MCQ Questions for Class 7 Maths Algebraic Expressions Quiz 6

What should be taken away from

$3x^2-4y^2+5xy+20$ to get

$-x^2-y^2+6xy+20$ .

0%
$x^2-y^2-xy$
0%
$x^3-3y^2-xy$
0%
$x^2-y^3-xy$
0%
$4x^2-3y^2-xy$

Explanation

Let

$p$ be the required expression. Then according the question, we have:

$(3x^{ 2 }−4y^{ 2 }+5xy+20)-p=−x^{ 2 }−y^{ 2 }+6xy+20\\ \Rightarrow p=(3x^{ 2 }−4y^{ 2 }+5xy+20)-(-x^{ 2 }−y^{ 2 }+6xy+20)\\ \Rightarrow p=3x^{ 2 }−4y^{ 2 }+5xy+20+x^{ 2 }+y^{ 2 }-6xy-20\\ \Rightarrow p=(3x^{ 2 }+x^{ 2 })+(−4y^{ 2 }+y^{ 2 })+(5xy-6xy)+(20-20)\quad \quad \quad \quad (Combining\quad like\quad terms)\\ \Rightarrow p=4x^{ 2 }−3y^{ 2 }-xy$

Hence,

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

Find the like terms from the given terms and add them:

$4xyz, 2x^2y^2, 3x^2y^2, 4x^2, 9x^2y^2, 18x^2, 6xyz, 10x^2, 7xyz$

0%
(1) 11xyz
(2) $12x^2y^2$
(3) $32x^2$
0%
(1) 17xyz
(2) $14x^2y^2$
(3) $32x^2$
0%
(1) -17xyz
(2) $14x^2y^2$
(3) $28x^2$
0%
none

Explanation

We know that like terms are terms whose variables (and their exponents such as the

$2$ in

$x^2$ ) are the same.

In the given terms

$4xyz,2x^2y^2,3x^2y^2,4x^2,9x^2y^2,18x^2,6xyz,10x^2,7xyz$ , the like terms are as follows:

(i)

$4xyz,6xyz,7xyz$

(ii)

$2x^2y^2,3x^2y^2,9x^2y^2$

(iii)

$4x^2,18x^2,10x^2$

Now, adding the like terms, we get:

(i)

$4xyz+6xyz+7xyz=17xyz$

(ii)

$2x^2y^2+3x^2y^2+9x^2y^2=14x^2y^2$

(iii)

$4x^2+18x^2+10x^2=32x^2$

Hence, after adding the like terms we get

$17xyz, 14x^2y^2,32x^2$ .

State True or False

$13m^2-2m^2 = 11m^2$

0%
True
0%
False

Explanation

Let us solve the given expression

$13m^2-2m^2$ by combining like terms as shown below:

$13m^2-2m^2=(13-2)m^2=11m^2$

Hence,

$13m^2-2m^2=11m^2$ .

Subtract

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

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

0%
$28+5p-18q+8pq-2q^2+p^2q-5pq^2$
0%
$8+11p-18q+8pq-2q^2+p^2q-5pq^2$
0%
$28-5p+18q-8pq+2q^2-p^2q+5pq^2$
0%
none

Explanation

We subtract the given expression

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

$18-3p-11q+5pq-2q^2+5p^2q$ as shown below:

$(18-3p-11q+5pq-2q^{ 2 }+5p^{ 2 }q)-(4p^{ 2 }q-3pq+5pq^{ 2 }-8p+7q-10)\\ =18-3p-11q+5pq-2q^{ 2 }+5p^{ 2 }q-4p^{ 2 }q+3pq-5pq^{ 2 }+8p-7q+10\\ =(18+10)+(-3p+8p)+(-11q-7q)+(5pq+3pq)-2q^{ 2 }+(5p^{ 2 }q-4p^{ 2 }q)-5pq^{ 2 }\\ (Combining\quad like\quad terms)\\ =28+5p-18q+8pq-2q^{ 2 }+p^{ 2 }q-5pq^{ 2 }$

Hence,

$(18-3p-11q+5pq-2q^{ 2 }+5p^{ 2 }q)-(4p^{ 2 }q-3pq+5pq^{ 2 }-8p+7q-10)=28+5p-18q+8pq-2q^{ 2 }+p^{ 2 }q-5pq^{ 2 }$ .

Simplify:

$\left( { x }^{ 2 }-{ y }^{ 2 }+2xy+1 \right) -\left( { x }^{ 2 }-{ y }^{ 2 }+4xy-5 \right)$

0%
$-2xy+6$
0%
$2xy+6$
0%
$2x{ y }^{ 2 }-2xy+6$
0%
$-2{ y }^{ 2 }-2xy-6$

Explanation

Given:

$(x^2-y^2+2xy+1)-(x^2-y^2+4xy-5)$

$=x^2-y^2+2xy+1-x^2+y^2-4xy+5$

$= x^2-x^2-y^2+y^2+2xy-4xy+1+5$

$=-2xy+6$

$\therefore$

$(x^2-y^2+2xy+1)-(x^2-y^2+4xy-5)=-2xy+6$

If A =

$10{w}^{3} + 20{w}^{2} - 55w + 60$ ,
B =

$-25{w}^{2} + 15w - 10$ and
C =

$5{w}^{2} - 10w + 20$ ,
then A + B - C is equal to ______.

0%
$10{w}^{3} + 10{w}^{2} + 30w + 30$
0%
$10{w}^{3} + 10{w}^{2} - 30w + 30$
0%
$10{w}^{3} - 10{w}^{2} - 30w + 30$
0%
None of these

Explanation

So,

$A+B-C=(10W^3+20W^2-55W+60)+(-25W^2+15W-10)-(5W^2-10W+20)$

Note- Polynomials with same order will be operated with each other.

So,

$A+B-C=10W^3+(20W^2-25W^2-5W^2)+(-55W+15W+10W)+(60-10-20)$

So,

$A+B-C=10W^3-10W^2-30W+30$

Simplify :
(3x + 2y - 9) (2x - 6y + 2) - [(4x - 9y - 1) + (-3x + 8y + 7)]

0%
$6{x}^{2} - 14xy - 12{y}^{2} - 13x + 59y - 24$
0%
$6{x}^{2} - 12xy - 189 - 17x + 61y - 29$
0%
$8{x}^{2} - 14xy - 12{y}^{2} - 13x + 57y- 24$
0%
$8{x}^{2} - 14xy - 12{y}^{2} - 17x + 61y - 29$

Explanation

Given expression is

$(3x+2y-9)(2x-6y+2)-[(4x-9y-1)+(-3x+8y+7)]$

We can perform the given operations to simplify the expression further.

$=(6x^2-18xy+6x+4xy+4y-12y^2-18x+54y-18)-[4x-9y-1-3x+8y+7]$

$\Rightarrow 6x^2-12y^2-14xy-12x+58y-18-x+y-6$

$\Rightarrow 6x^2-12y^2-14xy-13x+59y-24$

Hence,

$(3x+2y-9)(2x-6y+2)-[(4x-9y-1)+(-3x+8y+7)]=6x^2-12y^2-14xy-13x+59y-24$

By how much is

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

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

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

Explanation

We have two expressions

$a^{4}-4a^{2}b^{2}+b^{2}$ ............(1)

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

Subtract (2) from (1), we get

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

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

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

$= 4a^{2}b^{2}$

What must be subtracted from

$3a^{2} - 6ab - 3b^{2} - 1$ to get

$4a^{2} - 7ab - 4b^{2} + 1$ ?

0%
$a^{2} - ab - b^{2} + 2$
0%
$6a^{2} + 2ab + 8b^{2} - 7$
0%
$-a^{2} + ab + b^{2} - 2$
0%
$3a^{2} - ab + b^{2} - 2$

Explanation

Let the polynomial to be subtracted be

$s$ . Then,

$(3a^2-6ab-3b^2-1)-s=(4a^2-7ab-4b^2+1)$

$\Rightarrow s = \left( {3{a^2} - 6ab - 3{b^2} - 1} \right) - \left( {4{a^2} - 7ab - 4{b^2} + 1} \right)$

$\Rightarrow s = 3{a^2} - 6ab - 3{b^2} - 1 - 4{a^2} + 7ab + 4{b^2} -1$

$\Rightarrow s = \left( {3{a^2} - 4{a^2}} \right) + \left( { - 6ab + 7ab} \right) + \left( { - 3{b^2} + 4{b^2}} \right) + \left( { - 1 - 1} \right)$

$\Rightarrow s = - {a^2} + ab - {b^2} - 2$

Therefore,

$-a^2+ab+b^2-2$ must be subtracted from

$3a^2-6ab-3b^2-1$ to get

$4a^2-7ab-4b^2+1.$

Hence, option

$C$ is correct.

Add

$5x^2-7x+3, -8x^2+2x-5$ and

$7x^2-x-2$

0%
$4x^2-7x+3$
0%
$4x^2-6x-4$
0%
$4x^2+7x-4$
0%
$4x^2-6x-3$

Explanation

To simplify an algebraic expression that consists of both like and unlike terms, we need to move the like terms together and then add or subtract their coefficients.

Here, the terms are

$5x^2-7x+3,-8x^2+2x-5$ and

$7x^2-x-2$ . We add these terms as shown below:

$(5x^{ 2 }-7x+3)+(-8x^{ 2 }+2x-5)+(7x^{ 2 }-x-2)\\ =\left( 5{ x }^{ 2 }-8{ x }^{ 2 }+7{ x }^{ 2 } \right) +\left( -7x+2x-x \right) +\left( 3-5-2 \right) \\ ={ x }^{ 2 }\left( 5-8+7 \right) +x\left( -7+2-1 \right) +\left( 3-5-2 \right) \\ =4{ x }^{ 2 }-6x-4$

Hence,

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

Do the following subtractions as directed.
Subtract

$(8p+9k-17)$ from

$(2pq+7p-8q+15)$

0%
$-p-8q-9k+2pq+32$
0%
$p+8q-9k+2pq+32$
0%
$-p-8q+9k+pq+32$
0%
$p-8q-9k+2pq-32$

Explanation

We know that Like terms are terms that contain the same variables raised to the same power.

We group together the like terms and subtract

$8p+9k-17$ from

$2pq+7p-8q+15$ as follows:

$(2pq+7p-8q+15)-(8p+9k-17)\\ =2pq+7p-8q+15-8p-9k+17\\ =(7p-8p)-8q-9k+2pq+(15+17)\\ =-p-8q-9k+2pq+32$

Hence,

$(2pq+7p-8q+15)-(8p+9k-17)=-p-8q-9k+2pq+32$ .

Do the following subtractions as directed.
Subtract

$t^4-3t^2+7$ from

$5t^3-9$

0%
$t^4-3t^2-5t^3+16$
0%
$-t^4-5t^3-3t^2+16$
0%
$-t^4+5t^3+3t^2-16$
0%
$t^4-5t^3-3t^2-16$

Explanation

We know that Like terms are terms that contain the same variables raised to the same power.

The only like term in the given expressions

$t^4-3t^2+7$ and

$5t^3-9$ are

$7$ and

$-9$ . So we group together the like terms and subtract

$t^4-3t^2+7$ from

$5t^3-9$ as follows:

$(5t^{ 3 }-9)-(t^{ 4 }-3t^{ 2 }+7)\\ =5t^{ 3 }-9-t^{ 4 }+3t^{ 2 }-7\\ =-t^{ 4 }+5t^{ 3 }+3t^{ 2 }+(-9-7)\\ =-t^{ 4 }+5t^{ 3 }+3t^{ 2 }-16$

Hence,

$(5t^{ 3 }-9)-(t^{ 4 }-3t^{ 2 }+7)=-t^{ 4 }+5t^{ 3 }+3t^{ 2 }-16$ .

Do the following subtractions as directed.
Subtract

$(3x^3-7x+4)$ from

$(5x^3+7x^2-2x)$

0%
$2x^3+4x^2+5x-2$
0%
$2x^3+7x^2+5x-2$
0%
$2x^3+7x^2+5x-4$
0%
$2x^3+x^2+5x-2$

Explanation

We know that Like terms are terms that contain the same variables raised to the same power.

We group together the like terms and subtract

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

$5x^3+7x^2-2x$ as follows:

$(5x^{ 3 }+7x^{ 2 }-2x)-(3x^{ 3 }-7x+4)\\ =5x^{ 3 }+7x^{ 2 }-2x-3x^{ 3 }+7x-4\\ =(5x^{ 3 }-3x^{ 3 })+7x^{ 2 }+(-2x+7x)-4\\ =2x^{ 3 }+7x^{ 2 }+5x-4$

Hence,

$(5x^{ 3 }+7x^{ 2 }-2x)-(3x^{ 3 }-7x+4)=2x^{ 3 }+7x^{ 2 }+5x-4$ .

Add the following :
i)

$5y^3 , 26y^3, 10y^3, -3y^3$
ii)

$3x^2, -10x^2, 4x^2$
iii)

$4x^2y, -3xy^2, -5xy^2, 5x^2y$

0%
(i) $38y^3$ (ii) $-3x^2$ (iii) $8x^2y-9xy^2$
0%
(i) $38y^3$ (ii) $-3x^2$ (iii) $9x^2y-8xy^2$
0%
(i) $41y^3$ (ii) $-3x^2$ (iii) $9x^2y-8xy^2$
0%
None

Explanation

$5y^3,26y^3,10y^3,−3y^3$

$5y^3+26y^3+10y^3−3y^3$

$y^3(5+26+10-3)$

$38y^3$

ii)

$3x^2,−10x^2,4x^2$

$3x^2−10x^2+4x^2$

$x^2(3-10+4)$

$-3x^2$

iii)

$4x^2y,−3xy^2,−5xy^2,5x^2y$

$4x^2y−3xy^2−5xy^2+5x^2y$

$4x^2y+5x^2y−3xy^2−5xy^2$

$9x^2y−8xy^2$

Subtract

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

$4x^{2}+x^{2}+x+6$ , then the resultant value is

0%
$6x^{3}+5x^{2}-2x+1$
0%
$2x^{3}+5x^{2}-2x+1$
0%
$2x^{3}-5x^{2}-2x+1$
0%
$2x^{3}-5x^{2}+2x-1$

What must be added to

$5x^{3}-2x^{2}+6x+7$ to make the sum

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

0%
$4x^{3}-5x^{2}+7x+6$
0%
$4x^{3}+5x^{2}-7x+6$
0%
$-4x^{3}+5x^{2}-7x-6$
0%
$None\ of\ these$

Explanation

$\left( {x}^{3} + 3 {x}^{2} - x + 1 \right) - \left( 5 {x}^{3} - 2 {x}^{2} + 6x + 7 \right)$

$= \left( 1 - 5 \right) {x}^{3} + \left( 3 + 2 \right) {x}^{2} - \left( 1 + 6 \right) x + 1 - 7$

$= -4 {x}^{3} + 5 {x}^{2} - 7x - 6$

Hence

$\left( -4 {x}^{3} + 5 {x}^{2} - 7x - 6 \right)$ must be added to

$\left( {x}^{3} + 3 {x}^{2} - x + 1 \right)$ to make the sum

$\left( 5 {x}^{3} - 2 {x}^{2} + 6x + 7 \right)$ .

Simplify combining like terms :

$p-(p-q)-q-(q-p)$

0%
$q-p$
0%
$q$
0%
$0$
0%
$p-q$

Explanation

$p-(p-q)-q-(q-p)$

$=p-p+q-q-q+p$

$=p-q$

Add:

$2u+3v,\,\,-2v+3w,\,\,3u-v+2w$

0%
$5u-5w$
0%
$5u+5w$
0%
$u-w$
0%
$u+w$

Explanation

given,

$\left(2u+3v+3u-v+2w\right)+\left(-2v+3w\right)$

$=5u+2v+2w-2v+3w$

$=5u+5w$

What must be added to

${ 7z }^{ 3 }-{ 11z }^{ 2 }-129$ to get

${ 5z }^{ 2 }+7z-92$ ?

0%
${ 7z }^{ 3 }+{ 16z }^{ 2 }+7z+37$
0%
$-{ 7z }^{ 3 }+{ 16z }^{ 2 }+7z+37$
0%
$-{ 7z }^{ 3 }+{ 16z }^{ 2 }+7z-37$
0%
$-{ 7z }^{ 3 }-{ 7z }^{ 2 }+7z-37$

Explanation

Let a must be added.

Then,

$7z^3 - 11z^2 -129 + a = 5z^2 +7z -92$

$\rightarrow a = -7z^3 + 16z^2 + 7z + 37$

Thus, B is the correct answer.

Simplify:

$a-[b-\{c-(a-b-c)\}-c]+a$

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

The sum of

$(6a+4b-c+3), (2b-3c+4), (11b-7a+2c-1)$ and

$(2c-5a-6)$ is

0%
$(4a-6b+2)$
0%
$(-3a+14b-3c+2)$
0%
$(-6a+17b)$
0%
$(-6a+6b+c-4)$

Explanation

$(6a+4b−c+3)+(2b−3c+4)+(11b−7a+2c−1)+(2c−5a−6)$

$\Rightarrow$

$6a+4b−c+3+2b−3c+4+11b−7a+2c−1+2c−5a−6$

Now arranging and combining the like terms,

$\Rightarrow$

$(6a-7a-5a)+(4b+2b+11b)+(-c-3c+2c+2c)+(3+4-1-6)$

$\Rightarrow$

$-6a+17b$

$\therefore$ The sum of

$(6a+4b−c+3),(2b−3c+4),(11b−7a+2c−1)$ and

$(2c−5a−6)$ is

$(-6a+17b)$

The length of a side of square is given as

$2x+3$ . Which expression represents the perimeter of the square.

0%
$2x+16$
0%
$6x+9$
0%
$8x+3$
0%
$8x+12$

Explanation

Apply the formula of perimeter of square
Since, the perimeter of a square

$=side+side+side+side=4 \times side$
The perimeter of the given square

$=4 \times (2x+3)$

$=4 \times 2 x+4 \times 3$

$\Rightarrow 4 \times(2 x+3)=8 x+12$
Therefore, the perimeter of the given square

$=8x+12$

$\textbf{Hence, the correct option is (d)}$

Number of terms in the expression

$3{ x }^{ 2 }{ y }-2{ y }^{ 2 }{ z }-{ z }^{ 2 }x+5$ is

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

Explanation

The terms in the given expression are

$3 x^{2} y,-2 y^{2} z,-z^{2} x$ and

$5$
Therefore, the number of terms in the given expression are

$4 .$

$\textbf{Hence, the correct option is (c)}$

The terms of the expression

$4{ x }^{ 2 }-3xy$ are:

0%
$4{ x }^{ 2 }$ and $-3xy$
0%
$4{ x }^{ 2 }$ and $3xy$
0%
$4{ x }^{ 2 }$ and $-xy$
0%
${ x }^{ 2 }$ and $xy$

Explanation

We know that terms are added in an expression.
Therefore terms in the expression

$4 x^{2}-3 x y$ are

$4 x^{2}$ and

$-3 x y$

$\textbf{Hence, the correct option is (a)}$

State true or false.

$(3a-b+3)-(a+b)$ is a binomial.

0%
True
0%
False

State true or false:

$7x$ has two terms,

$7$ and

$x$

0%
True
0%
False

Explanation

$7x$ is a multiplication of

$7$ and

$x$ .

Since,

$7x$ contains no addition and subtraction operations and therefore it has only one term

State the following statement is true or false.
Sum of

$x^2+x$ and

$y^2+y$ is

$2x^2+2y^2$

0%
True
0%
False

Explanation

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

Hence, sum of

$x^2+x$ and

$y^2+y$ is not

$2x^2+2y^2$

State the following statement is true or false.
If we subtract a monomial from a binomial, then the answer is at least a binomial.

0%
True
0%
False

If the perimeter of rectangle is given by

$2(a+b)$ , where

$a$ and

$b$ denotes the length and breadth of the rectangle respectively.. The variable used here are ____ and ____.

0%
$2$ and $a$
0%
$2$ and $b$
0%
$a$ and $b$
0%
$2, a$ and $b$

$a$ represent the side of a square, then perimeter of the square is ____.

0%
$2a$
0%
$4a$
0%
$6a$
0%
$8a$

Explanation

Given side of square:

$a$ units

We know

$\textit{Perimeter of square} = 4 \times side$

$\therefore$ the required expression here is

$4a$

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 6 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 6 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

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