NCERT Solution class 7 Mathematics EXERCISE- 1.2

NCERT Solution class 7 Mathematics EXERCISE- 1.2

EXERCISE- 1.2

Q1. Write down a pair of integer whose:

(a)  sum is -7

SOLUTION :-

=-4+(-3)

=  -4-3 ____[ (+X-=-)]

= -7

(B)   Difference is -10

SOLUTION:-

= -25-(-15)

= -25 + 15 ____[(-X-=+)]

= -10

(C)  Sum is 10

SOLUTION:-

= 4+(-4)

= 4-4

=0

Q2.  (A)  Write a pair of negative whose integer whose difference gives 8

SOLUTION :-

= (-5) – (-13)

= -5+ 13 ____[(-X-=+)]

= 8

(B) Write a negative integer and a positive integer whose sum is -5 .

SOLUTION:-

= -25 +20

= -5

(c)  Write a positive integer and a positive integer whose difference is -3 .

SOLUTION:-

= -6-(-3)

= -6+3____[(-X-=+)]

= -3

 

Q3. In a quiz team A scored 40,10,0 and team B scored

10, 0 , -40 in three successive rounds. Which team scored more ? Can we say that we can add integer in any order ?

SOLUTION:-

From the question, it is given that

Score of team A = -40 , 10 , 0

Total score obtained by team A = -40,+10,+0

=  -30

Score of team B = 10,0,-40

Total score obtained by team B = 10+ 0+(-40)

=10+0-40

= -30

Thus, the score of the  both A team and B team is same.

Yes, we can say that we can add integer in any order .

 

Q4.  Fill in the blanks to make the following statements true:

(i) (-5) + (-8) = (-8) + (____)

SOLUTION:-

Let us assume the missing integer be x,

Then,

= (-5) + (-8) =  (-8) + x

=-5 -8 = -8 +x

= -13 = -8 + x

By sending  -8 from the RHS to LHS  it becomes 8,

= -13+ 8 = x

=x= -5

Now substitute the value in the blanks place ,

(-5) + (-8) = (-8) +(-5)

[ This equation is in the form of commutative law of addition ]

 

(ii) -53 +  _________= -53

SOLUTION :-

Let us assume that missing integer be x,

Then,

= -53 + x = -53

By sending the -53 from LHS to RHS it becomes 53,

=x=-53 +53

=x=0

Now substitute the x value in the blank place.

= -53 + 0 =-53 ____ [This equation is in the form of closure property of addition ]

(iii) 17 + _______=0

SOLUTION :-

Let us assume the missing integer be x,

Then,

= 17 + x = 0

By sending 17 from LHS to RHS it becomes -17,

=x=0-17

=x=-17

Now substitute the x value in the blank place ,

=17 +(-17)=0 ________[ This equation is in the form of closure property of addition]

= 17-17=0

 

(iv) [13+(-12)]+(______)= 13 +[(-12)+(-7)]

SOLUTION :-

Let us assume the missing integer be x,

Then,

= [13+((-12)]+(x)=13+[(-12)+(-7)]

= [13-12] +(x)=13+[-12 -7]

= [1] +(x)=13+[-19]

= 1+(x) = 13-19

= 1+(x) = -6

By sending 1 from LHS to RHS it becomes -1,

=x=-6-1

=x=-7

Now substitute the x value in the blanks place ,

=[13+ (-12) ]+(-7) = 13+ [(-12) +(-7) ]____[This equation is in the form of associative property of addition ]

 

(v)  (-4) +[15+(-3) ]=[-4+15]+_______

SOLUTION:-

Let us assume that missing integer be x ,

Then,

= (-4)+[15+(-3)]=[-4+15]+x

= (-4) +[15-3)]=[-4+15 ] +x

= (-4) +[12]=[11]+x

=8=11+x

By sending 11 from RHS to LHS it becomes -11,

=8-11=x

=x= -3

Now substitute the x value in the blank place

= (-4) +[15+(-3) ]=[-4+15]+ (-3)______________[ thoid equation is in the form of associative property of addition ]

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