RD Sharma Solutions for Class 6 Maths Exercise 2.11 PDF
Each concept is explained in a comprehensive manner with a huge number of examples to make understanding easier for the students. The solutions can be used as a reference material by the students which helps them in performing well in the exam for Class 6. The students can self-analyse their performance and spend more time on their areas of weaknesses. The solutions of exercise-wise problems are available in PDF format, which can be downloaded for free by the students. RD Sharma Solutions for Class 6 Maths Chapter 2 Playing with Numbers Exercise 2.11 are provided here.
RD Sharma Solutions for Class 6 Maths Chapter 2: Playing with Numbers Exercise 2.11 Download PDF
Access answers to Maths RD Sharma Solutions for Class 6 Chapter 2: Playing with Numbers Exercise 2.11
1. For each of the following pairs of numbers, verify the property:
Product of the number = Product of their HCF and LCM
(i) 25, 65
(ii) 117, 221
(iii) 35, 40
(iv) 87, 145
(v) 490, 1155
Solution:
(i) 25, 65
Prime factorization of
25 = 5 × 5
65 = 5 × 13
So we get
HCF of 25, 65 = 5
LCM of 25 ,65 = 5 × 5 × 13 = 325
So we get the product of numbers = 25 × 65 = 1625
Product of HCF and LCM = 5 × 325 = 1625
Hence, it is verified that product of the number = product of their HCF and LCM.
(ii) 117, 221
Prime factorization of
117 = 3 × 3 × 13
221 = 13 × 17
So we get
HCF of 117, 221 = 13
LCM of 117, 221 = 3 × 3 × 13 × 17 = 1989
So we get the product of numbers = 117 × 221 = 25857
Product of HCF and LCM = 13 × 1989 = 25857
Hence, it is verified that product of the number = product of their HCF and LCM.
(iii) 35, 40
Prime factorization of
35 = 5 × 7
40 = 2 × 2 × 2 × 5
So we get
HCF of 35, 40 = 5
LCM of 35, 40 = 2 × 2 × 2 × 5 × 7 = 280
So we get the product of numbers = 35 × 40 = 1400
Product of HCF and LCM = 5 × 280 = 1400
Hence, it is verified that product of the number = product of their HCF and LCM.
(iv) 87, 145
Prime factorization of
87 = 3 × 29
145 = 5 × 29
So we get
HCF of 87, 145 = 29
LCM of 87, 145 = 3 × 5 × 29 = 435
So we get the product of numbers = 87 × 145 = 12615
Product of HCF and LCM = 29 × 435 = 12615
Hence, it is verified that product of the number = product of their HCF and LCM.
(v) 490, 1155
Prime factorization of
490 = 2 × 5 × 7 × 7
1155 = 3 × 5 × 7 × 11
So we get
HCF of 490, 1155 = 35
LCM of 490, 1155 = 2 × 3 × 3 × 5 × 7 × 7 × 11 = 16170
So we get the product of numbers = 490 × 1155 = 565950
Product of HCF and LCM = 35 × 16170 = 565950
Hence, it is verified that product of the number = product of their HCF and LCM.
2. Find the HCF and LCM of the following pairs of numbers:
(i) 117, 221
(ii) 234, 572
(iii) 145, 232
(iv) 861, 1353
Solution:
(i) 117, 221
Prime factorization of
117 = 3 × 3 × 13
221 = 13 × 17
So the required HCF = 13
Required LCM = 3 × 3 × 13 × 17 = 1989
(ii) 234, 572
Prime factorization of
234 = 2 × 3 × 3 × 13
572 = 2 × 2 × 11 × 13
So the required HCF = 26
Required LCM = 2 × 2 × 3 × 3 × 11 × 13 = 5148
(iii) 145, 232
Prime factorization of
145 = 5 × 29
232 = 2 × 2 × 2 × 29
So the required HCF = 29
Required LCM = 2 × 2 × 2 × 5 × 29 = 1160
(iv) 861, 1353
Prime factorization of
861 = 3 × 7 × 41
1353 = 3 × 11 × 41
So the required HCF = 123
Required LCM = 3 × 7 × 11 × 41 = 9471
3. The LCM and HCF of two numbers are 180 and 6 respectively. If one of the numbers is 30, find the other number.
Solution:
It is given that
LCM of two numbers = 180
HCF of two numbers = 6
One of the number = 30
We know that
Product of two numbers = Product of HCF and LCM
So we get
30 × other number = 6 × 180
On further calculation
Other number = (6 × 180)/ 30 = 36
Hence, the other number is 36.
4. The HCF of two numbers is 16 and their product is 3072. Find their LCM.
Solution:
It is given that
HCF of two numbers = 16
Product = 3072
We know that
Product of two numbers = Product of HCF and LCM
So we get
3072 = 16 × LCM
On further calculation
LCM = 3072/ 16 = 192
Therefore, the LCM of two numbers is 192.
5. The HCF of two numbers is 145, their LCM is 2175. If one number is 725, find the other.
Solution:
It is given that
HCF of two numbers = 145
LCM = 2175
We know that
Product of two numbers = Product of HCF and LCM
So we get
725 × other number = 145 × 2175
On further calculation
Other number = (145 × 2175)/ 725 = 435
Therefore, the other number is 435.
6. Can two numbers have 16 as their HCF and 380 as their LCM? Give reason.
Solution:
No. HCF of two numbers must exactly divide their LCM
We know that 16 does not divide 380 exactly
Therefore, no two numbers can exist with 16 as HCF and 380 as LCM.
RD Sharma Solutions for Class 6 Maths Chapter 2 – Playing with Numbers Exercise 2.11
RD Sharma Solutions Class 6 Maths Chapter 2 Playing with Numbers Exercise 2.11 explains the important properties of HCF and LCM of given numbers.
Key Features of RD Sharma Solutions for Class 6 Maths Chapter 2: Playing with Numbers Exercise 2.11
- The solutions created by subject experts help in obtaining an overall idea about the concepts.
- The self-study method strengthens among students, which helps them clear doubts by themselves, referring to the PDF.
- Important formulas, shortcuts and tips are highlighted in order to help students grasp the concepts faster.
- A huge number of examples before each exercise improves knowledge among students about the various methods used in solving problems.
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