RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3A

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3A.

Other Exercises

Question 1.
Solution:
After 30999, three whole numbers will be
30999 + 1 = 31000
31000 + 1 = 31001
31001 + 1 = 31002
i.e., 31000, 31001 and 31002

Question 2.
Solution:
Before 10001, three whole numbers will be 10001 – 1 = 10000
10000 – 1 = 9999
9999 – 1 = 9998
i.e., 10000, 9999, 9998

Question 3.
Solution:
Between 1032 and 1209, whole number are 1209 – 1031
= 178

Question 4.
Solution:
The smallest whole number is 0

Question 5.
Solution:
The successor of
(i) 2540801 is 2540801 + 1 = 2540802
(ii) 9999 is 9999 + 1 = 10000
(iii) 50904 is 50904 + 1 = 50905
(iv) 61639 is 61639 + 1 = 61640
(v) 687890 is 687890 + 1 = 687891
(vi) 5386700 is 5386700 + 1 = 5386701
(vii) 6475999 is 6475999 + 1 = 6476000
(viii) 9999999 is 9999999 + 1 = 10000000

Question 6.
Solution:
Predecessor of
(i) 97 is 97 – 1 = 96
(ii) 10000 is 10000 – 1 = 9999
(iii) 36900 is 36900 – 1 = 36899
(iv) 7684320 is 7684320 – 1 = 7684319
(v) 1566391 is 1566391 – 1 = 1566390
(vi) 2456800 is 2456800 – 1 = 2456799
(vii) 100000 is 100000 – 1 = 99999
(viii) 1000000 is 1000000 – 1 = 999999

Question 7.
Solution:
Three consecutive whole numbers just preceding 7510001 are (7510001 – 1), (7510001 – 2), (7510001 – 3)
i.e. 7510000, 7509999, 7509998.

Question 8.
Solution:
(i) Zero is not a natural number. (F)
(ii) Zero is the smallest whole number (T)
(iii) No, it is false, as zero is not a natural number but it is a whole number.
(iv) Yes, it is true, as set of natural numbers is a subset of whole numbers.
(v) False, zero is the smallest whole number.
(vi) The natural number 1 has no predecessor as 0 is the predecessor of 1 (T)
Which is not a natural number
(vii) The whole number 1 has no predecessor (F)
Predecessor of 1 is 0 which is a whole number
(viii) The whole number 0 has no predecessor (T)
(ix) The predecessor of a two-digit number is never a single-digit number (F)
As predecessor of two digit number say 99 is 99 – 1 = 98
Which is also a two-digit number and of 10 is 10 – 1 = 9
which is single-digit number
(x) The successor of a two-digit number is always a two-digit number (F)
The successor of a two-digit number 99 is 99 + 1 = 100 which is a three digit number
(xi) 500 is the predecessor of 499 (F)
As predecessor of 499 is 499 – 1 = 498 not 500 as 500 is the successor of 499
(xii) 7000 is the successor of 6999 (T)

 

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3A are helpful to complete your math homework.

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

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

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.

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 2E

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

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

Question 1.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 1.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 2.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 3.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 4.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 5.1
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:
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 6.1
∴ 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:
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 7.1
∴L.C.M. of given numbers = 2 x 2 x 3 x 3 x 5 x 7
= 36 x 35
= 1260

Question 8.
Solution:
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 8.1
∴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:
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 9.1
∴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 :
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 10.1
∴ 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 :
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 11.1
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 :
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 12.1
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 :
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 13.1
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 :
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 14.1
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 :
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 15.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 16.1
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 16.2
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 16.3

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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 21.1
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)
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 22.1
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 :
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 23.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 23.2
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 24.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 24.2
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 25.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 26.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 27.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 28.1

Question 29.
Solution:
The time after which both the devices will beep together = L.C.M. of 15 minutes and 20 minutes
Now,
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 29.1
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,
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2E 30.1
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 4 Integers Ex 4B

RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4B.

Other Exercises

Question 1.
Solution:
(i) On the number line we start from 0 and move 9 steps to the right to reach a point A. Now, starting from A, we move 6 steps to the left to reach a point B, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 1.1
Now, B represents the integer 3
9 + ( – 6) = 3
(ii) On the number line, we start from 0 and move 3 steps to the left to reach a point A. Now, starting from A, we move 7 steps to the right to reach a point B, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 1.2
And B represents the integer 4
( – 3) + 7 = 4
(iii) On the number line, we start from 0 and move 8 steps to the right to reach a point A. Now, starting from A, we move 8 steps to the left to reach a point B, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 1.3
And, B represents the integer 0.
8 + ( – 8) = 0
(iv) On the number line, we start from 0 and move 1 step the left to reach a point A. Now, starting from point A, we move 3 steps to the left to reach g. point B, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 1.4
And, B represents the integer – 4
( – 1) + ( – 3) = – 4.
(v) On the number line, we start from 0 and move 4 steps to the left to reach a point A. Now, starting from point A, we move 7 steps to the left to reach a point B, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 1.5
And, B represents the integer -11.
( – 4) + ( – 7) = – 11
(vi) On the number line we start from 0 and move 2 steps to the left to reach a point A. Now, starting from A, we move 8 steps to the left to reach a point B, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 1.6
And, B represents the integer – 10
( – 2) + ( – 8) = – 10
(vii) On the number line we start from 0 and move 3 steps to the right to reach a point A. Now, starting from A, we move 2 steps to the left to reach a point B and again starting from left to reach a point B and again starting from B, we move 4 steps to the left to reach a point C, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 1.7
And, C represents the integer – 3
3 + ( – 2) + ( – 4) = – 3
(viii) On the number line we start from 0 and move 1 step to the left to reach a point A. Now, starting from A, we move 2 steps to the left to reach a point B and again starting from B, we move 3 steps to the left to reach point C, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 1.8
And, C represents the integer – 6
( – 1) + ( – 2) + ( – 3) = – 6.
(ix) On the number line we start from 0 and move 5 steps to the right to reach a point A. Now, starting from A, we move 2 steps to the left to reach a point B and again starting from point B, we move 6 steps to the left to reach a point C, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 1.9
And, C represents the integer – 3.
5 + (- 2) + (- 6) = – 3

Question 2.
Solution:
(i) (- 3) + ( – 9) = – 12
(Using the rule for addition of integers having like signs)
(ii) ( – 7) + ( – 8) = – 15
(Using the rule for addition of integers having like signs)
(iii) ( – 9) + 16 = 7
(Using the rule for addition of integers having unlike signs)
(iv) ( – 13) + 25 = 12
(Using the rule for addition of integers having unlike signs)
(v) 8 + ( – 17) = – 9
(Using the rule for addition of integers having unlike signs)
(vi) 2 + ( – 12) = – 10
(Using the rule for addition of integers having unlike signs)

Question 3.
Solution:
(i) Using the rule for addition of integers with like signs, we get:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 3.1
(ii) Using the rule for addition of integers with like signs, we get :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 3.2
(iii) Using the rule for addition of integers with like signs, we get :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 3.3
(iv) Using the rule for addition of integers with like signs, we get:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 3.4

Question 4.
Solution:
(i) Using the rule for addition of integers with unlike signs, we get:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 4.1
(ii) Using the rule for addition of integers with unlike signs, we get:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 4.2
(iii) Using the rule for addition of integers with unlike signs, we have
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 4.3
(iv) Using the rule for addition of integers with unlike signs, we have
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 4.4

Question 5.
Solution:
(i) Using the rule for addition of integers with unlike signs, we get :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 5.1
(ii) Using-the rule for addition of integers with unlike signs, we get
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 5.2
(iii) Using the rule for addition of integers with unlike signs, we get :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 5.3
(iv) Using the rule for addition of integers with unlike signs, we get :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 5.4
(v) Using the rule for addition of integers with like signs, we get:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 5.5
(vi) Using the rule for addition of integers with unlike signs, we get :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 5.6
(vii) Using the rule for addition of integers with unlike signs, we get :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4B 5.7
(viii) We have, ( – 18) + 25 + ( – 37)
= [( – 18) + 25] + ( – 37)
= 7 + ( – 37)
= – 30
(ix) We have, – 312 + 39 + 192
= ( – 312) + (39 + 192)
= ( – 312) + 231
= – 81
(x) We have ( – 51) + ( – 203) + 36 + ( – 28)
= [( – 51) + ( – 203)] + [36 + ( – 28)]
= ( – 254) + 8
= – 246

Question 6.
Solution:
(i) The additive inverse of – 57 is 57
(ii) The additive inverse of 183 is – 183
(iii) The additive inverse of 0 is 0
(iv) The additive inverse of – 1001 is 1001
(v) The additive inverse of 2054 is – 2054

Question 7.
Solution:
(i) Successor of 201 = 201 + 1 = 202
(ii) Successor of 70 = 70 + 1 = 71
(iii) Successor of – 5 = – 5 + 1 = – 4
(iv) Successor of – 99 = – 99 + 1 = – 98
(v) Successor of – 500 = – 500 + 1 = – 499 Ans.

Question 8.
Solution:
(i) Predecessor of 120 = 120 – 1 = 119
(ii) Predecessor of 79 = 79 – 1 = 78
(iii) Predecessor of – 8 = – 8 – 1 = – 9
(iv) Predecessor of – 141 = – 141 – 1 = – 142
(v) Predecessor of – 300 = – 300 – 1 = – 301 Ans.

Question 9.
Solution:
(i) ( – 7) + ( – 9) + 12 + ( – 16)
= – 7 – 9 + 12 – 16
= – 7 – 9 – 16 + 12
= – 32 + 12
= – 20
(ii) 37 + ( – 23) + ( – 65) + 9 + ( – 12)
= 37 – 23 – 65 + 9 – 12
= 37 + 9 – 23 – 65 – 12
= 46 – 100
= – 54
(iii) ( – 145) + 79 + ( – 265) + ( – 41) + 2
= – 145 + 79 – 265 – 41 + 2
= 79 + 2 – 145 – 265 – 41
= 81 – 451
= – 370
(iv) 1056 + ( – 798) + ( – 38) + 44 + ( – 1)
= 1056 – 798 – 38 + 44 – 1
= 1056 + 44 – 798 – 38 – 1
= 1100 – 837
= 263 Ans.

Question 10.
Solution:
Distance travelled from Patna to its north = 60 km
Distance travelled from that place to south of it = 90 km
Distance of the final place to Patna = 60 – 90
= – 30 km
= 30 km south
Ans.

Question 11.
Solution:
Total amount of pencils purchased = Rs. 30 + Rs. 25
= Rs 55
Total amount of pens purchased = Rs. 90
Total cost price = Rs. 55 + Rs. 90
= Rs. 145
Total sale price of pencils and pens = Rs 20 + Rs. 70
= Rs. 90
Loss = cost price – selling price
= Rs. 145 – Rs. 90
= Rs. 55 Ans.

Question 12.
Solution:
(i) True.
(ii) False : As if positive integer is greater then it will be positive.
(iii) True : As ( – a + a = 0).
(iv) False : As the sum of three integers can be zero or non-zero.
(v) False : As | – 5 | = 5 and | – 3 | = 3 and 5 ≮ 3.
(vi) False : | 8 – 5 | = | 3 | = 3 and | 8 | + | – 5 | = 8 + 5 = 13.

Question 13.
Solution:
(i) a + 6 = 0
Subtracting 6 from both sides,
a + 6 – 6 = 0 – 6
=> a = – 6
a = – 6.
(ii) 5 + a = 0
Subtracting 5 from both sides,
5 + a – 5 = 0 – 5
=> a = – 5
a = – 5
(iii) a + ( – 4) = 0
Adding 4 to both sides,
a + ( – 4) + 4 = 0 + 4
=> a = 4
a = 4
(iv) – 8 + a = 0
Adding 8 to both sides,
– 8 + a + 8 = 0 + 8
=> a – 8
a = 8 Ans.

Hope given RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4B are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A

RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4A.

Other Exercises

Question 1.
Solution:
(i) A decrease of 8
(ii) A gain of Rs. 7
(iii) Loosing a weight of 5 kg
(iv) 10 km below sea level
(v) 5°C above the freezing point
(vi) A withdrawal of Rs. 100
(vii) Spending Rs. 500
(viii) Going 6 m to the west
(ix) – 24
(x) 34

Question 2.
Solution:
(i) + Rs. 600
(ii) – Rs. 800
(iii) – 7°C
(iv) – 9
(v) + 2 km
(vi) – 3 km
(vii) + Rs. 200
(viii) – Rs. 300

Question 3.
Solution:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 3.1

Question 4.
Solution:
(i) 0
(ii) – 3
(iii) 2
(iv) 8
(v) – 365
(vi) 8

Question 5.
Solution:
(i) – 7
(ii) – 1
(iii) – 27
(iv) – 26
(v) – 603
(vi) – 777

Question 6.
Solution:
(i) The integers between 0 and 6 are
1, 2, 3, 4, 5.
(ii) The integers between – 5 and 0 are
– 4, – 3, – 2, – 1.
(iii) The integers between – 3 and 3 are
– 2, – 1, 0, 1, 2.
(iv) The integer between – 7 and – 5 is – 6.

Question 7.
Solution:
(i) 0 < 7
(ii) 0 > – 3
(iii) – 5 < – 2
(iv) – 15 < 13
(v) – 231 < – 132
(vi) – 6 < 6

Question 8.
Solution:
(i) – 7, – 2, 0, 5, 8
(ii) – 100, – 23, – 6, – 1, 0, 12
(iii) – 501, – 363, – 17, 15, 165
(iv) – 106, – 81, – 16, – 2, 0, 16, 21.

Question 9.
Solution:
(i) 36, 7, 0, – 3, – 9, – 132
(ii) 51, 0, – 2, – 8, – 53
(iii) 36, 0, – 5, – 71, – 81
(iv) 413, 102, – 7, – 365, – 515.

Question 10.
Solution:
(i) We want to write an integer 4 more than 6. So, we start from 6 and proceed 4 steps to the right to obtain 10, as shown below:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 10.1
∴ 4 more than 6 is 10.
(ii) We want to write an integer 5 more than – 6. So, we start from – 6 and proceed 5 steps to the right to obtain – 1, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 10.2
∴ 5 more than – 6 is – 1.
(iii) We want to write an integer 6 less than 2. So we start from 2 and come back to the left by 6 steps to obtain – 4, as shown below:
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 10.3
∴ 6 less than 2 is – 4.
(iv) We want to write an integer 2 less than – 3. So we start from – 3 and come back to the left by 2 steps to obtain – 5, as shown below :
RS Aggarwal Class 6 Solutions Chapter 4 Integers Ex 4A 10.4
∴ 2 less than – 3 is – 5.

Question 11.
Solution:
(i) False, as zero is greater than every negative integer.
(ii) False, as zero is an integer.
(iii) True, as zero is neither positive nor negative.
(iv) False, as – 10 is to the left of – 6 on a number line.
(v) False, as absolute value of an integer is always equal to the integer.
(vi) True, as 0 is to right of every negative integer, on a number line.
(vii) False, as every natural number is positive. False, the successor is – 186
(viii) False, the predecessor is – 216

Question 12.
Solution:
(i) | – 9 | = 9
(ii) | 36 | = 36
(iii) | 0 | = 0
(iv) | 15 | = 15
(v) – | – 3 | = – 3
(vi) 7 + | – 3 | = 7 + 3 = 10
(vii) |7 – 4| = | 3 | = 3
(viii) 8 – | – 7| = 8 – 7 = 1

Question 13.
Solution:
The required integers are – 6, – 5, – 4, – 3, – 2.
The required integers are – 21, – 22, – 23, – 24, – 25.
The required integers are – 21, – 22, – 23, – 24, – 25.

 

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RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3F.

Other Exercises

Objective questions
Mark against the correct answer in each of the following :

Question 1.
Solution:
The smallest whole number is 0 (b)

Question 2.
Solution:
The least 4-digit number = 1000
On dividing 1000 by 9, we get
Remainder = 1
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 1.1
Least 4-digit number which is
Divisible by 9 = 1000 – 1 + 9
= 1000 + 8
= 1008 (d)

Question 3.
Solution:
The largest 6-digit number = 999999
On dividing by 16, we get
Remainder =15
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 3.1
The greatest 6-digit number divisible by 16
= 999999 – 15
= 999984 (c)

Question 4.
Solution:
On dividing 10004 by 12, we get remainder = 8
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 4.1
8 is to be subtracted from 10004 (c)

Question 5.
Solution:
On dividing 10056 by 23 We get remainder =12
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 5.1
The least number to be added = 23 – 5
= 18 (b)

Question 6.
Solution:
On dividing 457 by 11
We get remainder = 6
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3F 6.1
Which is greater than half of 11
The number nearest to 457 which is divisible 11 will be = 457 – 6 + 11
= 457 + 5
= 462 (d)

Question 7.
Solution:
Whole number between 1018 and 1203 are 1019 to 1202 are 1202 – 1018
= 184 (c)

Question 8.
Solution:
Divisor = 46
Quotient =11
Remainder =15
Number = Divisor x Quotient + Remainder
= 46 x 11 + 15
= 506 + 15
= 521 (b)

Question 9.
Solution:
Dividend = 199
Quotient =16
Remainder = 7
Divisor = \(\\ \frac { 199-7 }{ 16 } \) = \(\\ \frac { 192 }{ 16 } \)
= 12 (c)

Question 10.
Solution:
7589 – ? = 3434
Required number = 7589 – 3434
= 4155 (c)

Question 11.
Solution:
587 x 99 = 587 x (100 – 1)
= 587 x 100 – 587 x 1
= 58700 – 587
= 58113 (c)

Question 12.
Solution:
4 x 538 x 25 = 538 x 4 x 25
= 538 x 100
= 53800 (c)

Question 13.
Solution:
24679 x 92 + 24679 x 8
= 24679 x (92 + 8)
= 24679 x 100
= 2467900 (c)

Question 14.
Solution:
1625 x 1625 – 1625 x 625
= 1625 (1625 – 625)
= 1625 x 1000
= 1625000 (a)

Question 15.
Solution:
1568 x 185 – 1568 x 85
= 1568 (185 – 85)
= 1568 x 100
= 156800 (c)

Question 16.
Solution:
888 + 111 + 555 = 111 x ?
= 11 (8 + 7 + 5)
= 111 x 20 (c)

Question 17.
Solution:
The sum of two odd number is an even number (b)

Question 18.
Solution:
The product of two odd numbers is an odd number (a)

Question 19.
Solution:
If a is a whole number such that
a + a = a, then a = 0
as 0 + 0 = 0
None of these (d)

Question 20.
Solution:
The predecessor of 10000 is 10000 – 1
= 9999 (b)

Question 21.
Solution:
The successor of 1001 is 1001 + 1
= 1002 (b)

Question 22.
Solution:
The smallest even whole number is 2 (b)

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3F are helpful to complete your math homework.

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

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

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

Question 1.
Solution:
We have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 1.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 2.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 3.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 4.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 5.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 6.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 7.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 8.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 9.1
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 :
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 10.1
∴ H.C.F. of 58 and 70 = 2.

Question 11.
Solution:
By division method, we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 11.1
∴ H.C.F. of 399 and 437 = 19

Question 12.
Solution:
By division method, we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 12.1
∴ H.C.F. of 1045 and 1520 = 95.

Question 13.
Solution:
By division method, we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 13.1
∴ H.C.F. of 1965 and 2096 = 131

Question 14.
Solution:
By division method, we have
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 14.1
∴ H.C.F. of 2241 and 2324 = 83.

Question 15.
Solution:
First, we find the H.C.F. of 658 and 940
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 15.1
∴ H.C.F. of 658 and 940 is 94.
Now, we find the H.C.F. of 94 and 1128.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 15.2
∴ 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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 16.1
∴ H.C.F. of 754 and 1508 is 754
Now, we find the H.C.F. of 754 and 1972
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 16.2
∴ 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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 17.1
∴ H.C.F. of 391 and 425 is 17.
Now, we find the H.C.F. of 17 and 527
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 17.2
∴ 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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 18.1
H.C.F. of 1794 and 2346 is 138.
Now, we find the H.C.F. of 138 and 4761
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 18.2
∴ 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.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 19.1
∴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.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 20.1
∴ 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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 21.1
∴ 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.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 22.1
∴ 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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 23.1
∴ 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.
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 24.1
∴ 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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 25.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 26.1
∴ 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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 27.1
∴ H.C.F. of 441 and 567 is 63
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 27.2
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 28.1
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 28.2
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 28.3

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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 29.1
∴ H.C.F. of 42 and 49 = 7
Now we find the H.C.F. of 7 and 63
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 29.2
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 30.1
∴ 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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 31.1
∴ 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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 32.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 33.1
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
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 34.1
RS Aggarwal Class 6 Solutions Chapter 2 Factors and Multiples Ex 2D 34.2

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RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E.

Other Exercises

Question 1.
Solution:
(i) By actual division, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.1
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.2
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.3
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.4
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 1.5

Question 2.
Solution:
(i) By actual division, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.1
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.2
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.3
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.4
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.5
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 2.6

Question 3.
Solution:
(i) We know that any number (non-zero) divided by 1 gives the number itself
65007 ÷ 1 = 65007
(ii) We know that 0 divided by any natural number gives 0
0 ÷ 879 = 0
(iii) 981 + 5720 ÷ 10 = 981 + (5720 ÷ 10)
= 981 + 572
= 1553
(iv) 1507 – 625 ÷ 25 = 1507 – (625 ÷ 25)
= 1507 – 25
= 1482
(v) 32277 ÷ (648 – 39)
= 32277 ÷ 609
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 3.1
32277 ÷ (648 – 39) = 53
(vi) 1573 ÷ 1573 – 1573 ÷ 1573
= (1573 ÷ 1573) – (1573 ÷ 1573)
= 1 – 1
= 0

Question 4.
Solution:
We have n ÷ n = n
let n = 1, 1 ÷ 1 = 1
1 = 1
which is true
∴ Hence 1 is the required whole number.

Question 5.
Solution:
Product of two numbers = 504347
One number = 317
Other number = 504347 ÷ 317
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 5.1
∴ Other number = 1591

Question 6.
Solution:
Here Dividend = 59761, Quotient = 189
∴ Remainder = 37
We know that Dividend = Divisor x Quotient + Remainder
59761 = Divisor x 189 + 37
59761 – 37 = Divisor x 189
59724 = Divisor x 189
Divisor x 189 = 59724
∴ Divisor = 59724 ÷ 189
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 6.1
∴ Divisor = 59724 ÷ 189 = 316

Question 7.
Solution:
Here dividend = 55390,
Divisor = 299 and Remainder = 75
By division algorithm, we have
Dividend = Quotient x Divisor + Remainder
55390 = Quotient x 299 + 75
55390 – 75 = Quotient x 299
55315 = Quotient x 299
Quotient x 299 = 55315
Quotient = 55315 ÷ 299
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 7.1
∴ Required quotient = 185

Question 8.
Solution:
On dividing 13601 by 87, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 8.1
It is clear that if we subtract 29 from 13601, the resulting number will be exactly divisible by 87.
∴ The required least number = 29.

Question 9.
Solution:
Here dividend = 1056, Divisor = 23
By actual division, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 9.1
It is clear that if we add 2 to 21, it will become 23 which is divisible by 23.
∴ Required least number = 2.

Question 10.
Solution:
Greatest 4-digit number = 9999
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 10.1
On, dividing by 16, we get remainder as 15
∴ The required largest 4-digit number = 9999 – 15
= 9984

Question 11.
Solution:
Largest number of 5-digits = 99999
On dividing 99999 by 653, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 11.1
∴ Quotient = 153, Remainder = 90
Check : By division algorithm Dividend = Divisor x Quotient + Remainder
= 653 x 153 + 90
= 99909 + 90
= 99999

Question 12.
Solution:
The least 6-digit number = 100000
On dividing 100000 by 83, we have
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 12.1
It is clear that if we add 15 to 68, it will become 83 which is divisible by 83.
∴ Required least 6-digit number = 100000 + 15
= 100015

Question 13.
Solution:
Cost of 1 dozen bananas = Rs. 29
Bananas can be purchase in Rs. 1392
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 13.1
1392 ÷ 29
= 48 dozens

Question 14.
Solution:
Total number of trees = 19625
Total number of rows = 157
Number of trees in each row = 19625 ÷ 157
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 14.1
∴ Number of trees in each row = 125

Question 15.
Solution:
Total population of the town = 517530
Since there is one educated person out of 15
Total number of educated persons in the town = 517530 ÷ 15
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 15.1
∴ Total number of educated persons in the town = 34502.

Question 16.
Solution:
Cost of 23 colour TV sets = Rs. 570055
Cost of 1 colour TV set
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3E 16.1
∴ Cost of 1 color TV set
= Rs. 570055 ÷ 23
= Rs. 24785

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E are helpful to complete your math homework.

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RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3D

RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3D

These Solutions are part of RS Aggarwal Solutions Class 6. Here we have given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3D.

Other Exercises

Question 1.
Solution:
(i) 246 x 1 = 246
(By multiplicative property of 1)
(ii) 1369 x 0 = 0
(By multiplicative property of 0)
(iii) 593 x 188 = 188 x 593
(By commutative law of multiplication)
(iv) 286 x 753 = 753 x 286
(By commutative law of multiplication)
(v) 38 x (91 x 37) = 91 x (38 x 37)
(By associative law of multiplication)
(vi) 13 x 100 x 1000 = 1300000
(vii) 59 x 66 + 59 x 34 = 59 x (66 + 34)
(By distributive law of multiplication)
(vii) 68 x 95 = 68 x 100 – 68 x 5

Question 2.
Solution:
(i) Commutative law of multiplication
(ii) Closure property
(iii) Associative law of multiplication
(iv) Multiplicative property of 1
(v) Multiplicative property of 0
(vi) Distributive law of multiplication over addition in whole numbers
(vii) Distributive law of multiplication over subtraction in whole numbers

Question 3.
Solution:
Using the law of distribution over addition and subtraction
(i) 647 x 13 + 647 x 7
= 647 x (13 + 7)
= 647 x 20
= 12940
(ii) 8759 x 94 + 8759 x 6
= 8759 x (94 + 6)
= 8759 x 100
= 875900
(iii) 7459 x 999 + 7459
= 7459 x 999 + 7459 x 1
= 7459 x (999 + 1)
= 7459 x 1000
= 7459000
(iv) 9870 x 561 – 9870 x 461
= 9870 x (561 – 461)
= 9870 x 100
= 987000
(v) 569 x 17 + 569 x 13 + 569 x 70
= 569 x (17 + 13 + 70)
= 569 x 100
= 56900
(vi) 16825 x 16825 – 16825 x 6825
= 16825 x (16825 – 6825)
= 16825 x 10000
= 168250000

Question 4.
Solution:
(i) 2 x 1658 x 50 = 1658 x (2 x 50) (Associative law of multiplication)
= 1658 x 100
= 165800
(ii) 4 x 927 x 25 = 927 x (4 x 25) (Associative law of multiplication)
= 927 x 100
= 92700
(iii) 625 x 20 x 8 x 50
(By associative law of multiplication)
(625 x 8) x (20 x 50)
= 5000 x 1000
= 5000000
(iv) 574 x (625 x 16)
= 574 x 10000
= 5740000
(v) 250 x 60 x 50 x 8
= (250 x 8) x (60 x 50)
(By associative law)
= 2000 x 3000
= 6000000
(vi) 8 x 125 x 40 x 25
= (8 x 125) x (40 x 25)
= 1000 x 1000
= 1000000

Question 5.
Solution:
Using distributive law of multiplication over addition or subtraction,
(i) 740 x 105
= 740 x (100 + 5)
= 740 x 100 + 740 x 5
= 74000 + 3700
= 77700
(ii) 245 x 1008
= 245 x (1000 + 8)
= 245 x 1000 + 245 x 8
= 245000 + 1960
= 246960
(iii) 947 x 96
= 947 x (100 – 4)
= 947 x 100 – 947 x 4
= 94700 – 3788
= 90912
(iv) 996 x 367
= 367 x (1000 – 4)
= 367 x 1000 – 367 x 4
= 367000 – 1468
= 365532
(v) 472 x 1097
= 472 x (1100 – 3)
= 472 x 1100 – 472 x 3
= 519200 – 1416
= 517784
(vi) 580 x 64
= 580 x (60 + 4)
= 580 x 60 + 580 x 4
= 34800 + 2320
= 37120
(vii) 439 x 997
= 437 x (1000 – 3)
= 439 x 1000 – 439 x 3
= 439000 – 1317
= 437683
(viii) 1553 x 198
= 1553 x (200 – 2)
= 1553 x 200 – 1553 x 2
= 310600 – 3106
= 307494

Question 6.
Solution:
(i) 3576 x 9 = 3576 x (10 – 1)
= 3576 x 10 – 3576 x 1
= 35760 – 3576
= 32184
(ii) 847 x 99 = 84 x (100 – 1)
= 847 x 100 – 847 x 1
= 84700 – 847
= 83853
(iii) 2437 x 999 = 2437 x (1000 – 1)
= 2437 x 1000 – 2437 x 1
= 2437000 – 2437
= 2434563

Question 7.
Solution:
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3D 7.1

Question 8.
Solution:
Largest 3-digit number = 999
Largest 5-digit number = 99999
Required product = 99999 x 999
= 99999 x (1000 – 1)
= 99999 x 1000-99999 x 1
= 99999000 – 99999
= 9,98,99,001

Question 9.
Solution:
Speed of car = 75 km per hour
In 1 hour, distance covered by a car = 75 km
.’. In 98 hours, distance will be covered = 75 x 98
= 75 x (100 – 2)
= 75 x 100 – 75 x 2
= 7500 – 150
= 7350 km

Question 10.
Solution:
Cost of 1 set of VCR = Rs. 24350
Cost of 139 sets of VCR = Rs. 24350 x 139
= Rs. 3384650
RS Aggarwal Class 6 Solutions Chapter 3 Whole Numbers Ex 3D 10.1

Question 11.
Solution:
Cost of 1 house = Rs. 450000
Cost of 197 houses = Rs. 450000 x 197
= Rs. 450000 x (200 – 3)
= Rs. (450000 x 200 – 450000 x 3)
= Rs. (90000000 – 1350000)
= Rs. 88650000

Question 12.
Solution:
Cost of each chair = Rs. 1065
Cost of 50 chairs = Rs. 1065 x 50
= Rs. 53250
Cost of each blackboard = Rs. 1645
Cost of 30 blackboards = Rs. 1645 x 30
= Rs. 49350
Total cost of 50 chairs and 30 blackboards = Rs. 53250 + 49350
= Rs. 102600

Question 13.
Solution:
Number of students in 1 section = 45
Number of students in 6 sections = 45 x 6 = 270
Monthly charges of 1 student = Rs. 1650
Total monthly incomes from the class VI = Rs. 270 x 1650
= Rs. (300 – 30) x 1650
= Rs. (1650 x 300 – 1650 x 30)
= Rs. (495000 – 49500)
= Rs. 445500

Question 14.
Solution:
Since the product of two whole numbers is zero
.’. From multiplicative property of zero, we conclude that one of the whole numbers is zero.

Question 15.
Solution:
(i) Sum of two odd numbers is an even number.
(ii) Product of two odd numbers is an odd number.
(iii) a x a = a => a = 1 as 1 x 1 = 1

Hope given RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3D are helpful to complete your math homework.

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

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

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.

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

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

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 2A.

Other Exercises

Question 1.
Solution:
(i) A factor of a number is an exact divisor of that number.
Examples : 1. 2 is a factor of 8
2. 5 is a factor of 15
3. 9 is a factor of 27
4. 4 is a factor of 20
5. 3 is a factor of 12.
(ii) Multiple. A number is said to be a multiple of any of its factors.
Examples : 1. 15 is a multiple of 3
2. 8 is a multiple of 4
3. 10 is a multiple of 2
4. 25 is a multiple of 5
5.18 is a multiple of 9.

Question 2.
Solution:
(i) We know that
20 = 1 x 20, 20 = 2 x 10, 20 = 4 x 5
which shows that the numbers 1, 2, 4, 5, 10, 20 exactly divide 20.
1, 2, 4, 5, 10 and 20 are all factors of 20
(ii) We know that
36 = 1 x 36, 36 = 2 x 18, 36 = 3 x 12, 36 = 4 x 9, 36 = 6 x 6
This shows that each of the numbers 1, 2, 3, 4, 6, 9, 12, 18, 36 exactly divides 36.
1, 2, 3, 4, 6, 9, 12, 18, 36 are the factors of 36.
(iii) We know that
60 = 1 x 60, 60 = 2 x 30, 60 = 3 x 20, 60 = 4 x 15, 60 = 5 x 12, 60 = 6 x 10
This shows that each of the numbers 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 exactly divides 60.
1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 are all the factors of 60.
(iv) We know that
75 = 1 x 73, 75 = 3 x 25, 75 = 5 x 15
This shows that each of the numbers 1, 3, 5, 15, 25, 75 exactly divides 75.
1, 3, 5, 15, 25, 75 are all the factors of 75.

Question 3.
Solution:
(i) First five multiples of 17 are :
17 x 1 = 17
17 x 2 = 34
17 x 3 = 51
17 x 4 = 68
17 x 5 = 85
(ii) First five multiples of 23 are :
23 x 1 = 23
23 x 2 = 46
23 x 3 = 69
23 x 4 = 92
23 x 5 = 115
(iii) First five multiples of 65 are :
65 x 1 = 65
65 x 2 = 130
65 x 3 = 195
65 x 4 = 260
65 x 5 = 325
(iv) First five multiples of 70 are :
70 x 1 = 70
70 x 2 = 140
70 x 3 = 210
70 x 4 = 280
70 x 5 = 350

Question 4.
Solution:
(i) 32 is a multiple of 2, so it is an even number.
(ii) 37 is not a multiple of 2, so it is an odd number.
(iii) 50 is a multiple of 2, so it is an even number.
(iv) 58 is a multiple of 2, so it is an even number.
(v) 69 is not a multiple of 2, so it is an odd number.
(vi) 144 is a multiple of 2, so it is an even number.
(vii) 321 is not a multiple of 2, so it is an odd number.
(viii) 253 is not a multiple of 2, so it is an odd number.

Question 5.
Solution:
Prime Numbers. Each of the numbers which has exactly two factors, namely 1 and itself, is called a prime number.
Examples. The numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 are all prime numbers.

Question 6.
Solution:
(i) Prime numbers between 10 and 40 are: 11, 13, 17, 19, 23, 29, 31, 37.
(ii) Prime numbers between 80 and 100 are : 83, 89, 97.
(iii) Prime numbers between 40 and 80 are : 41, 43, 47, 53, 59, 61, 67, 71, 73, 79
(iv) Prime numbers between 30 and 40 are : 31, 37.

Question 7.
Solution:
(i) 2 is the smallest prime number.
(ii) 2 is the only even prime number.
(iii) 3 is the smallest odd prime number.

Question 8.
Solution:
(i) We know that
87 = 1 x 87, 87 = 3 x 29
This shows that 1, 3, 29, 87 are the factors of 87.
The number 87 is not a prime number as it has more than 2 factors.
(ii) We have 89 = 1 x 89
The number 89 is a prime number as it has only 2 factors.
(iii) We have 63 = 1 x 63, 63 = 3 x 21,
63 = 7 x 9
This shows that the number 63 has more than 2 factors namely 1, 3, 7, 9, 21,63. So, it is not a prime number.
(iv) We have 91 = 1 x 91, 91 = 7 x 13 This shows that the number 91 has more than 2 factors namely 1, 7, 13, 91.
So, it is not a prime number.

Question 9.
Solution:
From the Sieve of Eratosthenes, we see that the seven consecutive numbers are 90, 91, 92, 93, 94, 95 and 96

Question 10.
Solution:
(i) There is no counting number having no factor at all.
(ii) The number 1 has exactly one factor.
(iii) The numbers between 1 and 100 having exactly three factors are : 4, 9, 25, 49.

Question 11.
Solution:
Composite Numbers. Numbers having more than two factors are called composite numbers. A composite number can be an odd number. The smallest odd composite number is 9.

Question 12.
Solution:
Twin-primes. Two consecutive odd prime numbers are known as twin- primes.
The prime numbers between 50 and 100 are:
53, 59, 61, 67, 71, 73, 79, 83, 89, 97
From above pairs of twin-primes are (59, 61), (71, 73)

Question 13.
Solution:
Co-primes. Two numbers are said to
be co-prime if they do not have a common factor.
Examples. Five pairs of co-primes are:
(i) 2, 3
(ii) 3, 4
(iii) 4, 5
(iv) 8, 15
(v) 9, 16
Co-primes are not always prime.
Illustration. In the pair (3, 4) of co-primes, 3 is a prime number whereas 4 is a composite number.

Question 14.
Solution:
(i) 36 = 7 + 29
(ii) 42 = 5 + 37
(iii) 84= 17 + 67
(iv) 98 = 19 + 79

Question 15.
Solution:
(i) 31 = 5 + 7 + 19
(ii) 35 = 5 + 7 + 23
(iii) 49 = 3 + 5 + 41
(iv) 63 = 7+ 13 +43

Question 16.
Solution:
(i) 36 = 17 + 19
(ii) 84 = 41 + 43
(iii) 120 = 59 + 61
(iv) 144 = 71+73

Question 17.
Solution:
(i) to (iv). None of the given statements is true.

 

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

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