M.C.Q - Page 2 - MATHS STD 9 Questions - Vidyadip

Questions · Page 2 of 2

M.C.Q

Question 511 Mark

If $\sqrt{2}=1.414$ then $\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}}=?$

0.207
2.414
0.414
0.621

Answer

0.414

Solution:
$\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}}=\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}\times\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}-1\big)}}$
$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{\big(\sqrt{2}\big)^2-(1)^2}}$
$=\sqrt{\frac{\big(\sqrt{2}-1\big)^2}{1}}$
$=\sqrt{\big(\sqrt{2}-1\big)^2}$
$=\sqrt{2}-1$
$=1.414-1$
$=0.414$
Hence, the correct option is (c).

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Question 521 Mark

The simplest for of $0.\overline{32}$ is:

$\frac{16}{45}$
$\frac{32}{99}$
$\frac{29}{90}$
None of these.

Answer

$\frac{29}{90}$

Solutions:
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 531 Mark

The value of $x^{p-q}.x^{q-r}⋅x^{r-p}$ is equal to:

A
$0$
✓
$1$
C
$x$
D
$x^{pqr}$

Answer

Correct option: B.

$1$

$x^{p-q}.x^{q-r}⋅x^{r-p}$
$= x^{p-q+q-r+r-p}$
$= x^0$
$= 1$

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Question 541 Mark

If $\Big(\frac{2}{3}\Big)^{\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{81}{16}$ then x = ?

1
2
3
4

Answer

4

Solution:
$\Big(\frac{2}{3}\Big)^{\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{81}{16}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{3^4}{2^4}$
$\Rightarrow\Big(\frac{3}{2}\Big)^{-\text{x}+2\text{x}}=\Big(\frac{3}{2}\Big)^4$
$\Rightarrow\Big(\frac{3}{2}\Big)^{\text{x}}=\Big(\frac{3}{2}\Big)^4$
$\Rightarrow\text{x}=4$
Hence, the correct option is (d).

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Question 551 Mark

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=?$

$\sqrt{2}$
2
4
8

Answer

4

Solution:
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{4\times8}+\sqrt{4\times12}}{\sqrt{8}+\sqrt{12}}$
$=\frac{2\sqrt{8}+2\sqrt{12}}{\sqrt{8}+\sqrt{12}}=\frac{2\big(\sqrt{8}+\sqrt{12}\big)}{\sqrt{8}+\sqrt{12}}=2$
Hence, the correct answer is option (b).

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Question 561 Mark

An irrational number between 5 and 6 is:

$\frac{1}{2}(5+6)$
$\sqrt{5+6}$
$\sqrt{5\times6}$
None of these.

Answer

$\sqrt{5\times6}$

Solution:
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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Question 571 Mark

If $\sqrt{2}=1.41$ then $\frac{1}{\sqrt{2}}=?$

0.075
0.75
0.705
7.05

Answer

0.705

Solution:
$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}$
$=\frac{\sqrt{2}}{2}$
$=\frac{1.41}{2}$
$=0.705$
Hence, the correct option is (c).

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Question 581 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:

$\frac{1}{5}$
$\frac{2}{9}$
$\frac{2}{5}$
$\frac{1}{8}$

Answer

$\frac{2}{9}$

Solution:
$\frac{2}{9}=0.2222222...=0.\overline{2}$
Hence, the correct option is (b).

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Question 591 Mark

The value of $\sqrt[4]{(64)^{-2}}$ is:

$\frac{1}{8}$
$\frac{1}{2}$
$8$
$\frac{1}{64}$

Answer

$\frac{1}{8}$

Solution:
$\sqrt[4]{(64)^{-2}}=\sqrt[4]{(8^2)^{-2}}=8^{-4\times\frac{1}{4}}=8^{-1}=\frac{1}{8}$
Hence, the correct answer is option (a).

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Question 601 Mark

The simplest for of $0.\overline{54}$ is:

$\frac{27}{50}$
$\frac{6}{11}$
$\frac{4}{7}$
None of these.

Answer

$\frac{6}{11}$

Solutions:
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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Question 611 Mark

If $\text{x}=\frac{\sqrt{7}}{5}$ and $\frac{5}{\text{x}}=\text{p}\sqrt{7}$ then the value of p is:

$\frac{7}{25}$
$\frac{25}{7}$
$\frac{7}{15}$
$\frac{15}{7}$

Answer

$\frac{25}{7}$

Solution:
$\text{x}=\frac{\sqrt{7}}{5}$ 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 621 Mark

If $(3^3)^2 = 9^x$ then $5^x =$ ?

A
$1$
B
$5$
C
$25$
✓
$125$

Answer

Correct option: D.

$125$

$(3^3)^2 = 9^x$
$\Rightarrow (3^2)^3 = (3^2)^x$
$\Rightarrow x = 3$
Then $5^x = 5^3 = 125$

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Question 631 Mark

$\big(6+\sqrt{27}\big)-\big(3+\sqrt{3}\big)+\big(1-2\sqrt{3}\big)$ when simplified is:

positive and irrational.
positive and rational.
negative and irrational.
negative and rational.

Answer

positive and rational.

Solution:
$\big(6+\sqrt{27}\big)-\big(3+\sqrt{3}\big)+\big(1-2\sqrt{3}\big)$
$=6+\sqrt{27}-3-\sqrt{3}-2\sqrt{3}$
$=6-3+3\sqrt{3}-\sqrt{3}-2\sqrt{3}$
$=3$
which is positive and rational number.
Hence, the correct answer is option (b).

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Question 641 Mark

The simplest for of $1.\overline{6}$ is:

$\frac{833}{500}$
$\frac{8}{5}$
$\frac{5}{3}$
None of these.

Answer

$\frac{5}{3}$

Solution:
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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Question 651 Mark

The value of $\sqrt{5+2\sqrt{6}}$ is:

$\sqrt{5}+\sqrt{6}$
$\sqrt{5}-\sqrt{6}$
$\sqrt{3}+\sqrt{2}$
$\sqrt{3}-\sqrt{2}$

Answer

$\sqrt{3}+\sqrt{2}$

Solution:
$\sqrt{5+2\sqrt{2}}=\sqrt{\big(\sqrt{2}\big)^2+\big(\sqrt{3}\big)^2+2\times\sqrt{2}\times\sqrt{3}}$
$=\sqrt{\big(\sqrt{2}+\sqrt{3}\big)^2}$
$=\sqrt{2}+\sqrt{3}$
Hence, the correct option is (c).

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Question 661 Mark

The value of $\Big(\frac{256\text{x}^{16}}{81\text{y}^4}\Big)^{-\frac{1}{4}}$ is:

$\frac{3\text{y}}{8\text{x}^4}$
$\frac{3\text{y}}{4\text{x}^4}$
$\frac{4\text{y}}{5\text{x}^4}$
$\frac{4\text{x}^4}{3\text{y}}$

Answer

$\frac{3\text{y}}{4\text{x}^4}$

Solution:
$\Big(\frac{256\text{x}^{16}}{81\text{y}^4}\Big)^{-\frac{1}{4}}=\Big(\frac{81\text{y}^{4}}{256\text{x}^{16}}\Big)^{\frac{1}{4}}$
$=\Big(\frac{3^4\text{y}^{4}}{4^4(\text{x}^4)^4}\Big)^{\frac{1}{4}}=\Big(\frac{3\text{y}}{4\text{x}^4}\Big)^{4\times\frac{1}{4}}=\frac{3\text{y}}{4\text{x}^4}$
Hence, the correct option is (b).

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Question 671 Mark

Simplified value of $(25)^{\frac{1}{3}}\times5^{\frac{1}{3}}$ is:

25
3
1
5

Answer

5

Solution:
$(25)^{\frac{1}{3}}\times5^{\frac{1}{3}}=5^{2\times\frac{1}{3}}\times5^\frac{1}{3}$
$=5^\frac{2}{3}\times5^\frac{1}{3}=5^{\frac{2}{3}+\frac{1}{3}}=5^{\frac{3}{3}}=5$

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Question 681 Mark

Which of the following is a rational number?

$1+\sqrt{3}$
$\pi$
$2\sqrt{3}$
$0$

Answer

$0$

Solution:
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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Question 691 Mark

The value of $\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$ is:

3
-3
9
$\frac{1}{3}$

Answer

3

Solution:
$\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}=\Big[\big(9^2\big)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$
$=9^{2\times\frac{1}{2}\times\frac{1}{2}}=9^{\frac{1}{2}}=3^{2\times\frac{1}{2}}=3$
Hence, the correct option is (a).

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Question 701 Mark

$\sqrt{10}\times\sqrt{15}=?$

$\sqrt{25}$
$5\sqrt{6}$
$6\sqrt{5}$
None of these.

Answer

$5\sqrt{6}$

Solution:
$\sqrt{10}\times\sqrt{15}=\sqrt{5\times2}\times\sqrt{5\times3}$
$=\sqrt{5}\times\sqrt{2}\times\sqrt{5}\times\sqrt{3}=5\sqrt{6}$
Hence, the correct answer is option (b).

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Question 711 Mark

$\big(-2-\sqrt{3}\big)\big(-2+\sqrt{3}\big)$ when simplified is:

positive and irrational.
positive and rational.
negative and irrational.
negative and rational.

Answer

positive and rational.

Solution:
$\big(-2-\sqrt{3}\big)\big(-2+\sqrt{3}\big)$
$\big(-2\big)^2-\big(\sqrt{3}\big)^2$
$=4-3=1,$ which is positive and rational number.
Hence, the correct answer is option (b).

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Question 721 Mark

The value of $\sqrt{3-2\sqrt{2}}$ is:

$\sqrt{3}+\sqrt{2}$
$\sqrt{3}-\sqrt{2}$
$\sqrt{2}+1$
$\sqrt{2}-1$

Answer

$\sqrt{2}-1$

Solution:
$\sqrt{3-2\sqrt{2}}=\sqrt{\big(\sqrt{2}\big)^2+(1)^2-2\times\sqrt{2}\times1}$
$=\sqrt{\big(\sqrt{2}-1\big)^2}$
$=\sqrt{2}-1$
Hence, the correct option is (d).

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Question 731 Mark

If $\text{x}=3+\sqrt{8}$ then $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=?$

34
56
28
63

Answer

34

Solution:
$\text{x}=3+\sqrt{8}$
$\Rightarrow\text{x}^2=\big(3+\sqrt{8}\big)^2=9+8+6\sqrt{8}=17+6\sqrt{8}$
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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Question 741 Mark

$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}=?$

$2$
$\sqrt{2}$
$2\sqrt{2}$
$4\sqrt{2}$

Answer

$2$

Solution:
$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}$
$=\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{2^5}$
$=2^\frac{1}{3}\times2^\frac{1}{4}\times2^{5\times\frac{1}{12}}$
$=2^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$
$=2^{\frac{4+3+5}{12}}$
$=2^{\frac{12}{12}}$
$=2$
Hence, the correct option is (a).

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Question 751 Mark

Which of the following is a true statment?

$\pi$ and $\frac{22}{7}$ are both rationals.
$\pi$ and $\frac{22}{7}$ are both irrationals
$\pi$ is rational and $\frac{22}{7}$ is irrational.
$\pi$ is irrational and $\frac{22}{7}$ is rational.

Answer

$\pi$ is irrational and $\frac{22}{7}$ is rational.

Solution:
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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M.C.Q - Page 2 - MATHS STD 9 Questions - Vidyadip