MCQ Questions for Class 7 Maths Algebraic Expressions Quiz 4

An expression is taken away from

$3x^{2} - 4y^{2} + 5xy + 20$ to obtain

$-x^{2} - y^{2} + 6xy + 20$ , then the expression is __________.

0%
$4x^{2} - 3y^{2} - xy$
0%
$2x^{2} - 5y^{2} + xy + 40$
0%
$3y^{2} - xy - 4x^{2}$
0%
$4x^{2} + 3y^{2} + xy$

Explanation

We are given two expression

$3x^{2}-4y^{2}+5xy+20$ ................................(1)

$-x^{2}-y^{2}+6xy+20$ ................................(2)

An expression is taken away from (1) to obtain (2) is given by

$3x^{2}-4y^{2}+5xy+20 - (-x^{2}-y^{2}+6xy+20)$

$= 3x^{2}-4y^{2}+5xy+20 +x^{2}+y^{2}-6xy-20)$

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

Simplify :

$(a^{3} - 2a^{2} + 4a - 5) - (-a^{3} - 8a + 2a^{2} + 5)$ .

0%
$2a^{3} + 7a^{2} + 6a - 10$
0%
$2a^{3} + 7a^{2} + 12a - 10$
0%
$2a^{3} - 4a^{2} + 12a - 10$
0%
$2a^{3} - 4a^{2} + 6a - 10$

Explanation

We need to simplify

$(a^{3}-2a^{2}+4a-5) - (-a^{3}-8a+2a^{2}+5)$

$=a^{3}-2a^{2}+4a-5 +a^{3}+8a-2a^{2}-5$

$=(a^{3}+a^{3})+(-2a^{2}-2a^{2})+(4a+8a)-10$

$=2a^{3}-4a^{2}+12a-10$

By how much is

${a}^{4} + 4{a}^{2}{b}^{2} + {b}^{4}$ more than

${a}^{4} - 8{a}^{2}{b}^{2} + {b}^{4}$ ?

0%
$12{a}^{2}{b}^{2}$
0%
$-12{a}^{2}{b}^{2}$
0%
$2{a}^{4} + 2{b}^{4}$
0%
$10{a}^{2}{b}^{2}$

Explanation

Required expression

$= ({a}^{4} + 4{a}^{2}{b}^{2} + {b}^{4}) - ({a}^{4} - 8{a}^{2}{b}^{2} + {b}^{4}$ )

$={a}^{4} + 4{a}^{2}{b}^{2} + {b}^{4} - {a}^{4} + 8{a}^{2}{b}^{2} - {b}^{4}$

$=12{a}^{2}{b}^{2}$

Subtract

$(2a - 3b + 4c)$ from the sum of

$(a + 3b - 4c), (4a - b + 9c)$ and

$(-2b + 3c - a)$ .

0%
$3a + 2b + 4c$
0%
$2a - 2b + 4c$
0%
$3a - 4b - 2c$
0%
$2a + 3b + 4c$

Explanation

It is instructed to subtract

$(2a - 3b + 4c)$ from the sum of

$(a + 3b - 4c), (4a - b + 9c)$ and

$(-2b + 3c - a)$ .

So, First we will do the sum of the three given polynomials,

Sum

$= (a + 3b - 4c) + (4a - b + 9c) + (-2b + 3c - a)$

$= (a + 4a - a)+ (3b - b - 2b) + (-4c + 9c + 3c)$

$= 4a + 8c$

Now, we can perform the subtraction,

$\therefore$ Required difference

$= (4a + 8c) - (2a - 3b + 4c)$

$= 4a + 8c - 2a + 3b - 4c$

$= 2a + 3b + 4c$

Add the following:

$2{p}^{2}{q}^{2}-3pq+4, 5+7pq-3{p}^{2}{q}^{2}$

0%
$-p^2q^2-4pq+9$
0%
$-p^2q^2+4pq+9$
0%
$-p^2q^2+2pq-9$
0%
None of these

Explanation

$2{p}^{2}{q}^{2}-3pq+4+5+7pq-3{p}^{2}{q}^{2}$

$=2p^2q^2-3p^2q^2-3pq+7pq+9$

$=-p^2q^2+4pq+9$

add

$2x$ and

$3x$ :-

0%
$9x$
0%
$3x$
0%
$5x$
0%
$7x$

Explanation

$2x+3x=(2+3)x=5x$

State true or false

We can add or subtract like terms in an algebraic expression.

0%
True
0%
False

Explanation

We can always subtract like terms in an algebraic expression and same is true for addition.

Ex: Let

$ax^2-bx^2+cx+dx+e= 0$ is a algebraic expression then this can be writtens as

$x^2(a-b)+x(c+d)+e=0$

Answer :

$(A)$

If we subtract a monomial from a binomial, then answer is atleast a binomial

0%
True
0%
False

Explanation

Consider a binomial

$-2xy+z$ and a monomial

$-z$ .
If we add

$-2xy+z$ and

$-z$ , we get

$-2xy+z+-z=-2xy$ , which is a monomial.
So the result need not be a binomial.

$\textbf{Hence, the given statement is False.}$

Sum of

${x}^{2}+x$ and

$y+{y}^{2}$ is

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

0%
True
0%
False

Explanation

The sum of

$x^{2}+x$ and

$y+y^{2}$ is given by

$\left(x^{2}+x\right)+\left(y+y^{2}\right)=x^{2}+y^{2}+x+y$
Therefore, Sum of

$x^{2}+x$ and

$y+y^{2}$ is not

$2 x^{2}+2 y^{2}$ .

$\textbf{Hence, the given statement is False.}$

State True or False:When we add a monomial and a trinomial, then answer can be a monomial.

0%
True
0%
False

Explanation

Consider a trinomial

$3xy+x+5$ and monomial

$-x$ and

$3xy$

$\bullet$ If we subtract

$3xy$ from

$3xy+x+5$ , we get

$x+5$ which is a binomial

$\bullet$ if we subtract

$-x$ from

$3xy+x+5$ , we get

$3xy+2x+5$ which is a trinomial..
We know that, a binomial or a trinomial is a polynomial.
Therefore, when we subtract a monomial from a trinomial, then answer will be a polynomial.

$\textbf{Hence, given statement is True.}$

The sum of

$-7pq$ and

$2pq$ is

0%
$-9pq$
0%
$\ 9pq$
0%
$\ 5pq$
0%
$- 5pq$

Explanation

Guven monomials are

$-7pq$ and

$2pq$

$\therefore$ Required sum

$=(- 7pq) + (2pq )$

$-7pq+2q$

$= (-7+2) pq\qquad[\because -7pq$ and

$2pq$ are like terms

$]$

$= -5pq$

Hence, the correct answer is option (D).

If we subtract

$-3x^{2}y^{2}$ from

$x^{2}y^{2}$ , then we get

0%
$- 2x^{2}y^{2}$
0%
$-4x^{2}y^{2}$
0%
$2x^{2}y^{2}$
0%
$4x^{2}y^{2}$

Explanation

Given monomials are :

$-3x^{2}y^{2}$ from

$x^{2}y^{2}$

Required difference

$=(x^2y^2)-(-3x^2y^2)$

$=x^2y^2+3x^2y^2$

$=(1+3)x^2y^2$ [Since the given monomials are like terms]

$=4x^{2}y^{2}$

Hence, option (D) is the correct answer.

Sum of

$a-b + ab$ ,

$b +c-bc$ and

$c -a-ac$ is

0%
$2c+ab-ac-bc$
0%
$2c - ab-ac-bc$
0%
$2c+ab+ac+bc$
0%
$2c -ab+ac + bc$

Explanation

Sum of

$a-b + ab$ ,

$b +c-bc$ and

$c -a-ac$

$=(a-b+ab)+(b-c-bc)+(c-a-ac)$

$=a-b+ab+b+c-bc+c-a-ac$

Rearranging and collecting like terms,

$=a-a-b+b+c+c+ab-bc-ac$

$=2c+ab-ac-bc$

Hence, option (A) is correct.

$-xy-(-5xy)$ is equal to

0%
$-6xy$
0%
$6xy$
0%
$-4xy$
0%
$4xy$

Explanation

$\begin{aligned}{} - xy - ( - 5xy) &= - xy + 5xy\\& = ( - 1 + 5)xy\\ &= 4xy\end{aligned}$

State the following statement is true or false.
Sum of

$2$ and

$p$ is

$2p$ .

0%
True
0%
False

Explanation

Sum of

$2$ and

$p$ is

$2+p$ not

$2p$ .

Simplify:

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

0%
$\displaystyle { 4x }^{ 2 }y+{ 2xy }^{ 2 }$
0%
$\displaystyle { 2x }^{ 2 }y-8xy$
0%
$-xy^2$
0%
$-xy-2x+y^2$

Explanation

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

$\displaystyle =\left( { 2x }^{ 2 }y+{ 2x }^{ 2 }y-{ 4x }^{ 2 }y \right) +\left( -{ 3xy }^{ 2 }+{ 2xy }^{ 2 } \right) ,$ (Arranging like terms together)

$\displaystyle =0-{ xy }^{ 2 }=-{ xy }^{ 2 }$
Hence simplified form of the given expression is

$-xy^2$

From the sum of

$z^3+3z^2+5z+8$ and

$4z^3+2z^2-7z-2$ subtract

$2z^3-3z^2+z-4$ .

0%
$0$
0%
$8z^2-3z$
0%
$3z^3+8z^2-3z+10$
0%
$2-z$

Explanation

The sum of

$z^3+3z^2+5z+8$ and

$4z^3+2z^2-7z-2$ is,

$=z^3+3z^2+5z+8 + 4z^3+2z^2-7z-2$

$=5z^3+5z^2-2z+6$

Now,

Subtracting

$2z^3-3z^2+z-4$ from

$5z^3+5z^2-2z+6$ we get,

$5z^3+5z^2-2z+6$

$-(2z^3-3z^2+z-4)$

$=5z^3+5z^2-2z+6$

$-2z^3+3z^2-z+4$

$=3z^3+8z^2-3z+10$

$If \:x^2 + 2x = 45, what \: is \: the\: value \: of \: x^4 + 4x^3 + 4x^2 - 13?$

0%
2013
0%
1986
0%
2012
0%
32

Subtract the second expression from the first:

$5x^2-6xy+2$ and

$3x^2+10xy-8$ .

0%
$x^2-16xy-10$
0%
$2x^2+86xy+10$
0%
$x^2-8xy- 50$
0%
$2x^2-16xy+10$

Explanation

$(5x^2-6xy+2) - (3x^2+10xy-8)$

$=5x^2-6xy+2-3x^2-10xy+8$

$=2x^2-16xy+10$

Subtract

$4a -7ab + 3b + 12$ from

$12a- 9ab + 5b- 3$ .

0%
$8a+2ab + 2b+15$
0%
$8a + 2b+15$
0%
$8a-2ab + 2b+15$
0%
$8a-2ab + 2b-15$

Explanation

subtract

$4a - 7ab + 3b + 12$ from

$12a- 9ab + 5b - 3$

$=12a- 9ab + 5b - 3-(4a - 7ab + 3b+12)$

$=12a- 9ab + 5b - 3-4a +7ab - 3b-12$

$=8a-2ab+2b-15$

From

$\displaystyle 8-y+{ 2y }^{ 2 }$ take away

$\displaystyle \left( { y }^{ 2 }-7-2y \right)$ .

0%
$y^2+y+15$
0%
$5y^2-1$
0%
$3y-7y^2+11$
0%
None of these

Explanation

Here we have to just subtract

$y^2-7-2y$ from

$8-y+2y^2$

$\therefore \displaystyle \left( 8-y+{ 2y }^{ 2 } \right) -\left( { y }^{ 2 }-7-2y \right) =8-y+{ 2y }^{ 2 }-{ y }^{ 2 }+7+2y$

$=\displaystyle \left( 8+7 \right) +\left( -y+2y \right) +\left( { 2y }^{ 2 }-{ y }^{ 2 } \right)$

$\displaystyle =15+y+{ y }^{ 2 }$

Hence, required value is

$y^2+y+15$ .

What should be subtracted from

$2a+6b-5$ to get

$-3a+2b+3$ ?

0%
$5+4b-8$
0%
$5a+4b-8$
0%
$5a+4ab-8$
0%
$5a+4b-10$

Explanation

Let

$X=-3a+2b+3$ and

$Y=2a+6b-5$
Let

$Z$ be the required expression
Now,

$X=Y-Z$

$=>Z=Y-X$
Thus,

$2a+6b-5-(-3a+2b+3)$

$=2a+6b-5+3a-2b-3$

$=5a+4b-8$

Find the sum of

$\displaystyle { x }^{ 2 }-{ 2y }^{ 2 },{ 2x }^{ 2 }-4xy+{ 5y }^{ 2 },{ 6y }^{ 2 }+11xy-{ 6x }^{ 2 }$

0%
$x^2-5xy+2y^2$
0%
$5y^2+2xy-2x^2$
0%
$9y^2+7xy-3x^2$
0%
$2x^2-7xy+y^2$

Explanation

Required sum

$=\displaystyle { (x }^{ 2 }-{ 2y }^{ 2 })+{ (2x }^{ 2 }-4xy+{ 5y }^{ 2 })+{ (6y }^{ 2 }+11xy-{ 6x }^{ 2 })$

$=\displaystyle \left( { x }^{ 2 }+{ 2x }^{ 2 }-{ 6x }^{ 2 } \right) +\left( { -2y }^{ 2 }+{ 5y }^{ 2 }+{ 6y }^{ 2 } \right) +\left( -4xy+11xy \right)$ (Combining like terms)

$\displaystyle= -{ 3 }x^{ 2 }+{ 9y }^{ 2 }+7xy$ .

What should be added to

$5x^2+2xy+y^2$ to get

$3x^2+4xy$ ?

0%
$-2x^2+2xy-y^2$
0%
$x^2+2y-y^2$
0%
$-2x^2+2y-xy^2$
0%
$x^2+2xy-y^2$

Explanation

Let

$A=3x^2+4xy$ and

$B=5x^2+2xy+y^2$
Let

$C$ be the required expression
Now,

$A=B+C$

$=>C=A-B$
Thus,

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

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

$=-2x^2+2xy-y^2$

Subtract

$(5x^{2}-5x-7)$ from the sum of

$(x^{2}-3),\:(5-4x)$ and

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

0%
$x+18$
0%
$x^{2}+x+10$
0%
$-x+24$
0%
$-9x+18$

Explanation

$(x^2-3)+(5-4x)+(9+4x^2)=5x^2-4x+11$
Now,

$(5x^2-4x+11)-(5x^2-5x-7)=x+18$
Option A is correct.

Simplify

$\displaystyle \left ( a^{3}-2a^{2}+4a-5 \right )-\left ( -a^{3}-8a+2a^{2}+5 \right )$

0%
$\displaystyle 2a^{3}+7a^{2}+6a-10$
0%
$\displaystyle 2a^{3}+7a^{2}+12a-10$
0%
$\displaystyle 2a^{3}-4a^{2}+12a-10$
0%
$\displaystyle 2a^{3}+4a^{2}+6a-10$

Explanation

$\displaystyle 3^{3}-2a^{2}+4a-5$

$\displaystyle (-)-a^{3}+2a^{2}-8a+5$
+ - + -

$\displaystyle \underline{\overline{2a^{3}-4a^{2}+12a-10}}$

Simplify :

$\displaystyle 5x-\left[ 4x-\left\{ \left( 2x-5 \right) -3\left( 3x-4 \right) \right\} \right]$

0%
$\displaystyle -7-6x$
0%
$\displaystyle -7+6x$
0%
$\displaystyle 7+6x$
0%
$\displaystyle 7-6x$

Explanation

$\displaystyle 5x-\left[ 4x-\left\{ \left( 2x-5 \right) -3\left( 3x-4 \right) \right\} \right]$

$=5x-4x+[(2x-5)-3(3x-4)]$

$=5x-4x+2x-5-9x+12$

$=7-6x$

$\displaystyle -a-[a+\{ { a+b-2a-(a-2b)\} }-b]$ is equal to

0%
$\displaystyle -4a+b$
0%
$\displaystyle 4a-2b$
0%
$0$
0%
$\displaystyle -2b$

Explanation

Given,

$\displaystyle -a-[a+\{ { a+b-2a-(a-2b)\} }-b]$
=

$\displaystyle -a-[a+\{ { a+b-2a-a+2b\} }-b]$

$\displaystyle =-a-[a+\{ { -2a+3b\} }-b]$

$\displaystyle =-a-\left[ a-2a+3b-b \right]$

$\displaystyle =-a-\left[ -a+2b \right]$

$\displaystyle =-a+a-2b$

$\displaystyle =-2b$ , which is simplified form of the given expression.

By how much is

$\displaystyle x^{4}-4x^{2}y^{2}+y^{4}$ less than

$\displaystyle x^{4}+8x^{2}y^{2}+y^{4}$ ?

0%
$\displaystyle -12x^{2}y^{2}$
0%
$\displaystyle 12x^{2}y^{2}$
0%
$\displaystyle -12xy$
0%
$\displaystyle 12xy$

Explanation

We have to find solution of

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

$-(x^{4}-4x^{2}y^{2}+y^{4})$
Separating like terms and unlike terms, we get

$=$

$x^4 - x^4 +y^4 - y^4 +8x^2y^2-(-4x^2y^2)$

$=$

$8x^2y^2 + 4x^2y^2$

$=$

$12x^2y^2$

Subtract the sum of

$\displaystyle \left( { 5x }^{ 2 }-7x+4 \right)$ and

$\displaystyle \left( 2x-{ 5x }^{ 3 }+1 \right)$ from

$\displaystyle \left( { 3x }^{ 2 }-1+5x \right)$ .

0%
$\displaystyle { 5x }^{ 3 }+{ 11x }^{ 2 }-5x+3$
0%
$5x^3-2x^2+10x-6$
0%
$\displaystyle { 3x }^{ 3 }+{ 11x }^{ 2 }+3x+5$
0%
$\displaystyle { 11x }^{ 3 }+{ 3x }^{ 2 }+5x-3$

Explanation

$3x^2 - 1 + 5x - [(5x^2-7x+4) + (2x-5x^3+1)]$
=

$3x^2 - 1 +5x - [5x^2-5x+5- 5x^3]$
=

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

$-2x^2 + 10x -6 + 5x^3$
=

$5x^3 -2x^2 + 10x - 6$

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 4 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 4 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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