MCQ Questions for Class 7 Maths Algebraic Expressions Quiz 5

The sum of three algebric expressions is

$\displaystyle x^{2}+y^{2}+z^{2}$ . If two of them are

$\displaystyle 4x^{2}-5y^{2}+3z^{2}$ and

$\displaystyle -3x^{2}+4y^{2}-2z^{2}$ then the third expression is

0%
$\displaystyle 2x^{2}+2z^{2}$
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$\displaystyle 2y^{2}$
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$\displaystyle 2x^{2}+2y^{2}-z^{2}$
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$\displaystyle 2y^{2}+2z^{2}$

Explanation

Third exp =

$\displaystyle x^{2}+y^{2}+z^{2}-\left \{ \left ( 4x^{2}-5y^{2}+3z^{2} \right )+\left ( -3x^{2}+4y^{2}-2z^{2} \right ) \right \}$

$x^2+y^2+z^2 - ( x^2 - y^2 + z^2)$

$x^2 + y^2 + z^2 - x^2 + y^2 - z^2$

$2y^2$

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

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

$P - Q + R$ is equal to

0%
$\displaystyle 13x-20y+16z$
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$0$
0%
$x + y + z$
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$\displaystyle 2x-4y+3z$

Explanation

Given,

$P=3x-4y-8z, Q=-10y+7x+11z, R=19z-6y+4x$

We need to find value of

$P - Q + R$

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

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

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

$=0+0+0=0$

$\displaystyle 5a-\left[ 3b-\left\{ a-3\left( 2a-b \right) \right\} \right]$ is equal to

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

Explanation

Given,

$\displaystyle 5a-\left[ 3b-\left\{ a-3\left( 2a-b \right) \right\} \right]$

$\displaystyle =5a-\left[ 3b-\left\{ a-6a+3b \right\} \right]$

$\displaystyle =5a-\left[ 3b-\left\{ -5a+3b \right\} \right]$

$\displaystyle =5a-\left[ 3b+5a-3b \right]$

$\displaystyle =5a-5a=0$
Hence given expression vanishes.

Simplify:

$\displaystyle x^{2}y^{3}-1.5x^{2}y^{3}+1.4x^{2}y^{3}$

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$\displaystyle 0.9x^{2}y^{3}$
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$\displaystyle -0.9x^{2}y^{3}$
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$0.9$
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$-0.9$

Explanation

$\displaystyle x^{2}y^{3}-1.5x^{2}y^{3}+1.4x^{2}y^{3}$
=

$\displaystyle x^{2}y^{3}(1-1.5+1.4)$

$\displaystyle =0.9x^{2}y^{3}$

What must be added to

$x^2\,+\, 5x\,-\,6$ to get

$x^3\, -\,x^2\,+\, 3x\, -\, 2$ ?

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$x^3\,-\,2x^2\,-\,2x\,-\,4$
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$x^3\,+\,2x^2\,-\,2x\,+\,4$
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$x^3\,-\,2x^2\,-\,2x\,+\,4$
0%
None of these

Explanation

Let the polynomial to be added be

$p(x)$

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

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

$\therefore p(x) = x^3 -2x^2-2x+4$

What must be subtracted from

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

$x^3\,+\, x^2\,+\, x\, -\,1$ ?

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$x^4\, -\,x^3\,+ x^2\, -\, 4x\, +\, 8$
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$x^3\,+\, x^2\, -\, 4x\, +\, 8$
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$x^4\, -\,x^3\, +\, x^2\,+\, 4x\, -\,8$
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$x^4\, -\,x^3\, -\,x^2\,+\, 4x\, -\,8$

Explanation

Let the polynomial to be subtracted be

$p(x)$

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

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

$\therefore p(x) = x^4 - x^3 + x^2 - 4x + 8$

What must be subtracted from

$x^3 \,-\, 3x^2\, + 5x\, -\, 1$ to get

$2x^3\, +\, x^2\, - \,4x\, +\, 2$ ?

0%
$-x^3\, +\, 4x^2\, -\, 9x\, +\, 3$
0%
$x^3\, +\, 4x^2\, -\, 9x\, +\, 3$
0%
$x^3\, -\, 4x^2\, +\, 9x\, -\, 3$
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$-x^3\, -\, 4x^2\, +\, 9x\, -\, 3$

Explanation

Let the polynomial to be subtracted be

$p(x)$ .

$\therefore x^3 - 3x^2 + 5x -1 - p(x) = 2x^3 + x^2 - 4x + 2$

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

$\therefore p(x) = -x^3 - 4x^2 + 9x -3$

What must be added to

$x^3\, +\, 3x\, -\, 8$ to get

$3x^3 \,+ \,x^2\, +\, 6$ ?

0%
$2x^3\, +\, x^2\, - \,3x\, +\, 14$
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$2x^2\, + \, x^2\, +\, 14$
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$2x^3\, +\, x^2\, - \,6x\, -\,14$
0%
None of these

Explanation

Let the polynomial to be added be

$p$ .

$\therefore x^3 + 3x - 8 + p = 3x^3 + x^2 + 6$

$\therefore p = 3x^3 + x^2 + 6 - x^3 - 3x + 8$

$\therefore p = 2x^3 + x^2- 3x + 14$

Hence, Option A is correct.

What must be added to the sum of

$\displaystyle 2a^{2}-3a+7,-5a^{2}-2a-11$ and

$\displaystyle 3a^{2}+5a-8$ to get

$0$ ?

0%
$-12$
0%
$12$
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$\displaystyle a^{2}+a$
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$\displaystyle a-1$

Explanation

Let '

$x$ ' be added to these polynomial to get

$0$ .

$\Rightarrow (2a^2-3a+7)+(-5a^2-2a-11)+(3a^2+5a-8)+x=0$

$\Rightarrow (2a^2-5a^2+3a^2)+(-3a-2a+5a)+(7-11-8)+x=0$

$\Rightarrow 0+0+(-12)+x=0$

$\Rightarrow x=12$
Option B is correct.

Which of the following expressions are exactly equal in value?

${(3x-y)}^{2}-({5x}^{2}-2xy)$

${(2x-y)}^{2}$

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

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

0%
$1$ and $2$ only
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$1,2$ and $3$ only
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$2$ and $4$ only
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$1,2$ and $4$ only

Explanation

From

$1^{st}$ expression

${(3x-y)}^{2}-({5x}^{2}-2xy)$

$=9{x}^{2}-6xy+{y}^{2}-5{x}^{2}+2xy$

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

$2^{nd}$ expression

${(2x-y)}^{2} =4{x}^{2}-4xy+{y}^{2}$
From

$3^{rd}$ expression

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

$=4{ x }^{ 2 }+2xy+{ y }^{ 2 }\quad$
From

$4^{th}$ expression

${(2x+3y)}^{2}-8y(2x+y)=4{ x }^{ 2 }+12xy+{ 9y }^{ 2 }-16xy-8{ y }^{ 2 }$

$=4{ x }^{ 2 }-4xy+{ y }^{ 2 }$
Hence the value of

$1,2$ and

$4$ expression will be same

Subtract:

$-6p+ q+3r+ 8$ from

$p -2q - 5r -8$

0%
$7p-3q+8r-16$
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$7p-3q-8r-16$
0%
$p-3q-8r+16$
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$p-3q-8r-16$

Explanation

$\begin{aligned}{}p - 2q - 5r - 8 - ( - 6p + q + 3r + 8) &= p - 2q - 5r - 8 + 6p - q - 3r - 8\\ &= p + 6p - 2q - q - 5r - 3r - 8 - 8\\& = 7p - 3q - 8r - 16\end{aligned}$

Hence, option

$B$ is correct.

What should be added to the product of

$({x}^{2}+xy-{y}^{2})$ and

$({x}^{2}-xy+{y}^{2})$ to get

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

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

Explanation

The given expressions are

$({x}^{2}+xy-{y}^{2})$ and

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

There product

$= ({x}^{2}+xy-{y}^{2})({x}^{2}-xy+{y}^{2})$

$= {x}^{2}({x}^{2}-xy+{y}^{2})+xy({x}^{2}-xy+{y}^{2})-{y}^{2}({x}^{2}-xy+{y}^{2})$

$= \left ( x^4-x^3y+x^2y^2 \right )+\left(x^3y-x^2y^2+xy^3 \right )+\left (-x^2y^2+xy^3-y^4 \right )$

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

let we added

$a$ then

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

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

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

Subtract

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

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

0%
$8a - 2ab + 2b -15$
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$a - ab + 2b -25$
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$a - 2ab + 2b -5$
0%
$8a - ab + 2b -10$

Explanation

To subtract

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

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

Rearranging the terms of the given expressions, changing the sign of each term of the expression to be subtracted and adding the two expressions, we get.

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

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

$-$

$+$

$-$

$-$
------------------------------

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

Hence,

$Op-A$ is correct .

Subtract:

$5y^4-3y^3+2y^2+y-1$ from

$4y^4- 2y^3 -6y^2-y + 5$

0%
$-y^4 + y^3-8y^2-2y + 6$
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$y^4 + y^3-8y^2-2y + 4$
0%
$-y^4 + y^3-8y^2-2y - 4$
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$y^4 + y^3-8y^2+2y + 4$

Explanation

After subtracting

$5y^4-3y^3+2y^2+y-1$ from,

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

$4y^4- 2y^3- 6y^2-y + 5-(5y^4-3y^3+2y^2+y-1)$

$=4y^4- 2y^3- 6y^2-y + 5-5y^4+3y^3-2y^2-y+1$

$=-y^4 + y^3-8y^2-2y + 6$

$(5x^{2} + 6x - 3) + (2x^{2} - 7x - 9)$

0%
$7x^{2} - x - 12$
0%
$7x^{2} - 2x - 12$
0%
$7x^{2} - 3x - 12$
0%
None of the above

Explanation

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

$+\underline { 2x^{2} - 7x - 9}$

$7x^{2} - x - 12$

The total terms in

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

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

Explanation

We have a polynomial given by

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

The total number of terms is given by 4, since we have 4 terms with either an addition or subtraction operated between them.

Those 4 terms are

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

What is to be added to

$2a^3+4ab^2-5b^3-6a^2b$ to get

$3a^3-4a^2b+11ab^2+b^3$ ?

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

Explanation

$3a^3-4a^2b+11ab^2+b^3$

$\underline {-2a^3+6a^2b-4ab^2+5b^3}$

$a^3+2a^2b+7ab^2+6b^3$

What must be subtracted from

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

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

0%
$-a^2 + ab + b^3-2$
0%
$-a^2 + ab + b^2-2$
0%
$a^2 + ab + b^2-2$
0%
$a^2 + ab + b^3-2$

Explanation

Let

$X$ be subtracted from

$3a^2-6ab-3b^2$
Then,

$3a^2-6ab-3b^2-1 -X= 4a^2-7ab -4b^2 + 1$

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

$x=3a^2-6ab-3b^2-1-4a^2+7ab +4b^2 - 1$

$x=-a^2+ab+b^2-2$

How many terms are there in the expression

$5x^3+3xy+y^2$ ?

0%
$1$
0%
$2$
0%
$3$
0%
$5$

Explanation

There are 3 terms. The terms are

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

If we add two binomials

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

$7x^2+y^2$ , what will be the result?

0%
monomial
0%
binomial
0%
trinomial
0%
cannot be determined

Explanation

$(4x^2+5y^2)+(7x^2+y^2)=11x^2+6y^2$ is a binomial.

What is the measure of the third side of a triangle given that its two sides are

$a^{2} - 2a + 1$ and

$3a^{2} - 5a + 3$ and has a perimeter

$6a^{2} - 4a + 9$ ?

0%
$2a^{2} - 3a - 5$
0%
$2a^{2} + 3a - 5$
0%
$2a^{2} + 3a + 5$
0%
$2a^{2} - 3a + 5$

Explanation

Let the third side of the triangle be

$x.$

The perimeter of a triangle is equal to the sum of its sides so,

$\begin{aligned}{}x + \left( {{a^2} - 2a + 1} \right) + \left( {3{a^2} - 5a + 3} \right) &= 6{a^2} - 4a + 9\\x + \left( {{a^2} + 3{a^2}} \right) + \left( { - 2a - 5a} \right) + \left( {1 + 3} \right)& = 6{a^2} - 4a + 9\\x + 4{a^2} - 7a + 4 &= 6{a^2} - 4a + 9\\x &= \left( {6{a^2} - 4{a^2}} \right) + \left( { - 4a + 7a} \right) + \left( {9 - 4} \right)\\& = 2{a^2} + 3a + 5\end{aligned}$

Hence, option

$C$ is correct.

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

$5x^{2} - 2x - 6$
Which of the following is the sum of the two polynomials shown above?

0%
$8x^{2} - 7x - 4$
0%
$8x^{2} + 7x - 4$
0%
$8x^{4} - 7x^{2} - 4$
0%
$8x^{4} + 7x^{2} - 4$

Explanation

Given polynomials are

$3x^2-5x+2$ and

$5x^2-2x-6$

The sum of the given polynomials is given by:

$(3x^2-5x+2)+(5x^2-2x-6)$

$=(3x^2+5x^2)+(-5x-2x)+(2-6)$

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

$=8x^2-7x-4$

$P=3x-4y-8z$ ,

$Q=-10y+7x+11z$ and

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

$P-Q+R$ is equal to

0%
$13x-20y+16z$
0%
$0$
0%
$x+y+z$
0%
$2x-4y+3z$

Explanation

$P=3x-4y-8z$

$Q=-10y+7x+11z$

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

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

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

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

Simplify the polynomial and write it in standard form:

$-3({x}^{3}-{x}^{2}-2x-5)-(4{x}^{3}-7x-1)$

0%
$-7{x}^{3}+3{x}^{2}+13x+16$
0%
$7{x}^{3}+2{x}^{2}+11x+16$
0%
$-6{x}^{3}+3{x}^{2}+12x+14$
0%
$-4{x}^{3}+3{x}^{2}+11x+15$

Explanation

Solve the polynomial as follows:

$-3(x^3-x^2-2x-5)-(4x^3-7x-1)$

$=-3x^3+3x^2+6x+15-4x^3+7x+1$

$=-7x^3+3x^2+13x+16$

The sum of three expressions is

${ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }$ . If two of them are

$4{ x }^{ 2 }-5{ y }^{ 2 }+3{ z }^{ 2 }$ and

$-3{ x }^{ 2 }+4{ y }^{ 2 }+2{ z }^{ 2 }$ , the third expression is

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

Explanation

Add two expressions

$\Rightarrow \left ( 4x^{2}-5y^{2}+3z^{2} \right )+(-3x^{2}+4y^{2}+2z^{2})$

$=\left ( 4x^{2}-5y^{2}+3z^{2}-3x^{2}+4y^{2}+2z^{2} \right )$

$=4x^{2}-3x^{2}-5y^{2}+4y^{2}+3z^{2}+2z^{2}$

$=x^{2}-y^{2}+5z^{2}$

But

$x^{2}-y^{2}+5z^{2}+P(x)=x^{2}+y^{2}+z^{2}$ , where

$P(x)$ is the third expression

$\therefore P(x)=\left ( x^{2}+y^{2}+z^{2} \right )-\left ( x^{2}-y^{2}+5z^{2} \right )$

$=x^{2}+y^{2}+z^{2}-x^{2}+y^{2}-5z^{2}=2y^{2}-4z^{2}$

Add

$(12x^2+4x+5y)$ and

$(3x^2-2x+3y)$ .

0%
$2x^2-2x+8y$
0%
$2x^2+15x+8y$
0%
$15x^2-2x+8y$
0%
$15x^2+2x+8y$

Explanation

Adding

$(12x^2+4x+5y)+(3x^2-2x+3y)$

$\Rightarrow 12x^2+4x+5y+3x^2-2x+3y$

$\Rightarrow 15x^2+2x+8y$

$x+y-\left( z-x-\left[ y+z-\left( x+y-\left\{ z+x-\left( y+z+x \right) \right\} \right) \right] \right)$ is equal to

0%
$3x$
0%
$2y$
0%
$x$
0%
$0$

Explanation

$x+y-\left (z-x-\left [ y+z-(x+y-\left \{ z+x-(y+z+x) \right \}) \right ] \right )$

$=x+y-\left ( z-x-\left [ y+z-(x+y-\left \{ z+x-y-z-x \right \}) \right ] \right )$

$=x+y-\left (z-x-\left [ y+z-(x+y-z-x+y+z+x) \right ] \right )$

$=x+y-\left ( z-x-\left [ y+z-x-y+z+x-y-z-x \right ] \right )$

$=x+y-\left (z-x-y-z+x+y-z-x+y+z+x \right )$

$=x+y-z+x+y+z-x-y+z+x-y-z-x$

$=x$

Simplify:

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

0%
$5{ x }^{ 2 }-3x+5$
0%
$6{ x }^{ 2 }-4x+1$
0%
$8{ x }^{ 2 }-8x+1$
0%
$8{ x }^{ 2 }-9x+5$

Explanation

The value of

$(5{x}^{2}-3x+2)-(3{x}^{2}+5x-1)+(6{x}^{2}-2)$

$= (5{x}^{2}-3{x}^{2}+6{x}^{2})+(-3x-5x)+(2+1-2)$

$= 8{x}^{2}-8x+1$

What is the simplified result of following the steps in order?

$(1)$ add

$5y$ to

$2x$

$(2)$ multiply the sum by

$3$

$(3)$ subtract x

$+$ y from the product

0%
$5x+14y$
0%
$5x+16y$
0%
$5x+5y$
0%
$6x+4y$
0%
$3x+12y$

$\left( 4p{ x }^{ 2 }+5{ q }^{ 2 }y-9rz \right) -\left( -3{ q }^{ 2 }y+7p{ x }^{ 2 }-rz \right) =$

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

Explanation

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

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

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

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

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 5 - MCQExams.com

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Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 5 - MCQExams.com

0:0:1

Practice Class 7 Maths Quiz Questions and Answers

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