Step-by-step explanation:
Q-4)
Given numbers are 51 and 83
The number of whole numbers between them
= 83-51 +1
= 84-51
Q-5)
a) Given that : 1783+198+347+602
= (1783+347)+(198+602)
(By Commutative Property)
= 2130+800
= 2930
Therefore, 1783+198+347+602 = 2930
b) Given that : 125×25×8×4
= (125×8)×(25×4)
= 1000×100
= 100000
Therefore, 125×25×8×4 = 100000
c) Given that : 1002×268
= (1000+2)×268
= (1000×268)+(2×268)
(By Distributive Property)
= 268000+536
= 268536
Therefore, 1002×268 = 268536
d) Given that : 68×101-68
= (68×101)-(68×1)
= 68×(101-1)
= 68×100
= 6800
Therefore, 68×101-68 = 6800
Let a , b be any two integers , if a×b = b×a is called Commutative Property under multiplication .
Let a , b ,c be any three non zero integers , if a×(b+c)= (a×b)+(a×c) is called Distributive Property under multiplication over addition.
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Verified answer
Step-by-step explanation:
Solution:-
Q-4)
Given numbers are 51 and 83
The number of whole numbers between them
= 83-51 +1
= 84-51
= 33
Q-5)
a) Given that : 1783+198+347+602
= (1783+347)+(198+602)
(By Commutative Property)
= 2130+800
= 2930
Therefore, 1783+198+347+602 = 2930
b) Given that : 125×25×8×4
= (125×8)×(25×4)
(By Commutative Property)
= 1000×100
= 100000
Therefore, 125×25×8×4 = 100000
c) Given that : 1002×268
= (1000+2)×268
= (1000×268)+(2×268)
(By Distributive Property)
= 268000+536
= 268536
Therefore, 1002×268 = 268536
d) Given that : 68×101-68
= (68×101)-(68×1)
= 68×(101-1)
(By Distributive Property)
= 68×100
= 6800
Therefore, 68×101-68 = 6800
Used Properties:-
Commutative Property:-
Let a , b be any two integers , if a×b = b×a is called Commutative Property under multiplication .
Distributive Property:-
Let a , b ,c be any three non zero integers , if a×(b+c)= (a×b)+(a×c) is called Distributive Property under multiplication over addition.