MCQ - Maths STD 7 Questions - Vidyadip

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24 questions · auto-graded multiple-choice test.

MCQ 11 Mark

$\Big\{\Big(\frac{1}{3}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}+\Big(\frac{1}{4}\Big)^{-3}\Big\}$

✓
$\frac{19}{64}$
B
$\frac{64}{19}$
C
$\frac{27}{16}$
D
$\frac{-19}{64}$

Answer

Correct option: A.

$\frac{19}{64}$

$\Big\{\Big(\frac{1}{3}\Big)^{-3}-\Big(\frac{1}{2}\Big)^{-3}\Big\}\div\Big(\frac{1}{4}\Big)^{-3}$
$=\Big\{\Big(\frac{3}{1}\Big)^{3}-\Big(\frac{2}{1}\Big)^{3}\Big\}\div\Big(\frac{4}{1}\Big)^{3}$ $\text{As, }\text{x}^{-1}=\frac{1}{\text{x}}$
$=\{3^3-2^3\}\div4^3$
$=\{27-8\}\div64$
$=19\div64$
$=\frac{19}{64}$
Hence, the correct alternative is option $(a).$

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MCQ 21 Mark

$(1+3+5+7+9+11)^{\frac{3}{2}}=$

A
$36$
✓
$216$
C
$256$
D
None of these.

Answer

Correct option: B.

$216$

$(1+3+5+7+9+11)^{\frac{3}{2}}$
$=(36)^{\frac{3}{2}}$
$=(6^2)^{\frac{3}{2}}$
$=6^{2\times\frac{3}{2}}$
$[\text{As,}(\text{x}^{\text{m}})=\text{x}^{\text{mn}}]$
$=6^3$
$=216$
Hence, the correct alternative is option $(b).$

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MCQ 31 Mark

If $\Big(\frac{5}{3}\Big)^{-5}\times\Big(\frac{5}{3}\Big)^{11}=\Big(\frac{5}{3}\Big)^{8\text{x}}, \text{then x}=?$

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

Answer

Correct option: C.

$\frac{3}{4}$

$\Big(\frac{5}{3}\Big)^{-5}\times\Big(\frac{5}{3}\Big)^{11}=\Big(\frac{5}{3}\Big)^{8\text{x}}$
$\Rightarrow\Big(\frac{5}{3}\Big)^{-5+11}=\Big(\frac{5}{3}\Big)^{8\text{x}}$ $\Big(\text{As, }\text{x}^{\text{m}\times\text{n}}=\text{x}^{\text{m+n}}\Big)$
$\Rightarrow\Big(\frac{5}{3}\Big)^{6}=\Big(\frac{5}{3}\Big)^{8\text{x}}$
Comparing the exponent of both the sides, we get:
$8\text{x}=6$
$\Rightarrow\text{x}=\frac{6}{8}$
$\therefore\text{x}=\frac{3}{4}$
Hence, the correct alternative is option $(c).$

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MCQ 41 Mark

If $3^{ x }=6561$, then $3^{x-3}=$

A
$81$
✓
$243$
C
$729$
D
$27$

Answer

Correct option: B.

$243$

$3^x=6561$
$\Rightarrow 3^x=3^8$
Comparing the exponent of both the sides, we get:
$x=8$
Now, $3^{x-3}=3^{8-3}$
$=3^5=243$
Hence, the correct alternative is option $(b).$

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MCQ 51 Mark

If $x y z=0$, then find the value of $\left(a^x\right)^{y z}+\left(a^y\right)^{z x}+\left(a^z\right)^{x y}=$

✓
$1. 3$
B
$2. 2$
C
$3. 1$
D
$4. 0$

Answer

Correct option: A.

$1. 3$

Since,
$\left(a^x\right)^{y z}+\left(a^y\right)^{z x}+\left(a^z\right)^{x y}$
$=a^{x y z}+a^{y z x}+a^{z x y}\left[\text { As, }\left(x^m\right)^n=x^{m n}\right]$
$=a^{x y z}+a^{x y z}+a^{x y z}$
$=a^0+a^0+a^0$
$=1+1+1\left(\text { As, } x^0=1\right)$
$=3$
Hence, the correct option is $(a).$

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MCQ 61 Mark

$2^{3^{2}}=$

A
$64$
B
$32$
C
$256$
✓
$512$

Answer

Correct option: D.

$512$

Since, $2^{3^{2}}=2^9=512$
Hence, the correct alternative is option $(d).$

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MCQ 71 Mark

$(8^4+8^2)^{\frac{1}{2}}=$

A
$84$
B
$8\sqrt{77}$
C
$72$
✓
$8\sqrt{65}$

Answer

Correct option: D.

$8\sqrt{65}$

$(8^4+8^2)^{\frac{1}{2}}$
$=(8^{2+2}+8^2)^{\frac{1}{2}}$
$=(8^2\times8^2+8^2)^{\frac{1}{2}}$ $(\text{As},\text{x}^{\text{m+n}}=\text{x}^{\text{m}}\times\text{x}^{\text{n}})$
$=[8^2\times(8^2+1)]^{\frac{1}{2}}$ $[\text{As}, \text{ab+ac}=\text{a}\times(\text{a+c})]$
$=(8^2)^{\frac{1}{2}}\times(8^2+1)^{\frac{1}{2}}$ $[\text{As, }(\text{ab})^{\text{m}}=\text{a}^{\text{m}}\times\text{b}^{\text{m}}]$
$=8^{2\times\frac{1}{2}}\times(64+1)^{\frac{1}{2}}$
$=8\times65^{\frac{1}{2}}$
$=8\sqrt{65}$
Hence, the correct alternative is option $(d).$

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MCQ 81 Mark

The number $4,70,394$ is standard form is written as:

✓
$4.70394 \times 10^5$
B
$4.70394 \times 10^4$
C
$47.0394 \times 10^4$
D
$4703.94 \times 10^2$

Answer

Correct option: A.

$4.70394 \times 10^5$

Since, $4,70,394=4.70394 \times 100000=4.70394 \times 10^5$
So, the number $4,70,394$ in standard form is written as $4.70394 \times 10^5$
Hence, the correct alternative is option $(a).$

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MCQ 91 Mark

$(6^{-1} - 8^{-1})^{-1 }=$

A
$\frac{1}{24}$
✓
$24$
C
$-24$
D
$-\frac{1}{24}$

Answer

Correct option: B.

$24$

$(6^{-1}-8^{-1})^{-1}=$
$=\Big(\frac{1}{6}-\frac{1}{8}\Big)^{-1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}}\Big)$
$=\Big(\frac{4-3}{24}\Big)^{-1}$
$=\Big(\frac{1}{24}\Big)^{-1}$
$=\frac{24}{1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}})$
$=24$
Hence, the correct alternative is option $(b).$

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MCQ 101 Mark

If $2^n=4096$, then $2^{n-5}=$

✓
$128$
B
$64$
C
$256$
D
$32$

Answer

Correct option: A.

$128$

Since,
As, $2^n=4096$
$\Rightarrow 2^n=2^{12}$
Comparing the exponent of both the sides, we get:
$n =12$
Now, $2^{n-5}=2^{12-5}$
$=2^7$
$=128$
Hence, the correct alternative is option $(a).$

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MCQ 111 Mark

$(2^{3^{4}})=$

A
$2^{4^{3}}$
B
$2^{3^{4}}$
✓
$\big(2^4\big)^3$
D
None of these.

Answer

Correct option: C.

$\big(2^4\big)^3$

$\big(2^3\big)^4$
$=2^{3\times4}$ $[\text{As}, (\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=2^{4\times3}$
$=(2^{4})^{3}$
Hence, the correct alternative is option $(c).$

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MCQ 121 Mark

If $a = 25$, then $\text{a}^{25^{0}}+\text{a}^{0^{25}}$

A
$25$
✓
$26$
C
$24$
D
$0$

Answer

Correct option: B.

$26$

$\text{a}^{25^{0}}+\text{a}^{0^{25}}$
$=\text{a}^{1}+\text{a}^{0}$ $\Big(\text{As, }25^{0}=1\text{ and }0^{25}=0\Big)$
$=\text{a}+1$ $\Big(\text{As, }\text{a}^{0}=1\Big)$
$=25+1$
$=26$
Hence, the correct alternative is option $(b).$

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MCQ 131 Mark

$\bigg[\Big\{\big(-\frac{1}{3}\big)\Big\}^{-2}\bigg]^{-1}=$

✓
$\frac{1}{81}$
B
$\frac{1}{9}$
C
$-\frac{1}{81}$
D
$-\frac{1}{9}$

Answer

Correct option: A.

$\frac{1}{81}$

$\bigg[\Big\{\big(-\frac{1}{3}\big)\Big\}^{-2}\bigg]^{-1}$
$\bigg[\Big\{\big(-\frac{1}{3}\big)^{2}\Big\}^{(-2)\times(-1)}\bigg]$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}\big]$
$=\Big(\frac{-1}{3}\Big)^{2\times2}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}\big]$
$=\Big(\frac{-1}{3}\Big)^4$
$=\frac{(-1)^4}{3^4}$ $\Big[\text{As, }\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{m}}=\frac{\text{x}^{\text{m}}}{\text{y}^{\text{m}}}\Big]$
$=\frac{1}{81}$
Hence, the correct alternative is option $(a).$

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MCQ 141 Mark

$\text{If abc} = 0,$ $\text{then}\frac{\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}}{\Big\{(\text{x}^{\text{b}})^{\text{c}}\Big\}}=$

A
$3$
B
$0$
C
$-1$
✓
$1$

Answer

Correct option: D.

$1$

$\frac{\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}}{\Big\{(\text{x}^{\text{b}})^{\text{c}}\Big\}}$
$=\frac{(\text{x}^{\text{a}})^{\text{bc}}}{(\text{x}^{\text{b}})^{\text{ac}}}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\frac{\text{x}^{\text{abc}}}{\text{x}^{\text{abc}}}$ $[\text{As,}(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\frac{\text{x}^{0}}{\text{x}^{0}}$ $[\text{As, abc}=0]$
$=\frac{1}{1}$ $[\text{As, }\text{x}^{0}=1]$
$=1$
Hence, the correct alternative is option $(d).$

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MCQ 151 Mark

$(3^{-1}\times5^{-1})^{-1}=$

A
$\frac{1}{15}$
B
$-\frac{1}{15}$
✓
$15$
D
$-15$

Answer

Correct option: C.

$15$

$(3^{-1}\times5^{-1})^{-1}$
$=\Big(\frac{1}{3}\times\frac{1}{5}\Big)^{-1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}}\Big)$
$=\Big(\frac{1}{15}\Big)^{-1}$
$=\frac{15}{1}$ $\Big(\text{As},\text{x}^{-1}=\frac{1}{\text{x}}\Big)$
$=15$
Hence, the correct alternative is option $(c).$

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MCQ 161 Mark

$\Big(-\frac{3}{5}\Big)^{-1}=$

A
$\frac{3}{5}$
B
$\frac{5}{3}$
✓
$-\frac{5}{3}$
D
$-\frac{3}{5}$

Answer

Correct option: C.

$-\frac{5}{3}$

Since,
$\Big(-\frac{3}{5}\Big)^{-1}=-\frac{5}{3}$ $\Big(\text{As, }\text{x}^{-1}=\frac{\text{1}}{\text{x}}\Big)$
Hence, the correct alternative is option $(c).$

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MCQ 171 Mark

If $a =3^{-3}-3^3$ and $b =3^3-3^{-3}$, then $\frac{ a }{ b }-\frac{ b }{ a }=$

✓
$0$
B
$1$
C
$-1$
D
$2$

Answer

Correct option: A.

$0$

Since,
$\text{a}=3^{-3}-3^3$
$=\frac{1}{3^3}-3^3$ $\Big(\text{As},\text{x}^{-\text{m}}=\frac{1}{\text{x}^{\text{m}}}\Big)$
$=\frac{1}{27}-\frac{27}{1}$
$=\frac{1}{27}-\frac{27\times27}{1\times27}$
$=\frac{1}{27}-\frac{729}{27}$
$=\frac{1-729}{27}$
$=\frac{-728}{27}$
Also, $\text{b}=3^3-3^{-3}$
$=3^3-\frac{1}{3^3}$ $\Big(\text{As, }\text{x}^{-\text{m}}=\frac{1}{\text{x}^{\text{m}}}\Big)$
$=\frac{27}{1}-\frac{1}{27}$
$=\frac{27\times27}{1\times27}-\frac{1}{27}$
$=\frac{729}{27}-\frac{1}{27}$
$=\frac{729-1}{27}$
$=\frac{728}{27}$
Now,
$\frac{\text{a}}{\text{b}}-\frac{\text{b}}{\text{a}}$
$=\frac{\Big(\frac{-728}{27}\Big)}{\Big(\frac{728}{27}\Big)}-\frac{\Big(\frac{728}{27}\Big)}{\Big(\frac{-728}{27}\Big)}$
$=(-1)-(-1)$
$=-1+1$
$=0$
Hence, the correct alternative is option $(a).$

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MCQ 181 Mark

The number $2.35 \times 10^4$ in the usual form is written as:

A
$2.35 \times 10^3$
✓
$23500$
C
$2350000$
D
$235 \times 10^4$

Answer

Correct option: B.

$23500$

Since, $2.35 \times 10^4=2.35 \times 10000=23500$
So, the number $2.35 \times 10^4$ in the usual form is written as $23500 .$
Hence, the correct alternative is option $(b).$

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MCQ 191 Mark

$\frac{(144)^{\frac{1}{2}}+(256)^{\frac{1}{2}}}{3^2-2}=$

A
$8$
✓
$4$
C
$-4$
D
$-8$

Answer

Correct option: B.

$4$

$\frac{(144)^{\frac{1}{2}}+(256)^{\frac{1}{2}}}{3^2-2}$
$=\frac{(12^2)^{\frac{1}{2}}+(16^2)^{\frac{1}{2}}}{9-2}$
$=\frac{12^{2\times\frac{1}{2}}+16^{2\times\frac{1}{2}}}{7}$ $[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\frac{12+16}{7}$
$=\frac{28}{7}$
$=4$
Hence, the correct alternative is option $(b).$

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MCQ 201 Mark

$(-1)^{301}+(-1)^{302}+(-1)^{303}+\dots+(-1)^{400}$

A
$1$
B
$101$
C
$100$
✓
$0$

Answer

Correct option: D.

$0$

Since,
$(-1)^{301}+(-1)^{302}+(-1)^{303}+\dots+(-1)^{400}$
$=(-1)+(1)+(-1)+\dots+(1)$ $\Big[\text{As, }(-1)^{\text{odd}}=-1\text{ and }(-1)^{\text{even}}=1\Big]$
$=-50+50$ $[\text{As, there are} 50(-1)'\text{s and 50 (1)}'\text{s}]$
$=0$
Hence, the correct alternative is option is $(d).$

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MCQ 211 Mark

If $2^n=1024$, then $2^{\left(\frac{n}{2}+2\right)}=$

A
$64$
✓
$128$
C
$256$
D
$512$

Answer

Correct option: B.

$128$

As, $2^n=1024$
$\Rightarrow 2^n=2^{10}$
Comparing the exponent of both the sides, we get:
$n = 10$
Now,
$2^{\Big(\frac{\text{n}}{2}+2\Big)}$
$=2^{\Big(\frac{\text{10}}{2}+2\Big)}$
$=2^{(5+2)}$
$=2^7$
$=128$
Hence, the correct alternative is option $(b).$

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MCQ 221 Mark

What should be multiplied to $6^{-2}$ so that the product may be equal to $216$ ?

A
$6^4$
✓
$6^5$
C
$6^3$
D
$6$

Answer

Correct option: B.

$6^5$

Since,
Let $6^{ m }$ should be multiplied to $6^{-2}$
$\text{A.T.Q.}$
$6^{-2} \times 6^m=216$
$\Rightarrow 6^{-2+m}=6^3$
Comparing th exponent of both the sides, we get
$-2+m=3$
$\Rightarrow m=3+2$
$\therefore m=5$
So, $6^5$ should be multiplied.
Hence, the correct alternative is option $(b).$

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MCQ 231 Mark

If $abc = 0$, then find the value of $\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}$

✓
$1$
B
$a$
C
$b$
D
$c$

Answer

Correct option: A.

$1$

Since,
$\Big\{(\text{x}^{\text{a}})^{\text{b}}\Big\}^{\text{c}}$
$=(\text{x}^{\text{a}})^{\text{bc}}$
$[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\text{x}^{\text{abc}}$
$[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=\text{x}^{0}$
$[\text{As, abc}=0]$
$=1$
$[\text{As, }\text{x}^{0}=1]$
Hence, the correct alternative is option $(a).$

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MCQ 241 Mark

$\Big\{(33)^2-(31)^2\Big\}^{\frac{5}{7}}$

A
$64$
B
$16$
✓
$32$
D
$4$

Answer

Correct option: C.

$32$

Since,
$\Big\{(33)^2-(31)^2\Big\}^{\frac{5}{7}}$
$=\{1089-961\}^{\frac{5}{7}}$
$=\{128\}^{\frac{5}{7}}$
$=\{2^7\}^{\frac{5}{7}}$
$=2^{7\times\frac{5}{7}}$$[\text{As, }(\text{x}^{\text{m}})^{\text{n}}=\text{x}^{\text{mn}}]$
$=2^5$
$=32$
Hence, the correct alternative is option $(c).$

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