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

 

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

EXERCISE 2E

Find the LCM of the numbers given below: 

1. 42, 63         2. 60, 75         3. 12, 18, 20     4. 36. 60. 72 

5. 36, 40, 126 6. 16, 28, 40, 77 7. 28, 36, 45, 60 8. 144. 180. 384 

9. 48, 64, 72, 96, 108 

Solution (i) 

The given numbers are 42 and 63.

We have:

742,6336,9    32,3     22,1         1,1$\begin{array}{l}7\overline{)42,63}\\ 3\overline{)6,9}\\ 3\overline{)\text{2,3 }}\\ 2\overline{)\text{2,1 }}\\ 1,1\end{array}$

∴ LCM =7 × 3 × 3 × 2 × 1
            =126

Solution (ii) 

The given numbers are 60 and 75.

We have:

360,75520,2554,5    24,1    22,1        1,1 $\begin{array}{l}3\overline{)60,75}\\ 5\overline{)20,25}\\ 5\overline{)4,5}\\ 2\overline{)4,1}\\ 2\overline{)2,1}\\ \text{ 1,1 }\end{array}$

∴  LCM = 3 × 5× 5 × 2 × 2
              = 300

Solution (iii) 

The given numbers are 12, 18 and 20.
We have:

212,18,202 6,9,10   33,9,5      31,3,5       51,1,5              1,1,1 $\begin{array}{l}2\overline{)12,18,20}\\ 2\overline{)\text{ 6,9,10 }}\\ 3\overline{)\text{3,9,5 }}\\ 3\overline{)\text{1,3,5 }}\\ 5\overline{)\text{1,1,5 }}\\ \text{ 1,1,1 }\end{array}$
∴  LCM = 2 × 2 × 3× 3 × 5
              = 180

Solution (iv) 

The given numbers are 36, 60 and 72.

We have:

236,60,72218,30,36 39,15,18   33,5,6       51,5,2        21,1,2             1,1,1$\begin{array}{l}2\overline{)36,60,72}\\ 2\overline{)\text{18,30,36 }}\\ 3\overline{)\text{9,15,18 }}\\ 3\overline{)3,5,6}\\ 5\overline{)1,5,2}\\ 2\overline{)1,1,2}\\ \text{ 1,1,1}\\ \end{array}$
∴  LCM = 2 × 2 × 2 × 3 × 3 × 5
             = 360

Solution (v) 

The given numbers are 36, 40 and 126.

We have:

236,40,126318,20,63  36,20,21  22,20,7       21,10,7       51,5,7        71,1,7             1,1,1$\begin{array}{l}2\overline{)36,40,126}\\ 3\overline{)\text{18,20,63 }}\\ 3\overline{)\text{6,20,21 }}\\ 2\overline{)\text{2,20,7 }}\\ 2\overline{)\text{1,10,7 }}\\ 5\overline{)1,5,7}\\ \text{ 7}\overline{)1,1,7}\\ \text{ 1,1,1}\\ \\ \end{array}$
∴  LCM = 2 × 3 × 3 ×2 × 2 × 5 × 7
                   = 2520

Solution (vi) 

The given numbers are 16, 28, 40 and 77.
We have:

 216,28,40,77 78,14,20,77    28,2,20,11      24,1,10,11       22,1,5,11        51,1,5,11      111,1,1,11          1,1,1,1      $$\begin{gathered} \begin{array}{l} 2\overline{)16,28,40,77}\\ 7\overline{)\text{8,14,20,77 }}\\ 2\overline{)\text{8,2,20,11 }}\\ 2\overline{)\text{4,1,10,11 }}\\ 2\overline{)\text{2,1,5,11 }}\end{array} \\ 5\overline{)\text{1,1,5,11 }} \\ 11\overline{)\text{1,1,1,11 }} \\ \overline{)\text{1,1,1,1 }} \end{gathered}$$

∴  LCM = 2 × 7 × 2 × 2 × 2 × 5 × 11
             = 6160

Solution (vii) 

The given numbers are 28, 36, 45 and 60.

We have:

228,36,45,602 14,18,45,3037,9,45,15     37,3,15,5       57,1,5,5          77,1,1,1               1,1,1,1    $\begin{array}{l}2\overline{)28,36,45,60}\\ 2\overline{)\text{ 14,18,45,30}}\\ 3\overline{)\text{7,9,45,15 }}\\ 3\overline{)\text{7,3,15,5 }}\\ 5\overline{)\text{7,1,5,5 }}\\ 7\overline{)\text{7,1,1,1 }}\\ \text{ 1,1,1,1 }\end{array}$

∴  LCM = 2 × 2 × 3 × 3 × 5 × 7
            = 1260

Solution (viii) 

The given numbers are 144, 180 and 384.
We have:
2144,180,3842 72,90,192   236,45,96      218,45,48       39,45,24         33,15,8            21,5,8              21,5,4              21,5,2              51,5,1                  1,1,1                 $\begin{array}{l}2\overline{)144,180,384}\\ 2\overline{)\text{ 72,90,192 }}\\ 2\overline{)\text{36,45,96 }}\\ 2\overline{)\text{18,45,48 }}\\ 3\overline{)\text{9,45,24 }}\\ 3\overline{)\text{3,15,8 }}\\ 2\overline{)\text{1,5,8 }}\\ 2\overline{)\text{1,5,4 }}\\ 2\overline{)\text{1,5,2 }}\\ 5\overline{)\text{1,5,1 }}\\ \overline{)\text{1,1,1 }}\\ \end{array}$

∴  LCM = 27 × 32 × 5
              = 5760

Solution (ix)

The given numbers are 48, 64, 72, 96 and 108.
We have:

248,64,72,96,108224,32,36,48,54 212,16,18,24,27  26,8,9,12,27      33,4,9,6,27       21,4,3,2,9       21,2,3,1,9       31,1,3,1,9      31,1,1,1,3           1,1,1,1,1$\begin{array}{l}2\overline{)48,64,72,96,108}\\ 2\overline{)24,32,36,48,54}\\ 2\overline{)12,16,18,24,27}\\ 2\overline{)6,8,9,12,27}\\ 3\overline{)3,4,9,6,27}\\ 2\overline{)1,4,3,2,9}\\ 2\overline{)1,2,3,1,9}\\ 3\overline{)1,1,3,1,9}\\ 3\overline{)1,1,1,1,3}\\ \text{ 1,1,1,1,1}\end{array}$

∴  LCM = 26 × 33
              = 1728

Find the HCF and L.CM of 

10. 117, 221 11. 234, 572 12. 693, 1078 13 145. 232 

14. 861, 1353 15. 2923, 3239 

Solution (10)

The given numbers are 117 and 221.

We have:

  3117  339 1313       1$\begin{array}{l} 3\overline{)117}\\ 3\overline{)\text{39 }}\\ 13\overline{)13}\\ \text{ 1}\end{array}$    132211717     1$$\begin{gathered} \begin{array}{l}13\overline{)221}\\ 17\overline{)17}\end{array} \\ \overline{)1} \end{gathered}$$

Now,
117 = 3 × 3 × 13
221 = 13 × 17

∴  HCF = 13 × 1

Now, LCM = 13 × 17 × 3 × 3
                  = 1989

Solution (11)

The given numbers are 234 and 572.

We have:

   2234 3117 339 1313      1$\begin{array}{l} 2\overline{)234}\\ \text{ 3}\overline{)117}\\ \text{ 3}\overline{)39}\\ \text{13}\overline{)\text{13 }}\\ \text{ 1}\end{array}$      2572  228613143 1111      1$\begin{array}{l} 2\overline{)572}\\ \text{ 2}\overline{)286}\\ \text{13}\overline{)\text{143 }}\\ \text{11}\overline{)\text{11 }}\\ \text{ 1}\end{array}$

Now, we have:

234 = 2 × 3 × 3 × 13
572 = 2 × 2 × 1 3 × 11

∴  LCM = 13 × 2  × 2 × 11 × 9
              = 5148
Also, HCF = 13 × 2 = 26

Solution (12)

The given numbers are 693 and 1078.

We have:

 3693 3231 777 1111      1$\begin{array}{l} 3\overline{)693}\\ 3\overline{)231}\\ 7\overline{)\text{77 }}\\ \text{11}\overline{)\text{11 }}\\ \text{ 1}\end{array}$  21078  7539  777 1111     1$\begin{array}{l} 2\overline{)1078}\\ 7\overline{)539}\\ 7\overline{)\text{77 }}\\ \text{11}\overline{)\text{11 }}\\ \text{ 1}\end{array}$

Now, we have:

693 = 3 × 3 ×7 × 11
1078 = 2 × 7× 7 × 11

∴  HCF = 7 × 11= 77
Also, LCM = 2 × 3 × 3 ×7 × 7 × 11 = 9702

Solution (13) 

The given numbers are 145 and 232.
We have:
  51452929     1$\begin{array}{l} 5\overline{)145}\\ 29\overline{)29}\\ \text{ 1}\end{array}$             2232  2116  258 2929       1  $\begin{array}{l} 2\overline{)232}\\ 2\overline{)116}\\ 2\overline{)58}\\ 29\overline{)\text{29 }}\\ \text{ 1 }\end{array}$

Now, we have:

145 = 5 × 29
232 = 2 ×2 × 2 × 29

∴  HCF = 29
Also, LCM = 29 × 2 × 2 × 2 × 5 = 1160

Solution (14)

The given numbers are 861 and 1353.

We have:

  3861  72874141      1$$\begin{gathered} \begin{array}{l} 3\overline{)861}\\ \text{ 7}\overline{)287}\\ \text{41}\overline{)41}\end{array} \\ \overline{)1} \end{gathered}$$  31353114514141      1$$\begin{gathered} \begin{array}{l} 3\overline{)1353}\\ \text{11}\overline{)451}\\ \text{41}\overline{)41}\end{array} \\ \overline{)1} \end{gathered}$$

Now, we have:

861 = 3 × 41 × 7
1353 = 41 × 11 × 3

∴ HCF = 41 × 3 = 123
Also, LCM = 41 × 3 × 11 × 7 = 9471

Solution (15)

HCF of 2923 and 3239:

∴ HCF = 79

We know that product of two numbers = HCF × LCM

⇒ LCM = Product of two numbersHCF ⇒ LCM = 2923×323979∴ LCM = 119843

16. For each pair of numbers, verify that their product = (HCF x LCM). 

    (i) 87, 145         (ii) 186, 403     (iii) 490, 1155 

Solution (i)

(i) 87 and 145


We have:
87 = 3 × 29
145 = 5 × 29

HCF = 29
LCM = 29 × 15 × 1 = 435

Now, 

HCF × LCM = Product of the two numbers

29 × 435       =   87 × 145   

12615            =    12615

∴ HCF × LCM = Product of the two numbers
Verified.

Solution (ii)

(ii)186 and 403

186 = 2 × 3 × 31
403 = 31 × 13

HCF = 31
LCM = 31 × 13 × 6 = 2418

Now, 

HCF × LCM = Product of the two numbers

31 × 2418       =   186 × 403   

74958            =    74958

∴ HCF × LCM = Product of the two numbers
Verified.

Solution (iii)

(iii) 490 and 1155

490 = 7 × 7 × 2 × 5
1155 = 5 × 7 ×3 × 11

HCF = 7 × 5 = 35
LCM = 7 × 5 ×7 × 2 × 3 × 11 = 16170

Now, 

HCF × LCM = Product of the two numbers

35 × 16170       =  490 ×  1155   

565950            =   565950

∴ HCF × LCM = Product of the two numbers
Verified.

17. The product of two numbers is 2160 and their HCF is 12. Find their LCM. 

Solution: 

Product of the two numbers = 2160

HCF = 12

We know that LCM × HCF = Product of the two numbers
       
∴ LCM = 216012$\frac{2160}{12}$ = 180

18 The product of two numbers is 2560 and their LCM is 320. Find their HCF 

Solution:

Product of the two numbers = 2560
LCM = 320

We know that

LCM × HCF = Product of the two numbers

∴ HCF = 2560320$\frac{2560}{320}$ = 8      

   

19. The HCF of two numbers is 145 and their LCM is 2175. If one of the numbers is 725, find the other. 

Solution:

HCF = 145
LCM = 2175
One of the number = 725

We know that
HCF × LCM = Product of two numbers
∴ Other number = 145×2175725$\frac{145\times 2175}{725}$ = 435

20 The HCF and LCM of two numbers are 131 and 8253 respectively.If one of the numbers is 917, find the other. 

Solution:

HCF = 131
LCM = 8253
One of the number = 917

We know that
LCM × HCF = Product of two numbers
Other number = 8253×131917$\frac{8253\times 131}{917}$

∴ The other number is 1179.

21. Find the least number divisible by 15, 20, 24, 32 and 36. 

Solution:

The given numbers are 15, 20, 24, 32 and 36.

The smallest number divisible by the numbers given above will be their LCM.

215,20,24,32,36315,10,12,16,18 55,10,4,16,6  21,2,4,16,6   21,1,2,8,3       21,1,1,4,3      21,1,1,2,3       31,1,1,1,3           1,1,1,1,1  $\begin{array}{l}2\overline{)15,20,24,32,36}\\ 3\overline{)15,10,12,16,18}\\ 5\overline{)5,10,4,16,6}\\ 2\overline{)1,2,4,16,6}\\ 2\overline{)1,1,2,8,3}\\ 2\overline{)1,1,1,4,3}\\ 2\overline{)1,1,1,2,3}\\ 3\overline{)1,1,1,1,3}\\ 1,1,1,1,1 \\ \end{array}$

LCM = 25 × 32 × 5 
            =  1440
∴ The least number divisible by 15, 20, 24, 32 and 36 is 1440.
 

22. Find the least number which when divided by 25, 40 and 60 leaves 9 as the remainder in each case. 

Solution:

25, 40 and 60 exactly divides the least number that is equal to their LCM.
So, the required number that leaves 9 as a remainder will be LCM + 9.

Finding the LCM:

225,40,60225,20,30 225,10,15  325,5,15   525,5,5       55, 1, 1        1, 1, 1    $\begin{array}{l}2\overline{)25,40,60}\\ 2\overline{)25,20,30}\\ 2\overline{)25,10,15}\\ 3\overline{)25,5,15}\\ 5\overline{)25,5,5}\\ 5\overline{)\text{5, 1, 1 }}\\ \text{ 1, 1, 1 }\end{array}$
LCM = 23 × 3 × 52 = 600
∴ Required number = 600 + 9 = 609 

23, Find the least number of five digits that is exactly divisible by 16. 18, 24 and 30. 

Solution:

LCM of 16, 18, 24 and 30:

216,18,24,3028,9,12,15    24,9,6,15    22,9,3,15      31,9,3,15      31,3,1,5        51,1,1,5             1,1,1,1$\begin{array}{l}2\overline{)16,18,24,30}\\ 2\overline{)8,9,12,15}\\ 2\overline{)4,9,6,15}\\ 2\overline{)2,9,3,15}\\ 3\overline{)1,9,3,15}\\ 3\overline{)1,3,1,5}\\ 5\overline{)1,1,1,5}\\ \text{ 1,1,1,1}\end{array}$


L.C.M of the given numbers = 2 x 2 x 2 x 3 x 2 x 3 x 5

    = 24 x 30 = 720

Now least number of five digits = 10000 Dividing 10000 by 720, we get

Clearly if we add 80 to 640, it will become 720 which is exactly divisible by 720.
Required least number of five digits = 10000 + 80 = 10080
Required greatest number of five digits
= 99999 – 279
= 99720

24. Find the greatest number of five digits exactly divisible by 9, 12. 15. 18 and 24. 

Solution: 

First, we will find the LCM of 9, 12, 15, 18 and 24.

L.C.M of the given numbers = 2 x 2 x 3 x 3 x 5 x 2 = 360
                                             
Now greatest number of five digits = 99999
Dividing 99999 by 360, we get

Required greatest number of five digits

= 99999 – 279

= 99720

25. Three bells toll at intervals of 9, 12, 15 minutes. If they start tolling together, after what time will they next toll together?

Solution: 

Three bells toll at intervals of 9, 12 and, 15 minutes.

So, we first find the LCM of 9, 12 and 15

 39,12,1533,4,5    51,4,5    21,4,1    21,2,1       1,1,1$\begin{array}{l}3\overline{)9,12,15}\\ 3\overline{)3,4,5}\\ 5\overline{)1,4,5}\\ 2\overline{)1,4,1}\\ 2\overline{)1,2,1}\\ \text{ 1,1,1}\end{array}$

Required time = 2 × 2 ×3 × 3 × 5
                          = 180 minutes
                          =3 h
If they start tolling together, they will toll together again after 3 h.

26. Three boys step off together from the same place. If their steps measure 36 cm, 48 cm and 54 cm, at what distance from the starting point will they again step together?

Solution: 

The least distance from the starting point where they will step together will be given by the LCM of 36, 48 and 54.

So, we first find the LCM of 36, 48 and 54.

236,48,54218,24,2739,12,2733,4,931,4,321,4,1 21,2,1     1,1,1$\begin{array}{l}2\overline{)36,48,54}\\ 2\overline{)18,24,27}\\ 3\overline{)9,12,27}\\ 3\overline{)3,4,9}\\ 3\overline{)1,4,3}\\ 2\overline{)1,4,1}\\ \text{ 2}\overline{)1,2,1}\\ \text{ 1,1,1}\end{array}$

The required distance = 2 × 2 ×3 × 3 × 3 × 2 × 2
                                          = 16 × 27
                                          = 432 cm
∴ They will step together again at a distance of 432 cm from the starting point.

27. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds. If they start changing simultaneously at 8 a.m., after how much time will they change again simultaneously? 

Solution: 

248,72,108224,36,54212,18,2726,9,2733,9,2731,3,931,1,3  1,1,1$\begin{array}{l}2\overline{)48,72,108}\\ 2\overline{)24,36,54}\\ 2\overline{)12,18,27}\\ 2\overline{)6,9,27}\\ 3\overline{)3,9,27}\\ 3\overline{)1,3,9}\\ 3\overline{)1,1,3}\\ \text{ 1,1,1}\end{array}$

L.C.M. of 48, 72, 108
= 2 x 2 x 2 x 2 x 3 x 3 x 3 sec

= 16 x 27 
= 432 sec

Required time = 432 sec.

= 43260

= 7 m in 12 sec

So, the lights will change simultaneously at 8:07:12 a.m.

28. Three measuring rods are 45 cm, 50 cm and 75 cm in length. What is the least length (in metres) of a rope that can be measured by the full length of each of these three rods? 

Solution: 

Lengths of three rods = 45 cm, 50 cm and 75 cm

Required least length of the rope = L.C.M. of 45 cm, 50 cm, 75 cm

We have

245,50,75345,25,75315,25,2555,25,2551,5,5   1,1,1$\begin{array}{l}2\overline{)45,50,75}\\ 3\overline{)45,25,75}\\ 3\overline{)15,25,25}\\ 5\overline{)5,25,25}\\ 5\overline{)1,5,5}\\ \text{ 1,1,1}\end{array}$

Required length = 3 × 3 ×5 × 5 × 2
                              = 450 cm

29. An electron device makes a beep after every 15 minutes. Another device makee after every 20 minutes. They beeped together at 6 a.m. At what time will they next beep together?

Solution: 

The time after which both the devices will beep together = L.C.M. of 15 minutes and 20 minutes

515,2033,421,421,2   1,1$$\begin{gathered} \begin{array}{l}5\overline{)15,20}\\ 3\overline{)3,4}\\ 2\overline{)1,4}\\ 2\overline{)1,2}\end{array} \\ \overline{)1,1} \end{gathered}$$

$\overline{)\mathrm{<span\; style="font-family:\; Arial,\; sans-serif;"><div>L.C.M.\; of\; 15\; minutes\; and\; 20\; minutes</div><div>=\; 5\; x\; 3\; x\; 4</div><div>=\; 60\; minutes</div><div>=\; 1\; hour</div><div>Both\; the\; devices\; will\; beep\; together\; after\; 1\; hour\; from\; 6\; a.m.</div><div>Required\; time\; =\; 6\; +\; 1</div><div>=\; 7\; a.m.</div><div><br></div><div><br></div></span>}}$

30. The circumference of four wheels are 50 cm, 60 cm, 75 cm and 100 cm. They start moving simultaneously. What least distance should they cover so that each wheel makes a complete number of revolutions?

Solution: 

The circumferences of four wheels = 50 cm, 60 cm, 75 cm and 100 cm

Required least distance = L.C.M. of 50 cm, 60 cm, 75 cm and 100 cm Now,

550,60,75,100510,12,15,2022,12,3,421,6,3,231,3,3,1   1,1,1,1 $\begin{array}{l}5\overline{)50,60,75,100}\\ 5\overline{)10,12,15,20}\\ 2\overline{)2,12,3,4}\\ 2\overline{)1,6,3,2}\\ 3\overline{)1,3,3,1}\\ \text{ 1,1,1,1 }\end{array}$
Required least distance = 5 × 5 ×2 × 2 × 3
                                             = 25 × 4 × 3
                                             = 300 cm = 3 m