Free Algebraic Expressions and Identities 01 Practice Test - 8th Grade

Question 1

When $7 x y + 5 y z - 3 z x,$ $4 y z + 9 z x - 4 y$ and $- 3 x z + 5 x - 2 x y$ are added, we get ___.

$5 x y + 9 y z + 3 z x + 5 x - 4 y$

$5 x y + 9 y z + 15 z x + 5 x - 4 y$

$5 x y + 9 y z - 3 z x + 5 x - 4 y$

$5 x y + 9 y z + 3 z x - 5 x - 4 y$

SOLUTION

Solution : A

$7 x y + 5 y z - 3 z x$
$4 y z + 9 z x - 4 y$
$- 2 x y - 3 x z + 5 x$ ________________________
$5 x y + 9 y z + 3 z x - 4 y + 5 x$

Question 2

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

$x^{3} - 2 x^{2} - 8 x + 7$

$7 x^{3} + 2 x^{2} - 8 x + 7$

$7 x^{3} - 2 x^{2} - 8 x + 3$

$7 x^{3} - 2 x^{2} - 8 x + 7$

SOLUTION

Solution : D

$4 x^{3} - 3 x + 5$
$- 3 x^{3} + 5 x - 2 + 2 x^{2}$
$(+) (-) (+) (-)$
________________________
$7 x^{3} - 8 x + 7 - 2 x^{2}$

$= 7 x^{3} - 2 x^{2} - 8 x + 7$

Question 3

Add $(3 + 2 y - 5 y^{2} + 6 y^{3}), (- 8 + 3 y + 7 y^{3})$ and $(5 - 6 y - 8 y^{3} + y^{2})$ .

$- 3 y - 4 y^{2} + 5 y^{3}$

$- y - 3 y^{2} + 5 y^{3}$

$- y - 4 y^{2} + 5 y^{3}$

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

SOLUTION

Solution : C

To add : $(3 + 2 y - 5 y^{2} + 6 y^{3}), (- 8 + 3 y + 7 y^{3})$ and $(5 - 6 y - 8 y^{3} + y^{2})$

$(- 8 + 3 y + 7 y^{3})$ does not have the term with $y^{2}$ . So, we add $0 y^{2}$ and hence, the expression will be $(- 8 + 3 y + 0 y^{2} + 7 y^{3})$ .

On adding there expressions, we get,
$3 + 2 y - 5 y^{2} + 6 y^{3}$
$- 8 + 3 y + 0 y^{2} + 7 y^{3}$
$5 - 6 y + y^{2} - 8 y^{3} - -------------------- -$
$- 1 y - 4 y^{2} + 5 y^{3}$

$(3 + 2 y - 5 y^{2} + 6 y^{3}) + (- 8 + 3 y + 7 y^{3}) + (5 - 6 y - 8 y^{3} + y^{2}) = (- y - 4 y^{2} + 5 y^{3})$

Question 4

${(1.05)}^{2} - {(0.95)}^{2} =$ ___

SOLUTION

Solution :
Using the identity
${(a)}^{2} - {(b)}^{2}$ $= (a + b) (a - b)$ ,
${(1.05)}^{2} - {(0.95)}^{2}$
$= (1.05 + 0.95) (1.05 - 0.95)$
$= (2) (0.1) = 0.2$

Question 5

${(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2} =$ ____________.

$44 p q^{2}$

$48 p^{2} q$

$48 p q^{2}$

$44 p^{2} q$

SOLUTION

Solution : C

${(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2}$

We have:

$(a + b)^{2} = a^{2} + 2 a b + b^{2} (a - b)^{2} = a^{2} - 2 a b + b^{2}$

So, ${(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2}$

$= [{(4 p q)}^{2} + {(3 q)}^{2} + 2 (4 p q) (3 q)] - [{(4 p q)}^{2} + {(3 q)}^{2} - 2 (4 p q) (3 q)]$

$= 24 p q^{2} + 24 p q^{2}$
$= 48 p q^{2}$

Question 6

Using an identity expand :
$(2 y + 5) (2 y + 5)$

$4 y^{2} + 10 y + 25$

$4 y^{2} + 20 y + 25$

$4 y^{2} + 20 y + 15$

$y^{2} + 20 y + 25$

SOLUTION

Solution : B

We know that,
$(a + b) \times (a + b) = (a + b)^{2}$

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

$(2 y + 5) (2 y + 5) = (2 y + 5)^{2}$
$= (2 y)^{2} + 2 (2 y) (5) + 5^{2}$
$= 4 y^{2} + 20 y + 25$

Question 7

A monomial multiplied by a monomial always gives a ________.

Monomial

Binomial

Trinomial

Constant

SOLUTION

Solution : A

When we multiply monomials, we first multiply the coefficients and then multiply the variables by adding the exponents. This will always give a monomial.

For example, $2 a b \times 2 b$ = $4 a b^{2}$ , which is a monomial.

Question 8

The numerical factor of a term is known as

Expression

Coefficient

Variable

Equation

SOLUTION

Solution : B

The numerical factor of a term is known as coefficient.

Question 9

Simplify $(3 x^{2} + 5 y^{2}) (4 x y - 5 y)$ .

$(6 x^{3} y - 15 x y + 20 x y^{3} - 5 y^{3})$

$(12 x^{3} y - 5 x^{2} y + 10 x y^{3} + 25 y^{3})$

(12 x^{3} y - 15 x^{2} y + 20 x y^{3} - 25 y^{3})

(6 x^{3} y - 15 x^{2} y + 10 x y^{3} - 5 y^{3})

SOLUTION

Solution : C

Given: $(3 x^{2} + 5 y^{2}) (4 x y - 5 y)$
$= 3 x^{2} (4 x y - 5 y) + 5 y^{2} (4 x y - 5 y)$
$= 12 x^{3} y - 15 x^{2} y + 20 x y^{3} - 25 y^{3}$

Question 10

Find the area of a rectangle whose length is $5 x y$ units and breadth is $8 x y^{2}$ units.

$40 x^{2}$ $y^{3}$ square units

$40 x^{2}$ $y^{2}$ square units

$40 x y^{3}$ square units

$40 x y$ square units

SOLUTION

Solution : A

Given:
Length of the rectangle = $5 x y$ units
Breadth of the rectangle = $8 x y^{2}$ units

Area of the rectangle = length $\times$ breadth
$= 5 x y \times 8 x y^{2}$
$= 40 x^{2} y^{3}$ square units

Hence, the area of the rectangle is $40 x^{2} y^{3}$ square units.

Free Algebraic Expressions and Identities 01 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

Related ExamsUnderstanding Quadrilaterals 01Direct and Inverse Proportions 01Squares and Square Roots 01

Related Exams
Understanding Quadrilaterals 01
Direct and Inverse Proportions 01
Squares and Square Roots 01