Question 3011 Mark
How many digits are there in the repeating block of digits in the decimal expansion of $\frac{17}{7}?$
Answer- 6
Solution:
$\frac{17}{7}=2.\overline{428571}$
Hence, the correct opion is (b).
View full question & answer→Question 3021 Mark
The value of $\sqrt[3]{1000}$ is:
Answer$(10)^3 = 1000$
So$, \sqrt[3]{1000}=1000^{\frac{1}{3}}=(10^3)^{\frac{1}{3}}$
$=10$
View full question & answer→Question 3031 Mark
$(125)^{-\frac{1}{3}}=?$
Answer- $\frac{1}{5}$
Solution:
$(125)^{-\frac{1}{3}}$
$=(5^3)^{-\frac{1}{3}}$
$=5^{-1}$
$=\frac{1}{5}$
View full question & answer→Question 3041 Mark
Which of the following is a rational number?
Answer- $\sqrt{225}$
Solution:
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.
$\sqrt{2},$ $\sqrt{23}$ and 0.1010010001... are irrational numbers, since they cannot be expressed in the form $\frac{\text{p}}{\text{q}}.$
$\sqrt{225}=15=\frac{15}{1}$ is a rational number.
Hence, the correct opion is (c).
View full question & answer→Question 3051 Mark
The rational number not lying between $-\frac{2}{5}$ and $-\frac{1}{5}$ is:
Answer- $0$
Solution:
Zero is greater than both numbers.
View full question & answer→Question 3061 Mark
If $\text{x}+\sqrt{15}=4,$ then $\text{x}+\frac{1}{\text{x}}=$
Answer- 8
Solution:
$\text{x}+\sqrt{15}=4$
$=\text{x}=4-\sqrt{15}\Rightarrow\frac{1}{\text{x}}=\frac{1}{4-\sqrt{15}}$
$\frac{1}{\text{x}}=\frac{1}{4-\sqrt{15}}\times\frac{4+\sqrt{15}}{4+\sqrt{15}}=\frac{4+\sqrt{15}}{(4)^2-\big(\sqrt{15}\big)^2}\\ \ =\frac{4+\sqrt{15}}{16-15}=4+\sqrt{15}$
Now, $\text{x}+\frac{1}{\text{x}}=4-\sqrt{15}+4+\sqrt{15}=8$
Hence, correct option is (c).
View full question & answer→Question 3071 Mark
A rational number lying between $\sqrt{2}$ and $\sqrt{3}$ is:
Answer- 1.6
Solution:
$\sqrt{2}=1. 414213562...$ and $\sqrt{3}=1. 7320508075...$
Hence, 1.6 lies between $\sqrt{2}$ and $\sqrt{3}$
Hence, the correct option is (c).
View full question & answer→Question 3081 Mark
If $\sqrt{7}=2.646$ then $\frac{1}{\sqrt{7}}=?$
Answer- 0.378
Solution:
$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}$
$=\frac{\sqrt{7}}{7}$
$=\frac{1}{7}\times\sqrt{7}$
$=\frac{1}{7}\times2.646$
$=0.378$
Hence, the correct option is (b).
View full question & answer→Question 3091 Mark
$\sqrt{10}\times\sqrt{15}$ is equal to :
Answer- $5\sqrt6$
Solution :
10 = 5 × 2
15 = 5 × 3
$\therefore\sqrt{10}\times\sqrt{15}=\sqrt{5\times2}\times\sqrt{5\times3}$
$=\sqrt{5}\times\sqrt{2}\times\sqrt{5}\times\sqrt3$
$=\big(\sqrt5\times\sqrt5\big)\times\sqrt2\times\sqrt3$
$=5\sqrt6$
Hence, correct option is (a).
View full question & answer→Question 3101 Mark
The number of consecutive zeros in $2^3 \times 3^4 \times 5^4 \times 7$, is:
Answer$5 \times 2 = 10$
$\Rightarrow$ one $5$ and one $2$ make one zero, so $5 \times 2 \times 5 \times 2 = 100$
Numbers of pairs of $5$ and $2$ will be equal to the number of consecutive zeros in the given number.
In the given number, there are three $2\ 's$ and four$ 5\ 's.$
So number of pairs of $5$ and $2$ are only three.
So there will be three consecutive zeros in the given number
View full question & answer→Question 3111 Mark
The value of $\Big(\frac{256\text{x}^{16}}{81\text{y}^{4}}\Big)^{-\frac{1}{4}}$ is:
Answer- $\frac{3\text{y}}{4\text{x}^4}$
Solution:
$\Big(\frac{256\text{x}^{16}}{81\text{y}^{4}}\Big)^{-\frac{1}{4}}$
$=\Big(\frac{81\text{y}^{4}}{256\text{x}^{16}}\Big)^{\frac{1}{4}}$
$=\Big(\frac{3^4\text{y}^4}{4^4\text{x}^{16}}\Big)^{\frac{1}{4}}$
$=\big[\Big(\frac{3\text{y}}{4\text{x}^4}\Big)^4\big]^{\frac{1}{4}}$
$=\Big(\frac{3\text{y}}{4\text{x}^4}\Big)^{\frac{1}{4}\times4}$
$=\frac{3\text{y}}{4\text{x}^4}$
View full question & answer→Question 3121 Mark
When simplified $\Big(-\frac{1}{27}\Big)^{-\frac{2}{3}}$ is:
Answer- $9$
Solution:
We have to find the value of $\Big(-\frac{1}{27}\Big)^{\frac{-2}{3}}$
So,
$\Big(-\frac{1}{27}\Big)^{\frac{-2}{3}}=\Big(-\frac{1}{27}\Big)^{\frac{-2}{3}}$
$=\Big(-\frac{1}{3^3}\Big)^{\frac{-2}{3}}$
$=-\frac{1}{3^{3\times\frac{-2}{3}}}$
$\Big(-\frac{1}{27}\Big)^{\frac{-2}{3}}=-\frac{1}{3^{-2}}$
$=-\frac{1}{\frac{1}{3^2}}$
$=\frac{1}{\frac{1}{9}}$
$=9$
Hence the correct choice is a.
View full question & answer→Question 3131 Mark
Between any two rational numbers there.
Answer- Are many rational numbers.
Solution:
Between any two rational number there are many rational number,
Example - 4 and 8
We have 5, 6, 7, 7.5....... and many more.
View full question & answer→Question 3141 Mark
$\sqrt{8}$ is an:
Answer- Irrational number.
Solution:
$\sqrt{8}$ is in irrational number
$\because\sqrt{4\times2}=2\sqrt{2}$
View full question & answer→Question 3151 Mark
The simplest rationalising factor of $\sqrt3+\sqrt5,$ is:
Answer- $\sqrt{3}-\sqrt5$
Solution:
Rationalising factor of any number of kind $\sqrt{\text{a}}\pm\sqrt{\text{b}}$ is $\sqrt{\text{a}}\mp\sqrt{\text{b}}$
So. for given number $\sqrt3+\sqrt5.$ Rationalising factor would be $\sqrt3-\sqrt5.$
Hence, correct option is (c).
View full question & answer→Question 3161 Mark
Which of the following numbers can be represented as non-terminating, repeating decimals?
Answer- $\frac{3}{11}$
Solution:
- $\frac{39}{24}=1.625=$ Terminating Decimal
- $\frac{3}{16}=0.1875=$ Terminating Decimal
- $\frac{3}{11}=0.27272727 \ ...=$ Non-Terminating Decimal
- $\frac{137}{25}=5.48=$ Terminating Decimal
Hence, option (c) is correct.
View full question & answer→Question 3171 Mark
Write the correct answer in the following:
Decimal representation of a rational number cannot be.
Answer- Non - terminating non - repeating.
Solution:
Decimal representation of a rational number cannot be non - terminating non-repeating because the decimal expansion of rational number is either terminating or non - terminating recurring (repeating).
View full question & answer→Question 3181 Mark
The decimal representation of a rational number is:
Answer- either terminating or repeating.
Solution:
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.
Decimal representation of a rational number is either terminating or a repeating decimal, since every decimal of this form can be expressed in the form $\frac{\text{p}}{\text{q}}.$
Hence, the correct opion is (b).
View full question & answer→Question 3191 Mark
Which of the following is an irrational number:
Answer- 3.141141114
Solution:
This is an irrational number because there is no repetition of numbers after the decimal.
View full question & answer→Question 3201 Mark
If $\text{x}=2+\sqrt{3}$ then $\text{x}+\frac{1}{\text{x}}=$
Answer- 4
Solution:
$\text{x}+\frac{1}{\text{x}}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}$
Now put, $\text{x}=2+\sqrt{3}$
We have,
$\frac{(2+\sqrt{3})^2+1}{2+\sqrt{3}}$
$\Rightarrow\frac{4+3+2(2\sqrt{3})+1}{2+\sqrt{3}}$
$\Rightarrow\frac{8+4\sqrt{3}}{2+\sqrt{3}}$
$\Rightarrow\frac{4(2+\sqrt{3})}{2+\sqrt{3}}$
$=4$
View full question & answer→Question 3211 Mark
An irrational number between 5 and 6 is:
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).
View full question & answer→Question 3221 Mark
Which of the following is a correct statement?
Answer- $5\text{a}^2\text{bc}^2$
Solution:
Find value of $\sqrt[5]{3125\text{a}^{10}\text{b}^{5}\text{c}^{10}}.$
$\sqrt[5]{3125\text{a}^{10}\text{b}^{5}\text{c}^{10}}=\sqrt[5]{5^5\text{a}^{10}\text{b}^{5}\text{c}^{10}}$
$=5^{5\times\frac{1}{5}}\text{a}^{10\times\frac{1}{2}}\text{b}^{5\times\frac{1}{5}}\text{c}^{10\times\frac{1}{5}}$
$\sqrt[5]{3125\text{a}^{10}\text{b}^5\text{c}^{10}}=5\text{a}^2\text{bc}^2$
Hence the correct choice is a.
View full question & answer→Question 3231 Mark
If $\text{x}=3+\sqrt{8}$ then $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=?$
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).
View full question & answer→Question 3241 Mark
The decimal expansion of $\sqrt{2}$ is :
Answer- nonterminating, nonrecurring.
Solution:
The decimal expansion of $\sqrt{2}=1.41421356...,$ which is non-terminating, nonrecurring.
Hence, the correct opion is (d).
View full question & answer→Question 3251 Mark
Which of the following is an irrational number.
Answer- $\sqrt{23}$
Solution:
It doesn't represent the form of $\frac{\text{p}}{\text{q}}$ and $\text{q}\not=0$
View full question & answer→Question 3261 Mark
The value of $x - y^{x-y}$ when$ x = 2$ and $y = -2$ is :
AnswerGiven $\text{x}-\text{y}^{\text{x}-\text{y}}$
Here $\text{x}=\text{2},\ \text{y}=-2$
By substituting in $\text{x}-\text{y}^{\text{x}-\text{y}}$ we get
$\text{x}-\text{y}^{\text{x}-\text{y}}=2-(-2)^{2-(-2)}$
$=2-(-2)^{2+2}$
$=2-(-2)^4$
$=2-(16)$
$=-14$
The value of $\text{x}-\text{y}^{\text{x}-\text{y}}$ is $-14$
View full question & answer→Question 3271 Mark
The Number $1.\overline{27}$ in the from $\frac{\text{p}}{\text{q}} ,$ where p and q are integers and $\text{q}\not=0,$ is:
Answer- $\frac{14}{11}$
Solution:
Let $\text{x}=1.\overline{27}=1.272727.....(\text{i})$
Multiplying equation (i) with 100 [since there is bar over 2 & 7, (two digits)]
100x = 100 × (1.272727...)
100x = 127.2727 .......(ii)
Subtracting (i) from (ii) i.e. (ii) - (i)
100x - x = 127.2727 ... - 1.272727...
99x = 126.000
$\text{x}=\frac{126}{99}$
$\text{x}=\frac{14}{11}$
Thus, $1.\overline{27}=\frac{14}{11}$
View full question & answer→Question 3281 Mark
If $(3^3)^2 = 9^x$ then $5^x =$ ?
Answer$(3^3)^2 = 9^x$
$\Rightarrow (3^2)^3 = (3^2)^x$
$\Rightarrow x = 3$
Then $5^x = 5^3 = 125$
View full question & answer→Question 3291 Mark
If $2^{-\text{m}}\times\frac{1}{2^{\text{m}}}=\frac{1}{4},$ then $\frac{1}{14}\bigg\{(4^\text{m})^{\frac{1}{2}}\Big(\frac{1}{5^{\text{m}}}\Big)^{-1}\bigg\}$ is equal to:
Answer- $\frac{1}{2}$
Solution:
We have to find the value of $\frac{1}{14}\bigg\{(4^\text{m})^{\frac{1}{2}}\Big(\frac{1}{5^{\text{m}}}\Big)^{-1}\bigg\}$ provided $2^{-\text{m}}\times\frac{1}{2^{\text{m}}}=\frac{1}{4},$
Consider,
$2^{-\text{m}}\times\frac{1}{2^{\text{m}}}=\frac{1}{4},$
$=\frac{1}{2^{\text{m}}}\times\frac{1}{2^\text{m}}$
$=\frac{1}{2^\text{m}\times2^\text{m}}$
$\frac{1}{2^{2\text{m}}}=\frac{1}{2^2}$
Equating the power of exponents we get
$2\text{m} = 2$
$\text{m}=\frac{2}{2}$
$\text{m}=1$
By substituting $\frac{1}{14}\bigg\{(4^\text{m})^{\frac{1}{2}}\Big(\frac{1}{5^{\text{m}}}\Big)^{-1}\bigg\}$ we get
$\frac{1}{14}\bigg\{(4^\text{m})^{\frac{1}{2}}\Big(\frac{1}{5^{\text{m}}}\Big)^{-1}\bigg\}=\frac{1}{14}\bigg\{4^{1\times\frac{1}{2}}+\Big(\frac{1}{5^1}\Big)^{-1}\bigg\}$
$=\frac{1}{14}\Big\{2^{2\times\frac{1}{2}}+\frac{1}{5^{-1}}\Big\}$
$=\frac{1}{14}\Big\{2+1\times\frac{5}{1}\Big\}$
$\frac{1}{14}\bigg\{(4^\text{m})^{\frac{1}{2}}\Big(\frac{1}{5^{\text{m}}}\Big)^{-1}\bigg\}=\frac{1}{14}\{2+5\}$
$=\frac{1}{14}(7)$
$=\frac{1}{14}\times7$
$=\frac{1}{2}$
Hence the correct choice is a.
View full question & answer→Question 3301 Mark
The decimal form of $\frac{2}{11}$ is:
Answer- $0.\overline{18}$
Solution:
When we divide 2 by 11
We have value = 0.181818
Which is $0.\overline{18}$
View full question & answer→Question 3311 Mark
The rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ is:
Answer- $\sqrt{12}+\sqrt{5}$
Solution:
$\frac{1}{2\sqrt{3}-\sqrt{5}}$
$=(2\sqrt{3}-\sqrt{5})(2\sqrt{3}+\sqrt{5})$
$=12-5$
$=7$
Rational number
$(2\sqrt{3}+\sqrt{5})=(\sqrt{4\times3}+\sqrt{5})=\sqrt{12}+\sqrt{5}$
View full question & answer→Question 3321 Mark
The value of $\sqrt{3-2\sqrt2},$ is:
Answer- $\sqrt2-1$
Solution:
$\sqrt{3-2\sqrt2}$
$=\sqrt{2+1-2\sqrt2}$
$=\sqrt{\big(\sqrt2\big)^2+(1)^2-2\big(\sqrt2\big)(1)}$
$=\sqrt{\big(\sqrt2-1\big)^2}$
$=\sqrt2-1$
Hence, correct option is (a).
View full question & answer→Question 3331 Mark
$8\sqrt{15}\div2\sqrt{3}$
Answer- $4\sqrt{5}$
Solution:
$\frac{8\sqrt{5}\times\sqrt{3}}{2\sqrt{3}}$
$\Rightarrow\frac{8\sqrt{5}\times\sqrt{3}}{2\sqrt{3}}$
$\Rightarrow\frac{8\sqrt{5}}{2}$
$\Rightarrow4\sqrt{5}$
View full question & answer→Question 3341 Mark
The value of $\sqrt{20}\times\sqrt{5}$ is :
Answer- $10$
Solution:
$\sqrt{20}\times\sqrt{5}$
$=2\sqrt{5}\times\sqrt{5}$
$=2\times5$
$=10$
View full question & answer→Question 3351 Mark
The value of $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}},$ is:
Answer- $\frac{4}{3}$
Solution:
$\sqrt{48}=\sqrt{16\times3}=4\sqrt3$
$\sqrt{32}=\sqrt{16\times2}=4\sqrt2$
$\sqrt{27}=\sqrt{9\times3}=3\sqrt3$
$\sqrt{18}=\sqrt{9\times2}=3\sqrt2$
Now, $\frac{\sqrt{48}+\sqrt{32}}{\sqrt{27}+\sqrt{18}}=\frac{4\sqrt3+4\sqrt2}{3\sqrt3+3\sqrt2}$
$=\frac{4\big(\sqrt{\not3}+\sqrt{\not2}\big)}{3\big(\sqrt{\not3}+\sqrt{\not2}\big)}$
$=\frac{4}{3}$
Hence, correct option is (a).
View full question & answer→Question 3361 Mark
If $\sqrt{3}=1.732$ and$\sqrt{3}=1.732$ then the value of $\frac{1}{\sqrt{3}-\sqrt{2}}$ is:
Answer- 3.146
Solution:
$\frac{1}{\sqrt{3}-\sqrt{2}}$
$\Rightarrow\frac{1}{\sqrt{3}-\sqrt{2}}\times\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\Rightarrow\frac{\sqrt{3}+\sqrt{2}}{3-2}=\sqrt{3}+\sqrt{2}$
$\Rightarrow1.732+1.414$
$\Rightarrow3.146$
View full question & answer→Question 3371 Mark
The rationalisation factor of $\sqrt3,$ is:
Answer- $\frac{1}{\sqrt3}$
Solution:
Rationalisation factor of any number like $\sqrt{\text{a}}$ is $\frac{1}{\text{a}}$ or $\frac{1}{\text{a}}$ is $\sqrt{\text{a}}.$
So. Rationalisation factor of $\sqrt3$ is $\frac{1}{\sqrt3}.$
Hence, correct option is (b).
View full question & answer→Question 3381 Mark
The simplest form of $0.12\overline{3}$ is:
Answer- None of these.
Solution:
Let $\text{x}=0.12\overline{3}$
Then, $\text{x}=0.12333 \ ...(\text{i})$
$\therefore100\text{x}=12.333... \ (\text{ii})$
and $1000\text{x}=123.333... \ (\text{iii})$
On subtracting (ii) from (iii), we get
$900\text{x}=111$
$\Rightarrow\text{x}=\frac{111}{900}=\frac{37}{300}$
View full question & answer→Question 3391 Mark
The simplest rationalisation factor of $\sqrt[3]{500}$ is :
Answer- $\sqrt[3]{2}$
Solution :
$\sqrt[3]{500}=500^{\frac{1}{3}}=\Big(\frac{500\times2}{2}\Big)^{\frac{1}{3}}$ $=\Big(\frac{1000}{2}\Big)^{\frac{1}{3}}=\frac{10^{3\times\frac{1}{3}}}{2^{\frac{1}{3}}}=\frac{10}{\sqrt[3]{2}}$
Thus, the simplest rationalisation factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}.$
Hence, the correct option is (d).
View full question & answer→Question 3401 Mark
Directions : In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following :
Assertion : $( 2+\sqrt2)^2=6+4\sqrt2$
Reason : $ (\text{a}+\text{b})^2=\text{a}^2+\text{b}^2+2\text{ab}$
Answer - Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
View full question & answer→