MCQ 11 Mark
The value of $\sqrt[3]{1000}$ is:
Answer- 10
Solution:
$(10)^3=1000$
So, $\sqrt[3]{1000}=1000^{\frac{1}{3}}=(10^3)^{\frac{1}{3}}$
View full question & answer→MCQ 21 Mark
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to :
Answer- 2
Solution :
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$
$\Rightarrow\frac{\sqrt{16\times2}+\sqrt{16\times3}}{\sqrt{4\times2}+\sqrt{4\times3}}$
$\Rightarrow\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}}$
$\Rightarrow\frac{4}{2}(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}})$
$\Rightarrow2$
View full question & answer→MCQ 31 Mark
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}=$
Answer- 10
Solution:
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$\Rightarrow\frac{(\sqrt{3}+\sqrt{2})^2+(\sqrt{3}-\sqrt{2})^2}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}$
$\Rightarrow\frac{(3+2+2\sqrt{6})+3+2-2\sqrt{6}}{3-2}$
$\Rightarrow10$
View full question & answer→MCQ 41 Mark
$\sqrt{10}\times\sqrt{15}=?$
- A
$\sqrt{25}$
- B
$5\sqrt{6}$
- C
$6\sqrt{5}$
- D
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).
View full question & answer→MCQ 51 Mark
Answer- 486
Solution:
93 + (-3)3 - 63
= 729 - 27 - 216
= 729 - 243
= 486
View full question & answer→MCQ 61 Mark
Which of the following is irrational?
Answer- 0.5015001500015.
Solution:
- $0.15=\frac{15}{100}=$ Rational number
- $0.1516=\frac{1516}{100000}=$ Rational number
- $0.\overline{1516}$ is a Non-terminating Repeating number = Rational number.
- 0.5015001500015. is a Non-terminating, Non-Repeating decimal number, So is a irrational number.
Hence, option (d) is correct.
View full question & answer→MCQ 71 Mark
If $\text{x}=\sqrt[3]{2+\sqrt3},$ then $\text{x}^3+\frac{1}{\text{x}^3}=$:
Answer- 4
Solution:
$\text{x}=\sqrt[3]{2+\sqrt3}+\big(2+\sqrt3\big)^{\frac{1}{3}}$
$\text{x}^3=\big\{\big({2+\sqrt3}\big)^{\frac{1}{\not3}}\big\}^{\not3}=\big(2+\sqrt3\big)$
$\Rightarrow\frac{1}{\text{x}^3}=\frac{1}{2+\sqrt3}=\frac{1}{2+\sqrt3}\times\frac{2-\sqrt3}{2-\sqrt3}\\ \ =\frac{2-\sqrt3}{4-3}=2-\sqrt3$
Now, $\text{x}^3+\frac{1}{\text{x}^3}=2+\sqrt3+2-\sqrt3=4$
Hence, correct option is (b).
View full question & answer→MCQ 81 Mark
Simplified value of $(16)^{-\frac{1}{4}}\times\sqrt[4]{16}$ is:
Answer- 1
Solution:
$(16)^{-\frac{1}{4}}\times\sqrt[4]{16}=(2^4)^{-\frac{1}{4}}\times(2^4)^\frac{1}{4}\\=2^{-1}\times2^1=\frac{1}{2}\times2=1$
Hence, the correct answer is option (a).
View full question & answer→MCQ 91 Mark
When simplified $\big(\text{x}^{-1}+\text{y}^{-1}\big)^{-1}$ is equal to:
Answer- $\frac{\text{xy}}{\text{x}+\text{y}}$
Solution:
We have to simplify $\big(\text{x}^{-1}+\text{y}^{-1}\big)^{-1}$
So,
$\big(\text{x}^{-1}+\text{y}^{-1}\big)^{-1}=\Big(\frac{1}{\text{x}}+\frac{1}{\text{y}}\Big)^{-1}$
$=\frac{1}{\frac{1}{\text{x}}+\frac{1}{\text{y}}}$
$=\frac{1}{\frac{1\times\text{y}}{\text{x}\times\text{y}}+\frac{1\times\text{x}}{\text{y}\times\text{x}}}$
$=\frac{1}{\frac{\text{y}}{\text{xy}}+\frac{\text{x}}{\text{xy}}}$
$\big(\text{x}^{-1}+\text{y}^{-1}\big)^{-1}=\frac{1}{\frac{\text{y}+\text{x}}{\text{xy}}}$
$=\frac{\text{xy}}{\text{y}+\text{x}}$
The value of $\big(\text{x}^{-1}+\text{y}^{-1}\big)^{-1}$ is $\frac{\text{xy}}{\text{y}+\text{x}}$
Hence the correct choice is c.
View full question & answer→MCQ 101 Mark
If a, b, c are positive real numbers, then $\sqrt{\text{a}^{-1}\text{b}}\times\sqrt{\text{b}^{-1}\text{c}}\times\sqrt{\text{c}^{-1}\text{a}}$ is equal to
- A
$1$
- B
$\text{abc}$
- C
$\sqrt{\text{abc}}$
- D
$\frac{1}{\text{abc}}$
Answer- $1$
Solution:
We have to find the value of $\sqrt{\text{a}^{-1}\text{b}}\times\sqrt{\text{b}^{-1}\text{c}}\times\sqrt{\text{c}^{-1}\text{a}}$ when a, b, c are positive real numbers.
So,
$\sqrt{\text{a}^{-1}\text{b}}\times\sqrt{\text{b}^{-1}\text{c}}\times\sqrt{\text{c}^{-1}\text{a}}$
$=\sqrt{\frac{1}{\text{a}}\times\text{b}}\times\sqrt{\frac{1}{\text{b}}\times\text{c}}\times\sqrt{\frac{1}{\text{c}}\times}\text{a}$
$=\sqrt{\frac{\text{b}}{\text{a}}}\times\sqrt{\frac{\text{c}}{\text{b}}}\times\sqrt{\frac{\text{a}}{\text{c}}}$
Taking square root as common we get
$\sqrt{\text{a}^{-1}\text{b}}\times\sqrt{\text{b}^{-1}\text{c}}\times\sqrt{\text{c}^{-1}\text{a}}=\sqrt{\frac{\text{b}}{\text{a}}\times\frac{\text{c}}{\text{b}}\times\frac{\text{a}}{\text{c}}}$
$\sqrt{\text{a}^{-1}\text{b}}\times\sqrt{\text{b}^{-1}\text{c}}\times\sqrt{\text{c}^{-1}\text{a}}=1$
Hence the correct alternative is a.
View full question & answer→MCQ 111 Mark
If $4\text{x}-4\text{x}^{-1}=24,$ then (2x)x equals:
- A
$5\sqrt{5}$
- B
$\sqrt{5}$
- C
$25\sqrt{5}$
- D
$125$
Answer- $25\sqrt{5}$
Solution:
We have to find the value of $(2\text{x})^\text{x}$ if $4\text{x}-4^{\text{x}-1}=24$
So,
Taking 4x as common factor we get
$4\text{x}(1-4^{-1})=24$
$4\text{x}\Big(1-\frac{1}{4}\Big)=24$
$4\text{x}\Big(\frac{1\times4}{1\times4}-\frac{1}{4}\Big)=24$
$4^4\Big(\frac{4-1}{4}\Big)=24$
$4^\text{x}\times\frac{3}{4}=24$
$4^\text{x}=24\times\frac{4}{3}$
$4^\text{x}=32$
$2^{2\text{x}}=2^{5}$
By equating powers of exponents we get
$2\text{x}=5$
$\text{x}=\frac{5}{2}$
By substituting $\text{x}=\frac{5}{2}$ in $(2\text{x})^\text{x}$ we get
$(2\text{x})^\text{x}=\Big(2\times\frac{5}{2}\Big)^{\frac{5}{2}}$
$=\Big(2\times\frac{5}{2}\Big)^{\frac{5}{2}}$
$=5^{\frac{5}{2}}$
$=5^{5\times\frac{1}{2}}$
$(2\text{x})^\text{x}=\sqrt[2]{5^5}$
$=\sqrt[2]{5\times5\times5\times5}$
$=5\times5\times^\sqrt[2]{5}$
$=25\sqrt{5}$
Hence the correct choice is c.
View full question & answer→MCQ 121 Mark
The value of m for which $\Bigg[\bigg\{\Big(\frac{1}{7^2}\Big)^{-2}\bigg\}^{-\frac{1}{3}}\Bigg]^{\frac{1}{4}}=7^{\text{m}},$ is:
- A
$-\frac{1}{3}$
- B
$\frac{1}{4}$
- C
$-3$
- D
$2$
Answer- $-\frac{1}{3}$
Solution:
We have to find the value of m for $\Bigg[\bigg\{\Big(\frac{1}{7^2}\Big)^{-2}\bigg\}^{-\frac{1}{3}}\Bigg]^{\frac{1}{4}}=7^{\text{m}},$
$\Rightarrow\bigg[\Big\{\frac{1}{7^{2\times-2}}^{-2}\Big\}^{-\frac{1}{3}}\bigg]^{\frac{1}{4}}=7^{\text{m}}$
$\Rightarrow\bigg[\Big\{\frac{1}{7^{-4}}\Big\}^{-\frac{1}{3}}\bigg]^{\frac{1}{4}}=7^{\text{m}}$
$\Rightarrow\Bigg[\bigg\{\frac{1}{7^{-4\times\frac{-1}{3}}}\bigg\}^{-\frac{1}{3}}\Bigg]^{\frac{1}{4}}=7^{\text{m}}$
$\Rightarrow\Bigg[\bigg\{\frac{1}{7^{\frac{4}{3}}}\bigg\}\Bigg]^{\frac{1}{4}}=7^{\text{m}}$
$\Rightarrow\Bigg[\frac{1}{7^{\frac{4}{3}\times\frac{1}{4}}}\Bigg]=7^{\text{m}}$
$\Rightarrow\Bigg[\frac{1}{7^{\frac{1}{3}}}\Bigg]=7^{\text{m}}$
By using rational exponents $\frac{1}{\text{a}^{\text{n}}}=\text{a}^{-\text{n}}$
$7^{\frac{-1}{3}}=7^{\text{m}}$
Equating power of exponents we get $-\frac{1}{3}=\text{m}$
Hence the correct choice is a.
View full question & answer→MCQ 131 Mark
If a, m, n are positive ingegers, then $\big\{\sqrt[\text{m}]{\sqrt[\text{n}]{\text{a}}}\big\}^{\text{mn}}$ is equal to
Answer- $\text{a}$
Solution:
Find the value of $\big\{\sqrt[\text{m}]{\sqrt[\text{n}]{\text{a}}}\big\}^{\text{mn}}.$
So,
$\big\{\sqrt[\text{m}]{\sqrt[\text{n}]{\text{a}}}\big\}^{\text{mn}}=\bigg\{\sqrt[\text{m}]{\text{a}^{\frac{1}{\text{n}}}}\bigg\}^{\text{mn}}$
$=\bigg\{\text{a}^{\frac{1}{\text{n}}\times\frac{1}{\text{m}}}\bigg\}^{\text{mn}}$
$=\bigg\{\text{a}^{\frac{1}{\text{n}}\times\frac{1}{\text{m}}}\times\text{m}\times\text{n}\bigg\}$
$\Rightarrow\big\{\sqrt[\text{m}]{\sqrt[\text{n}]{\text{a}}}\big\}=\bigg\{\text{a}^{\frac{1}{\text{n}}\times\frac{1}{\text{m}}}\times\text{m}\times\text{n}\bigg\}$
$\Rightarrow\big\{\sqrt[\text{m}]{\sqrt[\text{n}]{\text{a}}}\big\}=\text{a}$
Hence the correct choice is b.
View full question & answer→MCQ 141 Mark
The value of $\Big\{\big(23+2^2\big)^{\frac{2}{3}}+\big(140-19\big)^{\frac{1}{2}}\Big\}^2,$ is:
Answer- 400
Solution:
We have to find the value of $\Big\{\big(23+2^2\big)^{\frac{2}{3}}+\big(140-19\big)^{\frac{1}{2}}\Big\}^2,$
$\Big\{\big(23+2^2\big)^{\frac{2}{3}}+\big(140-19\big)^{\frac{1}{2}}\Big\}^2$
$=\Big\{\big(23+4\big)^{\frac{2}{3}}+\big(121\big)^{\frac{1}{2}}\Big\}^2$
$=\Big\{\big(27\big)^{\frac{2}{3}}+\big(121\big)^{\frac{1}{2}}\Big\}^2$
$=\Big\{\big(3^3\big)^{\frac{2}{3}}+\big(11^2\big)^{\frac{1}{2}}\Big\}^2$
$\Big\{\big(23+2^2\big)^{\frac{2}{3}}+\big(140-19\big)^{\frac{1}{2}}\Big\}^2\\=\Big\{3^{3\times\frac{2}{3}}+11^{2\times\frac{2}{3}}\Big\}$
$=\big\{3^2+11\big\}^2$
$\Rightarrow\Big\{\big(23+2^2\big)^{\frac{2}{3}}+\big(140-19\big)^{\frac{1}{2}}\Big\}^2=\{9+11\}^2$
By using the identity $(\text{a}+\text{b})^2=\text{a}^\text{2}+2\text{ab}+\text{b}^2$ we get,
$=9\times9+2\times9\times11+11\times11$
$=81+198+121$
$=400$
Hence correct choice is d.
View full question & answer→MCQ 151 Mark
If n is a natural number, then $\sqrt{\text{n}}$ is:
- A
- B
always an irrational number.
- C
always an irrational number.
- D
sometimes a natural number and sometimes an irrational number.
Answer- sometimes a natural number and sometimes an irrational number.
Solution:
- Is incorrect, because $\sqrt{\text{n}}$ can not be always a natural number
i.e. if $\text{n}=2, \ \sqrt{\text{n}}=\sqrt{2}$ (not a natural no.)
- Is incorrect, similiarly, if n = 2, 5, …. Or any odd no. or not perfect square, $\sqrt{\text{n}}=\sqrt{2},\sqrt{5},\sqrt{7}$ are Non-terminating and non-repeating, So irrational in nature, So, not always a rational number.
- Is also incorrect, $\sqrt{\text{n}}$ can aslo be rational or say a natural number. If n = 4, 9, 16... or any perfect square number then $\sqrt{\text{n}}=2,3,4...$ natural numbers.
- Is fully correct because if n is any odd number or non-perfect square number then $\sqrt{\text{n}}$ would be irrational, but if n is a perfect square number $\sqrt{\text{n}}$ then will be a natural number.
If n = 2, 3, 5, 8 ... $\sqrt{\text{n}}=\sqrt{2},\sqrt{3},\sqrt{8}...$ (irrational)
If n = 4, 9, 16 ... = 2, 3, 4 ... (Natural number)
So, correct option is (d).
View full question & answer→MCQ 161 Mark
Two rational numbers between $\frac{2}{3}$ and $\frac{5}{3}$ are :
- A
$\frac{1}{6}$ and $\frac{2}{6}$
- B
$\frac{1}{2}$ and $\frac{2}{1}$
- C
$\frac{5}{6}$ and $\frac{7}{6}$
- D
$\frac{2}{3}$ and $\frac{4}{3}$
Answer- $\frac{5}{6}$ and $\frac{7}{6}$
Solution:
We have,
$\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{5}{3}=\frac{5\times2}{3\times2}=\frac{10}{6}$
And, $\frac{1}{2}=\frac{1\times3}{2\times3}=\frac{3}{6}$ and $\frac{2}{1}=\frac{2\times6}{1\times6}=\frac{12}{6}$
Also, $\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{4}{3}=\frac{4\times2}{3\times2}=\frac{8}{6}$
Since, $\frac{1}{6}<\frac{2}{6}<\frac{3}{6}\Big(\frac{1}{2}\Big)<\frac{4}{6}\Big(=\frac{2}{3}\Big)<\frac{5}{6}<\frac{7}{6}\\<\frac{8}{6}\Big(=\frac{4}{3}\Big)<\frac{10}{6}\Big(=\frac{5}{3}\Big)<\frac{12}{6}\Big(=\frac{2}{1}\Big)$
So, the two rational numbers between $\frac{2}{3}$ and $\frac{5}{3}$ are $\frac{5}{6}$ and $\frac{7}{6}.$
Hence, the correct opion is (c).
View full question & answer→MCQ 171 Mark
The value of $(0.00032)^{\frac{-2}{5}}$ is:
Answer- 25
Solution:
$(0.00032)^{\frac{-2}{5}}$
$=(\frac{32}{100000})^{\frac{-2}{5}}$
$=(\frac{2}{10})^{5\times\frac{-2}{5}}$
$=(\frac{1}{5})^{-1}=25$
View full question & answer→MCQ 181 Mark
$\sqrt[4]{\sqrt[3]{2^2}}$ is equal to:
- A
$2^{-6}$
- B
$2^{\frac{1}{6}}$
- C
$2^{6}$
- D
$2^{-\frac{1}{6}}$
Answer- $2^{\frac{1}{6}}$
Solution:
$\sqrt[4]{\sqrt[3]{2^2}}=\sqrt[4]{(2)^{\frac{2}{3}}}$
$=(2)^{\frac{2}{3}\times\frac{1}{4}}$
$=(2)^{\frac{1}{6}}$
View full question & answer→MCQ 191 Mark
Which one of the following is not equal to $\Big(\frac{100}{9}\Big)^{-\frac{3}{2}}?$
- A
$\Big(\frac{9}{100}\Big)^{\frac{3}{2}}$
- B
$\bigg(\frac{1}{\frac{100}{9}}\bigg)^{\frac{3}{2}}$
- C
$\frac{3}{10}\times\frac{3}{10}\times\frac{3}{10}$
- D
$\sqrt{\frac{100}{9}}\times\sqrt{\frac{100}{9}}\times\sqrt{\frac{100}{9}}$
Answer- $\sqrt{\frac{100}{9}}\times\sqrt{\frac{100}{9}}\times\sqrt{\frac{100}{9}}$
Solution:
We have to find the value of $\Big(\frac{100}{9}\Big)^{-\frac{3}{2}}$
So,
$\Big(\frac{100}{9}\Big)^{-\frac{3}{2}}=\Big(\frac{10^2}{3^2}\Big)^{-\frac{3}{2}}$
$=\frac{10^{2\times\frac{3}{2}}}{3^{2\times\frac{3}{2}}}$
$=\frac{10^{-3}}{3^{-3}}$
$\Big(\frac{100}{9}\Big)^{-\frac{3}{2}}=\frac{\frac{1}{10^3}}{\frac{1}{3^3}}$
$=\frac{1}{10\times10\times10}\times\frac{3\times3\times3}{1}$
$=\frac{3\times3\times3}{10\times10\times10}$
Since, $\Big(\frac{100}{9}\Big)^{-\frac{3}{2}}$ is equal to $\Big(\frac{100}{9}\Big)^{\frac{3}{2}},\ \frac{1}{\Big(\frac{100}{9}\Big)^{\frac{3}{2}}},\ \frac{3\times3\times3}{10\times10\times10}.$
Hence the correct choice is d.
View full question & answer→MCQ 201 Mark
$(6+\sqrt{27})-(3+\sqrt{3})+(1-2\sqrt{3})$ when simplified is:
Answer- Positive and rational.
Solution:
$(6+\sqrt{27})-(3+\sqrt{3})+(1-2\sqrt{3})$
$=6+3\sqrt{3}-3-\sqrt{3}+1-2\sqrt{3}$
$=6+1-3$
$=4$
Positive and rational.
View full question & answer→MCQ 211 Mark
The simplest from of $0.5\bar{7}$ is:
- A
$\frac{57}{99}$
- B
$\frac{26}{45}$
- C
$\frac{57}{100}$
- D
Answer- $\frac{26}{45}$
Solution:
$0.\overline{57}=\frac{57-5}{90}$
$=\frac{52}{90}=\frac{26}{45}$
View full question & answer→MCQ 221 Mark
Which of the following is the value $(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})?$
- A
$\sqrt{7}$
- B
$\sqrt{11}$
- C
- D
Answer- 4
Solution:
$(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})$
$=(\sqrt{11})^2-(\sqrt{7})^2$
$=11-7$
$=4$
View full question & answer→MCQ 231 Mark
The value of $(2+\sqrt{3})(2-\sqrt{3})$ in.
Answer- 1
Solution:
We know the formula a2 - b2 = (a - b) (a - b)
Here put $\text{a}=2$ and $\text{b}=\sqrt{3}$
So, $2^2-(\sqrt{3})^2=4-3=1$
View full question & answer→MCQ 241 Mark
Write the correct answer in the following:
Between two rational numbers.
- A
There is no rational number.
- B
There is exactly one rational number.
- C
There are infinitely many rational numbers.
- D
There are only rational numbers and no irrational numbers.
Answer- There are infinitely many rational numbers.
Solution:
There are infinitely many rational numbers Between two rational numbers there are infinitely many rational number.
For example between 4 and 5 there are 4.1, 4.2.4.22, 4.223 ............
Hence, (C) is the correct answer.
View full question & answer→MCQ 251 Mark
If $\sqrt2=1.414,$ then the value of $\sqrt6-\sqrt3$ upto three place of decimal is :
Answer- 0.707
Solution :
$\sqrt6-\sqrt3$
$=\sqrt3\big(\sqrt2-1\big)$
Now, $\sqrt3=1.732$
$\sqrt2=1.414$
$\therefore\sqrt6-\sqrt3=1.732(1.414-1)\\ \ =1.732(0.414)=0.717$ (upto 3 decimal places)
Hence, correct option is (b).
View full question & answer→MCQ 261 Mark
Write the correct answer in the following:
If $\sqrt{2}=1.4142,$ then $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$ is equal to
Answer- 0.4142
Solution:
$\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\sqrt{\frac{\sqrt{2}-1}{{\sqrt{2}+1}}\cdot\frac{\sqrt{2}-1}{\sqrt{2}-1}}$
$[$Inside the root, multiplying numerator and denominator by $(\sqrt{2}-1)]$
$=\sqrt{\frac{(\sqrt{2}-1)^2}{(\sqrt{2})^2-(1)^2}}$ $ [\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$
$=\sqrt{\frac{(\sqrt{2}-1)^2}{2-1}}=\sqrt{2}-1=(1.4142...)-1=0.4142...$
View full question & answer→MCQ 271 Mark
If $\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$ and $\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},$ then $\text{x}^2+\text{xy}+\text{y}^2=$
Answer- 99
Solution:
$\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
$\therefore\text{x}=\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}\times\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2}\\ \ =\frac{\big(\sqrt3-\sqrt2\big)^2}{3-2}=5-2\sqrt6$
$\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2},$
$\therefore\text{y}=\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}\times\frac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}\\ \ =\frac{\big(\sqrt3+\sqrt{2}\big)^2}{3-2}=5+2\sqrt6$
Now, $\text{x}^2+\text{xy}+\text{y}^2$
$=\big(5 -2\sqrt6\big)^2+\big(5-2\sqrt6\big)\big(5+2\sqrt6\big)+\big(5+2\sqrt6\big)^2$
$=\big(25+24-20\sqrt6\big)+(25-24)+\big(25+24+20\sqrt6\big)$
$=49-20\sqrt6+1+49+20\sqrt6$
$=99$
Hence, correct option is (b).
View full question & answer→MCQ 281 Mark
The sum of two irrational numbers is.
- A
- B
- C
- D
Either irrational or rational.
Answer- Either irrational or rational.
Solution:
The sum of two irrational numbers, in some cases, will be irrational. However, if the irrational parts of the numbers have a zero sum (cancel each other out), the sum will be rational.
View full question & answer→MCQ 291 Mark
Choose the rational number which does not lie between $-\frac{2}{3}$ and $-\frac{1}{5}.$
- A
$-\frac{3}{10}$
- B
$\frac{3}{10}$
- C
$-\frac{1}{4}$
- D
$-\frac{7}{20}$
Answer- $\frac{3}{10}$
Solution:
Given two rational numbers are negative and $\frac{3}{10}$ is a positive rational number.
So, it does not lie between $-\frac{2}{3}$ and $-\frac{1}{5}.$
Hence, the correct opion is (b).
View full question & answer→MCQ 301 Mark
Answer- 27
Solution:
We have to find the value of 3x provided (23)2 = 4x
So,
23×2 = 22x
26 = 22x
By equating the exponents we get
6 = 2x
$\frac{6}{2}=\text{x}$
3 = x
By substituting in 3x we get
3x = 33
= 27
The value of 3x is 27
Hence the correct choice is d.
View full question & answer→MCQ 311 Mark
If x = 2 and y = 4, then $\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{x}-\text{y}}+\Big(\frac{\text{y}}{\text{x}}\Big)^{\text{y}-\text{x}}=$
Answer- 8
Solution:
We have to find the value of $\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{x}-\text{y}}+\Big(\frac{\text{y}}{\text{x}}\Big)^{\text{y}-\text{x}}$ if x = 2, y = 4
Substitute x = 2, y = 4, in $\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{x}-\text{y}}+\Big(\frac{\text{y}}{\text{x}}\Big)^{\text{y}-\text{x}}$ to get,
$\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{x}-\text{y}}+\Big(\frac{\text{y}}{\text{x}}\Big)^{\text{y}-\text{x}}$
$=\Big(\frac{2}{4}\Big)^{2-4}+\Big(\frac{4}{2}\Big)^{4-2}$
$=\Big(\frac{2}{4}\Big)^{-2}+\Big(\frac{4}{2}\Big)^{2}$
$=\Big(\frac{1}{2}\Big)^{-2}+(2)^2$
$=\Big(\frac{1}{2^{-2}}\Big)+4$
$\Big(\frac{\text{x}}{\text{y}}\Big)^{\text{x}-\text{y}}+\Big(\frac{\text{y}}{\text{x}}\Big)^{\text{y}-\text{x}}$
$=\frac{1}{\frac{1}{2^2}}+4$
$=\frac{1}{\frac{1}{4}}+4$
$=1\times\frac{4}{1}+4$
$=4+4$
$=8$
Hence the correct choice is b.
View full question & answer→MCQ 321 Mark
If $\text{x}=\frac{\sqrt{7}}{5}$ and $\frac{5}{\text{x}}=\text{p}\sqrt{7}$ then the value of p is:
- A
$\frac{7}{25}$
- B
$\frac{25}{7}$
- C
$\frac{7}{15}$
- D
$\frac{15}{7}$
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).
View full question & answer→MCQ 331 Mark
The product of a nonzero rational number with an irrational number is always a/ an.
Answer- Irrational number.
Solution:
The product of a non-zero rational number with an irrational number is always an irrational number.
View full question & answer→MCQ 341 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: -25 is not a rational number.
Reason : -25 can not be written in in the form of $\frac{\text{p}}{\text{q}}.$
- A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
- B
Both Assertion and Reason are correct and Reason is not the correct explanation for Assertion.
- C
Assertion is true but the reason is false.
- D
Both assertion and reason are false.
Answer - Both assertion and reason are false.
View full question & answer→MCQ 351 Mark
Select the correct statement from the following.
- A
$\frac{7}{9}>\frac{4}{5}$
- B
$\frac{2}{6}>\frac{3}{9}$
- C
$\frac{-5}{7}<\frac{-3}{4}$
- D
$\frac{-2}{3}<\frac{-4}{5}$
Answer- $\frac{-2}{3}<\frac{-4}{5}$
Solution:
$\frac{-2}{3}<\frac{-4}{5}$
Taking LCM of 3 and 5,
LCM = 15,
So, $\frac{-2\times5}{3\times5},\frac{-4\times3}{5\times3}$
$\Rightarrow\frac{-10}{15},\frac{-12}{15}$
Now, since both the denominator is equal so, we compare its numerator,
and since, -10 < -12
So, $\frac{-10}{15}>\frac{-12}{15}$
thus, $\frac{-2}{3}>\frac{-4}{5}$
View full question & answer→MCQ 361 Mark
$23.\overline{43}$ when expressed in the form $\frac{\text{p}}{\text{q}}$ $\big($p, q are integers and $\text{q}\neq0\big),$ is:
- A
$\frac{2320}{99}$
- B
$\frac{2343}{100}$
- C
$\frac{2343}{999}$
- D
$\frac{2320}{199}$
Answer- $\frac{2320}{99}$
Solution:
Let $\text{x}=23.\overline{43}=23.434343...(1)$
Now, $100\text{x}=2343.43333...(2)$
Subtracting equation (1) from (2), we get
$99\text{x}=2320$
$\Rightarrow\text{x}=\frac{2320}{99}$
Hence, option (a) is correct.
View full question & answer→MCQ 371 Mark
An irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is:
- A
$\Big(\frac{1}{7}\times\frac{2}{7}\Big)$
- B
$\frac{1}{2}\Big(\frac{1}{7}+\frac{2}{7}\Big)$
- C
$\sqrt{\frac{1}{7}\times\frac{2}{7}}$
- D
Answer- $\sqrt{\frac{1}{7}\times\frac{2}{7}}$
Solution:
An irrational number between a and b is given by $\sqrt{\text{ab}}.$
So, an irrational number between $\frac{1}{7}$ and $\frac{2}{7}$ is $\sqrt{\frac{1}{7}\times\frac{2}{7}}.$
View full question & answer→MCQ 381 Mark
Decimal representation of a rational number cannot be.
- A
Non-terminating non-repeating.
- B
Non-terminating repeating.
- C
- D
Answer- Non-terminating non-repeating.
Solution:
Decimal representation of a rational number cannot be non-terminating non-repeating.
It is always be terminating or non terminating repeating.
View full question & answer→MCQ 391 Mark
A rational number between $\sqrt{3}$ and $\sqrt{5}$ is:
Answer- 2.1
Solution:
$\sqrt{3}=1.73$ and $\sqrt{5}=2.236$
2.1 lies in between these two numbers.
View full question & answer→MCQ 401 Mark
The number 0.318564318564318564 ........ is:
Answer- a rational number
Solution:
$0.318564318564318564 \ ...=0.\overline{318564}$ is a Non-terminating repeating Number.
Hence, it is a rational number.
So, correct option is (c).
View full question & answer→MCQ 411 Mark
Simplified value of $(25)^{\frac{1}{3}}\times5^{\frac{1}{3}}$ is:
Answer- 5
Solution:
$(25)^{\frac{1}{3}}\times5^{\frac{1}{3}}=5^{2\times\frac{1}{3}}\times5^\frac{1}{3}$
$=5^\frac{2}{3}\times5^\frac{1}{3}=5^{\frac{2}{3}+\frac{1}{3}}=5^{\frac{3}{3}}=5$
View full question & answer→MCQ 421 Mark
Which of the following is a rational number:
- A
$\sqrt{180}$
- B
$\sqrt{31}$
- C
$\sqrt{196}$
- D
Answer- $\sqrt{196}$
Solution:
Because it is the square of 14 and can be written in the form of $\frac{\text{p}}{\text{q}}.$
View full question & answer→MCQ 431 Mark
Which of the following is rational?
- A
$\sqrt{3}$
- B
$\pi$
- C
$\frac{4}{0}$
- D
$\frac{0}{4}$
Answerd. $\frac{0}{4}$
Solution:- $\sqrt{3}=1.732 \ ...=$ Non-terminating and non-repeating number, hence irrational
- $\pi=3.14 \ ...$ also can not be terminated to $\frac{\text{p}}{\text{q}}$ form, and is non-terminating and non-repeating in nature. Hence, irrational.
- $\frac{4}{0}$ is not a rational number because this is in the form $\frac{\text{p}}{\text{q}}$ where p and q are integers but q = 0
- $\frac{0}{4}$ follows the defination of rational number.
Hence, correct option is (d).
View full question & answer→MCQ 441 Mark
The value of $\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$ is:
Answer- 1
Solution:
$\sqrt{\text{p}^{-1}\text{q}}.\sqrt{\text{q}^{-1}\text{r}}.\sqrt{\text{r}^{-1}\text{p}}$
$=\sqrt{\frac{\text{q}}{\text{p}}}.\sqrt{\frac{\text{r}}{\text{q}}}.\sqrt{\frac{\text{p}}{\text{r}}}$
$=\sqrt{\frac{\text{q}}{\text{p}}\times\frac{\text{r}}{\text{q}}\times\frac{\text{p}}{\text{r}}}$
$=\sqrt{1}$
$=1$
Hence, the correct option is (c).
View full question & answer→MCQ 451 Mark
If $\text{x}=4-\sqrt{15},$ then the value of $(\text{x}+\frac{1}{\text{x}})$ is:
Answer- 8
Solution:
$\text{x}+\frac{1}{\text{x}}=\frac{\text{x}^2+1}{\text{x}}$
Now, Put $\text{x}=4-\sqrt{15}$
$\Rightarrow\frac{(4-\sqrt{15})^2+1}{4-\sqrt{15}}$
$\Rightarrow\frac{16+15-8\sqrt{15}+1}{4-\sqrt{15}}$
$\Rightarrow\frac{32-8\sqrt{15}}{4-\sqrt{15}}$
$\Rightarrow8$
View full question & answer→MCQ 461 Mark
Which of the following is a true statment?
- A
The sum of two irrational numbers is an irrational number.
- B
The product of two irrational numbers is an irrational number.
- C
Every real number is always rational.
- D
Every real number is either rational or irrational.
Answer- Every real number is either rational or irrational.
Solution:
Consider, $\big(2+\sqrt{3}\big)$ and $\big(2-\sqrt{3}\big)$ which are two irrational numbers.
$\big(2+\sqrt{3}\big)+\big(2-\sqrt{3}\big)=4,$ which is a rational number.
Consider, $\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ which are two irrational numbers.
$\sqrt{3}\times\frac{1}{\sqrt{3}}=1,$ which is a rational number.
Every real number can either be a rational number or an irrational number.
Hence, the correct opion is (d).
View full question & answer→MCQ 471 Mark
The value of $(\frac{81}{16})^{\frac{-3}{4}}\times\{(\frac{25}{9})^{\frac{-3}{2}}\div(\frac{5}{2})^{-3}\}$ is:
Answer- 1
Solution:
$(\frac{81}{16})^{\frac{-3}{4}}\times\{(\frac{25}{9})^{\frac{-3}{2}}\div(\frac{5}{2})^{-3}\}$
$\Rightarrow(\frac{3}{2})^{4\times\frac{-3}{4}}\times\{(\frac{5}{3})^{2\times\frac{-3}{2}}\div(\frac{5}{2})^{-3}\}$
$\Rightarrow(\frac{3}{2})^{-3}\times\{(\frac{5}{3})^{-3}\div(\frac{5}{2})^{-3}\}$
$\Rightarrow(\frac{3}{2})^{-3}\times(\frac{5}{3}\times\frac{2}{5})^{-3}$
$\Rightarrow(\frac{3}{2})^{-3}\times(\frac{2}{3})^{-3}$
$\Rightarrow(\frac{3}{2}\times\frac{2}{3})^{-3}$
$\Rightarrow(1)^{-3}=1$
View full question & answer→MCQ 481 Mark
The value of xa-b ×b-c xc-a is:
Answer- 1
Solution:
xa-b ×b-c xc-a
⇒ xa-b+c+c-a
⇒ x0
= 1
View full question & answer→MCQ 491 Mark
If $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16,\ \frac{3^\text{p}}{2^\text{n}}=81$ and $\text{a}=2^{\frac{1}{10}},$ then $\frac{\text{a}^{2\text{m}+\text{n}-\text{p}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}=$
- A
$2$
- B
$\frac{1}{4}$
- C
$9$
- D
$\frac{1}{8}$
Answer- $2$
Solution:
Given: $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16,\ \frac{3^\text{p}}{2^\text{n}}=81$ and $\text{a}=2^{\frac{1}{10}}$
To find: $\frac{\text{a}^{2{\text{m}+\text{n}-\text{p}}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}$
Find: $\frac{2^{\text{m}+\text{n}}}{2^{\text{n}-\text{m}}}=16$
By using rational components $\frac{\text{a}^\text{m}}{\text{a}^\text{n}}=\text{a}^{\text{m}-\text{n}}$ We get
$2^{\text{m}+\text{n}-\text{n}+\text{m}}=16$
$2^{2\text{m}}=2^4$
By equating rational exponents we get
$2\text{m}=4$
$\text{m}=\frac{4}{2}$
$\text{m}=2$
Now, $\frac{\text{a}^{2\text{m}+\text{n}-\text{p}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})}=(\text{a}^{2\text{m}+\text{n}-\text{p}}).(\text{a}^{\text{m}-2\text{n}+2\text{p})}$ we get
$=\text{a}^{2\text{mn}+\text{n}+\text{p}+\text{m}-2\text{n}+2\text{p}}$
$=\text{a}^{3\text{m}-\text{n}+\text{p}}$
Now putting value of $\text{a}=2^\frac{1}{10}$ we get,
$=2^{\frac{3\text{m}-\text{n}+\text{p}}{10}}$
$=2^{\frac{6-\text{n}+\text{p}}{10}}$
Also, $\frac{3^\text{p}}{3^\text{n}}=81$
$3^{\text{p}-\text{n}}=3^4$
On comparing LHS and RHS we get, p - n = 4.
Now,
$\frac{\text{a}^{2{\text{m}+\text{n}-\text{p}}}}{(\text{a}^{\text{m}-2\text{n}+2\text{p}})^{-1}}=\text{a}^{3\text{m}-\text{n}+\text{p}}$
$=2^{\frac{6+(\text{p}-\text{n})}{10}}$
$=2^{\frac{6+4}{10}}$
$=2^{\frac{10}{10}}=2^1$
$=2$
So, option (a) is the correct answer.
View full question & answer→MCQ 501 Mark
The value of $\Big(\frac{12^{\frac{1}{5}}}{27^{\frac{1}{5}}}\Big)^{\frac{5}{2}}.$
- A
$\frac{12}{27}$
- B
$\frac{4}{9}$
- C
$\frac{2}{3}$
- D
Answer- $\frac{2}{3}$
Solution:
$\Big(\frac{12^{\frac{1}{5}}}{27^{\frac{1}{5}}}\Big)^{\frac{5}{2}}.$
$\Rightarrow\Big(\frac{12}{27}\Big)^{\frac{1}{5}\times\frac{5}{2}}$
$\Rightarrow\Big(\frac{12}{27}\Big)^{\frac{1}{2}}$
$\Rightarrow\frac{2\sqrt{3}}{3\sqrt{3}}$
$\Rightarrow\frac{2}{3}$
View full question & answer→