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 ]