GSEB Class 6 Maths Solutions Chapter 6 Integers Exercise 6.3

Question 1. Find
(a) \( 35 - (20) \)
(b) \( 72 - (90) \)
(c) \( (-15)-(-18) \)
(d) \( (-20) – (13) \)
(e) \( 23 - (-12) \)
(f) \( (-32) - (-40) \)
Answer:
(a) \( 35 - (20) \):
As, \( 35 = 20 + 15 \)
\( 35-20 = 20 + 15 – 20 = 20 – 20 + 15 \)
\( = 0 + 15 = 15 \)
Therefore, \( 35 - (20) = 15 \)

(b) \( 72 – (90) \):
Method 1: Since, \( – 90 = ( − 72) + (−18) \)
\( 72 – 90 = 72 + (− 72) + ( − 18) = 0 + (- 18) \)
\( = -18 \)
Method 2: \( 72 – 90 \)
\( = 72 + \) [Additive inverse of \( 90 \)]
\( = 72 + [- 90] = − 18 \)

(c) \( (-15) - (-18) \):
\( = (- 15) + \) [Additive inverse of \( (-18) \)]
\( = (-15) + [18] = − 15 + [15 + 3] \)
\( =-15 + 15 + 3 = 0 + 3 = 3 \)
Therefore, \( -15-(-18) = 3 \)

(d) \( (-20) – (13) \):
\( = (-20) + \) [Additive inverse of \( 13 \)]
\( = (-20) + [- 13] \)
\( = − [20 + 13] \)
\( = [33] \)
\( (-20) – (13) = − 33 \)

(e) \( 23 - (-12) \):
\( = 23 + \) [Additive inverse of \( (- 12) \)]
\( = 23 + [12] = 35 \)
So, \( 23- (-12) = 35 \)

(f) \( (-32) - (-40) \):
\( (-32) − ( − 40) = (- 32) + \) [Additive inverse of \( (-40) \)]
\( = (-32) + [+ 40] \)
\( = - 32 + 40 \)
\( = − 32 + 32 + 8 \)
\( = 0 + 8 = 8 \)
In simple words: To find the difference between integers, change subtraction to addition and use the additive inverse of the second number. Then combine the numbers according to their signs.

Exam Tip: Remember that subtracting a negative number is the same as adding a positive number. For example, \( a - (-b) = a + b \).

Question 2. Fill in the blanks with >, < or = sign.
(a) \( (-3) + (-6) \) .......... \( (-3)-(-6) \)
(b) \( (-21) - (-10) \) .......... \( (-31) + (-11) \)
(c) \( 45 - (-11) \) .......... \( 57 + (-4) \)
(d) \( (-25) - (-42) \) .......... \( (-42) - (-25) \)
Answer:
(a) For \( (-3) + (-6) \):
\( (-3) + (-6) = -9 \)
For \( (-3) - (-6) \):
\( (-3) - (- 6) = (- 3) + (6) = 3 \)
Here, \( (-9) < 3 \)
So, \( (-3) + (-6) < (-3) - (-6) \)

(b) For \( (-21) - (-10) \):
\( (-21) - (-10) = -21 + (+10) = -11 \)
For \( (-31) + (-11) \):
\( (-31) + (-11) = -42 \)
Here, \( (-11) > (-42) \)
So, \( (-21) - (-10) > (-31) + (-11) \)

(c) For \( 45 - (-11) \):
\( 45 - (-11) = 45 + \) [Additive inverse of \( (-11) \)]
\( = 45 + [11] = 56 \)
For \( 57 + (-4) \):
\( 57 + (-4) = 53 \)
Here, \( 56 > 53 \)
So, \( 45 - (-11) > 57 + (-4) \)

(d) For \( (-25) - (-42) \):
\( = (-25) + \) [Additive inverse of \( ( – 42) \)]
\( = (-25) + (42) \)
\( = − 25 + 25 + 17 \)
\( = 0 + 17 = 17 \)
For \( (-42) - (-25) \):
\( = (-42) + \) [Additive inverse of \( ( – 25) \)]
\( = (-42) + (25) \)
\( = – 42 + 25 \)
\( = -17 \)
Here, \( 17 > -17 \)
So, \( (-25) - (-42) > (-42) - (-25) \)
In simple words: To compare expressions with integers, first work out the value of each side. Then use the greater than (>), less than (<), or equals (=) sign to show which side is bigger or if they are the same.

Exam Tip: Pay close attention to the signs when performing operations, especially when dealing with double negatives, as \( -(-a) \) becomes \( +a \).

Question 3. Fill in the blanks.
(a) \( (-8)+ \)............ \( = 0 \)
(b) \( 13+ \)............ \( = 0 \)
(c) \( 12 + \)............ \( = 0 \)
(d) \( (− 4) + \)............ \( = − 12 \)
(e) ............ \( - 15 = - 10 \)
Answer:
(a) \( (-8) + 8 = 0 \). (The additive inverse of \( -8 \) is \( 8 \)).

(b) \( 13 + (-13) = 0 \). (The additive inverse of \( 13 \) is \( -13 \)).

(c) \( 12 + (-12) = 0 \). (Since \( 12 \) and \( -12 \) are additive inverses of each other).

(d) We have the expression: \( (− 4) + \) [Blank] \( = − 12 \).
Let the blank be \( x \). So, \( -4 + x = -12 \).
To find \( x \), we add \( 4 \) to both sides: \( x = -12 + 4 \).
\( x = -8 \).
Thus, \( (-4) + (-8) = -12 \).

(e) We have the expression: [Blank] \( - 15 = - 10 \).
Let the blank be \( y \). So, \( y - 15 = -10 \).
To find \( y \), we add \( 15 \) to both sides: \( y = -10 + 15 \).
\( y = 5 \).
Thus, \( 5 - 15 = -10 \).
In simple words: To fill in the blanks, think about what number makes the equation true. For questions asking for zero, use the opposite sign of the given number. For others, work backwards by doing the opposite operation.

Exam Tip: Remember that the additive inverse of a number is the number that, when added to it, results in zero. For example, the additive inverse of \( a \) is \( -a \).

Question 4. Find
(a) \( (-7) - 8 – ( – 25) \)
(b) \( (-13) + 32-8-1 \)
(c) \( (-7) + (-8) + (−90) \)
(d) \( 50 – ( – 40) – ( – 2) \)
Answer:
(a) \( (-7) - 8- ( − 25) \)
\( = (-7) + (-8) + (25) \) [We change subtraction to addition by taking the additive inverse of \( 8 \) and \( -25 \)]
\( = [-15] + 25 \)
\( = (-15) + 15 + 10 \)
\( = 0 + 10 = 10 \)

(b) \( (-13) + 32 – 8 – 1 \)
First, calculate \( (-13) - 8 - 1 \):
\( (-13) - 8 = -21 \)
\( -21 - 1 = -22 \)
So, the expression becomes: \( (-13) + 32 + (-8) + (-1) \)
\( = [(-13) + (-8) + (-1)] + 32 \) (Grouping similar terms)
\( = [-22] + 32 \)
\( = -22 + 22 + 10 = 0 + 10 = 10 \)

(c) \( (-7) + (-8) + (−90) \)
The given integers all have the same sign (negative).
\( = − [ 7 + 8 + 90] \)
\( = - [105] = -105 \)

(d) \( 50- (-40) – (- 2) \)
\( = 50 + \) [Additive inverse of \( ( – 40) \)] \( + \) [Additive inverse of \( (- 2) \)]
\( = 50 + [ + 40] + [ + 2] \)
\( = 50 + 40 + 2 = 92 \)
In simple words: When you have several integers with different operations, change all subtractions to additions using the additive inverse. Then, group the positive and negative numbers and add them up, taking care of their signs.

Exam Tip: Always convert all subtraction operations to addition of the additive inverse (e.g., \( a - b = a + (-b) \)) before proceeding with calculations, as this helps avoid common sign errors.