Question 2511 Mark
If $\sqrt{2}=1.414$ then $\sqrt{\frac{\big(\sqrt{2}-1\big)}{\big(\sqrt{2}+1\big)}}=?$
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).
View full question & answer→Question 2521 Mark
If 'm' is a positive integer which is not a perfect square, then $\sqrt{\text{m}}$ is:
Answer- An irrational number.
Solution:
In $\sqrt{\text{m}}$ ,if m is not perfect square ,then the value of $\sqrt{\text{m}}$ will be $\text{m}^\frac{1}{2}.$
But if it is perfect square then value of $\sqrt{\text{m}}=$ some integer,
Example $\sqrt{4}=2$
View full question & answer→Question 2531 Mark
Answer- Always a whole number.
Solution:
All natural number together with zero form the collection of all whole number.
So,
- Every natural number is a whole number.
- Zero is a whole number which is not a natural number.
View full question & answer→Question 2541 Mark
Value of $\sqrt[4]{(81)^{-2}}$ is:
Answer- $\frac{1}{9}$
Solution:
$\sqrt[4]{(81)^{-2}}$
$=\sqrt[4]{\frac{1}{(81)^2}}$
$=\sqrt[4]{\frac{1}{(9^2)^2}}$
$=\sqrt[4]{\frac{1}{9^4}}$
$=\Big(\frac{1}{9}\Big)^{4\times\frac{1}{4}}$
$=\frac{1}{9}$
View full question & answer→Question 2551 Mark
The decimal expansion that a rational number cannot have is:
Answer- $0.5030030003$
Solution:
0.5030030003
The decimal expansion that a rational number cannot have is non terminating, non repeating.
View full question & answer→Question 2561 Mark
How many digits are there in the repeating block of digits in the decimal expansion of $\frac{17}{7}$ is:
Answer- 6
Solution:
$\frac{17}{7}=2.\overline{428571}$
View full question & answer→Question 2571 Mark
The $\frac{\text{p}}{\text{q}}$ form of the number 0.8 is:
Answer- $\frac{8}{10}$
Solution:
$\frac{8}{10}$ or $\frac{4}{5}$
View full question & answer→Question 2581 Mark
The value of $\frac{3\sqrt{12}}{6\sqrt{27}}$ is:
Answer- $\frac{1}{3}$
Solution:
$\frac{3\sqrt{12}}{6\sqrt{27}}$
$=\frac{3\sqrt{4\times3}}{6\sqrt{9\times3}}\Leftrightarrow\frac{6\sqrt{3}}{18\sqrt{3}}$
$=\frac{1}{3}$
View full question & answer→Question 2591 Mark
The value of $\frac{2^0+7^0}{5^0}$ is:
Answer- 2
Solution:
$\frac{2^0+7^0}{5^0}$
$=\frac{1+1}{1}$
$=2$
View full question & answer→Question 2601 Mark
Choose the rational number which does not lie between $-\frac{2}{3}$ and $-\frac{1}{5}.$
Answer- $\frac{3}{10}$
Solution:
Since $\frac{3}{10}>-\frac{2}{3}$ and $\frac{3}{10}>-\frac{1}{5}$
View full question & answer→Question 2611 Mark
A rational number equivalent to a rational number $\frac{7}{19}$ is:
Answer- $\frac{21}{57}$
Solution:
Divide numerator and denominator by 3.
View full question & answer→Question 2621 Mark
The simplest from of $0.\overline{36}$ is:
Answer- $\frac{4}{11}$
Solution:
$0.\overline{36}=\frac{36}{36}=\frac{4\times9}{11\times9}$
$\frac{4}{11}$
View full question & answer→Question 2631 Mark
The value of $\frac{2}{\sqrt{5}-\sqrt{3}}$ is:
Answer- $\sqrt{5}+\sqrt{30}$
Solution:
$\frac{2}{\sqrt{5}-\sqrt{3}}$
multiplying nu nominator and denominator by
$\sqrt{5}+\sqrt{3},$ we get
$\frac{2(\sqrt{5}+\sqrt{3})}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}$
$=\frac{2(\sqrt{5}+\sqrt{3})}{5-3}=\sqrt{5}+\sqrt{3}$
View full question & answer→Question 2641 Mark
Which of the following is equal to 'x'?
Answer- $(\sqrt{\text{x}^3})^{\frac{2}{3}}$
Solution:
$(\sqrt{\text{x}^3})^{\frac{2}{3}}$
$=(\text{x}\frac{3}{2})^{\frac{2}{3}}$
$=\text{x}$
View full question & answer→Question 2651 Mark
$\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^\text{n}-2\times5^{\text{n}+1}}$ is equal to:
AnswerWe have to simplify $\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^\text{n}-2\times5^{\text{n}+1}}$
Taking $5^n$ as a common factor we get
$\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^\text{n}-2\times5^{\text{n}+1}}=\frac{5^\text{n}(5^2-6\times5^1)}{5^\text{n}(13-2\times5^1)}$
$=\frac{5^\text{n}(25-30)}{5^\text{n}(13-10)}$
$=\frac{-5}{3}$
View full question & answer→Question 2661 Mark
The decimal expansion of the number $\sqrt{2}$ is:
Answer- Non-terminating non-recurring.
Solution:
As $\sqrt{2}$ is an irrational number, so its decimal representation will be non terminating, non recurring.
View full question & answer→Question 2671 Mark
$\pi$ is:
Answer- An irrational number.
Solution:
$\pi$ = 3.14159265359..., which is non-terminating non-recurring.
Hence, it is an irrational number.
View full question & answer→Question 2681 Mark
$\big(6+\sqrt{27}\big)-\big(3+\sqrt{3}\big)+\big(1-2\sqrt{3}\big)$ when simplified is:
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).
View full question & answer→Question 2691 Mark
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is:
Answer- $6^{\frac{1}{4}}$
Solution:
$\sqrt{2}$ and $\sqrt{3}$
$=2^{\frac{1}{2}}$ and $3^{\frac{1}{2}}$
$=2^{\frac{2}{4}}$ and $3^{\frac{2}{4}}$
$=4^{\frac{1}{4}}$ and $9^{\frac{1}{4}}$
Irrational between $\sqrt{2}$ and $\sqrt{3}$ is $6^{\frac{1}{4}}$
View full question & answer→Question 2701 Mark
$(\frac{125}{216})^{\frac{-1}{3}}=$
Answer- $\frac{6}{5}$
Solution:
$(\frac{125}{216})^{\frac{-1}{3}}$
$\Rightarrow(\frac{5}{6})^3\times\frac{-1}{3}$
$\Rightarrow(\frac{5}{6})^{-1}$
$\Rightarrow\frac{6}{5}$
View full question & answer→Question 2711 Mark
If $\sqrt{7}=2.646$ then $\frac{1}{\sqrt{7}}=?$
Answer- 0.378
Solution:
$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times\frac{\sqrt{7}}{\sqrt{7}}$
$=\frac{\sqrt{7}}{7}$
$=\frac{1}{7}\times\sqrt{7}$
$=\frac{1}{7}\times2.646$
$=0.378$
View full question & answer→Question 2721 Mark
Every point on a number line represents:
Answer- a unique number.
Solution:
Every point on a number line represents a unique number.
Hence, the correct opion is (d).
View full question & answer→Question 2731 Mark
Every rational number is:
Answer- A real number.
Solution:
Every rational number $(1,4.5,10,\frac{1}{2},-27,\frac{75}{5},0)$ is a real number.
However, not every real number is a rational number.
Although some numbers that appear to be irrational are actually rational because they can be reduced i.e. $\sqrt{25}=5$
View full question & answer→Question 2741 Mark
If $\text{a}=7-4\sqrt{3},$ then the value of $\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}$ is :
Answerd. 4
Solution:
Let $\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}=\text{x}$
Then, squaring both side, we get
$\text{a}+\frac{1}{\text{a}}+2=\text{x}^2$
$\Rightarrow\frac{\text{a}^2+1}{\text{a}}+2=\text{x}^2$
Now, put the value of a,
$\frac{(7-4\sqrt{3})^2+1}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow\frac{49+48-56\sqrt{3}+1}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow\frac{98-56\sqrt{3}}{7-4\sqrt{3}}+2=\text{x}^2$
$\Rightarrow14\Big(\frac{7-4\sqrt{3}}{7-4\sqrt{3}}\Big)+2=\text{x}^2$
$\Rightarrow16=\text{x}^2$
$\Rightarrow\text{x}=4$
So, $\text{x}=\sqrt{\text{a}}+\frac{1}{\sqrt{\text{a}}}=4$
View full question & answer→Question 2751 Mark
The value of $\frac{4\sqrt{12}}{12\sqrt{27}}$ is:
Answer- $\frac{2}{9}$
Solution:
$\frac{4\sqrt{12}}{12\sqrt{27}}$
$=\frac{4\sqrt{4\times3}}{12\sqrt{9\times3}}$
$=\frac{8\sqrt{3}}{36\sqrt{3}}$
$=\sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}$
$=\sqrt{(\sqrt{2}-1)^2}$
$=1.414-1$
$=0.414$
$=\frac{2}{9}$
View full question & answer→Question 2761 Mark
Write the correct answer in the following:
The number obtained on rationalising the denominator of $\frac{1}{\sqrt{7}-2}$ is
Answer- $\frac{\sqrt{7}+2}{3}$
Solution:
$\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2}\times\frac{\sqrt{7}+2}{\sqrt{7}+2}$
$\frac{\sqrt{7}+2}{7-4}=\frac{\sqrt{7}+2}{3}$
Hence, (a) is the correct answer.
View full question & answer→Question 2771 Mark
If $\text{x}^{\frac{1}{12}}=49^{\frac{1}{24}},$ then the value of x is:
Answer- 7
Solution:
$\text{x}^{\frac{1}{12}}=49^{\frac{1}{24}},$
$\Rightarrow\text{x}^{\frac{1}{12}}=7\frac{2}{24}=7^\frac{1}{12}$
Equating both, x = 7
View full question & answer→Question 2781 Mark
If $0<\text{y}<\text{x},$ which statement must be true?
Answer- $\sqrt{\text{xy}}=\sqrt{\text{x}}\sqrt{\text{y}}$
Solution:
We have to find which statement must be true?
Given $0<\text{y}<\text{x},$
Option (a):
Left hand side:
$\sqrt{\text{x}}-\sqrt{\text{y}}=\sqrt{\text{x}\text{}}=\sqrt{\text{y}}$
Right Hand side:
$\sqrt{\text{x}-\text{y}}=\sqrt{\text{x}-\text{y}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (b):
$\sqrt{\text{x}}+\sqrt{\text{x}}=\sqrt{2\text{x}}$
Left hand side:
$\sqrt{\text{x}}+\sqrt{\text{x}}=2\sqrt{\text{x}}$
Right Hand side:
$\sqrt{2\text{x}}=\sqrt{2\text{x}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (c):
$\text{x}\sqrt{\text{y}}=\text{y}\sqrt{\text{x}}$
Left hand side:
$\text{x}\sqrt{\text{y}}=\text{x}\sqrt{\text{y}}$
Right Hand side:
$\text{y}\sqrt{\text{x}}=\text{y}\sqrt{\text{x}}$
Left hand side is not equal to right hand side
The statement is wrong.
Option (d):
$\sqrt{\text{xy}}=\sqrt{\text{x}}\sqrt{\text{y}}$
Left hand side:
$\sqrt{\text{xy}}=\sqrt{\text{xy}}$
Right Hand side:
$\sqrt{\text{x}}\sqrt{\text{y}}=\sqrt{\text{x}}\times\sqrt{\text{y}}$
$=\sqrt{\text{xy}}$
Left hand side is equal to right hand side
The statement is true.
Hence the correct choice is d.
View full question & answer→Question 2791 Mark
Which of the following is a rational number?
Answer- 0.853853853...
Solution:
The decimal expansion of a rational number is either terminating or non-terminating recurring.
Hence, 0.853853853... is a rational number.
Hence, the correct option is (d).
View full question & answer→Question 2801 Mark
If $\text{x}=\sqrt6+\sqrt5,$ then $\text{x}^2+\frac{1}{\text{x}^2}-2=$
Answer- 20
Solution :
$\text{x}^2+\frac{1}{\text{x}^2}-2=\Big(\text{x}-\frac{1}{\text{x}}\Big)^2$
$\text{x}=\sqrt6+\sqrt5$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt6+\sqrt5}=\frac{1}{\sqrt6+\sqrt5}\times\frac{\sqrt6-\sqrt5}{\sqrt6-\sqrt5}\\ \ =\frac{\sqrt6-\sqrt5}{1}=\sqrt6-\sqrt5$
Now,
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^2=\big[\sqrt6+\sqrt5-\big(\sqrt6-\sqrt5\big]^2\\ \ =\big(2\sqrt5\big)^2=4\times5=20$
Hence, correct option is (d).
View full question & answer→Question 2811 Mark
Write the correct answer in the following:
$\sqrt{10}\times\sqrt{15}$ is equal to.
Answer- $5\sqrt{6}$
Solution :
$\sqrt{10}, \sqrt{15}=\sqrt{2.5}\sqrt{3.5}=\sqrt{2}\sqrt{5}\sqrt{3}\sqrt{5}=5\sqrt{6}$
View full question & answer→Question 2821 Mark
If $\text{x}=2+\sqrt{3}$ then $\Big(\text{x}+\frac{1}{\text{x}}\Big)$ equals :
Answer- 4
Solution :
$\text{x}=2+\sqrt{3}$
$\therefore\frac{1}{\text{x}}=\frac{1}{2+\sqrt{3}}$
$=\frac{1}{2+\sqrt{3}}\times\frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{2^2-\big(\sqrt{3}\big)^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
$\therefore\Big(\text{x}+\frac{1}{\text{x}}\Big)=\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4$
Hence, the correct option is (c).
View full question & answer→Question 2831 Mark
The value of $\Big(\frac{\text{x}^{\text{l}}}{\text{x}^{\text{m}}}\Big)^{\frac{1}{\text{lm}}}\times\Big(\frac{\text{x}^{\text{m}}}{\text{x}^{\text{n}}}\Big)^{\frac{1}{\text{mn}}}\times\Big(\frac{\text{x}^{\text{n}}}{\text{x}^{\text{l}}}\Big)^{\frac{1}{\text{nl}}}$ is :
Answer- 1
Solution :
$\Big(\frac{\text{x}^{\text{l}}}{\text{x}^{\text{m}}}\Big)^{\frac{1}{\text{lm}}}\times\Big(\frac{\text{x}^{\text{m}}}{\text{x}^{\text{n}}}\Big)^{\frac{1}{\text{mn}}}\times\Big(\frac{\text{x}^{\text{n}}}{\text{x}^{\text{l}}}\Big)^{\frac{1}{\text{nl}}}$
$\Rightarrow\text{x}^{\text{a}^{2}-\text{b}^2}\times\text{x}^{\text{b}^2-\text{c}^2}\times\text{x}^{\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^{\text{a}^2-\text{b}^2+\text{b}^2-\text{c}^2+\text{c}^2-\text{a}^2}$
$\Rightarrow\text{x}^0=1$
View full question & answer→Question 2841 Mark
The Product $\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}$ is equal to:
Answer- $2$
Solution:
$\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{32}$
$=\sqrt[3]{2}\times\sqrt[4]{2}\times\sqrt[12]{(2)^5}$
$=(2)^{\frac{1}{3}}\times(2)^{\frac{1}{4}}\times(2)^{\frac{5}{15}}$
$=(2)^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$
$=(2)^{\frac{4+3+5}{12}}$
$=(2)^{\frac{12}{12}}$
$=2$
View full question & answer→Question 2851 Mark
The value of $\frac{1}{\sqrt{8}-3\sqrt{2}}$ is:
Answer- $-\frac{1}{\sqrt{2}}$
Solution:
$\frac{1}{\sqrt{8}-3\sqrt{2}}$
$\Rightarrow\frac{1}{2\sqrt{2}-3\sqrt{2}}$
$\Rightarrow\frac{1}{-\sqrt{2}}\Leftrightarrow\frac{-1}{\sqrt{2}}$
View full question & answer→Question 2861 Mark
On simplification, the expression $\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^{\text{n}}-2\times5^{\text{n}+1}}$ equals :
Answer- $-\frac{5}{3}$
Solution :
$\frac{5^{\text{n}+2}-6\times5^{\text{n}+1}}{13\times5^{\text{n}}-2\times5^{\text{n}+1}}$
$=\frac{5^{\text{n}+1}(5-6)}{5^{\text{n}}(13-2\times5)}$
$=\frac{5^{\text{n}}\times5\times(-1)}{5^{\text{n}}(13-10)}$
$=-\frac{5}{3}$
Hence, the correct option is (b).
View full question & answer→Question 2871 Mark
The simplest of $1.\bar6$ is:
Answer- $\frac{5}{3}$
Solution:
Let x = 1.666... ---(i)
Multiply eq. (i) by 10, we get
10x = 16.666... ----(ii)
Subtract eq(i) from (ii) we get
9x = 15
$\text{x}=\frac{5}{3}$
View full question & answer→Question 2881 Mark
If $\frac{3-\sqrt{5}}{3+2\sqrt{5}}=\text{a}\sqrt{5}-\frac{19}{11}\text{b},$ than the value of 'b' is:
Answer- 1
Solution:
$\frac{3-\sqrt{5}}{3+2\sqrt{5}}=\text{a}\sqrt{5}-\frac{19}{11}\text{b},$
Taking LHS,
$\Rightarrow\frac{3-\sqrt{5}}{3+2\sqrt{5}}\times\frac{3-2\sqrt{5}}{3-2\sqrt{5}}$
$\Rightarrow\frac{3(3-2\sqrt{5})-\sqrt{5}(3-2\sqrt{5})}{9-20}$
$\Rightarrow\frac{9-6\sqrt{5}-3\sqrt{5}+10}{-11}$
$\Rightarrow\frac{19-9\sqrt{5}}{-11}$
$\Rightarrow\frac{-19}{11}+\frac{9\sqrt{5}}{11}$
Equating this with RHS,
We get,
$=\frac{-19}{11}\text{b}=-\frac{19}{11}$
$\Rightarrow\text{b}=1$
View full question & answer→Question 2891 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: $3\sqrt2$ is a irrational number.
Reason: If m is a positive integer which is not a perfect cube, then $3\sqrt{\text{m}}$ is irrational.
Answer - Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
View full question & answer→Question 2901 Mark
If $8^{x+1} = 64 ,$ what is the value of $3^{2x+1}$?
AnswerWe have to find the value of $3^{2x+1}$ provided $8^{x+1} = 64$
So,
$2^{3(x+1)} = 64$
$2^{3x+3} = 2^6$
Equating the exponents we get
$3x + 3 = 6$
$3x = 6 - 3$
$3x = 3$
$\text{x}=\frac{3}{3}$
$x = 1$
By substitute in $3^{2x+1}$ we get
$= 3^{2\times 1+1}$
$= 3^{2+1}$
$= 3^3$
$= 27$
The real value of $3^{2x+1}$ is $27$
View full question & answer→Question 2911 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 2921 Mark
The difference of two irrational numbers is.
Answer- Either irrational or rational.
Solution :
Difference of two irrationals need not be an irrational.
Example:- each one of $(5\div\sqrt{2})$ and $(5-\sqrt{2})$ is irrational,
But, $(5+\sqrt{2})-(3+\sqrt{2})=2,$ which is rational.
View full question & answer→Question 2931 Mark
The seventh root of x divided by the eighth root of x is:
Answer- $\sqrt[56]{\text{x}}$
Solution:
We have to find he seventh root of x divided by the eighth root of x, so let it be L. So,
$\text{L}=\frac{\sqrt[7]{\text{x}}}{\sqrt[8]{\text{x}}}$
$=\frac{\text{x}^{\frac{1}{7}}}{\text{x}^{\frac{1}{8}}}$
$=\text{x}^{\frac{1}{7}-\frac{1}{8}}$
$=\text{x}^{\frac{1\times8}{7\times8}-\frac{1\times7}{8\times7}}$
$=\text{L}=\text{x}^{\frac{8}{56}-\frac{7}{56}}$
$=\text{x}^{\frac{1}{56}}$
$=\sqrt[56]{\text{x}}$
The seventh root of x divided by the eighth root of x is $=\sqrt[56]{\text{x}}$
Hence the correct choice is c.
View full question & answer→Question 2941 Mark
The sum of $0.\overline{3}$ and $0.\overline{4}$ is:
Answer- ${\frac{7}{9}}$
Solution:
$0.\overline{3}+0.\overline{4}$
$=0.\overline{7}=\frac{7}{9}$
View full question & answer→Question 2951 Mark
$(27)^{-\frac{2}{3}}$ is Equal to:
Answer$27$ can be written as $(3)^3$
So$, (27)^{-\frac{2}{3}}=\{(3)^3\}^{-\frac{2}{3}}$
$=\frac{1}{(3)^2}=\frac{1}{9}$
View full question & answer→Question 2961 Mark
If $\frac{\text{x}}{\text{x}^{1.5}}=8\text{x}^{-1}$ then x =
Answer- $64$
Solution:
For $\frac{\text{x}}{\text{x}^{1.5}}=8\text{x}^{-1}$ we have to find the value of x.
So,
$\frac{\text{x}^1}{\text{x}^{1.5}}=8\text{x}^{-1}$
$\text{x}^{1-1.5}8\text{x}^{-1}$
$\text{x}^{-0.5}=2^3\text{x}^{-1}$
$\frac{\text{x}^{0.5}}{\text{x}^{-1}}=2^3$
$\frac{\text{x}^{-\frac{5}{10}}}{\text{x}^{-1}}=2^3$
$\text{x}^{-\frac{1}{2}+1}=2^3$
$\text{x}^{\frac{1}{2}+\frac{2}{2}}=2^3$
$\text{x}^{\frac{-1+2}{2}}=2^3$
$\text{x}^{\frac{1}{2}}=2^3$
By raising both sides to the power 2 we get
$\text{x}^{\frac{1}{2}\times2}=2^{3\times2}$
$\text{x}^{\frac{1}{2}\times2}=2^6$
$\text{x}^1=64$
The value of x is 64
Hence the correct alternative is d.
View full question & answer→Question 2971 Mark
Which of the following is irrational?
Answer- 0.4014001400014
Solution:
0.4014001400014 is the irrational number.
Because an irrational number is non terminating and non repeation.
View full question & answer→Question 2981 Mark
The sum of $0.\overline{3}$ and $0.\overline{4}$ is:
Answer- $\frac{7}{9}$
Solution:
Let $\text{x}=0.\overline{3}$
i.e., $\text{x}=0.3333 \ ...(\text{i})$
$\Rightarrow10\text{x}=3.3333 \ ...(\text{ii})$
On subtracting (i) and (ii), we get
$9\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{9}$
Let $\text{y}=0.\overline{4}$
i.e., $\text{y}=0.4444 \ ...(\text{i})$
$\Rightarrow10\text{y}=4.4444 \ ...(\text{ii})$
On subtracting (i) and (ii), we get
$9\text{y}=4$
$\Rightarrow\text{y}=\frac{4}{9}$
$\therefore0.\overline{3}+0.\overline{4}=\text{x}+\text{y}=\frac{\text{3}}{\text{9}}+\frac{\text{4}}{\text{9}}=\frac{\text{7}}{\text{9}}$
Hence, the correct answer is option (b).
View full question & answer→Question 2991 Mark
Which of the following is a true statement?
Answer- $\pi$ is irrational and $\frac{22}{7}$ is rational.
Solution:
$\pi$ is irrational because $\frac{22}{7}$ is not the exact value of $\pi.$
But, here $\frac{22}{7}$ is fraction so, it is rational.
View full question & answer→Question 3001 Mark
After rationalising the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}},$ we get the denominator as.
Answer- 19
Solution:
After rationalizing:
$\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{7}{3\sqrt{3}-2\sqrt{2}}\times\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2}$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{27-8}$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{19}$
View full question & answer→