Which integer is greater than –5?
An integer on the horizontal number line is greater than the number on its left and less than the number on its right.
Therefore, here 1 will be greater than 5.
If a x (b − c) is 8 for a = 2, b = 10 and c = 6 then a x b − a x c is equal to:
By Distributive law,
a x (b − c) = a x b  a x c
⇒ a x b  a x c = 8
Alternative method
a x b  a x c = 2 x 10  2 x 6
⇒ a x b  a x c = 20  12 = 8 [using BODMAS]
Thus, option D is the correct.
The product of three negative integers is:
The product of 3 negative integers will always be negative.
a * (a) = a^{2}
(a) * (a) = a^{2}
(a) * (a) * (a) = a^{2} * (a) = a^{3}
The additive identity for integers is:
The additive identity property says that if you add an integer to zero or add zero to the integer, then you get the same integer back i.e. a + 0 = 0 + a = a.
Thus, 0 is an additive identity for the integers.
According to Distributive Law of Multiplication over Addition, a x (b + c) must be equal to:
According to Distributive Law of Multiplication over Addition:
a x (b + c) = a x b + a x c
Which amongst the following is the largest?
I89I, 89, 21, I21I
x gives the absolute value of x i.e. the magnitude of the real number without regard to its sign.
The magnitude of a number is its size.
Therefore, I89I = 89, I21I = 21
Thus, option C is correct.
The multiplicative identity for integers is:
The multiplicative identity of any integer p is a number q which when multiplied with p, leaves it unchanged, i.e. p × q = p.
Since, p × 1 = p = 1 × p
Therefore, 1 is the multiplicative identity for integers.
What will be the additive inverse of 7?
The additive inverse of a number x is the number that, when added to x, gives zero i.e. x + (x) = 0.
This number is also known as the opposite number.
Thus, 7 is the additive inverse of 7.
How many integers are there between 8 and 2 (excluding 8 and 2)?
The integers between 8 & 2 are: 7 , 6 , 5 , 4 , 3 , 2 , 1 , 0 , 1
⇒ There are 9 integers between 8 & 2.
The value of [(4) x (9) x (25)] ÷ [(2) x (3) x (5)] is:
Firstly we will solve brackets according to BODMAS rule,
(4) x (9) x (25) = 900
(2) x (3) x (5) = 30
⇒ [(4) x (9) x (25)] ÷ [(2) x (3) x (5)] = (900)/(30) = 30
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