Free Algebraic Expressions and Identities 03 Practice Test - 8th Grade

Question 1

When we add $5 a^{2} b + 6 a b + 7 b c + 54$ and $14 a b + 8 b c + 16$ , we get:

$5 a^{2} b + 20 a b + 7 b c + 70$

$5 a^{2} b + 6 a b + 7 b c + 8 a c + 70$

$5 a^{2} b + 16 a b + 7 b c + 8 a c + 70$

$5 a^{2} b + 20 a b + 15 b c + 70$

SOLUTION

Solution : D

$5 a^{2} b + 6 a b + 7 b c + 54$
$14 a b + 8 b c + 16$
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$5 a^{2} b + 20 a b + 15 b c + 70$

Question 2

Subtract $(8 - 6 x + x^{2} - 7 x^{3} + x^{5})$ from $(x^{4} - 6 x^{3} + x^{2} - 3 x + 1)$ .

$- x^{5} + x^{4} + x^{3} + 3 x$

$- x^{5} + x^{4} + x^{3} - 7$

$- x^{5} + x^{4} + x^{3} + 3 x - 7$

$- x^{5} + x^{4} + 4 x^{3} + 3 x - 7$

SOLUTION

Solution : C

$\begin{matrix} x^{4} & - 6 x^{3} & + x^{2} & - 3 x & + 1 x^{5} & - 7 x^{3} & + x^{2} & - 6 x & + 8 (-) & (+) & (-) & (+) & (-) - x^{5} + x^{4} & + x^{3} & + 3 x & - 7 \end{matrix}$

$= (- x^{5} + x^{4} + x^{3} + 3 x - 7)$

Question 3

What must be subtracted from $(x^{3} - 3 x^{2} + 5 x - 1)$ to get $(2 x^{3} + x^{2} - 4 x + 2)$ ?

$- x^{3} + 4 x^{2} + 9 x - 3$

$- x^{3} - 4 x^{2} + 6 x - 3$

$- x^{3} - 4 x^{2} + 9 x - 9$

$- x^{3} - 4 x^{2} + 9 x - 3$

SOLUTION

Solution : D

Let the unknown be $= y$

$(x^{3} - 3 x^{2} + 5 x - 1) - y = 2 x^{3} + x^{2} - 4 x + 2$

$\Rightarrow y = (x^{3} - 3 x^{2} + 5 x - 1) - (2 x^{3} + x^{2} - 4 x + 2)$

$x^{3} - 3 x^{2} + 5 x - 1$
$2 x^{3} + x^{2} - 4 x + 2$
$(-) (-) (+) (-)$
____________________
$- x^{3} - 4 x^{2} + 9 x - 3$

So we have to subtract $(- x^{3} - 4 x^{2} + 9 x - 3)$ to get the required value.

Question 4

$997 \times 998 =$ ___

SOLUTION

Solution :
$997 \times 998 = (1000 - 3) (1000 - 2)$
$= 1000^{2} - 3 \times 1000 - 2 \times 1000 + (- 3) (- 2)$
$= [1000000 - 5000 + 6] = 995006$

Question 5

$(x + y) (x - y) (x^{2} + y^{2})$ = ___________.

$x^{3} - y^{3}$

$x^{3} + y^{3}$

$x^{4} + y^{4}$

$x^{4} - y^{4}$

SOLUTION

Solution : D

$To simplify : (x + y) (x - y) (x^{2} + y^{2})$

Using the identity,
$(a + b) (a - b) = a^{2} - b^{2}$

we get,
$(x + y) (x - y) (x^{2} + y^{2})$
$= (x^{2} - y^{2}) (x^{2} + y^{2})$

Using the same identity again,
$where a = x^{2} and b = y^{2}$ ,

we get,

$(x^{2} - y^{2}) (x^{2} + y^{2})$
$= {(x^{2})}^{2} - {(y^{2})}^{2}$
$= x^{4} - y^{4}$

Question 6

Simplify $(x + 3) (x + 5)$ .

$x^{2} + 6 x + 9$

$x^{2} + 8 x + 15$

$x^{2} + 9 x + 6$

$9 x^{2} + 6 x + 9$

SOLUTION

Solution : B

To simplify: $(x + 3) (x + 5)$

Using the identity $[(x + a) (x + b)^{2} = x^{2} + (a + b) x + a b]$
we get,

$(x + 3) (x + 5) = x^{2} + (3 + 5) x + 3 \times 5 = x^{2} + 8 x + 15$

Question 7

Which of the following are like terms to $9 p q^{2}$ ?

8 a b

5 b c

$5 q^{2} p$

$12 p q^{2}$

SOLUTION

Solution : C and D

Like terms are terms with the same variables raised to the same power. Coefficients of like terms need not be the same.
$5 q^{2} p = 5 p q^{2}$ .
Hence $5 q^{2} p$ and $12 p q^{2}$ are both correct answers.

Question 8

$3 x y^{2} - 5 y^{2} - 5 x - 15$ is an example of trinomial.

True

False

SOLUTION

Solution : B

Expressions that contain exactly three terms are called trinomials. The given expression has four terms. Hence,the given statement is false.

Question 9

The area of a rectangle having length and width as $5 x y$ and $9 y$ respectively is $45 x y$ .

True

False

SOLUTION

Solution : B

Area of a rectangle = length × width

Area = $5 x y \times 9 y$

= $45 x y^{2}$

Hence the given statement is false.

Question 10

The product of $(t + s^{2})$ and $(t^{2} - s)$ is

$t^{3} + s^{2} t^{2} - s t - s^{3}$

$t^{3} + s^{2} t^{2} - s t - s^{2}$

$t^{3} + s^{2} t^{2} - 2 s t - s^{3}$

$t^{3} + t^{2} - s t - s^{3}$

SOLUTION

Solution : A

$(t + s^{2}) (t^{2} - s) = [t (t^{2} - s) + s^{2} (t^{2} - s)]$

$= t^{3} - t s + s^{2} t^{2} - s^{3}$

$= t^{3} + s^{2} t^{2} - s t - s^{3}$

Free Algebraic Expressions and Identities 03 Practice Test - 8th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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