Subtract the sum of $$\displaystyle { 9b }^{ 2 }+{ 3c }^{ 2 }$$ and $$\displaystyle { 2b }^{ 2 }+bc+{ 2c }^{ 2 }$$ from the sum of $$\displaystyle { 2b }^{ 2 }-2bc-{ c }^{ 2 }$$ and $$\displaystyle { c }^{ 2 }+2bc-{ b }^{ 2 }$$.

$$\displaystyle { 13b }^{ 2 }-5bc+{ 5c }^{ 2 }$$

$$\displaystyle { -10b }^{ 2 }-5bc-{ 5c }^{ 2 }$$

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

$$\displaystyle { 2a }^{ 3 }+{ 7a }^{ 2 }+6a-10$$

$$\displaystyle { 2a }^{ 3 }+{ 7a }^{ 2 }+12a-10$$

$$\displaystyle { 2a }^{ 3 }-{ 4a }^{ 2 }+6a-10$$

What should be subtracted from $$\displaystyle { x }^{ 2 }-4xy-{ y }^{ 2 }$$ to get $$1$$?

$$\displaystyle { x }^{ 2 }-4xy-{ y }^{ 2 }-1$$

$$\displaystyle { -x }^{ 2 }+4xy+{ y }^{ 2 }+1$$

$$\displaystyle { -x }^{ 2 }+4xy+{ y }^{ 2 }-1$$

Subtract $$\left( j{ k }^{ 2 }+2{ j }^{ 2 }k+5{ j }^{ 2 } \right) $$ from $$\displaystyle \left( 5j{ k }^{ 2 }+5{ j }^{ 2 }-2{ j }^{ 2 }k \right) $$.

$$\displaystyle 4j{ k }^{ 2 }-4{ j }^{ 2 }k$$

$$\displaystyle -(4k{ j }^{ 2 }-4jk^2)$$

$$\displaystyle 5{ j }^{ 2 }{ k }^{ 4 }-10{ j }^{ 4 }k$$

$$\displaystyle 8{ j }^{ 2 }{ k }^{ 3 }+7{ j }^{ 2 }k-5{ j }^{ 2 }$$

Simplify the expression: $$\displaystyle { t }^{ 2 }-59t+54-82{ t }^{ 2 }+60t$$

$$\displaystyle -81{ t }^{ 2 }+{ t }+54$$

$$\displaystyle -26{ t }^{ 2 }$$

$$\displaystyle -26{ t }^{ 6 }$$

$$\displaystyle -81{ t }^{ 4 }+{ t }^{ 2 }+54$$

MCQ Questions for Class 7 Maths Algebraic Expressions Quiz 3

5 a - {3 a - (4 - a) - 4}

$\displaystyle 5a-\{ { 3a-(4-a)-4\} }$ is equal to

0%
$\displaystyle 3a+8$
0%
$\displaystyle a-8$
0%
$\displaystyle a+8$
0%
$\displaystyle 3a-8$

Explanation

Given,

5 a - {3 a - (4 - a) - 4}

$\displaystyle 5a-\{ { 3a-(4-a)-4\} }$

= 5 a - {3 a - 4 + a - 4}

$=\displaystyle 5a-\left\{ 3a-4+a-4 \right\}$

= 5 a - {4 a - 8}

$=\displaystyle 5a-\left\{ 4a-8 \right\}$

= 5 a - 4 a + 8

$\displaystyle= 5a-4a+8$

= a + 8

$=\displaystyle a+8$ , which is simplified form.

a^{2} - (- a^{2})

$\displaystyle { a }^{ 2 }-\left( -{ a }^{ 2 } \right)$ is equal to

0%
$\displaystyle { 2a }^{ 2 }$
0%
$\displaystyle { a }^{ 2 }$
0%
0
0%
$\displaystyle { -2a }^{ 2 }$

Explanation

Given,

a^{2} - ({- a}^{2}) = a^{2} + a^{2} = 2 a^{2}

$\displaystyle { a }^{ 2 }-\left( { -a }^{ 2 } \right) ={ a }^{ 2 }+{ a }^{ 2 }=2a^2$ .
Hence simplified form of the given expression is

2 a^{2}

$2a^2$ .

Subtract the sum of

{9 b}^{2} + {3 c}^{2}

$\displaystyle { 9b }^{ 2 }+{ 3c }^{ 2 }$ and

{2 b}^{2} + b c + {2 c}^{2}

$\displaystyle { 2b }^{ 2 }+bc+{ 2c }^{ 2 }$ from the sum of

{2 b}^{2} - 2 b c - c^{2}

$\displaystyle { 2b }^{ 2 }-2bc-{ c }^{ 2 }$ and

c^{2} + 2 b c - b^{2}

$\displaystyle { c }^{ 2 }+2bc-{ b }^{ 2 }$ .

0%
$\displaystyle { 13b }^{ 2 }-5bc+{ 5c }^{ 2 }$
0%
$\displaystyle { -10b }^{ 2 }-bc-{ 5c }^{ 2 }$
0%
$\displaystyle { 10b }^{ 2 }+bc-{ 5c }^{ 2 }$
0%
$\displaystyle { -10b }^{ 2 }-5bc-{ 5c }^{ 2 }$

Explanation

According to the question,

{({2 b}^{2} - 2 b c - c^{2}) + (c^{2} + 2 b c - b^{2})} - {({9 b}^{2} + {3 c}^{2}) + ({2 b}^{2} + b c + {2 c}^{2})}

$\displaystyle \left\{ \left( { 2b }^{ 2 }-2bc-{ c }^{ 2 } \right) +\left( { c }^{ 2 }+2bc-{ b }^{ 2 } \right) \right\} -\left\{ \left( { 9b }^{ 2 }+{ 3c }^{ 2 } \right) +\left( { 2b }^{ 2 }+bc + { 2c }^{ 2 } \right) \right\}$
=

{({2 b}^{2} - b^{2}) + (- 2 b c + 2 b c) + (c^{2} - c^{2})} - {({9 b}^{2} + {2 b}^{2}) + ({3 c}^{2} + {2 c}^{2}) + b c}

$\displaystyle \left\{ \left( { 2b }^{ 2 }-{ b }^{ 2 } \right) +\left( -2bc+2bc \right) +\left( { c }^{ 2 }-{ c }^{ 2 } \right) \right\} -\left\{ \left( { 9b }^{ 2 }+{ 2b }^{ 2 } \right) +\left( { 3c }^{ 2 }+{ 2c }^{ 2 } \right) +bc \right\}$

= (b^{2}) - ({11 b}^{2} + {5 c}^{2} + b c)

$\displaystyle =\left( { b }^{ 2 } \right) -\left( { 11b }^{ 2 }+{ 5c }^{ 2 }+bc \right)$

= b^{2} - {11 b}^{2} - {5 c}^{2} - b c

$\displaystyle ={ b }^{ 2 }-{ 11b }^{ 2 }-{ 5c }^{ 2 }-bc$

= - {10 b}^{2} - b c - {5 c}^{2}

$\displaystyle =-{ 10b }^{ 2 }-bc-{ 5c }^{ 2 }$ .

Simplify:

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

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

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

Given expression is:

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

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

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

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

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

$\displaystyle ={ 2a }^{ 3 }-{ 4a }^{ 2 }+12a-10$
Hence simplified form of the given expression is

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

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

Add the following:

a b - b c, b c - c a, c a - a b

$ab-bc,bc-ca,ca -ab$

0%
0
0%
1
0%
2
0%
-1

Explanation

Adding them

= a d - b c + b c - c a + c a - a b

$=ad-bc+bc-ca+ca-ab$

= 0

$=0$

What should be subtracted from

x^{2} - 4 x y - y^{2}

$\displaystyle { x }^{ 2 }-4xy-{ y }^{ 2 }$ to get

1

$1$ ?

0%
$\displaystyle { -x }^{ 2 }-4xy+{ y }^{ 2 }$
0%
$\displaystyle { -x }^{ 2 }+4xy+{ y }^{ 2 }+1$
0%
$\displaystyle { x }^{ 2 }-4xy-{ y }^{ 2 }-1$
0%
$\displaystyle { -x }^{ 2 }+4xy+{ y }^{ 2 }-1$

Explanation

Let the algebraic expression that has to be subtracted be

a

$a$ . Then,

x^{2} - 4 x y - y^{2} - a = 1

$\displaystyle { x }^{ 2 }-4xy-{ y }^{ 2 }-a=1$

a = x^{2} - 4 x y - y^{2} - 1

$\displaystyle a={ x }^{ 2 }-4xy-{ y }^{ 2 }-1$
which required algebraic expression.

Simplify:

(a^{4} - 3 a^{2} + 6 a - 8) - (a^{4} - 10 a + 4 a^{2} + 8)

$(a^4-3a^2 + 6a - 8) - (a^4 - 10a + 4a^2 + 8)$

0%
$-7a^2- 16a + 16$
0%
$7a^2- 16a + 16$
0%
$-7a^2 + 16a - 16$
0%
$7a^2 + 16a - 16$

Explanation

(a^{4} - 3 a^{2} + 6 a - 8) - (a^{4} - 10 a + 4 a^{2} + 8)

$(a^4-3a^2+6a-8)-(a^4-10a+4a^2+8)$

= a^{4} - 3 a^{2} + 6 a - 8 - a^{4} + 10 a - 4 a^{2} - 8

$=a^4-3a^2+6a-8-a^4+10a-4a^2-8$

= a^{4} - a^{4} - 3 a^{2} - 4 a^{2} + 6 a + 10 a - 8 - 8

$=a^4-a^4-3a^2-4a^2+6a+10a-8-8$

= - 7 a^{2} + 16 a - 16

$=-7a^2+16a-16$

Add the following:

3 a - 4 b + 4 c, 2 a + 3 b - 8 c, a - 6 b + c

$3a - 4b + 4c, 2a + 3b - 8c, a - 6b + c$

0%
$6a-7b-3c$
0%
$6a-7b+3c$
0%
$6a+7b-3c$
0%
$5a-7b-3c$

Explanation

After adding we get:

(3 a - 4 b + 4 c) + (2 a + 3 b - 8 c) + (a - 6 b + c)

$(3a - 4b + 4c) + (2a + 3b - 8c) +( a - 6b + c)$

= 3 a + 2 a + a - 4 b + 3 b - 6 b + 4 c - 8 c + c

$=3a + 2a+ a- 4b + 3b- 6b+ 4c- 8c + c$

= 6 a - 7 b - 3 c

$=6a - 7b - 3c$

Subtract

3 a^{2} b

$3a^2b$ from

- 5 a^{2} b

$-5a^2b$

0%
$-2a^2b$
0%
$2a^2b$
0%
$-8a^2b$
0%
$0$

Explanation

After Subtracting

3 a^{2} b

$3a^2b$ from

- 5 a^{2} b

$-5a^2b$ we get,

= - 5 a^{2} b - 3 a^{2} b

$=-5a^2b-3a^2b$

= - 8 a^{2} b

$=-8a^2b$

By how much is x

^{4}

$^4$ + 4x

^{2}

$^2$ y

^{2}

$^2$ + y

^{4}

$^4$ more than x

^{4}

$^4$ - 8x

^{2}

$^2$ y

^{2}

$^2$ + y

^{4}

$^4$ ?

0%
-12x $^2$ y $^2$
0%
12x $^2$ y $^2$
0%
2x $^4$ + 2y $^4$
0%
None of these

What must be added to

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

$(x^3 + 3x - 8)$ to get

(3 x^{3} + x^{2} + 6)

$(3x^3 + x^2 + 6)$ ?

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

Explanation

Let,

Y

$Y$ is added to

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

$(x^3 + 3x - 8)$ to get

(3 x^{3} + x^{2} + 6)

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

So,

(x^{3} + 3 x - 8) + Y = 3 x^{3} + x^{2} + 6

$(x^3 + 3x - 8) +Y= 3x^3 + x^2 + 6$

\Rightarrow Y = (3 x^{3} + x^{2} + 6) - (x^{3} + 3 x - 8)

$\Rightarrow Y=(3x^3 + x^2 + 6) -(x^3 + 3x - 8)$

\Rightarrow Y = 3 x^{3} + x^{2} + 6 - x^{3} - 3 x + 8

$\Rightarrow Y=3x^3 + x^2 + 6 -x^3 - 3x + 8$

\Rightarrow Y = 2 x^{3} + x^{2} - 3 x + 14

$\Rightarrow Y=2x^3+x^2-3x+14$

So, option

A

$A$ is correct.

If we add two monomials, what will be the result?

0%
Monomial
0%
Binomial
0%
Polynomial
0%
Either $A$ or $B$

Explanation

If two monomials are added we will get a result which is addition of two terms.

So in most of the cases, the result will be a binomial but in other cases, when both terms differ by coefficients, their addition will give another monomial.
For eg.

3 x

$3x$ ,

4 y

$4y$ when added, gives

3 x + 4 y

$3x+4y$ .

However,

5 x^{2} + 7 x^{2}

$5x^2+7x^2$ when added, gives

5 x^{2} + 7 x^{2} = 12 x^{2}

$5x^2+7x^2=12x^2$ .

Add the following:

5 x - 8 y + 2 z, 3 z - 4 y - 2 x, 6 y - z - x

$5x - 8y + 2z, 3z - 4y - 2x,6y - z - x$

0%
$2x-6y-4z$
0%
$2x-6y+4z$
0%
$2x+6y+4z$
0%
$-2x-6y+4z$

Explanation

After adding we get:

(5 x - 8 y + 2 z) + (3 z - 4 y - 2 x) + (6 y - z - x)

$(5x - 8y + 2z)+ (3z - 4y - 2x)+(6y - z - x)$

= 5 x - 2 x - x - 8 y - 4 y + 6 y + 2 z + 3 z - z

$=5x- 2x - x- 8y- 4y+6y + 2z+ 3z- z$

= 2 x - 6 y + 4 z

$=2x - 6y + 4z$

How many terms are there in the algebraic expression

7 x^{3} + 2 x y + z - 7 y

$7x^3+2xy+z-7y$ ?

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

There are 4 terms. The terms are

7 x^{3}, 2 x y, z, 7 y

$7x^3, 2xy, z, 7y$

Simplify the expression

(2 b^{3} c^{2} + b^{2} c - 4 b c) - (b^{3} c^{2} - b^{2} c - 4 b c)

$(2b^3c^2+b^2c-4bc)-(b^3c^2-b^2c-4bc)$ .

0%
0
0%
$b^3c^2$
0%
$b^3c^2+2b^2c$
0%
$b^3c^2+2b^2c-8bc$

Explanation

On simplification, we get

(2 b^{3} c^{2} + b^{2} c - 4 b c) - (b^{3} c^{2} - b^{2} c - 4 b c)

$(2b^3c^2+b^2c-4bc)-(b^3c^2-b^2c-4bc)$

\Rightarrow 2 b^{3} c^{2} + b^{2} c - 4 b c - b^{3} c^{2} + b^{2} c + 4 b c

$\Rightarrow 2b^3c^2+b^2c-4bc-b^3c^2+b^2c+4bc$

\Rightarrow b^{3} c^{2} + 2 b^{2} c

$\Rightarrow b^3c^2+2b^2c$

If two unlike terms are added , the result will be a

0%
monomial
0%
binomial
0%
trinomial
0%
polynomial

Simplify

(2 x^{2} + 4 x + 8) - (2 x^{2} - 4 x + 7)

$(2x^2+4x+8)-(2x^2-4x+7)$

0%
$4x^2+8x+15$
0%
$2x^2+x+1$
0%
$8x+1$
0%
$8x+15$

Explanation

On simplifying, we get

(2 x^{2} + 4 x + 8) - (2 x^{2} - 4 x + 7)

$(2x^2+4x+8)-(2x^2-4x+7)$

\Rightarrow 2 x^{2} + 4 x + 8 - 2 x^{2} + 4 x - 7

$\Rightarrow 2x^2+4x+8-2x^2+4x-7$

\Rightarrow 8 x + 1

$\Rightarrow 8x+1$

Subtract

(j k^{2} + 2 j^{2} k + 5 j^{2})

$\left( j{ k }^{ 2 }+2{ j }^{ 2 }k+5{ j }^{ 2 } \right)$ from

(5 j k^{2} + 5 j^{2} - 2 j^{2} k)

$\displaystyle \left( 5j{ k }^{ 2 }+5{ j }^{ 2 }-2{ j }^{ 2 }k \right)$ .

0%
$\displaystyle -(4k{ j }^{ 2 }-4jk^2)$
0%
$\displaystyle 4j{ k }^{ 2 }-4{ j }^{ 2 }k$
0%
$\displaystyle 5{ j }^{ 2 }{ k }^{ 4 }-10{ j }^{ 4 }k$
0%
$\displaystyle 8{ j }^{ 2 }{ k }^{ 3 }+7{ j }^{ 2 }k-5{ j }^{ 2 }$

Explanation

(5 j k^{2} + 5 j^{2} - 2 j^{2} k) - (j k^{2} + 2 j^{2} k + 5 j^{2})

$(5jk^2 + 5j^2 - 2j^2k) - (jk^2 + 2j^2k + 5j^2)$

= (5 j k^{2} - j k^{2}) + (5 j^{2} - 5 j^{2}) + (- 2 j^{2} k - 2 j^{2} k)

$= (5jk^2 - jk^2) + (5j^2 - 5j^2) + (-2j^2k - 2j^2k)$

= 4 j k^{2} - 4 j^{2} k = - (4 j^{2} k - 4 k^{2} j)

$= 4jk^2 - 4j^2k=-(4j^2k-4k^2j)$

5 x^{3} + 5 x^{2} + 4 x + 3 + 4 x^{2} + 5 x =

$5x^{3} + 5x^{2} + 4x + 3 + 4x^{2} + 5x =$

0%
$9x^{3} + 4x^{2} + 4x + 5$
0%
$5x^{3} + 5x^{2} + 9x + 4$
0%
$5x^{3} + 9x^{2} + 9x + 3$
0%
None of the above

Explanation

5 x^{3} + 5 x^{2} + 4 x + 3 + 4 x^{2} + 5 x

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

= 5 x^{3} + 9 x^{2} + 9 x + 3

$=5x^{3} + 9x^{2} + 9x + 3$

If we add

7 x

$7x$ and

5 y^{2} + z

$5y^2+z$ , what will be the result?

0%
binomial
0%
trinomial
0%
polynimial
0%
can"t be determind

Explanation

(7 x) + (5 y^{2} + z) = 7 x + 5 y^{2} + z

$(7x)+(5y^2+z) =7x+5y^2+z$ is a trinomial

If we add

3 + 7 x

$3+7x$ and

11 x

$11x$ , what will be the result?

0%
monomial
0%
binominal
0%
trinimial
0%
can"t be determined

Explanation

(3 + 7 x) + 11 x = 3 + 18 x

$(3+7x)+11x=3+18x$ is a binomial.

Find an equivalent simplified expression for

2 (4 x + 7) - 3 (2 x - 4)

$\displaystyle 2\left( 4x+7 \right) -3\left( 2x-4 \right)$ .

0%
$\displaystyle x+2$
0%
$\displaystyle 2x+2$
0%
$\displaystyle 2x+26$
0%
$\displaystyle 3x+10$
0%
$\displaystyle 3x+11$

Explanation

The value of

2 (4 x + 7) - 3 (2 x - 4)

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

\Rightarrow 8 x + 14 - 6 x + 12

$\Rightarrow 8x+14-6x+12$

\Rightarrow 2 x + 26

$\Rightarrow 2x+26$

The expression

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

$\left( { x }^{ 2 }y-3{ y }^{ 2 }+5x{ y }^{ 2 } \right) -\left( -{ x }^{ 2 }y+3x{ y }^{ 2 }-3{ y }^{ 2 } \right)$ is equivalent to which of the following expression?

0%
$4{ x }^{ 2 }{ y }^{ 2 }$
0%
$8x{ y }^{ 2 }-6{ y }^{ 2 }$
0%
$2{ x }^{ 2 }y+2x{ y }^{ 2 }$
0%
$2{ x }^{ 2 }y+8x{ y }^{ 2 }-6{ y }^{ 2 }$

Explanation

We have,

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

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

= x^{2} y - 3 y^{2} + 5 x y^{2} + x^{2} y - 3 x y^{2} + 3 y^{2}

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

= 2 x^{2} y + 2 x y^{2}

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

Therefore, option C is correct.

Simplify the expression:

t^{2} - 59 t + 54 - 82 t^{2} + 60 t

$\displaystyle { t }^{ 2 }-59t+54-82{ t }^{ 2 }+60t$

0%
$\displaystyle -26{ t }^{ 2 }$
0%
$\displaystyle -26{ t }^{ 6 }$
0%
$\displaystyle -81{ t }^{ 4 }+{ t }^{ 2 }+54$
0%
$\displaystyle -81{ t }^{ 2 }+{ t }+54$

Explanation

t^{2} - 59 t + 45 - 82 t^{2} + 60 t

$t^2-59t+45-82t^2+60t$

\Rightarrow t^{2} - 82 t^{2} + 60 t - 59 t + 54

$\Rightarrow t^2-82t^2+60t-59t+54$

\Rightarrow - 81 t^{2} + t + 54

$\Rightarrow -81t^2+t+54$

(a + 2 b + 3 c) - (4 a + 6 b - 5 c)

$\left( a+2b+3c \right) -\left( 4a+6b-5c \right)$ is equivalent to:

0%
$-4a-8b-2c$
0%
$-4a-4b+8c$
0%
$-3a+8b-2c$
0%
$-3a-4b-2c$
0%
$-3a-4b+8c$

Explanation

The value of

(a + 2 b + 3 c) - (4 a + 6 b - 5 c)

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

\Rightarrow a + 2 b + 3 c - 4 a - 6 b + 5 c

$\Rightarrow a+2b+3c-4a-6b+5c$

\Rightarrow - 3 a - 4 b + 8 c

$\Rightarrow -3a-4b+8c$

Simplify:

(- 3 p q^{2} + 4 p q) - (7 p q - 9 p q^{2})

$\left( -3p{ q }^{ 2 }+4pq \right) -\left( 7pq-9p{ q }^{ 2 } \right)$

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

Explanation

(- 3 p q^{2} + 4 p q) - (7 p q - 9 p q^{2})

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

= - 3 p q^{2} + 4 p q - 7 p q + 9 p q^{2}

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

= - 3 p q^{2} + 9 p q^{2} + 4 p q - 7 p q

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

= 6 p q^{2} - 3 p q

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

Simplify :

(- 8 p s t + 3 p r q - 14) - (7 p r q - 20 + 2 p s t) =

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

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

Explanation

(- 8 p s t + 3 p r q - 14) - (7 p r q - 20 + 2 p s t)

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

= - 8 p s t + 3 p r q - 14 - 7 p r q + 20 - 2 p s t

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

= - 8 p s t - 2 p s t + 3 p r q - 7 p r q - 14 + 20

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

= - 10 p s t - 4 p r q + 6

$=-10pst-4prq+6$

= 6 - 10 p s t - 4 p r q

$=6-10pst-4prq$

What must be subtracted from

3 x^{2} + 4 y^{2} - 5

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

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

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

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

Explanation

Subtract both expression to get the answer

\Rightarrow (3 x^{2} + 4 y^{2} - 5) - (2 x^{2} - 3 y^{2} + 5)

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

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

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

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

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

= x^{2} + 7 y^{2} - 10

$=x^{2}+7y^{2}-10$

The sum of

3

$3$ expression is

8+13a+7a^

$8+13a+7a^$ Two of them are

2 a^{2} + 3 a + 2

$2a^2+3a+2$ and

3 a^{2} - 4 a + 1

$3a^2-4a+1$ . Find the third expression.

0%
$a^2+a+5$
0%
$2a^2+a+5$
0%
$2a^3+14a+5$
0%
$2a^2+14a+5$

Explanation

Expression be x

\Rightarrow x + (2 a^{2} + 3 a + 2) + (3 a^{2} - 4 a + 1)

$\Rightarrow x+(2a^2+3a+2)+(3a^2-4a+1)$

= 8 + 13 a + 7 a^{2}

$=8+13a+7a^2$

= x + 5 a^{2} - a + 3

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

= 8 + 13 a + 7 a^{2}

$=8+13a+7a^2$

\Rightarrow x = 7 a^{2} + 13 a + 8 - 5 a^{2} + a - 3

$\Rightarrow x=7a^2+13a+8-5a^2+a-3$

x = 2 a^{2} + 14 a + 5

$x=2a^2+14a+5$ .

Simplify :

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

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

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

Explanation

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

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

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

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

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

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

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 3 - MCQExams.com

0:0:2

Practice Class 7 Maths Quiz Questions and Answers

CBSE Questions for Class 7 Maths Algebraic Expressions Quiz 3 - MCQExams.com

0:0:2

Practice Class 7 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.