RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers

Read RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers below, students should study RD Sharma class 6 Mathematics available on Studiestoday.com with solved questions and answers. These chapter wise answers for class 6 Mathematics have been prepared by teacher of Grade 6. These RD Sharma class 6 Solutions have been designed as per the latest NCERT syllabus for class 6 and if practiced thoroughly can help you to score good marks in standard 6 Mathematics class tests and examinations

Exercise 5.1

 

Question 1:  Write the opposite of each of the following:

(i) Increase in population

(ii) Depositing money in a bank

(iii) Earning money

(iv) Going North

(v) Gaining a weight of 4kg

(vi) A loss of Rs. 1000

(vii) 25

(viii) – 15 

Solution 1:

(i) Decrease in population is the opposite of Increase in population.

(ii) Withdrawing money from a bank is the opposite of depositing money in a bank

(iii) Spending money is the opposite of earning money

(iv) Going south is the opposite of Going North.

(v) Losing a weight of 4kg is the opposite of gaining a weight of 4kg

(vi) A gain of Rs. 1000 is the opposite of A loss of Rs. 1000. 

(vii) – 25 is the opposite of 25.

(viii) 15 is the opposite of -15.

 

Question 2:  Indicate the following by using integers:

(i)  25° above zero.

(ii) 5° below zero.

(iii) A profit of Rs. 800.

(iv) A deposit of Rs. 2500.

(v) 3 km above sea level.

(vi) 2 km below level. 

Solution 2:

(i) +25°.

(ii) – 5°.

(iii) + Rs. 800.

(iv) + Rs. 2500.

(v) + 3.

(vi) – 2.

 

Question 3:  Mark the following integers on a number line:

(i) 7

(ii) -4

(iii) 0 

Solution 3:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers


 

Question 4:  Which number in each of the following pairs is smaller?

(i) 0, -4

(ii) -3 , 12

(iii) 8, 13

(iv) – 15, -27 

Solution 4:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-

 

Question 5:  Which number in each of the following pairs is larger?

(i) 3, -4

(ii) – 12, – 8

(iii) 0, 7

(iv) 12, – 18 

Solution 5:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-2


Question 6:  Write all integers between:

(i) – 7 and 3

(ii) – 2 and 2

(iii) – 4 and 0

(iv) 0 and 3 

Solution 6:

(i) – 6, – 5, – 4, – 3, – 2, – 1, 0, 1, 2 is the integers between – 7 and 3.

(ii) -1, 0, 1 is the integers between – 2 and 2.

(iii) -3, -2, -1 is the integers between – 4 and 0.

(iv) 1, 2 is the integers between 0 and 3.

 

Question 7:  How many integers are between?

(i) – 4 and 3

(ii) 5 and 12

(iii) – 9 and – 2

(iv) 0 and 5

Solution 7:

(i) -3, -2, -1, 0, 1, 2 is the integers between – 4 and 3.

Hence, number of integers between – 4 and 3 are 6.

 

(ii) 6, 7, 8, 9, 10, 11 is the integers between 5 and 12.

Hence, number of integers between 5 and 12 are 6.

 

(iii) -8, -7, -6, -5, -4, -3 is the integers between – 9 and – 2.

Hence, number of integers between -9 and -2 are 6.

 

(iv) 1, 2, 3, 4 is the integers between 0 and 5 are

Hence, number of integers between 0 and 5 are 4.

 

Question 8:  Replace * in each of the following by < or > so that the statement is true:

(i) 2 * 5

(ii) 0 * 3

(iii) 0 * – 7

(iv) – 18 * 15

(v) – 235 * – 532

(vi) – 20 * 20 

Solution 8:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-3


Question 9:  Write the following integers in increasing order:

(i) – 8, 5, 0, -12, 1, -9, 15

(ii) – 106, 107, – 320, – 7, 185 

Solution 9:

(i) The increasing order of given numbers is – 12 < – 9 < – 8 < 0 < 1 < 5 < 15. 

(ii) The increasing order of given numbers is -320 < – 106 < – 7 < 107 < 185.

 

Question 10:  Write the following integers in decreasing order:

(i) – 15, 0, -2, -9, 7, 6, -5, 8

(ii) -154, 123, -205, -89, -74 

Solution 10:

(i) The decreasing order of given numbers is 8 > 7 > 6 > 0 > -2 > -5 > -9 > -15. 

(ii) The decreasing order of given numbers is 123 > – 74 > – 89 > – 154 > – 205.

 

Question 11:  Using the number line, write the integer which is:

(i) 2 more than 3

(ii) 5 less than 3

(iii) 4 more than – 9 

Solution 11:

 RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-4

 

Question 12:  Write the absolute value of each of the following:

(i) 14

(ii) – 25

(iii) 0

(iv) – 125

(v) – 248

(vi) a – 7, if a is greater than 7

(vii) a – 7, if a – 2 is less than 7

(viii) a + 4, if a is greater than -4

(ix) a + 4 if a is less than – 4

(x) |-3|

(xi) -|-5|

(xii) |12 – 5|

Solution 12:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-5

 

Question 13:  

(i) Write 4 negative integers less than – 10.

(ii) Write 6 negative integers just greater than – 12. 

Solution 13:

(i) – 11, – 12, – 13, – 14 are the 4 negative integers less than – 10. 

(ii) -11, – 10, – 9, – 8, – 7, – 6 are the 6 negative integers just greater than – 12.

 

Question 14:  Which of the following statements are true?

(i) The smallest integer is zero.

(ii) The opposite of zero is zero.

(iii) Zero is not an integer.

(iv) 0 is larger than every negative integer.

(v) The absolute value of an integer is greater than the integer.

(vi) A positive integer is greater than its opposite.

(vii) Every negative integer is less than every natural number.

(viii) 0 is the smallest positive integer.

Solution 14:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-6

 

Exercise 5.2 

Question 1:  Draw a number line and represent each of the following on it:

(i) 5 + (-2)

(ii) (-9) + 4

(iii) (-3) + (-5)

(iv) 6 + (-6)

(v) (-1) + (-2) + 2

(vi) (-2) + 7 + (-9) 

Solution 1:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-7

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-8

 

Question 2:  Find the sum of

(i) -557 and 488

(ii) -522 and -160

(iii) 2567 and – 325

(iv) -10025 and 139

(v) 2547 and -2548

(vi) 2884 and -2884 

Solution 2:

(i) – 557 and 488

= – 557 + 488 (we know that-,+,=,-)

= 557 – 488

= 69.

 

(ii) – 522 and – 160

= – 522 + (-160) (we know that +,-,=,-)

= – 522 – 160

= – 682

 

(iii) 2567 and – 325

= 2567 + (-325) (we know that  +,-,=,-)

= 2567 – 325

= 2242

 

(iv) -10025 and 139

= -10025 + 139 (we know that-,+,=,-)

= -10025 + 139

= -9886

 

(v) 2547 and -2548

= 2547 + (-2548) (we know that +,-,=,-)

= 2547 – 2548

= -1

 

(vi) 2884 and -2884

= 2884 + (-2884) (we know that +,-,=,-)

= 2884 – 2884

= 0

 

Exercise 5.3

 

Question 1:  Find the additive inverse of each of the following integers:

(i) 52

(ii) – 176

(iii) 0

(iv) 1 

Solution 1:

(i) – 52 is the additive inverse of 52.

(ii) 176 is the additive inverse of – 176.

(iii) 0 is the additive inverse of 0.

(iv) – 1 is the additive inverse of 1.

 

Solution 2:  Find the successor of each of the following integers:

(i) – 42

(ii) -1

(iii) 0

(iv) – 200

(v) -99 

Solution 2:

(i) For successor we should add 1 in each value.

Successor of – 42 is 

 

(ii) For successor we should add 1 in each value.

Successor of – 1 is 

 

(iii) For successor we should add 1 in each value.

Successor of 0 is 

 

(iv) For successor we should add 1 in each value.

Successor of – 200 is 

 

(v) For successor we should add 1 in each value.

Successor of – 99 is 

 

Question 3:  Find the predecessor of each of the following integers:

(i) 0

(ii) 1

(iii) – 1

(iv) – 125

(v) 1000 

Solution 3:

(i) For predecessor we should subtract 1 in each value.

Predecessor of 0 is

= 0 – 1

= – 1

 

(ii) For predecessor we should subtract 1 in each value.

Predecessor of 1 is

= 1 – 1

= 0

 

(iii) For predecessor we should subtract 1 in each value.

Predecessor of -1 is

= -1 – 1

= -2

 

(iv) For predecessor we should subtract 1 in each value.

Predecessor of – 125 is

= -125 – 1

= – 126

 

(v) For predecessor we should subtract 1 in each value.

Predecessor of 1000 is

= 1000 – 1

= 999

 

Question 4:  Which of the following statements are true?

(i) The sum of a number and its opposite is zero.

(ii) The sum of two negative integers is a positive integer.

(iii) The sum of a negative integer and a positive integer is always a negative integer.

(iv) The successor of – 1 is 1.

(v) The sum of three different integers can never be zero. 

Solution 4:

(i) True

Reason:- 1 – 1 = 0

 

(ii) False

Reason:- -1 – 1 = -2

 

(iii) False

Reason:- – 2 + 3 = 1

 

(iv) False

Reason:- Successor of – 1 is 0.

 

(v) False

Reason:- 1 + 2 – 3 = 0

 

Question 5:  Write all integers whose absolute values are less than 5. 

Solution 5:

-4, – 3, – 2, – 1, 0, 1, 2, 3, 4

Hence, above are the integers whose absolute values are less than 5:-

 

Question 6:   Which of the following is false:

(i) |4 + 2| = |4| + |2|

(ii) |2 – 4| = |2| + |4|

(iii) |4 – 2| = |4| – |2|

(iv) |(-2) + (-4)| = |-2| + |-4| 

Solution 6:

(i) True

Reason :-

|4+2|= |4|+|2

|6|=4+2

6  =  6

 

(ii) False

Reason :-

|2 - 4| = |2|+|4|

|-2| = 2+4

2 = 6 

 

(iii) True

Reason :- 

|4-2|= |4|-|2|

|2|=4 - 2

2  =  2

 

(iv) True

Reason :-

|(-2) + (-4)| = |-2| + |-4|

|-2-4| = 2 + 4

|-6| = 6

6 = 6

 

Question 7:  Complete the following table:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-9

From the above table:

(i) Write all the pairs of integers whose sum is 0.

(ii) Is (-4) + (-2) = (-2) + (-4)?

(iii) Is 0 + (-6) = -6? 

Solution 7:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-10

(i) (6, -6), (4, – 4), (2, – 2), (0, 0) are the pairs of integers whose sum is 0.

 

(ii) By using commutatively of addition

(-4) + (-2) = (-2) + (-4)

- 4 – 2 = - 2 – 4

- 6 = - 6

Hence, it is a correct statement.

 

(iii) By using additive identity

 0 + (-6)

= -6.

 

Question 8:  Find an integer x such that

(i) x + 1 = 0

(ii) x + 5 = 0

(iii) – 3 + x = 0

(iv) x + (-8) = 0

(v) 7 + x = 0

(vi) x + 0 = 0 

Solution 8:

(i) x + 1 = 0 (Subtracting 1 on both sides)

x + 1 – 1 = 0 – 1

x = -1

 

(ii) x + 5 = 0 (Subtracting 5 on both sides)

x + 5 – 5 = 0 – 5

x = -5

 

(iii) – 3 + x = 0 (Adding 3 on both sides)

-3 + x + 3 = 0 + 3

x = 3

 

(iv) x + (-8) = 0 (Adding 8 on both sides)

x – 8 + 8 = 0 + 8

x = 8

 

(v) 7 + x = 0 (Subtracting 7 on both sides)

7 + x – 7 = 0 – 7

x = – 7

 

(vi) x + 0 = 0

x = 0

Exercise 5.4

 

Question 1:  Subtract the first integer from the second in each of the following:

(i) 12, -5

(ii) – 12, 8

(iii) – 225, – 135

(iv) 1001, 101

(v) – 812, 3126

(vi) 7560, – 8

(vii) – 3978, – 4109

(viii) 0, – 1005 

Solution 1:

(i) 12, -5

By subtracting the first integer from the second we get,
= -5 – 12  (we know that -,- = +) 
 = – 17 
 
(ii) – 12, 8
By subtracting the first integer from the second we get,
= 8 – (-12) (we know that -,- = +)  
= 8 + 12  
= 20 
 
(iii) – 225, – 135
By subtracting the first integer from the second we get,
= -135 – (-225) (we know that -,- = +) 
= 225 – 135  
= 90 
 
(iv) 1001, 101
By subtracting the first integer from the second we get,
= 101 – 1001  
= – 900 
 
(v) – 812, 3126
By subtracting the first integer from the second we get,
= 3126 – (-812) (we know that -,- = +) 
= 3126 + 812  
= 3938 
 
(vi) 7560, – 8
By subtracting the first integer from the second
= - 8 – 7560 (we know that -,- = +) 
= – 7568 
 
(vii) – 3978, – 4109
By subtracting the first integer from the second
= - 4109 – (- 3978) (we know that -,- = +)
= – 4109 + 3978 
= - 131
 
(viii) 0, – 1005
By subtracting the first integer from the second
= -1005 – 0 (we know that -,- = +)
= – 1005
 

Question 2:  Find the value of:

(i) – 27 – (- 23)

(ii) – 17 – 18 – (-35)

(iii) – 12 – (-5) – (-125) + 270

(iv) 373 + (-245) + (-373) + 145 + 3000

(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900

(vi) (-1) + (-304) + 304 + 304 + (-304) + 1 

Solution 2:

(i) – 27 – (- 23) 
= – 27 + 23 (we know that -,+ = -)
= 23 – 27
= – 4
 
(ii) – 17 – 18 – (-35)
= – 35 + 35
= 0
 
(iii) – 12 – (-5) – (-125) + 270
= – 12 + 5 + 125 + 270
= 400 – 12
= 388
 
(iv) 373 + (-245) + (-373) + 145 + 3000
= 373 – 245 – 373 + 145 + 3000
= 3145 + 373 – 373 – 245
= 3145 – 245
= 2900
 
(v) 1 + (-475) + (-475) + (-475) + (-475) + 1900
= 1 – 950 – 950 + 1900
= 1900 + 1 – 1900
= 1
 
(vi) (-1) + (-304) + 304 + 304 + (-304) + 1
= – 1 + 1 – 304 + 304 – 304 + 304= 0


Question 3:   Subtract the sum of – 5020 and 2320 from – 709. 

Solution 3:   

By the sum of -5020 and 2320 we get,

= -5020 + 2320

= 2320 – 5020

= – 2700

- 2700 Subtract from - 709 we get,

= – 709 – (-2700)  (we know -,- = +)

= – 709 + 2700

= 1991

 

Question 4:    Subtract the sum of – 1250 and 1138 from the sum of 1136 and – 1272. 

Solution 4:     

By the sum of – 1250 and 1138 is

= -1250 + 1138

= 1138 – 1250

= – 112__________(1)

By the sum of 1136 and – 1272 is

= 1136 – 1272

= – 136_____________(2)

Subtraction of equation 1 from equation 2 we get,

= -136 – (-112)

= – 136 + 112

= -24

 

Question 5:  From the sum of 233 and – 147, subtract – 284. 

Solution  5:   

By the sum of 233 and – 147 we get,

= 233 – 147

= 86_________(1)

Subtract – 284 from equation (1)

= 86 – (-284)

= 86 + 284

= 370

 

Question 6:  The sum of two integers is 238. If one of the integers is – 122, determine the other.

Solution 6:

Sum of two integers = 238

First integer is = – 122

Let other integer is = x

238 = -122 + x

238 + 122 = x

360 = x

Hence, the other integer is 360. 

 

Question 7:  The sum of two integers is – 223. If one of the integers is 172, find the other.

Solution 7:

Sum of two integers = – 223

First integer is = 172

Let other integer is = x

– 223 = 172 + x

– 223 – 172 = x

– 395 = x

Hence, the other integer is -395. 

 

Question 8:  Evaluate the following:

(i) – 8 – 24 + 31 – 26 – 28 + 7 + 19 – 18 – 8 + 33

(ii) – 26 – 20 + 33 – (-33) + 21 + 24 – (-25) – 26 – 14 – 34 

Solution 8:

(i) – 8 – 24 + 31 – 26 – 28 + 7 + 19 – 18 – 8 + 33

= – 8 – 24 – 26 – 28 – 18 – 8 + 31 + 7 + 19 + 33

= – 32 – 26 – 28 – 26 + 38 + 19 + 33

= 38 – 32 – 26 – 28 + 33 – 26 + 19

= 6 – 26 – 28 + 7 + 19

= 6 – 28 – 26 + 26

= 6 – 28

= – 22

 

(ii) – 26 – 20 + 33 – (-33) + 21 + 24 – (-25) – 26 – 14 – 34

= – 46 + 33 + 33 + 21 + 24 + 25 – 26 – 14 – 34

= – 46 + 66 + 21 + 24 + 25 + (-74)

= – 46 + 66 + 70 – 74

= – 46 – 4 + 66

= – 50 + 66

= 66 – 50

= 16

 

Question 9:  Calculate

1 – 2 + 3 – 4 + 5 – 6 + ……… + 15 – 16 

Solution 9:

1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 + 11 – 12 + 13 – 14 + 15 – 16

By calculation we get,

= – 1 – 1 – 1 – 1 – 1 – 1 – 1 – 1

= – 8

 

Question 10:  Calculate the sum:

5 + (-5) + 5 + (-5) + …..

(i) if the number of terms is 10.

(ii) if the number of terms is 11. 

Solution 10:

(i) if the number of terms is 10

5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5)

By calculation we get,

= 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5

= 0

 

(ii) if the number of terms is 11

5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5 + (-5) + 5

By calculation we get,

= 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 + 5 – 5 + 5

= 5

 

Question 11:  Replace * by < or > in each of the following to make the statement true:

(i) (-6) + (-9) * (-6) – (-9)

(ii) (-12) – (-12) * (-12) + (-12)

(iii) (-20) – (-20) * 20 – (65)

(iv) 28 – (-10) * (-16) – (-76) 

Solution 11:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers


Question 12:  If △ is an operation on integers such that a △ b = – a + b – (-2) for all integers a, b. Find the value of

(i) 4 △ 3

(ii) (-2) △ (-3)

(iii) 6 △ (-5)

(iv) (-5) △ 6 

Solution 12:

(i) 4△3
It is given that a△ b = – a + b – (-2)
4△3 
= – 4 + 3 – (-2)
= – 4 + 3+2
= – 4 + 5
= 1
 
(ii) (-2)△(-3)
It is given that a △ b = – a + b – (-2)
(-2)△(-3) 
= – (-2) + (-3) – (-2) 
= 2 -3 +2
= 1
 
(iii) 6△(-5)
It is given that a △ b = – a + b – (-2)
6△(-5) 
= – 6 + (-5) – (-2) 
= – 6-5+2
= – 9
 
(iv) (-5) △ 6
It is given that a △ b = – a + b – (-2)
We get
(-5) △  6 
= – (-5) + 6 – (-2) 
= + 5+ 6+2
= 13
 

Question 13:  If a and b are two integers such that a is the predecessor of b. Find the value of a – b. 

Solution 13:

As given in question a is the predecessor of b

a = b – 1

a + 1 = b ________(1)

From Equation (1) we get,

a – b = – 1

 

Question 14:  If a and b are two integers such that a is the successor of b. Find the value of a – b. 

Solution 14:

As given in question a is the successor of b

a = b + 1

a – 1 = b____________(1)

from Equation (1) we get,

a – b = 1

 

Question 15:  Which of the following statements are true:

(i) – 13 > – 8 – (-2)

(ii) – 4 + (-2) < 2

(iii) The negative of a negative integer is positive.

(iv) If a and b are two integers such that a > b, then a – b is always a positive integer.

(v) The difference of two integers is an integer.

(vi) Additive inverse of a negative integer is negative.

(vii) Additive inverse of a positive integer is negative.

(viii) Additive inverse of a negative integer is positive. 

Solution 15:

(i) – 13 > – 8 – (-2)                                                                                                     False
(ii) – 4 + (-2) < 2                                                                                                        True
(iii) The negative of a negative integer is positive.                                                           True
(iv) If a and b are two integers such that a > b, then a – b is always a positive integer.      True
(v) The difference of two integers is an integer.                                                            True
(vi) Additive inverse of a negative integer is negative.                                                   False
(vii) Additive inverse of a positive integer is negative.                                                   True
(viii) Additive inverse of a negative integer is positive.                                                  True


Question 16:  Fill in the blanks:

(i) – 7 + ….. = 0

(ii) 29 + ….. = 0

(iii) 132 + (-132) = ….

(iv) – 14 + ….. = 22

(v) – 1256 + ….. = – 742

(vi) ….. – 1234 = – 4539 

Solution 16:

RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers-


Objective Type Questions ::->

Mark the correct alternative in each of the following:
 
Question 1:   Which of the following statement is true?
(a) − 7 > − 5             
(b) − 7 < − 5                 
(c) (− 7) + (− 5) > 0                       
(d) (− 7) − (− 5) > 0 
 
Solution 1: (b) 
Negative numbers get smaller the father they are from zero.
 
Question 2: 5 less than − 2 is
(a) 3                           
(b) − 3                       
(c) − 7                                 
(d) 7
 
Solution 2: (c)
5 less than − 2 
= (− 2) − (5) 
= − 2 − 5 
= − 7
 
Question 3:  6 more than − 7 is
(a) 1                           
(b) − 1                       
(c) 13                                 
(d) – 13 
 
Solution 3: (b)
6 more than − 7 
= (− 7) + 6 
= − (7 − 6) 
= − 1
 
Question 4:  If x is a positive integer, then
(a) x + |x| = 0             
(b) x − |x| = 0                   
(c) x + |x| = −2x               
(d) x = − |x| 
 
Solution 4: (b) 
If x is positive integer, then 
We know |x| = x
= x - |x|
= 0 
 
Question 5:  If x is a negative integer, then
(a) x + |x| = 0             
(b) x − |x| = 0                   
(c) x + |x| = 2x               
(d) x − |x| = − 2x 
 
Solution 5: (a) 
It is given that x is negative integer, then |x| = −x
= x + |x| 
= x + x 
= 2x
 
Question 6:  If x is greater than 2, then |2 − x| =
(a) 2 − x             
(b) x − 2                   
(c) 2 + x               
(d) − x – 2 
 
Solution 6: (b) 
We know that if a is negative integer, then |a| = − a
It is given that x is greater than 2 where 2 − x is negative
Hence, |2 − x| = − (2 − x) = − 2 + x = x − 2.
 
Question 7:  9 + |− 4| is equal to
(a) 5                       
(b) − 5                       
(c) 13                           
(d) −13
 
Solution 7: (c)
= 9 + |− 4| 
= 9 + 4 
= 13
 
Question 8: (− 35) + (− 32) is equal to
(a) 67                       
(b) − 67                       
(c) − 3                           
(d) 3
 
Solution 8: (b) 
(− 35) + (− 32) 
= − (35 + 32) 
= − 67
 
Question 9:  (− 29) + 5 is equal to
(a) 24                         
(b) 34                       
(c) − 34                           
(d) – 24
 
Solution 9: (d)
It can be written as (− 29) + 5 = − (29 − 5) = − 24
 
Question 10:  |− |− 7| − 3| is equal to
(a) − 7                       
(b) 7                             
(c) 10                                   
(d) – 10
 
Solution 10: (c)
|− |− 7| − 3| 
= |− 7 − 3| 
= |− 10| 
= 10
 
Question 11:  The successor of − 22 is
(a) − 23                     
(b) − 21                             
(c) 23                                   
(d) 21
 
Solution 11: (b) 
We know that if ‘a’ is an integer a + 1 is its successor.
So the successor of − 22 
= − 22 + 1 (we know that -,+ = -)
= − 21
 
Question 12:  The predecessor of – 14 is
(a) – 15                       
(b) 15                           
(c) 13                                   
(d) – 13
 
Solution 12: (a)
The predecessor of – 14 
14 – 1 = – 15.
 
Question 13:  If the sum of two integers is − 26 and one of them is 14, then the other integer is
(a) − 12                     
(b) 12                             
(c) − 40                               
(d) 40
 
Solution 13: (c)
The sum of two integers = − 26
One of them = 14
Let the other number be x
14+ x=-26 
x = - 26 - 14 
x = - (26 + 14) 
x = - 40
 
Question 14:  Which of the following pairs of integers have 5 as a difference?
(a) 10, 5                       
(b) − 10, − 5                         
(c) 15, − 20                               
(d) both (a) and (b)
 
Solution 14: (d) 
By Option (a) we get 10 − 5 = 5
By Option (b) we get (− 5) − (− 10) = − 5 + 10 = 5
By Option (c) we get 15 − (− 20) = 15 + 20 = 35
 
Question 15:  If the product of two integers is 72 and one of them is − 9, then the other integers is
(a) − 8                     
(b) 8                         
(c) 81                               
(d) 63
 
Solution 15: (a)
The product of two integers are = 72
One of them = − 9
Let the other number be x
-9 ×x=72  
x=-72/9   
x=-8 
 
Question 16:  On subtracting − 7 from − 14, we get
(a) − 12                       
(b) − 7                         
(c) −14                               
(d) 21
 
Solution 16: (b)
− 14 − (− 7) 
= − 14 + 7 
= − (14 − 7) 
= − 7
 
Question 17:  The largest number that divides 64 and 72 and leave the remainders 12 and 7 respectively, is
(a) 17                             
(b) 13                                       
(c) 14                                     
(d) 18
 
Solution 17: (b) 
Subtract the remainders from the number we get,
64 − 12 = 52 and 72 − 7 = 65
So the required number is the HCF of 52 and 65.
It can be written as
Prime factorisation of 52 is = 4 × 13 
Prime factorisation of 65 is = 5 × 13
HCF of 52 and 65 = 13
Hence, the largest number that divides 64 and 72 and leave the remainders 12 and 7 respectively, is 13.
 
Question 18:  The sum of two integers is − 23. If one of them is 18, then the other is
(a) −14                             
(b) 14                                   
(c) 41                                   
(d) −41
 
Solution 18: (d) 
Sum of two integers = − 23
One of them = 18
Let the other number be x
18+x=-23 
x = - 23 - 18 
x = - (23 + 18) 
x = - 41
Hence, the other number is − 41.
 
Question 19:  The sum of two integers is − 35. If one of them is 40, then the other is
(a) 5                             
(b) − 75                                   
(c) 75                                   
(d) – 5
 
Solution 19: (b) 
Sum of two integers = − 35
One of them = 40
Let the other number be x
40 + x=-35 
x = (- 35) - (40) 
x = - 35 - 40 
x = - (35 + 40) 
x = - 75
Hence, the other number is − 75.
 
Question 20:  On subtracting − 5 from 0, we get
(a) − 5                             
(b) 5                                   
(c) 50                                   
(d) 0
 
Solution 20: (b) 
0 − (− 5) 
= 0 + 5 
= 5
Hence, subtracting − 5 from 0, we obtain 5.
 
Question 21:  (− 16) + 14 − (− 13) is equal to
(a) − 11                             
(b) 12                                   
(c) 11                                   
(d) – 15
 
Solution 21: (c) 
(− 16) + 14 − (− 13) 
= (− 16) + 14 + 13 
= (− 16) + 27 
= 27 − 16 
= 11
 
Question 22:  (− 2) × (− 3) × 6 × (− 1) is equal to
(a) 36                             
(b) − 36                               
(c) 6                                   
(d) – 6
 
Solution 22: (b) 
(− 2) × (− 3) × 6 × (− 1) 
= − (36 × 1) 
= − 36
 
Question 23:  86 + (- 28) + 12 + (- 34) is equal to
(a) 36                             
(b) − 36                               
(c) 6                                   
(d) – 6
 
Solution 23: (a) 
86 + (−28) + 12 + (−34) 
= (86 − 28) − (34 − 12) 
= (86 − 28) − 22 
= 58 − 22 
= 36
 
Question 24:  (−12) × (−9) − 6 × (−8) is equal to
(a) 156                             
(b) 60                               
(c) −156                                   
(d) – 60
 
Solution 24: (a) 
(−12) × (−9) − 6 × (−8) 
= 108 + 6 × 8 
= 108 + 48 
= 156
 
RD Sharma Solutions Class 6 Maths
RD Sharma Solutions Class 6 Maths Chapter 1 Knowing our Numbers
RD Sharma Solutions Class 6 Maths Chapter 2 Playing with Numbers
RD Sharma Solutions Class 6 Maths Chapter 3 Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 4 Operations on Whole Numbers
RD Sharma Solutions Class 6 Maths Chapter 5 Negative Numbers and Integers
RD Sharma Solutions Class 6 Maths Chapter 6 Fractions
RD Sharma Solutions Class 6 Maths Chapter 7 Decimals
RD Sharma Solutions Class 6 Maths Chapter 8 Introduction to Algebra
RD Sharma Solutions Class 6 Maths Chapter 9 Ratio Proportion and Unitary Method
RD Sharma Solutions Class 6 Maths Chapter 10 Basic Geomatrical Concepts
RD Sharma Solutions Class 6 Maths Chapter 11 Angles
RD Sharma Solutions Class 6 Maths Chapter 12 Triangle
RD Sharma Solutions Class 6 Maths Chapter 13 Quadrilaterals
RD Sharma Solutions Class 6 Maths Chapter 14 Circles
RD Sharma Solutions Class 6 Maths Chapter 15 Pair of Lines and Transversal
RD Sharma Solutions Class 6 Maths Chapter 16 Understanding Three Dimensional Shapes
RD Sharma Solutions Class 6 Maths Chapter 17 Symmetry
RD Sharma Solutions Class 6 Maths Chapter 18 Basic Geometrical Tools
RD Sharma Solutions Class 6 Maths Chapter 19 Geometrical Constructions
RD Sharma Solutions Class 6 Maths Chapter 20 Mensuration
RD Sharma Solutions Class 6 Maths Chapter 21 Data Handling Presentation of Data
RD Sharma Solutions Class 6 Maths Chapter 22 Data Handling Pictographs
RD Sharma Solutions Class 6 Maths Chapter 23 Data Handling Bar Graphs