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

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2D.

**Other Exercises**

- RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2A
- RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2B
- RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2C
- RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2D
- RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2E
- RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2F

**Find the H.C.F. of the numbers in each of the following, using the prime factorization method :**

**Question 1.**

**Solution:**

We have

84 = 2 x 2 x 3 x 7

= 2^{2} x 3 x 7

98 = 2 x 7 x 7 = 2 x 7^{2}

∴H.C.F. =2 x 7 = 14.

**Question 2.**

**Solution:**

We have

So, 170 = 2 x 5 x 17

238 = 2 x 7 x 17

∴ H.C.F. of 170 and 238 = 2 x 17 = 34

**Question 3.**

**Solution:**

We have

So, 504 = 2 x 2 x 2 x 3 x 3 x 7 = 2^{3} x 3^{2} x 7

980 = 2 x 2 x 5 x 7 x 7 = 2^{2} x 5 x 7^{2}

∴ H.C.F. of 504 and 980 = 2^{2} x 7

= 4 x 7 = 28.

**Question 4.**

**Solution:**

We have

So, 72 = 2 x 2 x 2 x 3 x 3 = 2^{3} x 3^{2}

108 = 2 x 2 x 3 x 3 x 3 = 2^{2} x 3^{3}

180 = 2 x 2 x 3 x 3 x 5 = 2^{2} x 3^{2} x 5

∴ H.C.F. of 72, 108,

180 = 2^{2} x 3^{2}

= 4 x 9 = 36

**Question 5.**

**Solution:**

We have

84 = 2 x 2 x 3 x 7 = 2^{2} x 3 x 7

120 = 2 x 2 x 2 x 3 x 5 = 2^{3} x 3 x 5

138 = 2 x 3 x 23

∴ H.C.F. of 84, 120 and 138 = 2 x 3 = 6

**Question 6.**

**Solution:**

We have

106 = 2 x 53

159 = 3 x 53

371 = 7 x 53

∴ H.C.F. of 106, 159, 371 = 53

**Question 7.**

**Solution:**

We have

272 = 2 x 2 x 2 x 2 x 17 = 2^{4} x 17

425 = 5 x 5 x 17

= 5^{2} x 17

∴ H.C.F. of 272 and 425 = 17.

**Question 8.**

**Solution:**

We have

So, 144 = 2 x 2 x 2 x 2 x 3 x 3 = 2^{4} x 3^{2}

252 = 2 x 2 x 3 x 3 x 7 = 2^{2} x 3^{2} x 7

630 = 2 x 3 x 3 x 5 x 7 = 2 x 3^{2} x 5 x 7

∴ H.C.F. of 144, 252 and 630 = 2 x 3^{2}

= 2 x 9 = 18.

**Question 9.**

**Solution:**

We have

So, 1197 = 3 x 3 x 7 x 19 = 3^{2} x 7 x 19

5320 = 2 x 2 x 2 x 5 x 7 x 19 = 2^{3} x 5 x 7 x 19

4389 = 7 x 3 x 11 x 19

∴ H.C.F. of 1197, 5320,

4389 = 7 x 19 = 133.

**Find the H.C.F. of the numbers in each of the following using division method:**

**Question 10.**

**Solution:**

By division method, we have :

∴ H.C.F. of 58 and 70 = 2.

**Question 11.**

**Solution:**

By division method, we have

∴ H.C.F. of 399 and 437 = 19

**Question 12.**

**Solution:**

By division method, we have

∴ H.C.F. of 1045 and 1520 = 95.

**Question 13.**

**Solution:**

By division method, we have

∴ H.C.F. of 1965 and 2096 = 131

**Question 14.**

**Solution:**

By division method, we have

∴ H.C.F. of 2241 and 2324 = 83.

**Question 15.**

**Solution:**

First, we find the H.C.F. of 658 and 940

∴ H.C.F. of 658 and 940 is 94.

Now, we find the H.C.F. of 94 and 1128.

∴ H.C.F. of 94 and 1128 = 94

Hence, H.C.F. of 658, 940 and 1128 = 94.

**Question 16.**

**Solution:**

First, we find the H.C.F. of 754 and 1508

∴ H.C.F. of 754 and 1508 is 754

Now, we find the H.C.F. of 754 and 1972

∴ H.C.F. of 754 and 1972 = 58

Hence, the H.C.F. of 754,1508 and 1972 = 58.

**Question 17.**

**Solution:**

First, we find the H.C.F. of 391 and 425

∴ H.C.F. of 391 and 425 is 17.

Now, we find the H.C.F. of 17 and 527

∴ H.C.F. of 17 and 527 is 17 Hence, H.C.F. of 391, 425 and 527 = 17.

**Question 18.**

**Solution:**

First, we find the H.C.F. of 1794 and 2346

H.C.F. of 1794 and 2346 is 138.

Now, we find the H.C.F. of 138 and 4761

∴ H.C.F. of 138 and 4761 is 69.

Hence, the H.C.F. of 1794, 2346 and 4761 = 69.

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

**Question 19.**

**Solution:**

First, we find the H.C.F. of 59, 97.

∴H.C.F. of 59 and 97 is 1.

Hence 59 and 97 are co-prime.

**Question 20.**

**Solution:**

First, we find the H.C.F. of 161 and 192.

∴ H.C.F. of 161 and 192 is 1.

Hence 161 and 192 are co-prime.

**Question 21.**

**Solution:**

First, we find the H.C.F. of 343 and 432

∴ H.C.F. of 343 and 432 is 1.

Hence 343 and 432 are co-prime.

**Question 22.**

**Solution:**

First, we find the H.C.F. of 512 and 945.

∴ H.C.F. of 512 and 945 is 1.

Hence 512 and 945 are co-prime.

**Question 23.**

**Solution:**

First, we find the H.C.F. of385 and 621

∴ H.C.F. of 385 and 621 is 1.

Hence the numbers 385 and 621 are co-prime

**Question 24.**

**Solution:**

First, we find the H.C.F. of 847 and 1014.

∴ H.C.F. of 847 and 1014 is 1.

Hence 847 and 1014 are co-prime.

**Question 25.
**

**Solution:**

Clearly, we have to find the greatest number which divides (615 – 6) and (963 – 6) exactly.

So, the required number = H.C.F. of 615 – 6 = 609 and 963 – 6 = 957

The required greatest number = 87.

**Question 26.**

**Solution:**

Clearly, we have to find the greatest number which divides 2011 – 9 = 2002 and 2623 – 5 = 2618.

So, the required greatest number = H.C.F. of 2002 and 2618

∴ Required greatest number = 154.

**Question 27.**

**Solution:**

Clearly, we have to find the greatest number which divides (445 4), (572 – 5) and (699 – 6). So, the required number = H.C.F. of 441, 567 and 693. First we find the H.C.F. of 441 and 567

∴ H.C.F. of 441 and 567 is 63

So H.C.F. of 63 and 693 is 63

Hence the required number = 63.

**Question 28.**

**Solution:**

(i) The given fraction = \(\\ \frac { 161 }{ 207 } \)

First we find the H.C.F. of 161 and 207

**Question 29.**

**Solution:**

Lengths of three pieces of timber = 42 metres, 49 metres, 63 metres Greatest possible length of each plank = H.C.F. of 42 metres, 49 metres and 63 metres

First we find the H.C.F. of 42 and 49

∴ H.C.F. of 42 and 49 = 7

Now we find the H.C.F. of 7 and 63

So, the H.C.F. of 7 and 63 is 7

∴ H.C.F. of 42 metres, 49 metres of 63 metres = 7 metres

Hence required possible length of each plank = 7 metres.

**Question 30.**

**Solution:**

Quantity of milk in three different containers = 403 L, 434 L and 465 L Clearly, the maximum capacity of the required container = H.C.F. of 403 L, 434 L, 465 L, we have

∴ 403 = 13 x 31

434 = 2 x 7 x 31

465 = 5 x 3 x 31

So the H.C.F. of 403 L, 434 L and 465 L = 31 L

Maximum capacity of the required container = 31 L.

**Question 31.**

**Solution:**

The given fruits = 527 apples, 646 pears and 748 oranges

Clearly, the greatest number of fruits in each heap = H.C.F. of 527, 646 and 748 we have

∴ 527 = 17 x 31

646 = 2 x 17 x 19

748 = 2 x 2 x 11 x 17

So, the H.C.F. of 527, 646 and 748 = 17

∴ Required number of fruits in each heap = 17

Total number of fruits = 527 + 646 + 748 = 1921

Number of heaps = \(\frac { Total\quad number\quad of\quad fruits }{ Number\quad of\quad fruits\quad in\quad one\quad heap } \)

= \( \frac { 1921 }{ 17 } \)

=113

**Question 32.**

**Solution:**

The given lengths are :

7 metres = 7 x 100 cm

= 700 cm 3 metres 85 cm

= (3 x 100 + 85) cm

= 385 cm

12 metres 95 cm = (12 x 100 + 95) cm

= 1295 cm

Clearly, the length of the required longest tape = H.C.F. of 700 cm, 385 cm and 1295 cm

We have

So, 700 = 2 x 2 x 5 x 5 x 7

= 2^{2} x 5^{2} x 7

385 = 5 x 7 x 11

1295 = 5 x 7 x 37

∴ H.C.F. of 700, 385 and 1295 = 5 x 7 = 35

∴ The required length of the longest tape

= 35 cm.

**Question 33.**

**Solution:**

Length of the courtyard = 18 m 72 cm = (18 x 100 + 72) cm = 1872 cm

Breadth of the courtyard = 13 m 20 cm = (13 x 100 + 20) cm = 1320 cm

Greatest side of each of the square tiles = H.C.F. of 1872 cm and 1320 cm

Now

1872 = 2 x 2 x 2 x 2 x 9 x 13

= 2^{4} x 3^{2} x 13

1320 = 2 x 2 x 2 x 3 x 5 x 11

= 2^{3} x 3 x 5 x 11

So, the H.C.F. of 1872 and 1320

= 2^{3} x 3 = 8 x 3 = 24

Greatest side of the square tile = 24 cm

Now Area of the courtyard = Length x Breadth = 1872 x 1320 cm^{2}

Area of one square tile = Side x Side

= 24 x 24 cm^{2}

∴ Least possible number of such tiles

= \(\frac { Area\quad of\quad the\quad courtyard }{ Area\quad of\quad the\quad tile } \)

= \( \frac { 1872\times 1320 }{ 24\times 24 } \)

= 78 x 55

= 4290

**Question 34.**

**Solution:**

Let the two prime numbers be 13 and 17, we find the H.C.F. of 13 and 17 as under

Hope given RS Aggarwal Solutions Class 6 Chapter 2 Factors and Multiples Ex 2D are helpful to complete your math homework.

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