Question 1011 Mark
Which of the following is an irrational number?
Answer- $\sqrt{7}$
Solution:
Because it cannot be express in $\frac{\text{p}}{\text{q}}$ form Or fraction.
View full question & answer→Question 1021 Mark
Write the correct answer in the following:
The decimal expansion of the number $\sqrt{2}$ is.
Answer- Non - terminating non - recurring.
Solution:
The decimal expansion of the number $\sqrt{2}$ is non-terminating non - recurring. Because $\sqrt{2}$ is an irrational number.
Also, we know that an irrational number is non - terminating non - recurring.
View full question & answer→Question 1031 Mark
If $\text{x}=3+\sqrt{8},$ than the value of $\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)$ is:
Answer- 34
Solution:
Given $\text{x}=3+\sqrt{8},$
$\frac{1}{\text{x}}=\frac{1}{(3+\sqrt{8})}=\frac{1}{(3+\sqrt{8})}\times\frac{(3-\sqrt{8})}{(3-\sqrt{8})}$
$=\frac{(3-\sqrt{8})}{(3^2-(\sqrt{8})^2}=\frac{(3+\sqrt{8})}{(9-8)}=(3-\sqrt{8})$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)=(3+\sqrt{8})+(3-\sqrt{8})=6$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=6^2=36$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+2\times\text{x}\times\frac{1}{\text{x}}=36$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)+2=36$
$\Rightarrow\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big) =36-2=34$
View full question & answer→Question 1041 Mark
Which of the following is irrational?
Answer- 0.1014001400014...
Solution:
- $0.14=\frac{14}{100},$ which is a Rational number
- $0.14\overline{16}$ is non-terminating but repeating, hence a rational number
- $0.\overline{1416}$ is non-terminating but repeating, hence a rational number
- 0.1014001400014... is non-terminating as well as non-repeating number, which is irrational in nature.
Hence, correct option is (d).
View full question & answer→Question 1051 Mark
If $\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$ and $\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3},$ then $\text{x}+\text{y}+\text{xy}=$
Answer- 9
Solution:
$\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}$
$\therefore\text{x}=\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\frac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}\\ \ =\frac{\big(\sqrt5+\sqrt3\big)^2}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}=\frac{8+2\sqrt{15}}{2}=4+\sqrt{15}$
$\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}$
$\therefore\text{y}=\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\times\frac{\sqrt5-\sqrt3}{\sqrt5-\sqrt3}\\ \ =\frac{\big(\sqrt5-\sqrt3\big)^2}{\big(\sqrt5\big)^2-\big(\sqrt3\big)^2}=\frac{8-2\sqrt{15}}{2}=4-\sqrt{15}$
$\text{xy}=\big(4+\sqrt{15}\big)\big(4-\sqrt{15}\big)=16-15=1$
Now, $\text{x}+\text{y}+\text{xy}=4+\sqrt{15}+4-\sqrt{15}+1\\ \ =4+4+1=9$
Hence, correct option is (a).
View full question & answer→Question 1061 Mark
Write the correct answer in the following:
Value of $\sqrt[4]{(81)^{-2}}$ is.
Answer- $\frac{1}{9}$
Solution:
$\sqrt[4]{(81)^{-2}}=\sqrt[4]{\Big(\frac{1}{81}\Big)^2}=\sqrt[4]{\Big\{\Big(\frac{1}{9}\Big)^2\Big\}^2}=\sqrt[4]{\Big(\frac{1}{9}\Big)^4}=\Big(\frac{1}{9}\Big)^{4\times\frac{1}{4}}=\frac{1}{9}$
Hence, (a) is the correct answer.
View full question & answer→Question 1071 Mark
Which of the following statements is true?
Answer- Product of a rational and an irrational number is always irrational.
Solution:
- Is incorrect, Product of two irrational numbers is not always irrational, it can be also rational sometimes.
when an irrational number is multiplied to itself, or multiplied by another irrational, that product becomes a perfect square.
Example:
$\sqrt{2}\times\sqrt{2}=2$ (Rational)
$\sqrt{2}\times\sqrt{8}=\sqrt{16}=\pm4$ (Rational)
- Is correct, because when a rational number is multiplied to an irrational number, it can not make an irrational number terminating or Non-terminating Repeating. Product again becomes a Non-terminating Non-Repeating number.
as: $2\times\sqrt{3}=2\sqrt{3}$
$\frac{2}{3}\times\sqrt{3}=\frac{2}{\sqrt{3}}$
So, product of a rational number and an irrational number is always an irrational, because irrational number is just changed in magnitude not in properties.
- Is incorrect, Sum of two irrational numbers can be an irrational number. i.e. if we add $\sqrt{2}$ and $\sqrt{3},$ we will get $\sqrt{2}+\sqrt{3}$ which is also an irrational.
- Is incorrect, Sum of an integer and a rational number can be a integer. Because all integers are rational numbers and also we can say some rational numbers are integers. So their sum with integer would be a integer
i.e. 2 + 3 = 5
Hence, correct option is (b).
View full question & answer→Question 1081 Mark
If $a = -2, b = -1,$ than $ab - ba$ is equal to:
Answer$a^b - b^a$
$= (-2)^{-1} (-1)^{-2}$
$=\frac{1}{(-2)}-\frac{1}{(-1)^2}$
$=\frac{-1}{2}-1$
$=\frac{-3}{2}$
$=-1.5$
View full question & answer→Question 1091 Mark
Write the correct answer in the following:
Which of the following is irrational?
Answer- $\sqrt{7}$
Solution:
$\because\sqrt{\frac{4}{9}}=\frac{2}{3}\text{(rational)},$
$\frac{\sqrt{12}}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3}}=2\text{(rational)},$
$\sqrt{81}=9\text{(rational)}$
but $\sqrt{7}$ is an irrational number.
Hence, $\sqrt{7}$ is an irrational number.
View full question & answer→Question 1101 Mark
If $x$ is a positive real number and $x^2 = 2,$ then $x^3 =$
AnswerWe have to find $x^3$ provided $x^2 = 2$ So,
By raising both sides to the power $\frac{1}{2}$
$\text{x}^{2\times\frac{1}{2}}=2^{\frac{1}{2}}$
$\text{x}^{2\times\frac{1}{2}}=\sqrt{2}$
$\text{x}=\sqrt{2}$
By substituting $\text{x}=\sqrt{2}$ in $x^3$ we get
$\text{x}^3\big(\sqrt{2}\big)^3$
$=\sqrt{2}\times\sqrt{2}\times\sqrt{2}$
$=2\sqrt{2}$
The value of $x^3$ is $2\sqrt{2}$
Hence the correct choice is $b.$
View full question & answer→Question 1111 Mark
$\sqrt{10}\times\sqrt{15}$ is equal to:
Answer- $5\sqrt{6}$
Solution:
$\sqrt{10}\times\sqrt{15}$
$=\sqrt{5\times2\times5\times3}$
$=5\sqrt{6}$
View full question & answer→Question 1121 Mark
Every rational number is:
Answer- a real number.
Solution:
A number whose square is non-negative, is called a real number.
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, a rational number is a real number since p and q which form a rational number are integers.
Hence, the correct opion is (d).
View full question & answer→Question 1131 Mark
A rational number equivalent to $\frac{7}{19}$ is:
Answer- $\frac{21}{57}$
Solution:
$\frac{7}{17}=\frac{7\times3}{17\times3}=\frac{21}{57}$
Hence, the correct opion is (d).
View full question & answer→Question 1141 Mark
The decimal representation of an irrational number is:
Answer- neither terminating nor repeating.
Solution:
A number which can neither be expressed as a terminating decimal nor as a repeating decimal is called an irrational number.
So, decimal representation of an irrational number is neither terminating nor a repeating decimal.
Hence, the correct opion is (d).
View full question & answer→Question 1151 Mark
If $(16)^{2x+3} = (64)^{x+3},$ then $4^{2x-2} =$
AnswerWe have to find the value of $4^{2x-2}$ provided $(16)^{2x+3} = (64)^{x+3}$
So,
$(16)^{2x+3} = (64)^{x+3}$
$(2^4)^{2x+3} = (2^6)^{x+3}$
$2^{8x+12} = 2^{6x+18}$
Equating the power of exponents we get
$8x +12 = 6x +18$
$8x - 6x = 18 - 12$
$2x = 6$
$\text{x}=\frac{6}{2}$
$x = 3$
The value of $4^{2x-2}$ is
$= 4^{2x-2}$
$= 4^{2\times 3-2}$
$= 4^{6-2}$
$= 4^4$
$= 256$
Hence the correct alternative is $b.$
View full question & answer→Question 1161 Mark
A rational number equivalent to $\frac{5}{7}$ is.
View full question & answer→Question 1171 Mark
The value of 'x' in $3+2^\text{x}=(64)^{\frac{1}{2}}+(27)^{\frac{1}{3}}$ is:
Answer- 3
Solution:
$3+2^\text{x}=(64)^{\frac{1}{2}}+(27)^{\frac{1}{3}}$
$\Rightarrow3+2^\text{x}=\sqrt{64}+\sqrt[3]{27}$
$\Rightarrow2^\text{x}=8=2^3$
Equating both,
$\text{x}=3$
View full question & answer→Question 1181 Mark
The value of $0.\overline{23}+0.\overline{22}$ is:
Answer- $0.\overline{45}$
Solution:
Let $\text{x}=0.\overline{23}=0.232323...(1)$
Now, $\text{y}=0.\overline{22}=0.22222...(2)$
Adding equation (1) and (2), we get
$\text{x}+\text{y}=0.454545=0.\overline{45}$
$\Rightarrow0.\overline{23}+0.\overline{22}=0.\overline{45}$
Hence, option (a) is correct.
View full question & answer→Question 1191 Mark
If $m$ is a positive integer which is not a perfect cube, then $\sqrt[3]{\text{m}}$ is:
AnswerA natural number is called a perfect cube if it is the cube of some natural number.
i.e., if $m = n^3,$ then $m$ is a perfect cube where $m$ and $n$ are natural numbers.
If $m$ is not a perfect cube then the number will be an irrational number because the value will be $\sqrt[3]{\text{m}}$ which cannot be written as a fraction or rational number.
View full question & answer→Question 1201 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: If two irrational number whose quotient is a rational number.
Reason: If two irrational number whose product is irrational number.
Answer - Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
View full question & answer→Question 1211 Mark
The simplest from of $0.\overline{123}$ is:
Answer- None of these.
Solution:
Since $0.\overline{123}=\frac{111}{900}=\frac{37}{300}$
View full question & answer→Question 1221 Mark
$(625)^{0.16} \times (625)^{0.09}=$
Answer$(625)^{0.16} \times (625)^{0.09}$
$(625)^{0.16 + 09}$
$(625)^{0.25}$ or $(625)^{\frac{1}{4}}$
But $625 = 5^4$
So, $(5^4)^{\frac{1}{4}}=5$
View full question & answer→Question 1231 Mark
The value of $x^{p-q}.x^{q-r}⋅x^{r-p}$ is equal to:
Answer$x^{p-q}.x^{q-r}⋅x^{r-p}$
$= x^{p-q+q-r+r-p}$
$= x^0$
$= 1$
View full question & answer→Question 1241 Mark
On simplification $(3+\sqrt{3})(3-\sqrt{3})$ gives:
Answer- $\frac{1}{8}$
Solution:
We know that,
$(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2$
So, here
$(4+\sqrt{3})(3-\sqrt{3})=3^2-(\sqrt{3})^2$
$\Rightarrow9-3-6$
View full question & answer→Question 1251 Mark
The decimal form of $\frac{1}{999}$ is:
Answer- $0.\overline{001}$
Solution:
When we divide 1 by 999 it result is 0.001001001001001
View full question & answer→Question 1261 Mark
If $\Big(\frac{2}{3}\Big)^{\text{x}}\Big(\frac{3}{2}\Big)^\text{2x}=\frac{81}{16}$ then x = ?
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).
View full question & answer→Question 1271 Mark
The value of $7^{\frac{1}{2}}\times8^{\frac{1}{2}}$ is:
Answer- $(56)^{\frac{1}{2}}$
Solution:
$7^{\frac{1}{2}}\times8^{\frac{1}{2}}$
$=(7\times8)^{\frac{1}{2}}$
$(56)^{\frac{1}{2}}$
View full question & answer→Question 1281 Mark
Write the correct answer in the following:
A rational number between $\sqrt{2}$ and $\sqrt{3}$ is
Answer- $1.5$
Solution:
We know that
$\sqrt{2}.=1.4142135...$ and $\sqrt{3}.=1.732050807...$
We see that 1.5 is a rational number which lies between 1.4142135... and 1.732050807...
Hence, (c) is tthe correct answer.
View full question & answer→Question 1291 Mark
If $10^{2y} = 25,$ then $10^{-y}$ equals:
AnswerWe have to find the value of $10^{-y}$
Given that $10^{2y} = 25,$ therefore,
$10^{2y} = 25$
$(10^y) 2 = 5^2$
$(10^\text{y})^{2\times\frac{1}{2}}=5^{2\times\frac{1}{2}}$
$\frac{10^{\text{y}}}{1}=\frac{5}{1}$
$\frac{1}{5}=\frac{1}{10^\text{y}}$
$\frac{1}{5}=10^\text{y}$
View full question & answer→Question 1301 Mark
If $10^\text{x}=64,$ what is the value of $10^{\frac{\text{x}}{2}+1}?$
Answer- 80
Solution:
We have to find the value of $10^{\frac{\text{x}}{2}+1}$ provided $10^\text{x}=64$
So,
$10^{\frac{\text{x}}{2}+1}=10^{\text{x}\times\frac{1}{2}}\times10^1$
$=\sqrt[2]{10^\text{x}}\times10^1$
By substituting $10^\text{x}=64$ we get
$=\sqrt[2]{64}\times10^1$
$=\sqrt[2]{8\times8}\times10$
$=8\times10$
$=80$
Hence the correct choice is c.
View full question & answer→Question 1311 Mark
The value of $xp-q xq-r xr-p$ is equal to:
Answer$x^{p-q} x^{q-r} x^{r-p}$
$= x^{p-q+q-r+r-p}$
$= x^0$
$= 1$
View full question & answer→Question 1321 Mark
The value of $\{2-3(2-3)^3\}^3,$ is:
Answer- $125$
Solution:
We have to find the value of $\{2-3(2-3)^3\}^3.$ So,
$\{2-3(2-3)^3\}^3=\{2-3(-1)^3\}^3$
$=\{2(-3\times-1)^3\}^3$
$=\{2+3\}^3$
$=5^3=125$
The value of $\{2-3(2-3)^3\}^3$ is 125
Hence the correct choice is b.
View full question & answer→Question 1331 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:
Answer- $\frac{2}{9}$
Solution:
$\frac{2}{9}=0.2222222...=0.\overline{2}$
Hence, the correct option is (b).
View full question & answer→Question 1341 Mark
The value of $\sqrt[4]{(64)^{-2}}$ is:
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).
View full question & answer→Question 1351 Mark
The simplest for of $0.\overline{54}$ is:
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).
View full question & answer→Question 1361 Mark
If $\frac{5-\sqrt3}{2+\sqrt3}=\text{x}+\text{y}\sqrt3,$ then:
Answer- x = 13, y = -7
Solution:
$\frac{5-\sqrt3}{2+\sqrt3}$
$=\frac{5-\sqrt3}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}$
$=\frac{\big(5-\sqrt3\big)\big(2-\sqrt3\big)}{(2)^2-\big(\sqrt3\big)^2}$
$=\frac{10-5\sqrt3-2\sqrt3+3}{4-3}$
$=\frac{13-7\sqrt3}{1}$
$=13-7\sqrt3$
⇒ x = 13 and y = -7
Hence, correct option is (a).
View full question & answer→Question 1371 Mark
The value of $(32)^{\frac{1}{5}}+(-7)^0+(64)^{\frac{1}{2}}$ is:
Answer- 11
Solution:
$(32)^{\frac{1}{5}}+(-7)^0+(64)^{\frac{1}{2}}$
$=2+1+8$
$=11$
View full question & answer→Question 1381 Mark
The value of $\sqrt[4]{\sqrt[3]{2^2}}$ is:
Answer- $2^{\frac{1}{6}}$
Solution:
$\sqrt[4]{\sqrt[3]{2^2}}$
$=\sqrt[4]{\sqrt{2^{\frac{2}{3}}}}$
$=2^{\frac{2}{3\times4}}=2^{\frac{1}{6}}$
View full question & answer→Question 1391 Mark
The value of $(32)^{\frac{1}{5}}+(-7)^0+(64)^{\frac{1}{2}}$ is:
Answer- 11
Solution:
= 2 + 1 + 8
= 11
View full question & answer→Question 1401 Mark
$0.\overline{001}$ when expressed in the form $\frac{\text{p}}{\text{q}}$ $\big($p, q are integers and $\text{q}\neq0\big),$ is:
Answer- $\frac{1}{999}$
Solution:
Let $\text{x}=0.\overline{001}=0.001001001...(1)$
Now, $1000\text{x}=001.001001001...(2)$
Subtracting equation (1) from (2), we get
$999\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{999}$
Hence, option (d) is correct.
View full question & answer→Question 1411 Mark
The value of $\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$ is:
Answer- 3
Solution:
$\Big[(81)^{\frac{1}{2}}\Big]^{\frac{1}{2}}$
$=(3^4)^{\frac{1}{2}\times\frac{1}{2}}$
$=(3^4)^{\frac{1}{4}}$
$=3$
View full question & answer→Question 1421 Mark
Which of the following is a correct statement?
Answer- Sum of a rational and irrational number is always an irrational number.
Solution:
- Is incorrect, because sum of two irrational numbers is not an irrational number always. It can also be a rational number
i.e. if we add $2+\sqrt{3}$ and $2-\sqrt{3},$ sum comes out to be $2+\sqrt{\not\text{3}}+2-\sqrt{\not\text{3}}=4,$ which is a rational number.
- Is correct, if a rational number is added to an irrational number means to a Non- terminating-repeating number, the sum will also be non-terminating and Non-repeating number, i.e an irrational number.
Example: a rational number '2' and an irrational no $'\sqrt{3}'$ is added, sum $=2+\sqrt{3}$ which is again a non-terminating and non-repeating number, hence an irrational number always.
- Is incorrect, Square of an irrational number is not necessarily a rational number. Again it can be either a rational or irrational.
i.e $(\sqrt{2})^2=2$ (Rational)
$(2+\sqrt{3})^2=4+3+2\times2\sqrt{3}=7+4\sqrt{3}$ (irrational)
- Is incorrect, Sum of two rational numbers can be an integer and a rational number both.
i.e $\frac{1}{2}+\frac{1}{4}=\frac{3}{4}$ (Rational number)
Hence, correct option is (b).
View full question & answer→Question 1431 Mark
Which of the following is (are) not equal to $\Bigg\{\Big(\frac{5}{6}\Big)^{\frac{1}{5}}\Bigg\}^{-\frac{1}{6}}?$
Answer- $\Big(\frac{5}{6}\Big)^{\frac{1}{5}-\frac{1}{6}}$
Solution:
We have to find the value of $\Bigg\{\Big(\frac{5}{6}\Big)^{\frac{1}{5}}\Bigg\}^{-\frac{1}{6}}$
So,
$\Bigg\{\Big(\frac{5}{6}\Big)^{\frac{1}{5}}\Bigg\}^{-\frac{1}{6}}=\frac{5^{\frac{1}{5}\times\frac{-1}{6}}}{6^{\frac{1}{5}\times\frac{-1}{6}}}$
$=\frac{5^{-\frac{1}{30}}}{6^{\frac{-1}{30}}}$
$=\frac{\frac{1}{5^{\frac{1}{30}}}}{\frac{1}{6^{\frac{1}{30}}}}$
$\Bigg\{\Big(\frac{5}{6}\Big)^{\frac{1}{5}}\Bigg\}^{-\frac{1}{6}}=\frac{1}{5^\frac{1}{30}}\times\frac{6^\frac{1}{30}}{1}$
$=\frac{6^{\frac{1}{30}}}{5^{\frac{1}{30}}}$
$=\Big(\frac{5}{6}\Big)^{\frac{1}{30}}$
Hence the correct choice is a.
View full question & answer→Question 1441 Mark
A terminating decimal is:
Answer- A rational number.
Solution:
A rational number because it can be written in fraction.
View full question & answer→Question 1451 Mark
The simplest from of $0.\overline{54}$ is:
Answer- $\frac{6}{11}$
Solution:
0.5454...
$=\frac{54}{99}=\frac{6}{11}$
View full question & answer→Question 1461 Mark
From the choices given below mark the co-prime numbers.
Question 1471 Mark
The sum of rational and an irrational number:
Answer- Is always irrational.
Solution:
By definition, an irrational number in decimal form goes on forever without repeating (a non-repeating, non-terminating decimal). By definition, a rational number in the decimal form either terminates or repeats. If you add a non-repeating, non-terminating decimal to a terminating decimal, you still have a non-repeating, non-terminating decimal. If you add a non-repeating,
Non-terminating decimal to a repeating decimal, you will no longer have a repeating decimal.
$3+\sqrt{5}=3+2.236067977=5.236067977$ which is irreational.
$4\frac{1}{3}+\pi=4.33333333+3.141592653= 4.47925986$ which is also irrational
View full question & answer→Question 1481 Mark
Which of the following is irrational?
Answer- $\sqrt{7}$
Solution:
- Is incorrect, because $\sqrt{\frac{4}{9}}=\frac{\sqrt{4}}{\sqrt{9}}=\pm\frac{2}{3}$ (Rational)
- Is also incorrect, as $\frac{4}{5}$ is in the form of $\frac{\text{P}}{\text{Q}}(\text{Q}\neq0),$ (Rational)
- Is incorrect, because $\sqrt{7}$ is a non-terminating and Non-Repeating number.
- Is incorrect, because $\sqrt{81}=\pm9$ (Rational)
Hence, correct option is (c).
View full question & answer→Question 1491 Mark
The smallest rational number by which $\frac{1}{3}$ should be multiplied so that its decimal expansion terminates after one place of decimal, is:
Answer- $\frac{3}{10}$
Solution:
$\frac{1}{3}=0.33333...$ (a Non - termnating number)
Now, if we remove 3 from denominator it will terminate.
So, if we multiply by $\frac{3}{10}$
i.e. $\frac{1}{3}\times\frac{1}{3}=\frac{1}{10}=0.1$ ( terminates after one place of decimal)
By multiplying by $\frac{1}{10},$ 3 does not replaces.
By multiplying by 3, we get 1, which is not terminating after one place of decimal
And, by multiplying by 30, we get 10, again not terminating after one palce of decimal.
Hence, option (b) is correct.
View full question & answer→Question 1501 Mark
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is:
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).
View full question & answer→