Algebraic Expressions Test

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Algebraic Expressions Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

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

A
$$x^{2}+3y^{2}+5$$

B
$$x^2 - 4y^2 + 5$$

C
$$x^2 + 7y^2 - 10$$

D
$$x^2 - 7y^2 - 10$$

Solution

$$(3x^{2}+4y^{2}-5) - a = (2x^{2}-3y^{2}+5)$$
$$\therefore$$ $$a = 3x^2 + 4 y^2 - 5 - (2x^2 - 3y^2 + 5)$$
$$\therefore$$ $$a = 3x^2 + 4 y^2 - 5 - 2x^2 + 3y^2 - 5$$
$$\therefore$$ $$a = x^2 + 7y^2 - 10$$
Question 2
1 / -0

Subtract $$4p^2q - 3pq + 5pq^2 - 8p - 7q - 10$$ from $$18 - 3p + 11q + 5pq - 2pq^2 + 5p^2q$$

A
$$pq^2 - 7p^2q + 8pq + 18q + 5p + 28$$

B
$$p^2q - 7pq^2 + 8pq + 18p + 5q + 28$$

C
$$pq^2 - 7p^2q + 8pq + 18p + 5q + 28$$

D
$$p^2q - 7pq^2 + 8pq + 18q + 5p + 28$$

Solution

Subtract $$4p^2q - 3pq + 5pq^2 - 8p - 7q - 10$$ from $$18-3p+11q+5pq-2pq^{ 2 }+5p^{ 2 }q$$
$$=18-3p+11q+5pq-2pq^{ 2 }+5p^{ 2 }q-(4p^2q - 3pq + 5pq^2 - 8p - 7q - 10)$$
$$=18-3p+11q+5pq-2pq^{ 2 }+5p^{ 2 }q-4p^2q + 3pq - 5pq^2 + 8p + 7q + 10$$
$$=28+5p+18q+8pq-7pq^2+p^2q$$
Question 3
1 / -0

Simplify: $$-7p-4(-3p + 2 ) $$

A
$$-19p + 8$$

B
$$5p + 8$$

C
$$-19p - 8$$

D
$$5p - 8$$

Solution

Multiplying term by term then adding/subtracting like terms,
$$-7p-4(-3p+2)$$
$$=-7p-4(-3p)-4(2)$$
$$=-7p+12p-8$$
$$=5p-8$$
Question 4
1 / -0

Simplify :$$(-8pst + 3prq -14) - (7prq - 20 + 2pst) = $$

A
$$6-10pst + 10prq$$

B
$$6 - 10pst - 4prq$$

C
$$-6pst -10prq + 6$$

D
$$-10pst-4prq-34$$

Solution

$$(-8pst + 3prq -14) - (7prq - 20 + 2pst)$$ $$= (-8pst + 3prq -14 - 7prq + 20 - 2pst)$$
$$= (-8pst - 2pst + 3prq - 7prq + 20 - 14)$$
$$= (-10pst - 4prq + 6)$$
Question 5
1 / -0

Add the following polynomials
$$ 4 + 2y - 3y^{3}, -8 + 4y +7y^{3} $$ $$ and\ 5\ - 6y + 8y^{3} - 6y^{2} $$

A
$$12y^{3} - 6y^{2} + 1$$

B
$$12y^{3} +6y^{2} + 1$$

C
$$12y^{3} - 6y^{2} - 1$$

D
$$12y^{3} + 6y^{2} - 1$$

Solution

Let's add the given polynomials,

$$\therefore\ 4\ + 2y - 3y^{3} -8 + 4y +7y^{3} + 5\ - 6y + 8y^{3} - 6y^{2}\\$$
$$=(7+8-3)y^3+(-6)y^2+(2+4-6)y+(4-8+5)\\$$
$$=12y^{3} - 6y^{2} + 1$$
Question 6
1 / -0

If $$P = 3x -4y -8z, Q = -10y + 7x + 11z$$ and $$R = 19z - 6y + 4x$$, then $$P-Q + R$$ is equal to

A
$$13x -20y + 16z$$

B
0

C
$$x + y + z$$

D
$$2x - by + 3z$$

Solution

$$P - Q + R$$ $$= 3x - 4y - 8z - (7x - 10y + 11z) + (4x - 6y + 19z)$$
$$P - Q + R$$ $$= 3x - 4y - 8z - 7x + 10y - 11z + 4x - 6y + 19z \\ = (3 - 7 + 4)x + (-4 + 10 - 6)y + (-8 - 11 + 19)z \\ = 0$$
Question 7
1 / -0

$$(4px^{2}+5q^{2}y -9rz) - (-3q^{2}y + 7px^{2}-rz) = $$

A
$$11px^{2} + 9q^{2}y - 10rz$$

B
$$-3px^{2} + qy - 8rz$$

C
$$-3px^{2} + 8q^{2}y - 8rz$$

D
$$-11px^{2}-8q^{2}y-10rz$$

Solution

$$(4px^{2}+5q^{2}y -9rz) - (-3q^{2}y + 7px^{2}-rz)$$$$=(4px^{2}+5q^{2}y -9rz) + 3q^{2}y - 7px^{2} + rz $$
$$=4px^{2} - 7px^{2} + 5q^{2}y + 3q^{2}y - 9rz + rz $$
$$= (-3)px^{2} + 8q^{2}y - 8rz $$
Question 8
1 / -0

Simplify :$$(-3pq^{2}+ 4pq) - (7pq - 9pq^{2}).$$

A
$$6pq^{2} + 3pq$$

B
$$6pq^{2} - 3pq$$

C
$$-12pq^{2} + 3pq$$

D
$$-12pq^{2}-3pq$$

Solution

$$(-3pq^{2}+ 4pq) - (7pq - 9pq^{2})$$ $$=(-3pq^{2}+ 4pq - 7pq + 9pq^{2})$$
$$=(6pq^{2} - 3pq)$$
Question 9
1 / -0

Subtract $$4x+y+2$$ from the sum of $$3x-2y+7$$ and $$5x-3y-8$$.

A
$$4xy+6y-3$$

B
$$6x-y+2$$

C
$$x-2y-3$$

D
$$4x-6y-3$$

Solution

Sum of $$3x-2y+7$$ and $$5x-3y-8$$
$$=3x-2y+7+5x-3y-8$$
$$=8x-5y-1$$

Thus, $$8x-5y-1-(4x+y+2)$$
$$=8x-5y-1-4x-y-2$$
$$=4x-6y-3$$
Question 10
1 / -0

Which polynomial should be subtracted from $$y^3+2y^2+5y-1$$ to get $$2y^2+12$$?

A
$$y^3+5y-13$$

B
$$y^3+y-1$$

C
$$5y-3$$

D
$$y^3-y-14$$

Solution

The bigger polynomial is $$ y^3+2y^2+5y-1 $$
Let one polynomial be $$p$$.
Therefore,
$$ y^3+2y^2+5y-1 -p=2y^2+12$$
$$=>y^3+2y^2+5y-1-2y^2-12=p$$
$$=y^3+5y-13$$
Therefore,
The polynomial to be subtracted is $$y^3+5y-13$$

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Algebraic Expressions Test - 8

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Algebraic Expressions Test - 8

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Question 1

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

$$x^2 - 4y^2 + 5$$

$$x^2 + 7y^2 - 10$$

$$x^2 - 7y^2 - 10$$

Solution

Question 2

$$pq^2 - 7p^2q + 8pq + 18q + 5p + 28$$

$$p^2q - 7pq^2 + 8pq + 18p + 5q + 28$$

$$pq^2 - 7p^2q + 8pq + 18p + 5q + 28$$

$$p^2q - 7pq^2 + 8pq + 18q + 5p + 28$$

Solution

Question 3

$$-19p + 8$$

$$5p + 8$$

$$-19p - 8$$

$$5p - 8$$

Solution

Question 4

$$6-10pst + 10prq$$

$$6 - 10pst - 4prq$$

$$-6pst -10prq + 6$$

$$-10pst-4prq-34$$

Solution

Question 5

$$12y^{3} - 6y^{2} + 1$$

$$12y^{3} +6y^{2} + 1$$

$$12y^{3} - 6y^{2} - 1$$

$$12y^{3} + 6y^{2} - 1$$

Solution

Question 6

$$13x -20y + 16z$$

0

$$x + y + z$$

$$2x - by + 3z$$

Solution

Question 7

$$11px^{2} + 9q^{2}y - 10rz$$

$$-3px^{2} + qy - 8rz$$

$$-3px^{2} + 8q^{2}y - 8rz$$

$$-11px^{2}-8q^{2}y-10rz$$

Solution

Question 8

$$6pq^{2} + 3pq$$

$$6pq^{2} - 3pq$$

$$-12pq^{2} + 3pq$$

$$-12pq^{2}-3pq$$

Solution

Question 9

$$4xy+6y-3$$

$$6x-y+2$$

$$x-2y-3$$

$$4x-6y-3$$

Solution

Question 10

$$y^3+5y-13$$

$$y^3+y-1$$

$$5y-3$$

$$y^3-y-14$$

Solution

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