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

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 2E.

**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 L.C.M. of the numbers given below:**

**Question 1.**

**Solution:**

We have

42 = 2 x 3 x 7

63 = 3 x 3 x 7

= 3^{2} x 7

∴ L.C.M. of 42 and 63 = 2 x 3^{2} x 7

= 2 x 9 x 7

= 18 x 7

= 126

**Question 2.**

**Solution:**

We have

So, 60 = 2 x 2 x 3 x 5

= 2^{2} x 3 x 5

75 = 3 x 5 x 5 = 3 x 5^{2}

∴L.C.M. of 60 and 75 = 2^{2} x 3 x 5^{2}

= 4 x 3 x 25

= 4 x 75 = 300

**Question 3.**

**Solution:**

We have

So, 12 = 2 x 2 x 3 = 2^{2} x 3

18 = 2 x 3 x 3 = 2 x 3^{2}

20 = 2 x 2 x 5 = 2^{2} x 5

∴L.C.M. of 12, 18 and 20 = 2^{2} x 3^{2} x 5

=4 x 9 x 5

= 20 x 9

= 180

**Question 4.**

**Solution:**

We have

36 = 2 x 2 x 3 x 3 = 2^{2} x 3^{2}

60 = 2 x 2 x 3 x 5 = 2^{2} x 3 x 5

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

∴ L.C.M. of 36, 60 and 72 = 2^{3} x 3^{2} x 5

=8 x 9 x 5

= 40 x 9

= 360

**Question 5.**

**Solution:**

We have

36 = 2 x 2 x 3 x 3 = 2^{2} x 3^{2}

40 = 2 x 2 x 2 x 5 = 2^{3} x 5

126 = 2 x 3 x 3 x 7 = 2 x 3^{2} x 7

∴ L.C.M. of 36, 40 and 126 .

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

= 8 x 9 x 5 x 7

= 72 x 35

= 2520

**Question 6.**

**Solution:**

∴ L.C.M. of given numbers

= 2 x 2 x 2 x 7 x 2 x 5 x 11

= 8 x 14 x 55

= 112 x 55 = 6160

**Question 7.**

**Solution:**

∴L.C.M. of given numbers = 2 x 2 x 3 x 3 x 5 x 7

= 36 x 35

= 1260

**Question 8.**

**Solution:**

∴L.C.M. of given numbers

= 2 x 2 x 2 x 2 x 3 x 3 x 5 x 8

= 16 x 9 x 40

= 144 x 40

= 5760

**Question 9.**

**Solution:**

∴L.C.M. of given numbers = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 2 x 3

= 32 x 54

= 1728

**Find the H.C.F. and L.C.M. of :**

**Question 10.**

**Solution:**

First we find the H.C.F. of the given numbers as under :

∴ H.C.F. of 117 and 221 = 13

Now L.C.M. = \(\frac { product\quad of\quad numbers }{ their\quad H.C.F } \)

= \(\frac { 117\times 221 }{ 13 } \)

= 9 x 221 = 1989

∴ H.C.F. = 13 and L.C.M. = 1989

**Question 11.**

**Solution:**

First we find the H.C.F. of 234 and 572 as under :

H.C.F. of 234 and 572 = 26

Now L.C.M. = \(\frac { product\quad of\quad numbers }{ their\quad H.C.F } \)

= \(\frac { 234\times 572 }{ 26 } \)

= 9 x 572

= 5148

**Question 12.**

**Solution:**

First we find the H.C.F. of 693 and 1078 as under :

H.C.F. of 693 and 1078 = 77 Product of numbers

Now L.C.M. = \(\frac { product\quad of\quad numbers }{ their\quad H.C.F } \)

= \(\frac { 693\times 1078 }{ 77 } \)

= 9 x 1078

= 9702

H.C.F. = 77 and L.C.M. = 9702

**Question 13.**

**Solution:**

First we find the H.C.F. of 145 and 232 as under :

H.C.F. of 145 and 232 = 29

Now L.C.M. = \(\frac { product\quad of\quad numbers }{ their\quad H.C.F } \)

= \(\frac { 145\times 232 }{ 29 } \)

= 5 x 232 = 1160

H.C.F. = 29 and L.C.M. = 1160

**Question 14.**

**Solution:**

First we find the H.C.F. of 861 and 1353 as under :

H.C.F. of 861 and 1353 = 123

Now L.C.M. = \(\frac { product\quad of\quad numbers }{ their\quad H.C.F } \)

= \(\frac { 861\times 1353 }{ 123 } \)

= 7 x 1353 = 9471

H.C.F. = 123 and L.C.M. = 9471

**Question 15.**

**Solution:**

First we find the H.C.F. of 2923 and 3239 as under :

H.C.F. of 2923 and 3239 = 79

Now L.C.M. = \(\frac { product\quad of\quad numbers }{ their\quad H.C.F } \)

= \(\frac { 2923\times 3239 }{ 79 } \)

= 37 x 3239= 119843

H.C.F. = 79 and L.C.M. = 119843

**Question 16.**

**Solution:**

We have

**Question 17.**

**Solution:**

We know that

L.C.M = \(\frac { product\quad of\quad the\quad number }{ their\quad H.C.F } \)

= \(\\ \frac { 2160 }{ 12 } \)

= 180

**Question 18.**

**Solution:**

We know that

L.C.M = \(\frac { product\quad of\quad the\quad number }{ their\quad H.C.F } \)

= \(\\ \frac { 2560 }{ 320 } \)

= 8

**Question 19.**

**Solution:**

We know that

One number x The other number

= H.C.F. x L.C.M.

.’. The other number

= \(\\ \frac { H.C.F\times L.C.M }{ One\quad number } \)

= \(\\ \frac { 145\times 2175 }{ 725 } \)

= \(\\ \frac { 2175 }{ 5 } \)

= 435

Required number = 435

**Question 20.**

**Solution:**

We know that

One number x The other number

= H.C.F. x L.C.M.

The other number

= \(\\ \frac { H.C.F\times L.C.M }{ One\quad number } \)

= \(\\ \frac { 131\times 8253 }{ 917 } \)

= \(\\ \frac { 8253 }{ 7 } \)

Required number = 1179

**Question 21.**

**Solution:**

Required least number = L.C.M. of 15, 20, 24, 32 and 36

L.C.M. = 3 x 2 x 2 x 2 x 5 x 4 x 3

= 24 x 60

= 1440

Hence, required least number = 1440

**Question 22.**

**Solution:**

Clearly, required least number = (L.C.M. of the given numbers + 9)

L.C.M. of the given numbers

= 4 x 5 x 5 x 2 x 3

= 600

Required least number

= 600 + 9

= 609

**Question 23.**

**Solution:**

First we find the L.C.M. of the given numbers as under :

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

**Question 24.**

**Solution:**

The greatest number of five digits exactly divisible by the given numbers = The greatest number of five digits exactly divisible by the L.C.M. of given numbers.

Now

L.C.M. of 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

**Question 25.**

**Solution:**

Three bells will again toll together after an interval of time which is exactly divisible by 9, 12, 15 minutes.

Required time = L.C.M. of 9, 12, 15 minutes

L.C.M. of 9, 12, 15 minutes = 3 x 3 x 4 x 5 minutes

= 9 x 20 minutes

= 180 minutes

Required time = 180 minutes

= \(\\ \frac { 180 }{ 60 } \)

= 3 hours

**Question 26.**

**Solution:**

Required distance = L.C.M. of 36 cm, 48 cm and 54 cm

L.C.M. of 36 cm, 48 cm. 54 cm

= 2 x 2 x 3 x 3 x 4 x 3 cm

= 36 x 12 cm

= 432 cm

= 4 m 32 cm

Required distance = 4 m 32 cm

**Question 27.**

**Solution:**

Required time = L.C.M. of 48 seconds, 72 seconds and 108 seconds

L.C.M. of 48 sec., 72 sec. and 108 sec.

= 2 x 2 x 2 x 3 x 3 x 2 x 3 sec.

= 24 x 18 sec.

= 432 sec.

Required time = 432 sec.

= \(\\ \frac { 432 }{ 60 } \)

= 7 m in 12 sec

**Question 28.**

**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

**Question 29.**

**Solution:**

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

Now,

L.C.M. of 15 minutes and 20 minutes

= 5 x 3 x 4

= 60 minutes

= 1 hour

Both the devices will beep together after 1 hour from 6 a.m.

Required time = 6 + 1

= 7 a.m.

**Question 30.**

**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,

L.C.M. of 50 cm, 60 cm, 75 cm, 100 cm

= 2 x 2 x 3 x 5 x 5 cm

= 300 cm = 3 m

Required least distance = 3 m.

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

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