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Practice each question, then open the answer or worked solution.
Volume of a rectangular box (cuboid) with length = 2ab, breadth = 3ac and height = 2ac is
(a) 12a3bc2 (b) 12a3bc (c) 12a2bc (d) 2ab +3ac + 2ac
Solution:-
(a) 12a3bc2
We know that, volume of cuboid = length × breadth × height
Given, length = 2ab, breadth = 3ac, height = 2ac
= 2ab × 3ac × 2ac
= (2 × 3 × 2) × ab × ac × ac
= 12a3bc2
Product of 6a2 – 7b + 5ab and 2ab is
(a) 12a3b – 14ab2 + 10ab (b) 12a3b – 14ab2 + 10a2b2
(c) 6a2 – 7b + 7ab (d) 12a2b – 7ab2 + 10ab
Solution:-
(b) 12a3b – 14ab2 + 10a2b2
Now, we have to find the product of trinomial and monomial,
= (6a2 – 7b + 5ab) × 2ab
= (2ab × 6a2) – (2ab × 7b) + (2ab × 5ab)
= 12a3b – 14ab2 + 10a2b2
Square of 3x – 4y is
(a) 9x2 – 16y2 (b) 6x2 – 8y2
(c) 9x2 + 16y2 + 24xy (d) 9x2 + 16y2 – 24xy
Solution:-
(d) 9x2 + 16y2 – 24xy
As per the condition in the question, (3x – 4y)2
The standard identity = (a – b)2 = a2 – 2ab + b2
Where, a = 3x, b = 4y
Then,
(3x – 4y)2 = (3x)2 – (2 × 3x × 4y) + (4y)2
= 9x2 – 24xy + 16y2
Which of the following are like terms?
(a) 5xyz2, – 3xy2z (b) – 5xyz2, 7xyz2
(c) 5xyz2, 5x2yz (d) 5xyz2, x2y2z2
Solution:-
(b) – 5xyz2, 7xyz2
Like the terms are formed from the same variables, and the powers of these variables are also the same. But coefficients of like terms need not be the same.
Coefficient of y in the term –y/3 is
(a) – 1 (b) – 3 (c) -1/3 (d) 1/3
Solution:-
(c) -1/3
-y/3 can also be written as y × (-1/3)
So, the coefficient of y is -1/3
a2 – b2 is equal to
(a) (a – b)2 (b) (a – b) (a – b)
(c) (a + b) (a – b) (d) (a + b) (a + b)
Solution:-
(c) (a + b) (a – b)
(a2 – b2) = (a + b) (a – b) is one of the standard identities.
Common factor of 17abc, 34ab2, 51a2b is
(a) 17abc (b) 17ab (c) 17ac (d) 17a2b2c
Solution:-
(b) 17ab
The given factors can be written in expanded form as,
17abc = 17 × a × b × c
34ab2 = 2 × 17 × a × b × b
51a2b = 3 × 17 × a × a × b
So, common factors in the above are 17 × a × b
= 17ab
Square of 9x – 7xy is
(a) 81x2 + 49x2y2 (b) 81x2 – 49x2y2
(c) 81x2 + 49x2y2 –126x2y (d) 81x2 + 49x2y2 – 63x2y
Solution:-
(c) 81x2 + 49x2y2 –126x2y
As per the condition in the question, (9x – 7xy)2
The standard identity = (a – b)2 = a2 – 2ab + b2
Where, a = 9x, b = 7xy
Then,
(9x – 7xy)2 = (9x)2 – (2 × 9x × 7xy) + (7xy)2
= 81x2 – 126x2y + 49x2y2
Factorised form of 23xy – 46x + 54y – 108 is
(a) (23x + 54) (y – 2) (b) (23x + 54y) (y – 2)
(c) (23xy + 54y) (– 46x – 108) (d) (23x + 54) (y + 2)
Solution:-
(a) (23x + 54) (y – 2)
Factorised form of 23xy – 46x + 54y – 108 is = 23xy – (2 × 23x) + 54y – (2 × 54)
Take out the common factors,
= 23x (y – 2) + 54 (y – 2)
Again take out the common factor,
= (y – 2) (23x + 54)
Factorised form of r2 – 10r + 21 is
(a) (r – 1) (r – 4) (b) (r – 7) (r – 3)
(c) (r – 7) (r + 3) (d) (r + 7) (r + 3)
Solution:-
(b) (r – 7) (r – 3)
Factorised form of r2 – 10r + 21 is = r2 – 7r – 3r + 21
Take out the common factors,
= r (r – 7) – 3 (r – 7)
Again take out the common factor,
= (r – 7) (r – 3)