RS Aggarwal Solutions for Class 6 Chapter 2 Factors and Multiples Exercise 2D PDF are provided here. The exercise wise solutions designed by faculty at BYJUâ€™S help students in solving RS Aggarwal textbook problems. The solutions are in explanatory manner to improve conceptual knowledge among the students. The primary objective of preparing the solutions is to make the subject easier for students.

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**Find the HCF of the numbers in each of the following, using the prime factorization method:**

**1. 84, 98**

** Solution**

** **

The prime factors of 84 = 2 Ã— 2 Ã— 3 Ã— 7 = 2^{2 }Ã— 3 Ã— 7

The prime factors of 98 = 2 Ã— 7 Ã— 7 = 2 Ã— 7^{2}

^{ }âˆ´ HCF = 2Ã— 7 = 14

**2. 170, 238**

** Solution**

** **

** **

** **The prime factors of 170 = 2 Ã— 5 Ã— 17

The prime factors of 238 = 2 Ã— 7 Ã— 17

âˆ´ HCF = 2 Ã— 17 = 34

** 3. 504,980**

** Solution**

** **

** **

The prime factors of 504 = 2^{3 }Ã— 3^{2 }Ã— 7

The prime factors of 980 = 2^{2 }Ã— 5 Ã— 7^{2}

^{ }âˆ´ HCF = 2^{2 }Ã— 7 = 28

**72, 108, 180**

**Solution**

^{ }

^{ }

The prime factors of 72 = 2^{3 }Ã— 3^{2 }

The prime factors of 108 = 2^{3 }Ã—3

The prime factors of 180 = 2^{2 }Ã— 3^{2 }Ã—5

âˆ´ HCF = 2^{2 }Ã— 3^{2 }= 36

**84, 120, 138**

**Solution**

** **

** **

** **

** **

** **The prime factors of 84 = 2^{2 }Ã— 3 Ã—7

The prime factors of 120 = 2^{3 }Ã— 3 Ã— 5

The prime factors of 138 = 2 Ã— 3 Ã— 23

âˆ´ HCF = 2 Ã— 3 = 6

**106,159,371**

**Solution**

The prime factors of 106 = 2 Ã— 53

The prime factors of 159 = 3 Ã— 53

The prime factors of 371 = 7 Ã— 53

âˆ´ HCF = 53

**272, 425**

**Solution**

** **

** **

The prime factors of 272 = 2^{4} Ã— 17

The prime factors of 425 = 5^{2} Ã— 17

âˆ´ HCF = 17

**8**.**144, 252,630**

** Solution**

** **

** **

The prime factors of 144 = 2^{4 }Ã— 3^{2}

^{ }The prime factors of 252 = 2^{2 }Ã— 3^{2 }Ã— 7

The prime factors of 630 = 2 Ã— 3^{2 }Ã— 5

âˆ´ HCF = 18

**9. 1197, 5320, 4389**

** Solution**

** **

^{ }

^{ }

The prime factors of 1197 = 3^{2 }Ã— 7 Ã— 19

The prime factors of 5320 = 2^{3 }Ã— 5 Ã— 7 Ã— 19

The prime factors of 4389 = 3Ã— 7 Ã— 11 Ã— 19 ^{ }

âˆ´ HCF = 133

^{ }

** Find the HCF of the numbers in each of the following, using the division method:**

** 10. 58, 70**

** Solution**

** **

âˆ´ The common factor of 58, 70 =2

** 11. 399, 437**

** Solution**

** **

** **

** **âˆ´ The Common factor of 399, 437 = 19

** 12. 1045, 1520**

** Solution**

** **

** **âˆ´ The HCF of 1045, 1520 = 95

** 13. 1965, 2096**

** Solution**

** **

** **âˆ´ Common factor of 1965, 2096 = 131

** 14. 2241, 2324**

**Solution**

** **

** **

** **âˆ´ The Common factor of 2241, 2324 = 83

** 15. 658, 940, 1128**

**Solution**

** **

** **

** **

** **âˆ´ Common factor of 658, 940, 1128 = 94

**754, 1508, 1972**

**Solution**

** **

** **

** **

** **âˆ´ Common factor of 754, 1508, 1972 = 58

**17. 391, 425, 527**

**Solution**

âˆ´ Common factor of 391, 425, 527 = 17

**18. 1794, 2346, 4761**

**Solution**

** **âˆ´ Common factor of 1794, 2346, 4761 is 69

** Show that the following primes are co-primes:**

** 19. 59, 97**

** Solution**

59 = 59Ã— 1

97 = 97 Ã— 1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

** 20. 161,192**

** Solution**

161 = 7 Ã— 23

192 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã—3

= 2^{6 }Ã—3 Ã—1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

** 21.343, 432**

** Solution**

343 = 7Ã— 7 Ã— 7

= 7^{3 }Ã— 1

432 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3

= 2^{4 }Ã— 3^{3 }Ã— 1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

**22. 512, 945**

** Solution**

512 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

= 2^{9}

945 = 3 Ã— 3 Ã— 3 Ã— 5 Ã— 7

= 3^{3 }Ã— 5 Ã— 7

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

**23. 385, 621**

** Solution**

385 = 5 Ã— 7 Ã— 11 Ã— 1

621 = 3 Ã— 3 Ã— 3 Ã— 23 Ã— 1

= 3^{3 }Ã— 23

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

**24. 847, 1014**

** Solution**

847 = 7 Ã— 11 Ã— 11 Ã— 1

= 7 Ã— 11^{2 }Ã— 1

1014 = 2 Ã— 3 Ã— 13 Ã— 13 Ã— 1

= 2 Ã— 3 Ã— 13^{2 }Ã— 1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

**25. Find the greatest number which divides 615 and 963, leaving the remainder 6 in each case**

** Solution**

Since the remainder is 6 we have to find the number which exactly divides (615 -6) and (963 -6)

Required number = HCF of 609 and 957

Hence, 87 is the greatest number which divides 615 and 963 leaving the remainder 6

## RS Aggarwal Solutions for Class 6 Chapter 2 Factors and Multiples Exercise 2D

RS Aggarwal Solutions for Class 6 Chapter 2 Factors and Multiples Exercise 2D include the questions based on finding out the HCF and LCM using prime factorization and division method. Topics which are mentioned in other exercises are

- Finding prime and composite numbers from 1 to 100
- General properties of divisibility
- Prime factorization
- Properties of HCF and LCM of given numbers

Students can solve RS Aggarwal textbook using PDF of solutions to perform better in the exam. It helps students understand the other methods which can be used in solving the same problem in a shorter duration.