Class 6 Maths RS Aggarwal Solutions

RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2F

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 2F.

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

Objective Questions
Tick the correct answer in each of the following :

Question 1.
Solution:
(c) Because sum of its digits is 8 + 3 + 4 + 7 + 9 + 5 + 6 + 0 = 42 which is divisible by 3.

Question 2.
Solution:
(a) Because sum of its digits is 8 + 5 + 7 + 6 + 9 + 0 + 1 = 36 which is divisible by 9.

Question 3.
Solution:
(d) Because the number formed by tens and ones digits is divisible by 4 i.e. 32 ÷ 4 = 8.

Question 4.
Solution:
(b) Because the number formed by hundred, tens and ones digits is divisible by 8 i.e. 176 ÷ 8 = 22.

Question 5.
Solution:
(a) Because its one digit is divisible by 2 and sum of its digits is 8 + 7 + 9 + 0 + 4 + 3 + 2 = 33,
which is divisible by 3. Hence it is divisible by 6.

Question 6.
Solution:
(c) Because the difference of the sums of its odd places digits and of its even places digits is (2 + 2 + 2 + 2) – (2 + 2 + 2 + 2) i.e. 8 – 8 = 0, which is zero and is divisible by 11.

Question 7.
Solution:
(d) Because 97 has no factors other than 1 and itself.

Question 8.
Solution:
(c) Because 179 has no factors other than 1 and itself.

Question 9.
Solution:
(c) Because 263 has no factors other than 1 and itself.

Question 10.
Solution:
(a), (b) Because the common factors of 9 and 10 are none but 1.

Question 11.
Solution:
(c) Because 32 has factors which are 2, 2, 2, 2, 2.

Question 12.
Solution:
(d) Because 18 is the highest common factor of 144 and 198.

Question 13.
Solution:
(a) Because 12 is the highest common factors of these numbers 144, 180 and 192.

Question 14.
Solution:
(b) Because 161 and 192 have no common factor other than 1, i.e., HCF of 161 and 192 is 1.

Question 15.
Solution:
(d) Because HCF of 289 and 391 is 289
and \(\frac { 289\div 17 }{ 391\div 17 } \) = \(\\ \frac { 17 }{ 23 } \)

Question 16.
Solution:
(d) Because dividing 134 and 167 by 33 remainder is 2 in each case.

Question 17.
Solution:
(c) Because 360 is the least multiple of 24, 36 and 40.

Question 18.
Solution:
(d) Because 540 is the least multiple of 12, 15, 20 and 27

Question 19.
Solution:
(c) Because 1263 – 3 = 1260 is divisible by 14, 28, 36 and 45.

Question 20.
Solution:
(c) Because HCF of two co-prime number is always 1.

Question 21.
Solution:
(c) Because HCF of a and b, two co-primes is 1.
LCM = a x b = ab.

Question 22.
Solution:
(c) Because LCM of two numbers = Product of these number ÷ their HCF i.e 2160 ÷ 12 = 180.

Question 23.
Solution:
(b) Because second number
= \( \frac { LCM\times HCF }{ 1st\quad number } \)
i.e., \(\\ \frac { 145\times 2175 }{ 725 } \)
= 435

Question 24.
Solution:
(c) Because LCM of 15, 20, 24. 32 and 36 = 1440.

Question 25.
Solution:
(d) Because LCM of 9, 12, 15 is 180. 180
180 minutes = \(\\ \frac { 180 }{ 60 } \)
= 3 hours.

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

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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
= 3² x 7
∴ L.C.M. of 42 and 63 = 2 x 3² 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
= 2² x 3 x 5
75 = 3 x 5 x 5 = 3 x 5²
∴L.C.M. of 60 and 75 = 2² x 3 x 5²
= 4 x 3 x 25
= 4 x 75 = 300

Question 3.
Solution:
We have

So, 12 = 2 x 2 x 3 = 2² x 3
18 = 2 x 3 x 3 = 2 x 3²
20 = 2 x 2 x 5 = 2² x 5
∴L.C.M. of 12, 18 and 20 = 2² x 3² x 5
=4 x 9 x 5
= 20 x 9
= 180

Question 4.
Solution:
We have

36 = 2 x 2 x 3 x 3 = 2² x 3²
60 = 2 x 2 x 3 x 5 = 2² x 3 x 5
72 = 2 x 2 x 2 x 3 x 3 = 2³ x 3²
∴ L.C.M. of 36, 60 and 72 = 2³ x 3² x 5
=8 x 9 x 5
= 40 x 9
= 360

Question 5.
Solution:
We have

36 = 2 x 2 x 3 x 3 = 2² x 3²
40 = 2 x 2 x 2 x 5 = 2³ x 5
126 = 2 x 3 x 3 x 7 = 2 x 3² x 7
∴ L.C.M. of 36, 40 and 126 .
= 2³ x 3² 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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

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
= 2² x 3 x 7
98 = 2 x 7 x 7 = 2 x 7²
∴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 = 2³ x 3² x 7
980 = 2 x 2 x 5 x 7 x 7 = 2² x 5 x 7²
∴ H.C.F. of 504 and 980 = 2² x 7
= 4 x 7 = 28.

Question 4.
Solution:
We have

So, 72 = 2 x 2 x 2 x 3 x 3 = 2³ x 3²
108 = 2 x 2 x 3 x 3 x 3 = 2² x 3³
180 = 2 x 2 x 3 x 3 x 5 = 2² x 3² x 5
∴ H.C.F. of 72, 108,
180 = 2² x 3²
= 4 x 9 = 36

Question 5.
Solution:
We have

84 = 2 x 2 x 3 x 7 = 2² x 3 x 7
120 = 2 x 2 x 2 x 3 x 5 = 2³ 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 = 2⁴ x 17
425 = 5 x 5 x 17
= 5² 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 = 2⁴ x 3²
252 = 2 x 2 x 3 x 3 x 7 = 2² x 3² x 7
630 = 2 x 3 x 3 x 5 x 7 = 2 x 3² x 5 x 7
∴ H.C.F. of 144, 252 and 630 = 2 x 3²
= 2 x 9 = 18.

Question 9.
Solution:
We have

So, 1197 = 3 x 3 x 7 x 19 = 3² x 7 x 19
5320 = 2 x 2 x 2 x 5 x 7 x 19 = 2³ 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
= 2² x 5² 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
= 2⁴ x 3² x 13
1320 = 2 x 2 x 2 x 3 x 5 x 11
= 2³ x 3 x 5 x 11
So, the H.C.F. of 1872 and 1320
= 2³ 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 cm²
Area of one square tile = Side x Side
= 24 x 24 cm²
∴ 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.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2C

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 2C.

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

Give the prime factorization of each of the following numbers.

Question 1.
Solution:
We have

∴12 = 2 x 2 x 3 = 2² x 3.

Question 2.
Solution:
We have

∴18 = 2 x 3 x 3 = 2 x 3²

Question 3.
Solution:
We have

∴ 48 = 2 x 2 x 2 x 2 x 3
= 2⁴ x 3.

Question 4.
Solution:
We have

∴ 56 = 2 x 2 x 2 x 7
= 2³ x 7.

Question 5.
Solution:
We have

∴ 90 = 2 x 3 x 3 x 5
= 2 x 3² x 5.

Question 6.
Solution:
We have

136 = 2 x 2 x 2 x 17
= 2³ x 17.

Question 7.
Solution:
We have

∴ 252 = 2 x 2 x 3 x 3 x 7
= 2² x 3² x 7.

Question 8.
Solution:
We have

∴ 420 = 2 x 2 x 3 x 5 x 7
= 2² x 3 x 5 x 7.

Question 9.
Solution:
We have

∴ 637 = 7 x 7 x 13
= 7² x 13.

Question 10.
Solution:
We have

∴ 945 = 3 x 3 x 3 x 5 x 7
= 3³ x 5 x 7.

Question 11.
Solution:
We have

∴ 1224 = 2 x 2 x 2 x 3 x 3 x 17
= 2³ x 3² x 17.

Question 12.
Solution:
We have

∴ 1323 = 3 x 3 x 3 x 7 x 7
= 3³ x 7²

Question 13.
Solution:
We have

∴ 8712 = 2 x 2 x 2 x 3 x 3 x 11 x 11
= 2³ x 3² x 11².

Question 14.
Solution:
We have

∴9317 = 7 x 11 x 11 x 11
= 7 x 11³

Question 15.
Solution:
We have

∴ 1035 = 3 x 3 x 5 x 23
= 3² x 5 x 23

Question 16.
Solution:
We have

∴ 1197 = 3 x 3 x 7 x 19
= 3² x 7 x 19

Question 17.
Solution:
We have

∴ 4641 = 3 x 7 x 13 x 17.

Question 18.
Solution:
We have

∴ 4335 = 3 x 5 x 17 x 17
= 3 x 5 x 17²

Question 19.
Solution:
We have

∴ 2907 = 3 x 3 x 17 x 19
= 3² x 17 x 19.

Question 20.
Solution:
We have

∴ 13915 = 5 x 11 x 11 x 23 = 5 x 11² x 23

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2B

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 2B.

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

Question 1.
Solution:
(i) The given number = 2650
Digit at unit’s place = 0
It is divisible by 2.
(ii) The given number = 69435
Digit at unit’s place = 5
It is not divisible by 2.
(iii) The given number = 59628
Digit at unit’s place = 8
It is divisible by 2.
(iv) The given number = 789403
Digit at unit’s place = 3
It is not divisible by 2.
(v) The given number = 357986
Digit at unit’s place = 6
It is divisible by 2.
(vi) The given number = 367314
Digit at unit’s place = 4
It is divisible by 2.

Question 2.
Solution:
(i) The given number = 733
Sum of its digits = 7 + 3 + 3 = 13,
which is not divisible by 3.
∴ 733 is not divisible by 3.
(ii) The given number = 10038
Sum of its digits = 1 + 0 + 0 + 3 + 8 = 12,
which is divisible by 3
∴ 10038 is divisible by 3.
(iii) The given number = 20701
Sum of its digits = 2 + 0 + 7 + 0 + 1 = 10,
which is not divisible by 3
∴ 20701 is not divisible by 3.
(iv) The given number = 524781
Sum of its digits = 5 + 2 + 4 + 7 + 8 + 1 = 27,
which is divisible by 3
∴ 524781 is divisible by 3.
(v) The given number = 79124
Sum of its digits = 7 + 9 + 1 + 2 + 4 = 23,
which is not divisible by 3
∴ 79124 is not divisible by 3.
(vi) The given number = 872645
Sum of its digits = 8 + 7 + 2 + 6 + 4 + 5 = 32,
which is not divisible by 3
∴ 872645 is not divisible by 3.

Question 3.
Solution:
(i) The given number = 618
The number formed by ten’s and unit’s digits is 18, which is not divisible by 4.
∴ 618 is not divisible by 4.
(ii) The given number = 2314
The number formed by ten’s and unit’s digits is 14, which is not divisible by 4.
∴ 2314 is not divisible by 4.
(iii) The given number = 63712
The number formed by ten’s and unit’s digits is 12, which is divisible by 4
∴ 63712 is divisible by 4.
(iv) The given number = 35056
The number formed by ten’s and unit’s digits is 56, which is divisible by 4.
∴ 35056 is divisible by 4.
(v) The given number = 946126
The number formed by ten’s and unit’s digits is 26, which is not divisible by 4.
∴ 946126 is not divisible by 4.
(vi) The given number = 810524
The number formed by ten’s and unit’s digits is 24, which is divisible by 4.
∴ 810524 is divisible by 4.

Question 4.
Solution:
We know that a number is divisible by 5 if its ones digit is 0 or 5
(i) 4965, (ii) 23590 (iv) 723405 and (vi) 438750 are divisible by 5

Question 5.
Solution:
(i) The given number = 2070
Its unit’s digit = 0
So, it is divisible by 2
Sum of its digits = 2 + 0 + 7 + 0 = 9,
which is divisible by 3
∴ The given number is divisible by 3.
So, 2070 is divisible by both 2 and 3.
Hence it is divisible by 6.
(ii) The given number = 46523
Its unit’s digit = 3
So, it is not divisible by 2
Hence 46523 is not divisible by 6.
(iii) The given number = 71232
Its unit’s digit = 2
So, it is divisible by 2
Sum of its digits = 7 + 1 + 2 + 3 + 2
= 15, which is divisible by 3
∴ 71232 is divisible by both 2 and 3
Hence it is divisible by 6.
(iv) The given number = 934706
Its unit’s digit = 6 So,
it is divisible by 2
Sum of its digits = 9 + 3 + 4 + 7 + 0 + 6 = 29,
which is not divisible by 3
Hence 934706 is not divisible by 6.
(v) The given number = 251780
Its unit’s digit = 0
So, it is divisible by 2
Sum of its digits = 2 + 5 + 1 + 7 + 8 + 0 = 23,
which is not divisible by 3
251780 is not divisible 6
(vi) 872536 is not divisible by 6 as sum of its digits is 8 + 7 + 2 + 5 + 3 + 6 = 31 which is not divisible by 3

Question 6.
Solution:
We know that a number is divisible by 7 if the difference between twice the ones digit and the number formed by the other digits is either 0 or a multiple of 7
(i) 826, 6 x 2 = 12 and 82
Difference between 82 and 12 = 70
Which is divisible by 7
∴ 826 is divisible by 7
(ii) In 117, 7 x 2 = 14, 11
Difference between 14 and 11 = 14 – 11 = 3
Which is not divisible by 7
∴ 117 is not divisible by 7
(iii) In 2345, 5 x 2 = 10 and 234
Difference between 234 – 10 = 224
which is divisible by 7
∴ 2345 is divisible by 7
(iv) In 6021, 1 x 2 = 2, and 602
Difference between 602 and 2 = 600
which is not divisible by 7
∴ 6021 is not divisible by 7
(v) In 14126, 6 x 2 = 12 and 1412
Difference between 1412 – 12 = 1400
which 8 is divisible by 7
∴ 14126 is divisible by 7
(vi) In 25368, 8 x 2 = 16 and 2536
Difference between 2536 and 16 = 2520
which is divisible by 7
∴ 25368 is divisible by 7

Question 7.
Solution:
(i) The given number = 9364
The number formed by hundred’s, ten’s and unit’s digits is 364, which is not divisible by 8.
∴ 9364 is not divisible by 8.
(ii) The given number = 2138
The number formed by hundred’s, ten’s and unit’s digits is 138, which is not divisible by 8.
∴ 2138 is not divisible by 8.
(iii) The given number = 36792
The number formed by hundred’s, ten’s and unit’s digits is 792, which is divisible by 8.
∴ 36792 is divisible by 8.
(iv) The given number = 901674
The number formed by hundred’s, ten’s and unit’s digits is 674, which is not divisible by 8.
∴ 901674 is not divisible by 8.
(v) The given number = 136976
The number formed by hundred’s, ten’s and unit’s digits is 976, which is divisible by 8.
∴ 136976 is divisible by 8.
(vi) The given number = 1790184
The number formed by hundred’s, ten’s and unit’s digits is 184, which is divisible by 8.
∴ 1790184 is divisible by 8.

Question 8.
Solution:
We know that a number is divisible by 9, if the sum of its digits is divisible by 9
(i) In 2358
Sum of digits : 2 + 3 + 5 + 8 = 18
which is divisible by 9
∴ 2358 is divisible by 9
(ii) In 3333
Sum of digits 3 + 3 + 3 + 3 = 12
which is not divisible by 9
∴ 3333 is not divisible by 9
(iii) In 98712
Sum of digits = 9 + 8 + 7 + 1 + 2 = 27
Which is divisible by 9
∴ 98712 is divisible by 9
(iv) In 257106
Sum of digits = 2 + 5 + 7 + 1 + 0 + 6 = 21
which is not divisible by 9
∴ 257106 is not divisible by 9
(v) In 647514
Sum of digits = 6 + 4 + 7 + 5 + 1 + 4 = 27
which is divisible by 9
∴ 647514 is divisible by 9
(v) In 326999
Sum of digits = 3 + 2 + 6 + 9 + 9 + 9 = 38
which is not divisible by 9
∴ 326999 is divisible by 9

Question 9.
Solution:
We know that a number is divisible by 10 if its ones digit is 0
∴(i) 5790 is divisible by 10

Question 10.
Solution:
(i) The given number = 4334
Sum of its digits in odd places = 4 + 3 =7
Sum of its digits in even places = 3 + 4 = 7
Difference of the two sums = 7 – 7 = 0
∴4334 is divisible by 11.
(ii) The given number = 83721
Sum of its digits in odd places = 1 + 7 + 8 = 16
Sum of its digits in even places = 2 + 3 = 5
Difference of the two sums = 16 – 5 = 11,
which is multiple of 11.
∴ 83721 is divisible by 11.
(iii) The given number = 66311
Sum of its digits in odd places = 1 + 3 + 6 = 10
Sum of its digits in even places = 1 + 6 = 7
Difference of the two sums = 10 – 7 = 3,
which is not a multiple of 11.
∴ 66311 is not divisible by 11.
(iv) The given number = 137269
Sum of its digits in odd places = 9 + 2 + 3 = 14
Sum of its digits in even places = 6 + 7 + 1 = 14
Difference of the two sums = 14 – 14 = 0
∴ 137269 is divisible by 11.
(v) The given number = 901351
Sum of its digits in odd places = 1 + 3 + 0 = 4
Sum of its digits in even places = 5 + 1 + 9 = 15
Difference of the two sums = 15 – 4 = 11,
which is a multiple of 11.
∴ 901351 is divisible by 11.
(vi) The given number = 8790322
Sum of its digits in odd places = 2 + 3 + 9 + 8 = 22
Sum of its digits in even places = 2 + 0 + 7 = 9
Difference of the two sums = 22 – 9 = 13,
which is not a multiple of 11.
∴ 8790322 is not divisible by 11.

Question 11.
Solution:
(i) The given number = 27*4
Sum of its digits = 2 + 7 + 4 = 13
The number next to 13 which is divisible by 3 is 15.
∴ Required smallest number = 15 – 13
= 2.
(ii) The given number = 53*46
Sum of the given digits = 5 + 3 + 4 + 6 = 18,
which is divisible by 3.
∴ Required smallest number = 0.
(iii) The given number = 8*711
Sum of the given digits = 8 + 7 + 1 + 1 = 17
The number next to 17,
which is divisible by 3 is 18.
∴ Required smallest number =18 – 17 = 1
(iv) The given number = 62*35
Sum of the given digits = 6 + 2 + 3 + 5 = 16
The number next to 16,
which is divisible by 3 is 18.
∴ Required smallest number =18 – 16 = 2
(v) The given number = 234*17
Sum of the given digits = 2 + 3+ 4 + 1 + 7 = 17
The number next to 17, which is divisible by 3 is 18.
Required smallest number = 18 – 17 = 1.
(vi) The given number = 6* 1054
Sum of the given digits = 6 + 1+ 0 + 5 + 4 = 16
The number next to 16,
which is divisible by 3 is 18.
Required smallest number = 18 – 16 = 2.

Question 12.
Solution:
(i) The given number = 65*5
Sum of its given digits = 6 + 5 + 5 = 16
The number next to 16, which is divisible by 9 is 18.
∴ Required smallest number =18 – 16 = 2
(ii) The given number = 2*135
Sum of its given digits = 2 + 1 + 3 + 5
The number next to 11, which is divisible by 9 is 18.
∴ Required smallest number = 18 – 11 =7.
(iii) The given number = 6702*
Sum of its given digits = 6 + 7 + 0 + 2 = 15
The number next to 15, which is divisible by 9 is 18.
∴ Required smallest number = 18 – 15 = 3
(iv) The given number = 91*67
Sum of its given digits = 9 + 1 + 6 + 7 =23
The number next to 23, which is divisible by 9 is 27.
∴ Required smallest number = 27 – 23 = 4.
(v) The given number = 6678*1
Sum of its given digits= 6 + 6 + 7 + 8 + 1 = 28
The number next to 28, which is divisible by 9 is 36.
∴ Required smallest number = 36 – 28 = 8.
(vi) The given number = 835*86
Sum of its given digits = 8 + 3 + 5 + 8 + 6 = 30
The number next to 30, which is divisible by 9 is 36.
∴ Required smallest number = 36 – 30 = 6.

Question 13.
Solution:
(i) The given number = 26*5
Sum of its digits is odd places = 5 + 6 = 11
Sum of its digits in even places = * + 2
Difference of the two sums = 11 – (* + 2)
The given number will be divisible by 11 if the difference of the two sums = 0.
∴ 11 – (* + 2) = 0
11 = * + 2
11 – 2 = *
9 = *
Required smallest number = 9.
(ii) The given number = 39*43
Sum of its digits in odd places
=3 + * + 3 = * + 6
Sum of its digits in even places = 4 + 9 = 13
Difference of the two sums = * + 6 – 13 = * – 7
The given number will be divisible by 11, if the difference of the two sums = 0.
∴ * – 7 = 0
* = 7
∴ Required smallest number = 7.
(iii) The given number = 86*72
Sum of its digits in odd places
= 2 + * + 8 = * + 10
Sum of its digits in even places = 7 + 6 = 13
Difference of the two sums = * + 10 – 13 = * – 3
The given number will be divisible by 11, if the difference of the two sums = 0.
∴ * – 3 = 0
* = 3
∴ Required smallest number = 3.
(iv) The given number = 467*91
Sum of its digits in odd places = 1 + * + 6 = * + 7
Sum of its digits in even places = 9 + 7 + 4 = 20
Difference of the two sums
= 20 – (* + 7)
= 20 – * – 7 = 13 – *
Clearly the difference of the two sums will be multiple of 11 if 13 – * = 11
∴ 13 – 11 = *
2 = *
* = 2 .
∴ Required smallest number = 2.
(v) The given number = 1723*4
Sum of its digits in odd places = 4 + 3 + 7 = 14
Sum of its digits in even places = * + 2 + 1 = * + 3
Difference cf the two sums = * + 3 – 14 = * – 11
The given number will be divisible by 11,if *- 11 is a multiple of 11,which is possible if * = 0.
Required smallest number = 0.
(vi) The given number = 9*8071
Sum of its digits in odd places = 1 + 0 + * = 1 + *
Sum of its digits in even places = 7 + 8 + 9 = 24
Difference of the two sums = 24 – 1 – * = 23 – *
∴ The given number will be divisible by 11, if 23 – * is a multiple of 11, which is possible if * = 1.
∴ Required smallest number = 1.

Question 14.
Solution:
(i) The given number = 10000001
Sum of its digits in odd places
= 1 + 0 + 0 + 0 = 1
Sum of its digits in even places = 0 + 0 + 0 + 1 = 1
Difference of the two sums = 1 – 1 = 0
∴ The number 10000001 is divisible by 11.
(ii) The given number = 19083625
Sum of its digits in odd places
= 5 + 6 + 8 + 9 = 28
Sum of its digits in even places = 2 + 3 +0 + 1 = 6
Difference of the two sums = 28 – 6 = 22,
which is divisible by 11.
∴ The number 19083625 is divisible by 11.
(iii) The given number = 2134563
Sum of its digits
= 2 + 1 + 3 + 4 + 5 + 6 + 3 = 24,
which is not divisible by 9.
∴ The number 2134563 is not divisible by 9.
(iv) The given number = 10001001
Sum of its digits
= 1 + 0 + 0 + 0 + 1 + 0 + 0 + 1 = 3,
which is divisible by 3.
∴ The number 10001001 is divisible by 3.
(v) The given number = 10203574
The number formed by its ten Is and unit’s digits is 74, which is not divisible by 4.
The number 10203574 is not divisible by 4.
(vi) The given number = 12030624
The number formed by its hundred’s, ten’s and unit’s digits = 624,
which is divisible by 8.
∴ The number 12030624 is divisible by

Question 15.
Solution:
103, 137, 179, 277, 331, 397 are prime numbers.

Question 16.
Solution:
(i) 154
(ii) 612
(iii) 5112, 3816 etc.
(iv) 3426, 5142 etc.

Question 17.
Solution:
(i) False
(ii) True
(iii) False
(iv) True
(v) False
(vi) True
(vii) True
(viii) True

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

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.