M.C.Q. [1 Marks Each]

MCQ 11 Mark

For any two non-zero rational numbers $x$ and $y$, $x^5 \div y^5$ is equal to

A
$(x \div y)^1$
B
$(x \div y)^0$
✓
$(x \div y)^5$
D
$(x \div y)^{10}$

Answer

Correct option: C.

$(x \div y)^5$

MCQ 21 Mark

If $\frac{p}{q}=\left(\frac{5}{6}\right)^2 \div\left(\frac{5}{6}\right)^0$, then the value of $\left(\frac{p}{q}\right)^2$ is

A
$\frac{125}{1290}$
✓
$\frac{625}{1296}$
C
$\frac{164}{125}$
D
$\frac{169}{144}$

Answer

Correct option: B.

$\frac{625}{1296}$

(b) $\frac{625}{1296}$
$\quad \frac{p}{q}=\left(\frac{5}{6}\right)^2÷\left(\frac{5}{6}\right)^0$
$\Rightarrow \quad \frac{p}{q}=\left(\frac{5}{6}\right)^2 \quad\left[\because a^0=1\right]$
Now, $\left(\frac{p}{q}\right)^2=\left(\left(\frac{5}{6}\right)^2\right)^2$
$=\left(\frac{5}{6}\right)^4=\frac{625}{1296} \quad\left[\because\left(a^m\right)^n=a^{m n}\right]$

View full question & answer→

MCQ 31 Mark

$\left[\left\{\left(\frac{2}{-9}\right)^2\right\}^0\right]^2$ is equal to

A
2
B
$\frac{4}{81}$
C
$\frac{81}{4}$
✓
1

Answer

Correct option: D.

View full question & answer→

MCQ 41 Mark

If $\frac{a}{b}=\left(\frac{625}{81}\right) \div\left(\frac{5^4}{3^4}\right)$, then the value of $\left(\frac{a}{b}\right)^5$ is

A
$\left(\frac{5}{3}\right)^8$
B
$\left(\frac{3}{5}\right)^8$
✓
1
D
$\frac{3}{5}$

Answer

Correct option: C.

(c) 1
$\frac{a}{b}=\left(\frac{625}{81}\right)÷\left(\frac{5^4}{3^4}\right)=\left(\frac{5^4}{3^4}\right)÷\left(\frac{5^4}{3^4}\right)$
$\frac{a}{b}=1$
$\therefore\left(\frac{a}{b}\right)^5=(1)^5=1$

View full question & answer→

MCQ 51 Mark

If $\left(3^{102} \times 3^{101}\right) \div 3^{101}=k \cdot 3^{100}$, then the value of $k$ is

✓
9
B
10
C
11
D
12

Answer

Correct option: A.

(a) 9
$\left(3^{102} \times 3^{101}\right) \div 3^{101}=k \cdot 3^{100}$
$\Rightarrow \quad\left(3^{102+101}\right) \div 3^{101}=k \cdot 3^{100} \quad\left[\because a^m \times a^n=a^{m+n}\right]$
$\Rightarrow \quad 3^{203} \div 3^{101}=k \cdot 3^{100}$
$\Rightarrow \quad 3^{203-101}=k \cdot 3^{100} \quad\left[\because a^m÷a^n=a^{m-n}\right]$
$\Rightarrow \quad 3^{102}=k \cdot 3^{100}$
$\Rightarrow \quad 3^2 \cdot 3^{100}=k \cdot 3^{100}$
On comparing both sides, we get
$k=3^2=9$

View full question & answer→

MCQ 61 Mark

In standard form, the number 829030000 is written as $k \times 10^8$, where $k$ is equal to

A
82903
B
829.03
C
82.903
✓
8.2903

Answer

Correct option: D.

8.2903

View full question & answer→

MCQ 71 Mark

$(-4)^4 \times(-2)^0 \times(-1)^{202}$ is equal to

A
64
B
1
✓
256

Answer

Correct option: D.

256

View full question & answer→

MCQ 81 Mark

$\left(\frac{2}{3}\right)^3 \times\left(\frac{5}{7}\right)^3$ is equal to

A
$\left(\frac{10}{21}\right)^9$
B
$\left(\frac{10}{21}\right)^6$
✓
$\left(\frac{10}{21}\right)^3$
D
$\left(\frac{10}{21}\right)^0$

Answer

Correct option: C.

$\left(\frac{10}{21}\right)^3$

(c) $\left(\frac{10}{21}\right)^3$
$\left(\frac{2}{3}\right)^3 \times\left(\frac{5}{7}\right)^3$
$\Rightarrow \quad\left(\frac{2}{3} \times \frac{5}{7}\right)^3=\left(\frac{10}{21}\right)^3 \quad\left[\because a^m \times b^m=(a b)^m\right]$

View full question & answer→

MCQ 91 Mark

The value of $\frac{10^{22}+10^{20}}{10^{20}}$ is

A
10
B
$10^{42}$
✓
101
D
$10^{22}$

Answer

Correct option: C.

101

(c) $\frac{10^{22}+10^{20}}{10^{20}}=\frac{10^{20}\left[10^2+1\right]}{10^{20}}$
$=\frac{10^{20}[100+1]}{10^{20}}$
$=10^{20-20} \times 101=101$

View full question & answer→

MCQ 101 Mark

$\left(5^7÷5^2\right) \times\left(3^6÷3^2\right)$ is equal to

A
1426
B
1242
✓
253125
D
101962

Answer

Correct option: C.

253125

(c) 253125
$\left(5^7 \div 5^2\right) \times\left(3^6÷3^2\right)$
$\Rightarrow \quad\left(5^{7-2}\right) \times\left(3^{6-2}\right) \quad\left[\because a^m \div a^n=a^{m-n}\right]$
$\Rightarrow \quad 5^5 \times 3^4$
$\Rightarrow 3125 \times 81=253125$

View full question & answer→

MCQ 111 Mark

If $\left(\frac{5}{3}\right)^5 \times\left(\frac{5}{3}\right)^{11}=\left(\frac{5}{3}\right)^{8 x}$, then the value of $x$ is

A
3
B
$\frac{1}{2}$
C
1
✓
2

Answer

Correct option: D.

(d) 2
$\left(\frac{5}{3}\right)^5 \times\left(\frac{5}{3}\right)^{11}=\left(\frac{5}{3}\right)^{8 x}$
$\Rightarrow \quad\left(\frac{5}{3}\right)^{5+11}=\left(\frac{5}{3}\right)^{8 x} \quad\left[\because a^m \times a^n=a^{m+n}\right]$
$\Rightarrow \quad\left(\frac{5}{3}\right)^{16}=\left(\frac{5}{3}\right)^{8 x}$
On comparing both sides, we get
$\begin{aligned} 16 & =8 x \\ x & =\frac{16}{8} \\ x & =2\end{aligned}$

View full question & answer→

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

Take a timed test