MCQ Questions for Class 7 Maths Algebraic Expressions Quiz 1

State True or False:

Addition of

$5a+3b, \ a-2b, \ 3a+5b$ is

$9a+6b$ .

0%
True
0%
False

Explanation

$We\quad can\quad directly\quad add\quad the\quad coefficients\quad of\quad like\quad variables,\\ \Longrightarrow (5+1+3)a+(3-2+5)b=9a+6b$

State True or False:

$b^2y-9b^2y+2b^2y-5b^2y$ is equal to

$-11b^2y$ .

0%
True
0%
False

Explanation

$\quad We\quad can\quad directly\quad add\quad the\quad coefficients\quad of\quad like\quad variables,\\ \Longrightarrow (1-9+2-5)b^{ 2 }y=-11b^{ 2 }y$

State True or False:

$-7x^2+18x^2+3x^2-5x^2$ is equal to

$9x^2$

0%
True
0%
False

Explanation

$\quad We\quad can\quad directly\quad add\quad the\quad coefficients\quad of\quad like\quad variables,\\ \Longrightarrow (-7+18+3-5){ x }^{ 2 }\quad =9{ x }^{ 2 }$

$\displaystyle 4x-\left( -2y+5x \right)$ is equal to

0%
$\displaystyle 9x-2y$
0%
$\displaystyle 9x+2y$
0%
$\displaystyle x+2y$
0%
$\displaystyle -x+2y$

Explanation

Given,

$\displaystyle 4x-\left( -2y+5x \right)$

$\displaystyle =4x+2y-5x$

$\displaystyle =-x+2y$
Hence simplified form is

$-x+2y$

On subtracting 7x+5y-3 from 5y-3x-9 we get

0%
10x +6
0%
-10x-6
0%
10x+10y-12
0%
-10x-12

Explanation

$(5y-3x-9)-(7x+5y-3)$

$=5y-3x-9-7x-5y+3$

$=-10x-6$

The number of terms is

$\displaystyle 6x^{3}+5x^{2}-2x+3$

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Explanation

$6x^2-5x^2-2x+3$ has terms

$6x^3,5x^2,2x\ and\ 3$ , therefore four terms.

What must be subtracted from

$3x^{2}+4y^{2}-5$ to get

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

0%
$x^{2}+3y^{2}+5$
0%
$x^2 - 4y^2 + 5$
0%
$x^2 + 7y^2 - 10$
0%
$x^2 - 7y^2 - 10$

Explanation

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

$\therefore$

$a = 3x^2 + 4 y^2 - 5 - (2x^2 - 3y^2 + 5)$

$\therefore$

$a = 3x^2 + 4 y^2 - 5 - 2x^2 + 3y^2 - 5$

$\therefore$

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

State true or false.

After Subtracting

$(1 + x + 2x^{2} - 3x^{3})$ from

$(2x^{2} -4x^{3} +4x + 9)$ we get

$(-x^{3} + 3x +8)$ .

0%
True
0%
False

Explanation

$-(1 + x + 2x^{2} - 3x^{3})$ +

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

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

Evaluate :

$7x-9y+3-3x-5y+8$

0%
$4x+4y+11$
0%
$4x-14y+11$
0%
Cannot be determined
0%
None of these

Explanation

$\quad We\quad can\quad directly\quad add\quad the\quad coefficients\quad of\quad like\quad variables,\\ \Longrightarrow (7-3)x+(-9-5)y+(3+8)=4x-14y+11$

State True or False:
Addition of

$a-3b+3, \ 2a+5-3c, \ 6c-15+6b$ is

$3a+3b+3c-7$ .

0%
True
0%
False

Explanation

$We\quad can\quad directly\quad add\quad the\quad coefficients\quad of\quad like\quad variables,\\ \Longrightarrow (1+2)a+(-3+6)b+(-3+6)c+(3+5-15)=3a+3b+3c-7$

Subtract

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

$18 - 3p + 11q + 5pq - 2pq^2 + 5p^2q$

0%
$pq^2 - 7p^2q + 8pq + 18q + 5p + 28$
0%
$p^2q - 7pq^2 + 8pq + 18p + 5q + 28$
0%
$pq^2 - 7p^2q + 8pq + 18p + 5q + 28$
0%
$p^2q - 7pq^2 + 8pq + 18q + 5p + 28$

Explanation

Subtract

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

$18-3p+11q+5pq-2pq^{ 2 }+5p^{ 2 }q$

$=18-3p+11q+5pq-2pq^{ 2 }+5p^{ 2 }q-(4p^2q - 3pq + 5pq^2 - 8p - 7q - 10)$

$=18-3p+11q+5pq-2pq^{ 2 }+5p^{ 2 }q-4p^2q + 3pq - 5pq^2 + 8p + 7q + 10$

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

Simplify:

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

0%
$-19p + 8$
0%
$5p + 8$
0%
$-19p - 8$
0%
$5p - 8$

Explanation

Multiplying term by term then adding/subtracting like terms,

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

$=-7p-4(-3p)-4(2)$

$=-7p+12p-8$

$=5p-8$

Simplify :

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

0%
$6-10pst + 10prq$
0%
$6 - 10pst - 4prq$
0%
$-6pst -10prq + 6$
0%
$-10pst-4prq-34$

Explanation

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

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

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

$= (-10pst - 4prq + 6)$

Add the following polynomials

$4 + 2y - 3y^{3}, -8 + 4y +7y^{3}$

$and\ 5\ - 6y + 8y^{3} - 6y^{2}$

0%
$12y^{3} - 6y^{2} + 1$
0%
$12y^{3} +6y^{2} + 1$
0%
$12y^{3} - 6y^{2} - 1$
0%
$12y^{3} + 6y^{2} - 1$

Explanation

Let's add the given polynomials,

$\therefore\ 4\ + 2y - 3y^{3} -8 + 4y +7y^{3} + 5\ - 6y + 8y^{3} - 6y^{2}\\$

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

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

$P = 3x -4y -8z, Q = -10y + 7x + 11z$ and

$R = 19z - 6y + 4x$ , then

$P-Q + R$ is equal to

0%
$13x -20y + 16z$
0%
0
0%
$x + y + z$
0%
$2x - by + 3z$

Explanation

$P - Q + R$

$= 3x - 4y - 8z - (7x - 10y + 11z) + (4x - 6y + 19z)$

$P - Q + R$

$= 3x - 4y - 8z - 7x + 10y - 11z + 4x - 6y + 19z \\ = (3 - 7 + 4)x + (-4 + 10 - 6)y + (-8 - 11 + 19)z \\ = 0$

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

0%
$11px^{2} + 9q^{2}y - 10rz$
0%
$-3px^{2} + qy - 8rz$
0%
$-3px^{2} + 8q^{2}y - 8rz$
0%
$-11px^{2}-8q^{2}y-10rz$

Explanation

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

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

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

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

Simplify :

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

0%
$6pq^{2} + 3pq$
0%
$6pq^{2} - 3pq$
0%
$-12pq^{2} + 3pq$
0%
$-12pq^{2}-3pq$

Explanation

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

$=(-3pq^{2}+ 4pq - 7pq + 9pq^{2})$

$=(6pq^{2} - 3pq)$

Subtract

$4x+y+2$ from the sum of

$3x-2y+7$ and

$5x-3y-8$ .

0%
$4xy+6y-3$
0%
$6x-y+2$
0%
$x-2y-3$
0%
$4x-6y-3$

Explanation

Sum of

$3x-2y+7$ and

$5x-3y-8$

$=3x-2y+7+5x-3y-8$

$=8x-5y-1$

Thus,

$8x-5y-1-(4x+y+2)$

$=8x-5y-1-4x-y-2$

$=4x-6y-3$

State whether the following statement is true or false.

After Subtracting

$-x^{4} + 4x^{3} y - 8x^{2} y^{2} + 2 xy^{3} - 4y^{4}$ from

$-5x^{3}y + x^{2}y^{2} - 6y^{4}$ we get,

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

0%
True
0%
False

Explanation

$-(-x^{4} + 4x^{3} y - 8x^{2} y^{2} + 2 xy^{3} - 4y^{4})$ +

$(-5x^{3}y + x^{2}y^{2} - 6y^{4})$

$= x^{4} - 9x^{3}y + 9x^{2} y^{2} - 2xy^{3} -2y^{4}$

Which polynomial should be subtracted from

$y^3+2y^2+5y-1$ to get

$2y^2+12$ ?

0%
$y^3+5y-13$
0%
$y^3+y-1$
0%
$5y-3$
0%
$y^3-y-14$

Explanation

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

$y^3+5y-13$

Add the following expressions

$2a+b+7$ and

$4a+2b+3$ .

0%
$6a+2b+5$
0%
$3b-6a+5$
0%
$6ab+3b-10$
0%
$6a+3b+10$

Explanation

Adding

$(2a+b+7)$ and

$(4a+2b+3)$ ,

$2a+b+7$
+

$4a+2b+3$
___

$6a+3b+10$

What must be subtracted from

$x^{4} + 4x^{2} -3x + 7$ to get

$3x^{3} - x^{2} + 2x + 1$ ?

Answer:

$x^{4}-3x^{3}+5x^{2} - 5x +6$

0%
True
0%
False

Explanation

Let

$p(x)$ be the polynomial which must be subtracted from

$x^4+4x^2-3x+7$ to get

$3x^3-x^2+2x+1$ .

So,

$x^4+4x^2-3x+7$

$-p(x)$ =

${3x^3-x^2+2x+1}$ .

So,

$p(x)=x^4+4x^2-3x+7-(3x^3-x^2+2x+1)$ .

$\Rightarrow p(x)=x^4-3x^3+5x^2-5x+6$

So, given answer is

$True$

Which polynomial should be added to

$2x^4-3x^2+5x+8$ to get

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

0%
$x^3+5x^2-x$
0%
$2x^5+x^2-10x-9$
0%
$-2x^4+5x^2-1$
0%
$-2x^4+5x^2-10x-4$

Explanation

The sum of two polynomials=

$2x^2-5x+4$
Let one polynomial be

$p$ .
Therefore,

$2x^4-3x^2+5x+8 +p=2x^2-5x+4$

$=>p=2x^2-5x+4-2x^4+3x^2-5x-8$

$=>p=-2x^4+5x^2-10x-4$
The polynomial to be added is

$-2x^4+5x^2-10x-4$

Simplify :

$(3x^2-2x+1)+(x^2+5x-3)+(4x^2+8)$

0%
$8x^2+3x+6$
0%
$x^2+x$
0%
$x^2+x_6$
0%
$8x^2+x+9$

Explanation

$(3x^2-2x+1)+(x^2+5x-3)+(4x^2+8)$

$=3x^2-2x+1+x^2+5x-3+4x^2+8$

$=8x^2+3x+6$

Subtract the second polynomial from the first :

$x^4+x^2+x-1\, ; \, x^4-x^3-x^2+1$

0%
$2x^2+x-2$
0%
$x^3+2x^2+x-2$
0%
$x^3+2x^2+x-2$
0%
$x+2x^2-2$

Explanation

$\begin{aligned}{}\left( {{x^4} + {x^2} + x - 1} \right) - \left( {{x^4} - {x^3} - {x^2} + 1} \right) &= {x^4} + {x^2} + x - 1 - {x^4} + {x^3} + {x^2} - 1\\ &= {x^4} - {x^4} + {x^3} + {x^2} + {x^2} + x - 1 - 1\\ &= {x^3} + 2{x^2} + x - 2\end{aligned}$

Hence, option

$B$ is correct.

State True or False:

Addition of

$8x-3y+7z, \ -4x+5y-4z, \ -x-y-2z$ is

$3x + y+z$ .

0%
True
0%
False

Explanation

$We\quad can\quad directly\quad add\quad the\quad coefficients\quad of\quad like\quad variables,\\ \Longrightarrow (8-4-1)x+(-3+5-1)y+(7-4-2)z=3x+y+z$

Hence the given statement is true.

State True or False:

$abx-15abx-10abx+32abx$ is equal to

$8abx$ .

0%
True
0%
False

Explanation

$\quad We\quad can\quad directly\quad add\quad the\quad coefficients\quad of\quad like\quad variables,\\ \Longrightarrow (1-15-10+32)abx=8abx$

State True or False:

$3x^2+5xy-4y^2+x^2-8xy-5y^2$ is equal to

$4x^2-3xy-9y^2$ .

0%
True
0%
False

Explanation

$\quad We\quad can\quad directly\quad add\quad the\quad coefficients\quad of\quad like\quad variables,\\ \Longrightarrow (1+3){ x }^{ 2 }+(5-8)xy+(-4-5)y^{ 2 }=4{ x }^{ 2 }-3xy-9{ y }^{ 2 }$

State True or False:
Addition of

$3b- 7c+10, \ 5c-2b-15, \ 15+12c+b$ is

$2b+10c+10$ .

0%
True
0%
False

Explanation

$We\quad can\quad directly\quad add\quad the\quad coefficients\quad of\quad like\quad variables,\\ \Longrightarrow (3-2+1)b+(-7+5+12)c+(10-15+15)=2b+10c+10$

State True or False:
Total savings of a boy who saves Rs.

$(4x-6y)$ , Rs.

$(6x+2y)$ , Rs.

$(4y-x)$ and Rs.

$(y-2x)$ for four consecutive weeks is Rs.

$(7x+y)$ .

0%
True
0%
False

Explanation

$We\quad can\quad add\quad the\quad coefficient\quad of\quad like\quad variables\quad directly\quad to\quad get\quad the\quad savings,\\ Rs.\quad 4x-6y\quad +\quad 6x+2y\quad +\quad 4y-x\quad +\quad y-2x\\ =Rs.7x+y\\$

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 1 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 1 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.