M.C.Q. [1 Marks Each]

MCQ 11 Mark

The value of $\left(-27 x^2 y^2\right) \div(-9 x y)$ is:

A
$3xy$
B
$-3xy$
C
$-3x$
✓
$3x$

Answer

Correct option: D.

$3x$

We have, $\left(-27 x^2 y^2\right) \div(-9 x y)$
$=\frac{-27\text{x}^2\text{y}^2}{-9\text{xy}}$
$=\frac{27\times\text{x}\times\text{x}\times\text{y}}{9\times\text{x}\times\text{y}}$
$=\frac{27}{9}\text{x}$
$=3\text{x}$

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MCQ 21 Mark

Product of $6a^2- 7b + 5ab$ and $2ab$ is:

A
$ 12 a^3 b-14 a b^2+10 a b $
✓
$ 12 a^3 b-14 a b^2+10 a^2 b^2 $
C
$ 6 a^2-7 b+7 a b $
D
$ 12 a^2 b-7 a b^2+10 a b $

Answer

Correct option: B.

$ 12 a^3 b-14 a b^2+10 a^2 b^2 $

Required product $= 2ab \times (6a^2- 7b + 5ab)$
This is the product of a trinomial by a monomial,
so we multiply monomial with each term of the trinomial.
$2ab \times (6a^2- 7b + 5ab) $
$= 2ab \times 6a^2+ 2ab(-7b) + 2ab \times 5ab$
$= 12 a^3 b-14 a b^2+10 a^2 b^2 $

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MCQ 31 Mark

Sum of $a - b + ab, b + c - bc$ and $c - a - ac$ is:

✓
$2c + ab - ac - bc$
B
$2c - ab - ac - bc$
C
$2c + ab + ac + bc$
D
$2c - ab + ac + bc$

Answer

Correct option: A.

$2c + ab - ac - bc$

Required sum $= (a - b + ab) + (b + c - bc) + (c - a - ac)$
$= a - b + ab + b + c - bc + c - a - ac$
$= 2c + ab - ac - bc$

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MCQ 41 Mark

Coefficient of ? in the term $-\frac{\text{y}}{3}$ is:

A
$-1$
B
$-3$
✓
$-\frac{1}{3}$
D
$\frac{1}{3}$

Answer

Correct option: C.

$-\frac{1}{3}$

We can write $-\frac{\text{y}}{3}$ as $-\frac{1}{3}$xy
so, the cofficient of y is $-\frac{1}{3}$

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MCQ 51 Mark

Number of factors of $(a + b) 2$ is:

A
$4$
B
$3$
✓
$2$
D
$1$

Answer

Correct option: C.

$2$

We can write $(a + b)^2$ as, $(a + b)(a + b)$ and this cannot be factorised further.
Hence, number of factors of $(a + b)^2$ is $2.$

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MCQ 61 Mark

Square of $9x - 7xy$ is:

A
$81 x^2+49 x^2 y^2$
B
$81 x^2-49 x^2 y^2$
✓
$81 x^2+49 x^2 y^2-126 x^2 y$
D
$81 x^2+49 x^2 y^2-63 x^2 y$

Answer

Correct option: C.

$81 x^2+49 x^2 y^2-126 x^2 y$

Square of $(9 x-7 x y)=(9 x-7 x)^2$
Comparing with $(\mathrm{a}-\mathrm{b})^2$
we get $a = 9x$ and $b = 7xy$
$(9 x-7 x y)^2$
$=(9 x)^2-2 \cdot 9 x \cdot 7 x y+(7 x y)^2$
$=81 x^2-126 x^2 y+49 x^2 y^2$
$=81 x^2+49 x^2 y^2-126 x^2 y$

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MCQ 71 Mark

The value of $(2x^2+ 4) ÷ 2$ is:

A
$ 2 x^2+2 $
✓
$ x^2+2 $
C
$ x^2+4 $
D
$ 2 x^2+4 $

Answer

Correct option: B.

$ x^2+2 $

We have,
$(2x^2+ 4) ÷ 2$
$=\frac{2\text{x}^2+4}{2}$
$=\frac{2(\text{x}^2+4)}{2}$
$=\text{x}^2+2$

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MCQ 81 Mark

Square of $3x - 4y$ is:

A
$9 x^2-16 y^2$
B
$6 x^2-8 y^2$
C
$9 x^2+16 y^2+24 x y$
✓
$9 x^2+16 y^2-24 x y$

Answer

Correct option: D.

$9 x^2+16 y^2-24 x y$

Square of $(3 x-4 y)$ will be $(3 x-4 y)^2$
comparing $(3 x-4 y)^2$ with $(a-b)^2$
We get $a = 3x$ and $b = 4y$
now, using identity,$(a-b)^2=a^2-2 a b+b^2$
$(3 x-4 y)^2=(3 x)^2-2 \cdot 3 x \cdot 4 y+(4 y)^2$
$=9 x^2+16 y^2-24 x y$

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MCQ 91 Mark

On dividing $p(4p^2- 16)$ by $4p(p - 2)$, we get

A
$2p + 4$
B
$2p - 4$
✓
$p + 2$
D
$p - 2$

Answer

Correct option: C.

$p + 2$

We have,
$\frac{\text{p}(4\text{p}^2-16)}{4\text{p}(\text{p}-2)}=\frac{\text{p}\big[(2\text{p})^2-4^2\big]}{4\text{p}(\text{p}-2)}$
$=\frac{(2\text{p}-4)(2\text{p}+4)}{4(\text{p}-2)}$
$=\frac{2(\text{p}-2).2(\text{p}+2)}{4(\text{p}-2)}$
$=\frac{4(\text{p}-2)(\text{p}+2)}{4(\text{p}-2)}$
$=\text{p}+2$

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MCQ 101 Mark

The value of $(a + b)^2- (a - b)^2$ is:

✓
$4ab$
B
$-4ab$
C
$2a^2+ 2b^2$
D
$2a^2- 2b^2$

Answer

Correct option: A.

$4ab$

We have,
$ (a+b)^2-(a-b)^2$
$=a^2+b^2+2 a b-\left(a^2+b^2-2 a b\right) $
$ =a^2+b^2+2 a b-a^2-b^2+2 a b$
$=a^2-a^2+b^2-b^2+2 a b+2 a b $
$= 2ab + 2ab$
$= 4ab$

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MCQ 111 Mark

Like term as $4m^3n^2$ is:

✓
$ 4 m^2 n^2 $
B
$ -6 m^3 n^2 $
C
$ 6 p m^3 n^2 $
D
$ 4 m^3 n $

Answer

Correct option: A.

$ 4 m^2 n^2 $

We knoe that, the like terms contain the same literal factor. so, the like as $4m^3n^2, -6m^3n^2$, as it contains the same literal factor $m^3n^2$.

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MCQ 121 Mark

The value of $(a+b)^2+(a-b)^2$ is:

A
$2 a+2 b$
B
$2 a-2 b$
✓
$2 a^2+2 b^2$
D
$2 a^2-2 b^2$

Answer

Correct option: C.

$2 a^2+2 b^2$

We have,
$(a+b)^2+(a-b)^2$
$=\left(a^2+b^2+2 a b\right)+\left(a^2+b^2-2 a b\right) $
$=\left(a^2+a^2\right)+\left(b^2+b^2\right)+(2 a b-2 a b) $
$ =2 a^2+2 b^2$

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MCQ 131 Mark

The factorised form of $3x - 24$ is:

A
$3x \times 24$
✓
$3(x - 8)$
C
$24(x - 3)$
D
$3(x - 12)$

Answer

Correct option: B.

$3(x - 8)$

We have,
$3x - 24 = 3 × x - 3 × 8 = 3(x - 8)$
[taking $3$ as common]

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MCQ 141 Mark

The common factor of $3ab$ and $2cd$ is:

✓
$1$
B
$-1$
C
$a$
D
$c$

Answer

Correct option: A.

$1$

We have, monomials $3ab$ and $2cd$ Now, $3ab = 3 \times a \times b$ and $2cd = 2 \times c \times d$ Observing the monomials, we see that, there is no common factor (neither numerical nor literal) between them except $1.$

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MCQ 151 Mark

Product of the following monomials $4 p,-7 q^3,-7 p q$ is:

✓
$196 p^2 q^4$
B
$196 p q^4$
C
$-196 p^2 q^4$
D
$196 \mathrm{p}^2 \mathrm{q}^3$

Answer

Correct option: A.

$196 p^2 q^4$

Required Product $=4 p \times\left(-7 q^3\right) \times(-7 p q)$
$=4 \times(-7) \times(-7) p \times q^3 \times p q $
$=196 p^2 q^4$

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MCQ 161 Mark

Which of the following is a binomial?

A
$ 7 \times a+a$
B
$ 6 a^2+7 b+2 c $
C
$ 4 a \times 3 b \times 2 c $
✓
$ 6\left(a^2+b\right) $

Answer

Correct option: D.

$ 6\left(a^2+b\right) $

Binomials are algebraic consisting of two unlike terms.
$a. 7 \times a + a = 7a + a = 8a\ ($monomial$)$
$b. 6a^2+ 7b + 2c \ ($trinomial$)$
$c. 4a \times 3b \times 2c\ ($monomial$)$
$d. 6(a^2+ b) = 6a^2+ 6b \ ($binomial$)$

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MCQ 171 Mark

The value of $(3x^3+ 9x^2+ 27x) ÷ 3x$ is:

A
$x^2+9+27 x$
B
$3 x^3+3 x^2+27 x$
C
$3 x^3+9 x^2+9$
✓
$x^2+3 x+9$

Answer

Correct option: D.

$x^2+3 x+9$

$(3\text{x}^2+9\text{x}^2+27\text{x})$
$=\frac{3\text{x}^3+9\text{x}^2+27\text{x}}{3\text{x}}$
$=\frac{3\text{x}^3}{3\text{x}}+\frac{9\text{x}^2}{3\text{x}}+\frac{27\text{x}}{3\text{x}}$
$=\text{x}^2+3\text{x}+9$

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MCQ 181 Mark

On dividing $57p^2qr$ by $114pq,$ we get

A
$\frac{1}{4}\text{pr}$
B
$\frac{3}{4}\text{pr}$
✓
$\frac{1}{2}\text{pr}$
D
$2\text{pr}$

Answer

Correct option: C.

$\frac{1}{2}\text{pr}$

Required value = $\frac{57\text{p}^2\text{qr}}{114\text{pq}}$
$=\frac{57\times\text{p}\times\text{p}\times\text{q}\times\text{r}}{114\times\text{p}\times\text{q}}$
$=\frac{57}{114}\text{pr}$
$=\frac{1}{2}\text{pr}$

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MCQ 191 Mark

Factorised form of $23xy - 46x + 54y - 108$ is:

✓
$(23x + 54)(y - 2)$
B
$(23x + 54y)(y - 2)$
C
$(23xy + 54y)(-46x - 108)$
D
$(23x + 54)(y + 2)$

Answer

Correct option: A.

$(23x + 54)(y - 2)$

We have,
$23xy- 46x + 54y - 108 = 23xy - 2 × 23x + 54y - 2 × 54$
$= 23x(y - 2) + 54(y - 2)$
$= (y - 2)(23x + 54)$
$= (23x + 54)(y - 2)$

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MCQ 201 Mark

Common factor of $17abc, 34ab^2, 51a^2b$ is:

A
$17abc$
✓
$17ab$
C
$17ac$
D
$17a^2b^2c$

Answer

Correct option: B.

$17ab$

Given, $17??? = 17 \times a \times b\times c\ 34ab^2$
$= 2 \times 17 \times a \times b \times b\ 51a\ 2b $
$= 3 \times 17 \times a \times a \times a$
Now, collecting the common factors,
we get $17 \times a \times b $
$= 17ab$

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MCQ 211 Mark

Factorised form of $r^2- 10r + 21$ is:

A
$(r - 1)(r - 4)$
✓
$(r - 7)(r - 3)$
C
$(r - 7)(r + 3)$
D
$(r + 7)(r + 3)$

Answer

Correct option: B.

$(r - 7)(r - 3)$

We have,
$r^2- 10r + 21$
$= r^2- 7r - 3r + 21$
$= r(r - 7) - 3(r - 7)$
$= (r - 7)(r - 3)$

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MCQ 221 Mark

In a polynomial, the exponents of the variables are always:

A
Integers.
B
Positive integers.
✓
Non-negative integers.
D
Non-positive integers.

Answer

Correct option: C.

Non-negative integers.

In a polynomial, the exponents of the variables are either positive integers or $0.$ Constant term $C$ can be written as $C x^\circ .$ We do not consider the expressions as a polynomial which consist of the variables having negative/fractional exponent.

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MCQ 231 Mark

An irreducible factor of $24x^2y^2$ is:

A
$ x^2 $
B
$ y^2 $
✓
$ x $
D
$ 24 x $

Answer

Correct option: C.

$ x $

A factor is said to be irreducible, if it cannot be factorised further.
We have, $24x^2y^2= 2 \times 2 \times 2 \times 3 \times x \times x \times y \times y$
Hence, an irreducible factor of $24x^2y^2$ is $x.$

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MCQ 241 Mark

Volume of a rectangular box $($cuboid$)$ with length $= 2ab,$ breadth $= 3ac$ and height $= 2ac$ is:

✓
$ 12 a^3 b c^2 $
B
$ 12 a^3 b c $
C
$ 12 a^2 b c $
D
$ 2 a b+3 a c+2 a c $

Answer

Correct option: A.

$ 12 a^3 b c^2 $

We know that, volume of a cuboid $=$ Length $\times $ Breadth $\times $ Height
$= 2ab \times 3ac \times 2ac$
$= (2 \times 3 \times 2)ab \times ac \times ac$
$= 12 \times a \times a \times a \times b \times c \times c$
$= 12 a^3 b c^2 $

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MCQ 251 Mark

If we subtract $-3x^2y^2$ from $x^2y^2$, then we get:

A
$-4 x^2 y^2$
B
$-2 x^2 y^2$
C
$2 x^2 y^2$
✓
$4 x^2 y^2$

Answer

Correct option: D.

$4 x^2 y^2$

Given, Monomial are $-3x^2y^2$ from $x^2y^2$.
Now we have to substract the first one from the second one.
$ x^2 y^2-\left(3 x^2 y^2\right)$
$=x^2 y^2-(-3) x^2 y^2 $
$ =x^2 y^2+3 x^2 y^2 $
$ =(1+3) x^2 y^2 $
$ =4 x^2 y^2 $

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MCQ 261 Mark

The product of a monomial and a binomial is a:

A
Monomial.
✓
Binomial.
C
Trinomial.
D
None of these.

Answer

Correct option: B.

Binomial.

Monomial consist of only single term and binomial contains two terms. So, the multiplication of a binomial by a monomial will always produce a binomial, whose rst term is the product of monomial and the binomial's first term and second term is the product of monomial and the binomial's second term.

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MCQ 271 Mark

The sum of $-7pq$ and $2pq$ is:

A
$-9pq$
B
$9pq$
C
$5pq$
✓
$5pq$

Answer

Correct option: D.

$5pq$

Given, monomials are $-7pq$ and $2pq$
Their sum $= -7pq + 2pq$
$= (-7 + 2)pq$
$= -5pq$

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MCQ 281 Mark

Which of the following is correct?

A
$ (a-b)^2=a^2+2 a b-b^2 $
✓
$ (a-b)^2=a^2-2 a b+b^2 $
C
$ (a-b)^2=a^2-b^2 $
D
$ (a+b)^2=a^2+2 a b-b^2 $

Answer

Correct option: B.

$ (a-b)^2=a^2-2 a b+b^2 $

$ (a-b)^2$
$=(a-b)(a-b) $
$ =a(a-b)-b(a-b) $
$ =a \cdot a-a \cdot b-b \cdot a+b \cdot b $
$ =a^2-a b-a b+b^2 $
$ =a^2-2 a b+b^2\ \& \ (a+b)^2 $
$ =(a+b)(a+b) $
$ =a \cdot a+a \cdot b+b \cdot a+b \cdot b $
$ =a^2+2 a b+b^2 $

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Maths M.C.Q. [1 Marks Each]

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