CBSE Class 6

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

• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4A
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4B
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4C
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4D
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4E
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4F

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 :

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 :

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 :

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 :

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 :

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 :

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 :

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 :

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 :

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:

(ii) Using the rule for addition of integers with like signs, we get :

(iii) Using the rule for addition of integers with like signs, we get :

(iv) Using the rule for addition of integers with like signs, we get:

Question 4.
Solution:
(i) Using the rule for addition of integers with unlike signs, we get:

(ii) Using the rule for addition of integers with unlike signs, we get:

(iii) Using the rule for addition of integers with unlike signs, we have

(iv) Using the rule for addition of integers with unlike signs, we have

Question 5.
Solution:
(i) Using the rule for addition of integers with unlike signs, we get :

(ii) Using-the rule for addition of integers with unlike signs, we get

(iii) Using the rule for addition of integers with unlike signs, we get :

(iv) Using the rule for addition of integers with unlike signs, we get :

(v) Using the rule for addition of integers with like signs, we get:

(vi) Using the rule for addition of integers with unlike signs, we get :

(vii) Using the rule for addition of integers with unlike signs, we get :

(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.

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 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

• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4A
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4B
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4C
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4D
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4E
• RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4F

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:

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:

∴ 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 :

∴ 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:

∴ 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 :

∴ 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.

Hope given RS Aggarwal Solutions Class 6 Chapter 4 Integers Ex 4A 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 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

• RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3A
• RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3B
• RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3C
• RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3D
• RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3E
• RS Aggarwal Solutions Class 6 Chapter 3 Whole Numbers Ex 3F

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

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

The greatest 6-digit number divisible by 16
= 999999 – 15
= 999984 (c)

Question 4.
Solution:
On dividing 10004 by 12, we get remainder = 8

8 is to be subtracted from 10004 (c)

Question 5.
Solution:
On dividing 10056 by 23 We get remainder =12

The least number to be added = 23 – 5
= 18 (b)

Question 6.
Solution:
On dividing 457 by 11
We get remainder = 6

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.

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