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MATHS M.C.Q

Presentation of Data in Tabular Form · MP Board (MPBSE) STD 9 (Madhya Pradesh - English Medium). 75 questions.

Questions · Page 2 of 2

M.C.Q

MCQ 511 Mark
The simplest for of $0.\overline{32}$ is:
  • A
    $\frac{16}{45}$
  • B
    $\frac{32}{99}$
  • ✓
    $\frac{29}{90}$
  • D
    None of these.
Answer
Correct option: C.
$\frac{29}{90}$
Let $\text{x}=0.\overline{32}$
Then, $\text{x}=0.3222 \ ...(\text{i})$
$\therefore10\text{x}=3.222 \ ...(\text{ii})$
and $100\text{x}=32.222 \ ...(\text{iii})$
On subtracting $(ii)$ from $(iii),$ we get
$90\text{x}=29$
$\Rightarrow\text{x}=\frac{29}{90}$
Hence, the correct option is $(c).$
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MCQ 521 Mark
The value of $x^{p-q} \cdot x^{q-r} \cdot x^{r-p}$ is equal to:
  • A
    $0$
  • ✓
    $1$
  • C
    $x$
  • D
    $x^{\text {pqr }}$
Answer
Correct option: B.
$1$
$x^{p-q} \cdot x^{q-r} \cdot x^{r-p}$
$=x^{p-q+q-r+r-p}$
$=x^0$
$=1$
Hence, the correct option is $(b).$
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MCQ 531 Mark
If $\Big(\frac{2}{3}\Big)^{\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{81}{16}$ then $x = ?$
  • A
    $1$
  • B
    $2$
  • C
    $3$
  • ✓
    $4$
Answer
Correct option: D.
$4$
$\Big(\frac{2}{3}\Big)^{\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{81}{16}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{3^4}{2^4}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}+2\text{x}}=\Big(\frac{3}{2}\Big)^4$
$\Rightarrow\Big(\frac{3}{2}\Big)^{\text{x}}=\Big(\frac{3}{2}\Big)^4$
$\Rightarrow\text{x}=4$
Hence, the correct option is $(d).$
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MCQ 541 Mark
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=?$
  • A
    $\sqrt{2}$
  • ✓
    $2$
  • C
    $4$
  • D
    $8$
Answer
Correct option: B.
$2$
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{4\times8}+\sqrt{4\times12}}{\sqrt{8}+\sqrt{12}}$
$=\frac{2\sqrt{8}+2\sqrt{12}}{\sqrt{8}+\sqrt{12}}=\frac{2\big(\sqrt{8}+\sqrt{12}\big)}{\sqrt{8}+\sqrt{12}}=2$
Hence, the correct answer is option $(b).$
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MCQ 551 Mark
An irrational number between $5$ and $6$ is:
  • A
    $\frac{1}{2}(5+6)$
  • B
    $\sqrt{5+6}$
  • ✓
    $\sqrt{5\times6}$
  • D
    None of these.
Answer
Correct option: C.
$\sqrt{5\times6}$
An irrational number between $a$ and $b$ is given by $\sqrt{\text{ab}}$
So, an irrational number between $5$ and $6$ is given by $\sqrt{5\times6}$
Hence, the correct answer is option $(c).$
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MCQ 561 Mark
If $\sqrt{2}=1.41$ then $\frac{1}{\sqrt{2}}=?$
  • A
    $0.075$
  • B
    $0.75$
  • ✓
    $0.705$
  • D
    $7.05$
Answer
Correct option: C.
$0.705$
$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{\sqrt{2}}{2}$
$=\frac{1.41}{2}$
$=0.705$
Hence, the correct option is $(c).$
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MCQ 571 Mark
The value of $0.\overline{2}$ in the form $\frac{\text{p}}{\text{q}},$ where $p$ and $q$ are integers and $\text{q}\neq0,$ is:
  • A
    $\frac{1}{5}$
  • ✓
    $\frac{2}{9}$
  • C
    $\frac{2}{5}$
  • D
    $\frac{1}{8}$
Answer
Correct option: B.
$\frac{2}{9}$
$\frac{2}{9}=0.2222222...=0.\overline{2}$
Hence, the correct option is $(b).$
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MCQ 581 Mark
The value of $\sqrt[4]{(64)^{-2}}$ is:
  • ✓
    $\frac{1}{8}$
  • B
    $\frac{1}{2}$
  • C
    $8$
  • D
    $\frac{1}{64}$
Answer
Correct option: A.
$\frac{1}{8}$
$\sqrt[4]{(64)^{-2}}=\sqrt[4]{(8^2)^{-2}}=8^{-4\times\frac{1}{4}}=8^{-1}=\frac{1}{8}$
Hence, the correct answer is option $(a).$
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MCQ 591 Mark
The simplest for of $0.\overline{54}$ is:
  • A
    $\frac{27}{50}$
  • ✓
    $\frac{6}{11}$
  • C
    $\frac{4}{7}$
  • D
    None of these.
Answer
Correct option: B.
$\frac{6}{11}$
Let $\text{x}=0.\overline{54}$
Then, $\text{x}=54.5454 \ ...(\text{i})$
$\therefore100\text{x}=54.5454 \ ...(\text{ii})$
On subtracting $(i)$ from $(ii),$ we get
$99\text{x}=54$
$\Rightarrow\text{x}=\frac{54}{99}=\frac{6}{11}$
Hence, the correct option is $(c).$
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MCQ 601 Mark
If $\text{x}=\frac{\sqrt{7}}{5}$ and $\frac{5}{\text{x}}=\text{p}\sqrt{7}$ then the value of $p$ is:
  • A
    $\frac{7}{25}$
  • ✓
    $\frac{25}{7}$
  • C
    $\frac{7}{15}$
  • D
    $\frac{15}{7}$
Answer
Correct option: B.
$\frac{25}{7}$
$\text{x}=\frac{\sqrt{7}}{5}$ and $\frac{5}{\text{x}}=\text{p}\sqrt{7}$
$\Rightarrow\frac{5}{\sqrt{7}}=\text{p}\sqrt{7}$
$\Rightarrow\frac{25}{\sqrt{7}}=\text{p}\sqrt{7}$
$\Rightarrow\text{p}=\frac{25}{\sqrt{7}\times\sqrt{7}}=\frac{25}{7}$
Hence, the correct option is $(b).$
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MCQ 611 Mark
The value of $\Big(\frac{256\text{x}^{16}}{81\text{y}^4}\Big)^{-\frac{1}{4}}$ is:
  • A
    $\frac{3\text{y}}{8\text{x}^4}$
  • ✓
    $\frac{3\text{y}}{4\text{x}^4}$
  • C
    $\frac{4\text{y}}{5\text{x}^4}$
  • D
    $\frac{4\text{x}^4}{3\text{y}}$
Answer
Correct option: B.
$\frac{3\text{y}}{4\text{x}^4}$
$\Big(\frac{256\text{x}^{16}}{81\text{y}^4}\Big)^{-\frac{1}{4}}=\Big(\frac{81\text{y}^{4}}{256\text{x}^{16}}\Big)^{\frac{1}{4}}$
$=\Big(\frac{3^4\text{y}^{4}}{4^4(\text{x}^4)^4}\Big)^{\frac{1}{4}}=\Big(\frac{3\text{y}}{4\text{x}^4}\Big)^{4\times\frac{1}{4}}=\frac{3\text{y}}{4\text{x}^4}$
Hence, the correct option is $(b).$
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MCQ 621 Mark
If $\left(3^3\right)^2=9^x$ then $5^x=$ ?
 
  • A
    $1$
  • B
    $5$
  • C
    $25$
  • ✓
    $125$
Answer
Correct option: D.
$125$

$\left(3^3\right)^2=9^x$
$\Rightarrow\left(3^2\right)^3=\left(3^2\right)^x$
$\Rightarrow x=3$
Then $5^x=5^3=125$
Hence, the correct option is $(d).$

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MCQ 631 Mark
$\big(6+\sqrt{27}\big)-\big(3+\sqrt{3}\big)+\big(1-2\sqrt{3}\big)$ when simplified is:
  • A
    positive and irrational.
  • ✓
    positive and rational.
  • C
    negative and irrational.
  • D
    negative and rational.
Answer
Correct option: B.
positive and rational.
$\big(6+\sqrt{27}\big)-\big(3+\sqrt{3}\big)+\big(1-2\sqrt{3}\big)$
$=6+\sqrt{27}-3-\sqrt{3}-2\sqrt{3}$
$=6-3+3\sqrt{3}-\sqrt{3}-2\sqrt{3}$
$=3$
which is positive and rational number.
Hence, the correct answer is option $(b).$
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MCQ 641 Mark
The simplest for of $1.\overline{6}$ is:
  • A
    $\frac{833}{500}$
  • B
    $\frac{8}{5}$
  • ✓
    $\frac{5}{3}$
  • D
    None of these.
Answer
Correct option: C.
$\frac{5}{3}$
Let $\text{x}=1.\overline{6}$
$\Rightarrow\text{x}=1.666 \ ...(\text{i})$
$\therefore10\text{x}=16.666 \ ...(\text{ii})$
On subtracting $(i)$ from $(ii),$ we get
$9\text{x}=15$
$\Rightarrow\text{x}=\frac{15}{9}=\frac{5}{3}$
Hence, the correct option is $(c).$
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MCQ 651 Mark
The value of $\sqrt{5+2\sqrt{6}}$ is:
  • A
    $\sqrt{5}+\sqrt{6}$
  • B
    $\sqrt{5}-\sqrt{6}$
  • ✓
    $\sqrt{3}+\sqrt{2}$
  • D
    $\sqrt{3}-\sqrt{2}$
Answer
Correct option: C.
$\sqrt{3}+\sqrt{2}$
$\sqrt{5+2\sqrt{2}}=\sqrt{\big(\sqrt{2}\big)^2+\big(\sqrt{3}\big)^2+2\times\sqrt{2}\times\sqrt{3}}$
$=\sqrt{\big(\sqrt{2}+\sqrt{3}\big)^2}$
$=\sqrt{2}+\sqrt{3}$
Hence, the correct option is $(c).$
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MCQ 661 Mark
$(125)^{-\frac{1}{3}}=?$
  • A
    $5$
  • B
    $-5$
  • ✓
    $\frac{1}{5}$
  • D
    $-\frac{1}{5}$
Answer
Correct option: C.
$\frac{1}{5}$
$(125)^{-\frac{1}{3}}=\frac{1}{(125)^{\frac{1}{3}}}$
$=\frac{1}{5^{3-\frac{1}{3}}}=\frac{1}{5}$
Hence, the correct answer is option $(c).$
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MCQ 671 Mark
Simplified value of $(25)^{\frac{1}{3}}\times5^{\frac{1}{3}}$ is:
  • A
    $25$
  • B
    $3$
  • C
    $1$
  • ✓
    $5$
Answer
Correct option: D.
$5$
$(25)^{\frac{1}{3}}\times5^{\frac{1}{3}}=5^{2\times\frac{1}{3}}\times5^\frac{1}{3}$
$=5^\frac{2}{3}\times5^\frac{1}{3}=5^{\frac{2}{3}+\frac{1}{3}}=5^{\frac{3}{3}}=5$
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MCQ 681 Mark
Which of the following is a rational number$?$
  • A
    $1+\sqrt{3}$
  • B
    $\pi$
  • C
    $2\sqrt{3}$
  • ✓
    $0$
Answer
Correct option: D.
$0$
Since, the sum and product of a rational and an irrational is always irrational.
So, $1+\sqrt{3}$ and $2\sqrt{3}$ are irrational numbers.
Also, $\pi$ is an irrational number.
And, $0$ is an integer.
So, $0$ is a rational number.
Hence, the correct option is $(d).$
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MCQ 691 Mark
The value of $\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$ is:
  • ✓
    $3$
  • B
    $-3$
  • C
    $9$
  • D
    $\frac{1}{3}$
Answer
Correct option: A.
$3$
$\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}=\Big[\big(9^2\big)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$
$=9^{2\times\frac{1}{2}\times\frac{1}{2}}=9^{\frac{1}{2}}=3^{2\times\frac{1}{2}}=3$
Hence, the correct option is $(a).$
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MCQ 701 Mark
$\sqrt{10}\times\sqrt{15}=?$
  • A
    $\sqrt{25}$
  • ✓
    $5\sqrt{6}$
  • C
    $6\sqrt{5}$
  • D
    None of these.
Answer
Correct option: B.
$5\sqrt{6}$
$\sqrt{10}\times\sqrt{15}=\sqrt{5\times2}\times\sqrt{5\times3}$
$=\sqrt{5}\times\sqrt{2}\times\sqrt{5}\times\sqrt{3}=5\sqrt{6}$
Hence, the correct answer is option $(b).$
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MCQ 711 Mark
Which of the following is a true statment?
  • A
    $\pi$ and $\frac{22}{7}$ are both rationals.
  • B
    $\pi$ and $\frac{22}{7}$ are both irrationals
  • C
    $\pi$ is rational and $\frac{22}{7}$ is irrational.
  • ✓
    $\pi$ is irrational and $\frac{22}{7}$ is rational.
Answer
Correct option: D.
$\pi$ is irrational and $\frac{22}{7}$ is rational.
A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number.
So, $\pi=3.141592...$ is irrational.
The numbers of the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ are known as rational numbers.
So, $\frac{22}{7}$ is rational.
Hence, the correct option is $(d).$
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MCQ 721 Mark
$\big(-2-\sqrt{3}\big)\big(-2+\sqrt{3}\big)$ when simplified is:
  • A
    positive and irrational.
  • ✓
    positive and rational.
  • C
    negative and irrational.
  • D
    negative and rational.
Answer
Correct option: B.
positive and rational.
$\big(-2-\sqrt{3}\big)\big(-2+\sqrt{3}\big)$
$\big(-2\big)^2-\big(\sqrt{3}\big)^2$
$=4-3=1,$ which is positive and rational number.
Hence, the correct answer is option $(b).$
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MCQ 731 Mark
The value of $\sqrt{3-2\sqrt{2}}$ is:
  • A
    $\sqrt{3}+\sqrt{2}$
  • B
    $\sqrt{3}-\sqrt{2}$
  • C
    $\sqrt{2}+1$
  • ✓
    $\sqrt{2}-1$
Answer
Correct option: D.
$\sqrt{2}-1$
$\sqrt{3-2\sqrt{2}}=\sqrt{\big(\sqrt{2}\big)^2+(1)^2-2\times\sqrt{2}\times1}$
$=\sqrt{\big(\sqrt{2}-1\big)^2}$
$=\sqrt{2}-1$
Hence, the correct option is $(d).$
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MCQ 741 Mark
If $\text{x}=3+\sqrt{8}$ then $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=?$
  • ✓
    $34$
  • B
    $56$
  • C
    $28$
  • D
    $63$
Answer
Correct option: A.
$34$
$\text{x}=3+\sqrt{8}$
$\Rightarrow\text{x}^2=\big(3+\sqrt{8}\big)^2=9+8+6\sqrt{8}=17+6\sqrt{8}$
Now, $\frac{1}{\text{x}}=\frac{1}{3+\sqrt{8}}$
$=\frac{1}{3+\sqrt{8}}\times\frac{3-\sqrt{8}}{3-\sqrt{8}}$
$=\frac{3-\sqrt{8}}{3^2-\big(\sqrt{8}\big)^2}$
$=\frac{3-\sqrt{8}}{9-8}$
$=3-\sqrt{8}$
$\Rightarrow\Big(\frac{1}{\text{x}}\Big)^2=\big(3-\sqrt{8}\big)^2=9+8-6\sqrt{8}=17-6\sqrt{8}$
Then, $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=17+6\sqrt{8}+17-6\sqrt{8}=34$
Hence, the correct option is $(a).$
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MCQ 751 Mark
$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}=?$
  • ✓
    $2$
  • B
    $\sqrt{2}$
  • C
    $2\sqrt{2}$
  • D
    $4\sqrt{2}$
Answer
Correct option: A.
$2$
$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}$
$=\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{2^5}$
$=2^\frac{1}{3}\times2^\frac{1}{4}\times2^{5\times\frac{1}{12}}$
$=2^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$
$=2^{\frac{4+3+5}{12}}$
$=2^{\frac{12}{12}}$
$=2$
Hence, the correct option is $(a).$
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