RS Aggarwal Class 6 Maths Chapter 2 - Factors and Multiples EXERCISE 2D

 

RS Aggarwal 2021-2022 for Class 6 Maths Chapter 2 - Factors and Multiples 

 EXERCISE 2D

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

1. 84, 98                 2. 170, 238 3.  504, 980

4. 72, 108, 180         5. 84, 120, 138 6. 106. 159, 371

7. 272, 425         8. 144. 252. 630 9. 1197, 5320, 4389

Solution: 1. The given numbers are 84 and 98.

We do:

284        242     321      77           1$\begin{array}{l}2\overline{)\text{84 }}\\ 2\overline{)\text{42 }}\\ 3\overline{)\text{21 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\end{array}$298      749      77          1$\begin{array}{l}2\overline{)\text{98 }}\\ 7\overline{)\text{49 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\end{array}$
 
84 = 2 × 2 × 3 × 7 = 22 × 3 × 7
98 = 2 × 7 × 7 = 2 × 72
  
∴ H.C.F of the given numbers = 2 × 7 = 14

Solution: 2. The given numbers are 170 and 238.

We do:

  2170          585     1717           1$\begin{array}{l} 2\overline{)\text{170 }}\\ 5\overline{)\text{85 }}\\ 17\overline{)\text{17 }}\\ \text{ 1}\end{array}$  2238          7119     1717              1       $$\begin{gathered} \begin{array}{l} 2\overline{)\text{238 }}\\ 7\overline{)\text{119 }}\\ 17\overline{)\text{17 }}\end{array} \\ \overline{)\text{1 }} \end{gathered}$$

170 = 2 × 5 × 17
238 = 2 × 7 × 17

∴ H.C.F. of the given numbers = 2 × 17 = 34

Solution: 3. The given numbers are 504 and 980.

We do :

2504        2252     2126      363        321        77               1      $\begin{array}{l}2\overline{)\text{504 }}\\ 2\overline{)\text{252 }}\\ 2\overline{)\text{126 }}\\ 3\overline{)\text{63 }}\\ 3\overline{)\text{21 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$       2980        2490     5245      749        77               1      $\begin{array}{l}2\overline{)\text{980 }}\\ 2\overline{)\text{490 }}\\ 5\overline{)\text{245 }}\\ 7\overline{)\text{49 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$

    504 = 2 × 2 ×2 × 3 × 3 × 7 = 23 × 32 × 7
    980 = 2 × 2 × 5× 7 × 7 = 22 × 5 × 72
∴ H.C.F of the given numbers = 22 × 7 = 28

Solution: 4. The given numbers are 72, 108 and 180

We do:

272        236     218      39        33             1      $\begin{array}{l}2\overline{)\text{72 }}\\ 2\overline{)\text{36 }}\\ 2\overline{)\text{18 }}\\ 3\overline{)\text{9 }}\\ 3\overline{)\text{3 }}\\ \text{ 1}\\ \end{array}$   2108        254     327      39        33          1      $\begin{array}{l}2\overline{)\text{108 }}\\ 2\overline{)\text{54 }}\\ 3\overline{)\text{27 }}\\ 3\overline{)\text{9 }}\\ 3\overline{)\text{3 }}\\ \text{ 1}\\ \end{array}$    2180        2 90        345          315          5 5               1$\begin{array}{l}2\overline{)\text{180 }}\\ 2\overline{)\text{ 90 }}\\ 3\overline{)\text{45 }}\\ 3\overline{)\text{15 }}\\ 5\overline{)\text{ 5 }}\\ \text{ 1}\\ \end{array}$
Now, 72=2 × 2 × 2 × 3 × 3 = 23 × 32
      108 = 2 × 2 × 3 × 3 × 3 = 22 × 33
      180 = 2 × 2 × 3 ×3 × 5 = 22 × 32 × 5
 ∴ H.C.F =22×32 =36

Solution: 5. The given numbers are 84, 120 and 138.

We do:

284        242     321      77           1      $\begin{array}{l}2\overline{)\text{84 }}\\ 2\overline{)\text{42 }}\\ 3\overline{)\text{21 }}\\ 7\overline{)\text{7 }}\\ \text{ 1 }\end{array}$       2120        260     230      315        55             1      $\begin{array}{l}2\overline{)\text{120 }}\\ 2\overline{)\text{60 }}\\ 2\overline{)\text{30 }}\\ 3\overline{)\text{15 }}\\ 5\overline{)\text{5 }}\\ \text{ 1}\\ \end{array}$          2138          369     2323           1$\begin{array}{l} 2\overline{)\text{138 }}\\ 3\overline{)\text{69 }}\\ 23\overline{)\text{23 }}\\ \text{ 1}\end{array}$

Now, 84 = 2 × 2 ×3 × 7
       120 = 2 × 2× 2 ×3 × 5
       138 = 2 × 3 × 23
  ∴ H.C.F = 2 × 3 = 6

Solution: 6. The given numbers are 106, 159 and 371.

We do:

  2106        5353          1      $\begin{array}{l} 2\overline{)\text{106 }}\\ 53\overline{)\text{53 }}\\ \text{ 1}\\ \end{array}$  3159        5353          1      $\begin{array}{l} 3\overline{)\text{159 }}\\ 53\overline{)\text{53 }}\\ \text{ 1}\\ \end{array}$  7371        5353          1      $\begin{array}{l} 7\overline{)\text{371 }}\\ 53\overline{)\text{53 }}\\ \text{ 1}\\ \end{array}$
Now, 106 = 2 × 53
         159 = 3 × 53
         371 = 7 × 53
    ∴ H.C.F = 53

Solution 7. Given numbers are 272 and 425.

We do:

  2272          2136       268        234        1717             1      $\begin{array}{l} 2\overline{)\text{272 }}\\ 2\overline{)\text{136 }}\\ 2\overline{)\text{68 }}\\ 2\overline{)\text{34 }}\\ 17\overline{)\text{17 }}\\ \text{ 1}\\ \end{array}$          5425          585     1717           1      $\begin{array}{l} 5\overline{)\text{425 }}\\ 5\overline{)\text{85 }}\\ 17\overline{)\text{17 }}\\ \text{ 1}\\ \end{array}$
Now, 272 = 2 × 2× 2 × 2 × 17
         425 = 5 × 5 × 17
∴ The required H.C.F is 17.

Solution 8. The given numbers are 144, 252 and 630.

We do:

2144        272      236      218        39         33             1      $\begin{array}{l}2\overline{)\text{144 }}\\ 2\overline{)\text{72 }}\\ 2\overline{)\text{36 }}\\ 2\overline{)\text{18 }}\\ 3\overline{)\text{9 }}\\ 3\overline{)\text{3 }}\\ \text{ 1}\\ \end{array}$          2252        2126        363        321        77              1      $\begin{array}{l}2\overline{)\text{252 }}\\ 2\overline{)\text{126 }}\\ 3\overline{)\text{63 }}\\ 3\overline{)\text{21 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$        2630       3315      3105      535        77             1      $\begin{array}{l}2\overline{)\text{630 }}\\ 3\overline{)\text{315 }}\\ 3\overline{)\text{105 }}\\ 5\overline{)\text{35 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$   

Now, 144 = 2 × 2× 2 × 2× 3 × 3
         252 = 2 ×2 × 3× 3 × 7
         630 = 2 ×3 × 3× 5 × 7
     ∴ H.C.F = 2 × 3 × 3 =18

Solution 9. The given numbers are 1197, 5320 and 4389.

We do:

   31197           3399           7133         1919               1      $\begin{array}{l} 3\overline{)\text{1197 }}\\ 3\overline{)\text{399 }}\\ 7\overline{)\text{133 }}\\ 19\overline{)\text{19 }}\\ \text{ 1}\\ \end{array}$        25320          22660           21330           5665             7133          1919                1      $\begin{array}{l} 2\overline{)\text{5320 }}\\ 2\overline{)\text{2660 }}\\ 2\overline{)\text{1330 }}\\ 5\overline{)\text{665 }}\\ 7\overline{)\text{133 }}\\ 19\overline{)\text{19 }}\\ \text{ 1}\\ \end{array}$          34389         71463         19209         1111                 1 $\begin{array}{l} 3\overline{)\text{4389 }}\\ 7\overline{)\text{1463 }}\\ 19\overline{)\text{209 }}\\ 11\overline{)\text{11 }}\\ \text{ 1 }\end{array}$

Now, 1197 = 3 × 3× 7× 19 = 32 × 7 × 19
         5320 = 2 × 2 × 2× 5 × 7 × 19 = 23 × 5 × 7 × 19
         4389 = 3 ×7 × 19 × 11
∴ Required H.C.F = 19 × 7 = 133

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

10. 58, 70         11, 399, 437 12. 1045, 1520

13. 1965, 2096         14. 2241, 2324 15. 658, 940. 1128

16. 754, 1508, 1972 17. 391.425, 527 18. 1794, 2346, 4761

Solution : 10. We do:

∴ The H.C.F of 58 and 70 is 2.

Solution:11. The given numbers are 399 and 437.

    We do:

∴ The H.C.F of 399 and 437 is 19.

Solution:12.  The given numbers are 1045 and 1520.

    We do:

∴ The H.C.F of 1045 and 1520 is 95.

Solution: 13. The given numbers are 1965 and 2096.

We do:

∴ The H.C.F of 1965 and 2096 is 131.

Solution 14. The given numbers are 2241 and 2324.

We do:

∴ The H.C.F of 2241 and 2324 is 83.

Solution 15. The given numbers are 658, 940 and 1128.

We do:

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

Therefor, the H.C.F of 658 and 940 is 94.

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

Thus, the H.C.F of 94 and 1128 is 94.

∴ The H..CF of 658, 940 and 1128 is 94.

Solution 16. The given numbers are 754, 1508 and 1972.
 
We do: 

        First, we will find the HCF of 754 and 1508.

Therefor, the H.C.F of 754 and 1508 is 754.

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

 Thus, the H.C.F of 754 and 1172 is 58.

∴ The H.C.F of 754, 1508 and 1972 is 58.

Solution 17. The given numbers are 391.425, 527.

 
We do: 

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

Therefor, the H.C.F of 391 and 425 is 17.

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

Thus, the H.C.F of 17 and 527 is 17.

∴ The H.C.F of 391.425 and 527 is 17.

Solution 18. The given numbers are 1794, 2346 and 4761

We do: 

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

Therefor, the H.C.F of 1794 and 2346 is 138.

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

Thus, the HCF of 138 and 4761 is 69.

∴ The HCF of 1794, 2346 and 4761 is 69.

Show that the following pairs are co-primes:

19. 59, 97         20. 161. 192 21. 343, 432

 22, 512, 945         23. 385, 621 24. 847, 1014

    Hint. Two numbers are co-pemes if their HCF is 1.

Solution 19. The given numbers are 59 and 97. 59=59×1 97=97×1 ∴ H.C.F = 1 59 and 97 have only one common factor as 1, so, the two numbers are co-primes.

Solution 20. The given numbers are 161 and 192.

We do:

  7161  2323        1  $$\begin{gathered} \begin{array}{l} 7\overline{)161}\\ 23\overline{)23}\end{array} \\ \overline{)1} \end{gathered}$$ 2192      296        248        224        212        26               3$\begin{array}{l}2\overline{)192}\\ 2\overline{)\text{96 }}\\ 2\overline{)\text{48 }}\\ 2\overline{)\text{24 }}\\ 2\overline{)12}\\ 2\overline{)\text{6 }}\\ 3\\ \end{array}$
Now, 161 = 7 × 23 × 1

192 = 2 × 2× 2 ×2 × 2×  2 × 3 = 26  × 3 × 1

∴ H.C.F = 1

Hence, 161 and 192 are co-primes.

Solution 21. The given numbers are 343 and 432.

We do:

7343      749        77              1 $\begin{array}{l}7\overline{)\text{343 }}\\ 7\overline{)\text{49 }}\\ 7\overline{)\text{7 }}\\ \text{ 1}\\ \end{array}$2432      2216        2108        254        327        39           33             1           $\begin{array}{l}2\overline{)432}\\ 2\overline{)\text{216 }}\\ 2\overline{)\text{108 }}\\ 2\overline{)\text{54 }}\\ 3\overline{)27}\\ 3\overline{)\text{9 }}\\ 3\overline{)\text{3 }}\\ \overline{)\text{1 }}\end{array}$

Now, 343 = 7 × 7× 7 × 1 = 73 × 1

432 = 2 × 2× 2 ×2 × 3× 3 ×3 = 24 × 33 × 1

∴ H.C.F =1

Hence, 343 and 432 are co-primes.

Solution 22. Given numbers are 512 and 945.

We do:

2512      2256        2128        264        232        216           28           24           22                1    $\begin{array}{l}2\overline{)512}\\ 2\overline{)\text{256 }}\\ 2\overline{)\text{128 }}\\ 2\overline{)\text{64 }}\\ 2\overline{)32}\\ 2\overline{)\text{16 }}\\ 2\overline{)\text{8 }}\\ 2\overline{)\text{4 }}\\ 2\overline{)\text{2 }}\\ \text{ 1 }\\ \end{array}$
3945        3315      3105     535         $\begin{array}{l}3\overline{)\text{945 }}\\ 3\overline{)\text{315 }}\\ 3\overline{)\text{105 }}\\ 5\overline{)\text{35 }}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}\end{array}$
77       $\begin{array}{l}7\overline{)\text{7 }}\end{array}$
$\begin{array}{l}\end{array}$ 1         $\begin{array}{l}\overline{)\text{1 }}\\ \mathrm{}\end{array}$$\begin{array}{l}\end{array}$

512 = 2 × 2 ×2 × 2 × 2× 2 × 2× 2 × 2 = 29
 945 = 3 × 3 × 3 × 5 × 7 = 33 × 5 × 7
Thus, the H.C.F of 512 and 945 is 1.
∴ 512 and 945 are co-primes.

Solution 23. The given numbers are 385 and 621.

  5385          777        1111           1        $$\begin{gathered} \begin{array}{l} 5\overline{)\text{385 }}\\ 7\overline{)\text{77 }}\\ 11\overline{)\text{11 }}\end{array} \\ \overline{)\text{1 }} \end{gathered}$$   3621          3207          369        2323                1        $$\begin{gathered} \begin{array}{l} 3\overline{)\text{621 }}\\ \text{ 3}\overline{)\text{207 }}\\ 3\overline{)\text{69 }}\\ 23\overline{)\text{23 }}\end{array} \\ \overline{)\text{ 1 }} \end{gathered}$$

385 = 5 × 7 × 11 × 1
621 = 3 × 3 × 3 × 23 = 33 × 23 × 1
∴ HCF = 1

Hence, The given numbers 385 and 621 are co-primes.

Solution 24. The given numbers are 847 and 1014.

  7847        11121        1111            1        $$\begin{gathered} \begin{array}{l}7\overline{)\text{847 }}\\ 11\overline{)\text{121 }}\\ 11\overline{)\text{11 }}\end{array} \\ \overline{)\text{1 }} \end{gathered}$$  21014          3507        13169        1313           1        $$\begin{gathered} \begin{array}{l}2\overline{)\text{1014 }}\\ 3\overline{)\text{507 }}\\ 13\overline{)\text{169 }}\\ 13\overline{)\text{13 }}\end{array} \\ \overline{)\text{1 }} \end{gathered}$$

847 = 7 × 11 × 11 × 1 = 7 × 112 × 1
1014 = 2 × 3 × 13 × 13 × 1
∴ HCF = 1
Hence, 847 and 1014 are co-primes.

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

Solution : Because the remainder is 6, 

So, we do: 615 - 6 = 609 

           and 963 - 6 = 957

Now, Required number = HCF of 609 and 957
6091  957−609        348609(1           −348                       261348(1             −261––––           87261(3              −261––––                    0$\begin{array}{l}609\begin{array}{c}1\\ \overline{)\begin{array}{l}\text{ 957}\\ \underset{}{-609}\end{array}}\end{array}\\ \text{ 348}\overline{)609}(1\\ \underset{}{-348}\\ \text{ 261}\overline{)348}(1\\ -\underset{\_}{261}\\ \text{ 87}\overline{)261}(3\\ -\underset{\_}{261}\\ \text{ 0}\end{array}$

Therefore, the required number is 87. 

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

Solution : Because the remainder is 9 and 5

So, we do: 2011 - 9 = 2002 

           and 2623 - 5 = 2618

Now, Required number = HCF of 609 and 957

20021    2618−2002                   6162002(3                     −1848                                      154616(4                              −616                                   0

∴ The required number is 154.

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

Solution : Because the remainder is 4,.5 and 6, 

So, we do: 445 - 4 = 441 

                  572 - 5 = 567

                  699 - 6 = 693

Now, Required number = HCF of 441, 567 and 693

First, we will find the HCF of 441 and 567.

∴ HCF = 63

Now, we will find the H.C.F of 63 and 693.

∴ HCF = 63

Hence, the required number is 63.

28. Reduce each of the following fractions to the lowest terms:

    (i) `frac\{161}{207}` (ii) `frac\{517}{799}`         (iii) `frac\{296}{481}`

Solution: (i). To reduce the given fraction to its lowest term, 

 
First, we will find the HCF of 161 and 207.

∴ HCF = 23

Now, Divide the numerator and the denominator by the HCF, we get:

161÷23207÷23=79$\frac{161÷23}{207÷23}=\frac{7}{\mathrm{9 }}$

Solution:(ii)  517799$\frac{517}{799}$

To reduce the given fraction to its lowest term, 

First, we will find the HCF of 517 and 799.

∴ HCF = 47

Now, Divide the numerator and the denominator by the HCF, we get:

517÷47799÷47=1117$\frac{517÷47}{799÷47}=\frac{11}{17}$

Solution: (iii)  296481$\frac{296}{481}$

To reduce the given fraction to its lowest term, 

First, we will find the HCF of 296 and 481.

∴ HCF = 37

Now, Divide the numerator and the denominator by the HCF, we get:

296÷37481÷37=813

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

Solution: The lengths of the three pieces of timber are 42 m, 49 m and 63 m.

Firstly, we will  find the HCF of 42 and 49 by division method.

∴ The HCF of 42 and 49 is 7.
Now, we will find the HCF of 7 and 63.
​​

∴ The HCF of 7 and 63 is 7.
Therefore, HCF of all three numbers is 7
Hence, the greatest possible length of each plank is 7 m.

30. Three different containers 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: Three different containers contain 403 L, 434 L and 465 L of milk.

First, we will  find the HCF of  403 L and 434 L by division method.


∴ HCF = 31

Now, we will find the HCF of 31 and 465.


∴ HCF = 31

Hence, the capacity of the required container is 31 L.

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: Number of apples = 527

               Number of pears = 646
               Number of oranges = 748
The greatest number of fruits possible in each heap will be given by the HCF of 527, 646 and 748.

First, we will find the HCF of 527 and 646.


∴ HCF of 527, 646 and 748 = 17

So, the greatest number of fruits in each heap will be 17.

32. Determine the longest tape which can be used to measure exactly the lengths 7 m,

3 m 85 cm and 12 m 95 cm.

Hint. Convert all the lengths to em and then take the HCF

Solution: 7 m = 700 cm

            3 m 85 cm = 385 cm
            12 m 95 cm = 1295 cm

The required length of the tape that can measure the lengths 700 cm, 385 cm and 1295 cm will be given by the HCF of 700 cm, 385 cm and 1295 cm.

Evaluating the HCF of 700, 385 and 1295 using prime factorisation method, we have:
 
700 = 2 × 2 × 5 × 5 × 7 = 22 × 52 × 7

385 = 5 × 11 × 7

1295 = 5 × 7 × 37

∴ HCF = 5 ×7 = 35

Hence, the longest tape which can measure the lengths 7 m, 3 m 85 cm and 12 m 95 cm exactly is of 35 cm.

33. A rectangular courtyard is 18 m 72 cm long and 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: Length of the courtyard = 18 m 72 cm = 1872 cm

Breadth of the courtyard = 13 m 20 cm = 1320 cm

Now, First, we will find the HCF of 1872 cm and 1320 cm.

13201   1872−1320                    5521320(2                      −1104               _____                                      216552(2                              −432____                                 120216(1                                    −120_____                                         96120(1                                            −96____                                                 2496 (4                                                   −96___                                                        0$\begin{array}{l}1320\begin{array}{c}1\\ \overline{)\begin{array}{l}\text{ 1872}\\ \underset{}{-1320}\end{array}}\end{array}\\ \text{ 552}\overline{)1320}(2\\ \underset{ \_\_\_\_\_}{-1104}\\ \text{ 216}\overline{)552}(2\\ -\underset{\_\_\_\_}{432}\\ \text{ 120}\overline{)216}(1\\ -\underset{\_\_\_\_\_}{120}\\ \text{ 96}\overline{)120}(1\\ -\underset{\_\_\_\_}{96}\\ 24\overline{)96}(4\\ -\underset{\_\_\_}{96}\\ 0\end{array}$

HCF of 1872 and 1320 = 24
∴ maximum edge of the square tile = 24 cm

Required number of tiles = area of courtyardarea of each square tile=1872×132024×24=4290

 

34. Find the HCF of

(i) two prime numbers (ii) two consecutive numbers 

(iii) two co-primes (iv) 2 and an even number

Solution 

(i) 5 and 7 are two prime numbers.
Now, HCF of 5 and 7 is as follows:
5 = 5 × 1
7 = 7 × 1
∴ HCF = 1

Solution 

(ii) 6 and 7 are two consecutive numbers.
Now, HCF of 6 and 7 is as follows:
6 = 2 × 3 × 1
7 = 7 × 1
∴ HCF = 1

Solution 

(iii) 2 and 3 are two co-primes.
Now, HCF of 2 and 3 is as follows:
2 = 2 × 1
3 = 3 × 1
∴ HCF = 1

Solution 

(iv) 2 and 4 are two even numbers.
Now, HCF of 2 and 4 is as follows:​
  2 = 2 × 1
  4 = 2 × 2 × 1 
∴ HCF = 2 × 1 = 2