RS Aggarwal Class 6 Math Second Chapter Factors And Multiples Exercise 2D Solution

EXERCISE 2D

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

(1) 84, 98

Ans: All the factors of 84 are 2 x 2 x 3 x 7

All the factors of 98 are 2 x 7 x 7

Common factors of 84 and 98 are 2 x 7

Thus, the highest common factor of 84 and 98 is 7

Hence, HCF of 84 and 98= 7

(2) 170, 238

Ans: All the factors of 170 are 2 x 5 x 17

All the factors of 238 are 2 x 7 x 17

Common factors of 84 and 98 are 2 x 17

Thus, the highest common factor of 170 and 238 is 17

Hence, HCF of 84 and 98= 17

(3) 504, 980

Ans: All the factors of 504 are 2 x 2 x 2 x 3 x 3 x 7

All the factors of 980 are 2 x 2 x 5 x 7 x 7

Common factors of 504 and 980 are 2 x 2 x 7

Thus, the highest common factor of 504 and 980 is 7

Hence, HCF of 504 and 980= 7

(4) 72, 108, 180

Ans: All the factors of 72 are 2 x 2 x 2 x 3 x 3

All the factors of 108 are 2 x 2 x 3 x 3 x 3

All the factors of 180 are 2 x 2 x 3 x 3 x 5

Common factors of 72, 108 and 180 are 2 x 2 x 3 x 3

Thus, the highest common factor of 72, 108 and 180 is 3

Hence, HCF of 72, 108 and 180= 3

(5) 84, 120, 138

Ans: All the factors of 84 are 2 x 2 x 3 x 7

All the factors of 120 are 2 x 2 x 2 x 3 x 5

All the factors of 138 are 2 x 3 x 23

Common factors of 84, 120 and 138 are 2 x 3

Thus, the highest common factor of 84, 120 and 138 is 2 x 3

Hence, HCF of 84, 120 and 138= 3

(6) 106, 159, 371

Ans: All the factors of 106 are 2 x 53

All the factors of 159 are 3 x 53

All the factors of 371 are 7 x 53

Common factors of 106, 159 and 371 is 53

Thus, the highest common factor of 106, 159 and 371 is 53

Hence, HCF of 106, 159 and 371= 53

(7) 272, 425

Ans: All the factors of 272 are 2 x 2 x 2 x 2 x 17

All the factors of 425 are 5 x 3 x 17

Common factors of 272 and 425 is 17

Thus, the highest common factor of 272 and 425 is 17

Hence, HCF of 272 and 425= 17

(8) 144, 252, 630

Ans: All the factors of 144 are 2 x 2 x 2 x 2 x 3 x 3

All the factors of 252 are 2 x 2 x 3 x 3 x 7

All the factors of 630 are 2 x 3 x 3 x 5 x 7

Common factors of 144, 252 and 630 are 2 x 3 x 3

Thus, the highest common factor of 144, 252 and 630 is 18

Hence, HCF of 144, 252 and 630= 18

(9) 1197, 5320, 4389

Ans: All the factors of 1197 are 3 x 3 7 x 19

All the factors of 5320 are 2 x 2 x 2 x 5 x 7 x 19

All the factors of 4389 are 3 x 7 x 11 x 19

Common factors of 1197, 5320 and 4389 are 7 x 19

Thus, the highest common factor of 1197, 5320 and 4389 is 19

Hence, HCF of 1197, 5320 and 4389= 19

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

(10) 58, 70

Ans: We have:

Hence, the HCF of 58 and 70 is 2.

(11) 399, 437

Ans: We have

Hence, the HCF of 399 and 437 is 19.

(12) 1045, 1520

Ans: We have

Hence, the HCF of 1045 and 1520 is 95.

(13) 1965, 2096

Ans: We have

Hence, the HCF of 1965 and 2096 is 131.

(14) 2241, 2324

Ans: We have

Hence, the HCF of 2241 and 2324 is 83.

(15) 658, 940, 1128

Ans: We have

Hence, the HCF of 658, 940  and 1128 is 94.

(16) 754, 1508, 1972

Ans: By division method

We have, 1972 divide by 1508= 116

And 754 divide by 116

Hence, the HCF of 754, 1508 and 1972 is 58.

(17) 391, 425, 527

By division method

We have, 425 divide by 391= 17

And 527 divide by

Hence, the HCF of 391, 425 and 527 is 17

(18) 1794, 2346, 4761

Ans: By division method

We have, 4761 divide by 2346= 69

Hence, the HCF of 1794, 2346 and 4761 is 69.

Show that the following pairs are co-primes:

(19) 59, 97

Ans: 59, 97 are co-primes because their HCF is 1.

(20) 161, 192

Solution:

(21) 343, 432

Solution:

(22) 512, 945

Solution:

(23) 385, 621

Solution:

(24) 847, 1014

Solution: 

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

Ans:

(26) Find the greatest number which divides 2011 and 2623, leaving remainders 9 and 5 respectively.

Ans:

(27) Find the greatest number that will divide 445, 572 and 699, leaving remainders 4, 5, 6 respectively.

Ans:

(28) Reduce each of the following fractions to the lowest terms:

(29) Three piece of timber, 42-m, 49-m and 63-m long, to be divided into planks of the same length. What is the greatest possible length of each plank?

Solution:

(30) Three different contains contain 403 L, 434 L and 465 L of milk respectively. Find the capacity of a container which can measure the milk of all the containers in an exact number of times.

Solution:

(31) There are 527 apples, 646 pears and 748 oranges. These are to be arranged in heaps containing the same number of fruits. Find the greatest number of fruits possible in each heap. How many heaps are formed?

Solution:

(32) Determine the longest tape  which can be used to measure exactly the lengths 7 m, 3m 85 cm, and 12 m 95 cm.

Solution:

(33) A rectangular courtyard is 18 m 72 cm long 13 m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

Solution:

(34) Find the HCF of

(i) Two prime numbers= 1

(ii) Two consecutive numbers= 1

(iii) Two co-primes= 1

(iv) 2 and an even number= 2.