all the dimensions of the room in exact number of times?
Answer:
The HCF of 12, 15 and 18 will give us the length of the stick which can measure all the three dimensions of the room exactly.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 15
∴ HCF of 12, 15 and 18 is 3.
The length of stick is 3 m.

Question 4.
HCF of co-prime numbers 4 and 15 was found as follows by factorization. 4 = 2 × 2 and 15 = 3 × 5. Since there is no common prime factor, HCE of 4 and 15 is ‘0’. Is the answer correct? If not, what Is the correct HCF?
Answer:
The numbers which have only 1 as the common factor are called co-primes.
In the problem it is given that 4 and 15 are co-prime numbers.
∴ Their common factor is 1.
HCF of 4 and 15 is ‘0’ given in the problem. This is not correct.
The correct HCF is 1.

Question 5.
What is the capacity of the largest vessel which can empty the oil from three vessels containing 32 litres, 24 litres and 48 litres an exact number of times ?
Answer:
The quantity of oil in three vessels is 32 litres, 24 litres and 48 litres.
TS 6th Class Maths Solutions Chapter 3 Playing with Numbers Ex 3.4 16
∴ LCM = 2 × 2 × 2 × 3 × 2 × 2 = 96
The capacity 4 the largest vessel = 96 litres.

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