Question 511 Mark
Write the correct answer in the following :
The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to
Answer- 2
Solution:
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16\times2}+\sqrt{16\times3}}{\sqrt{4\times2}+\sqrt{4\times3}}$
$=\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}$
$=\frac{4}{2}=2$
Hence, (b) is correct answer.
View full question & answer→Question 521 Mark
The value of $\frac{\text{x}^{\text{a}(\text{c}-\text{c})}}{\text{x}^{\text{b}(\text{a}-\text{c})}}\div\Big(\frac{\text{x}^\text{b}}{\text{x}^{\text{a}}}\Big)^{\text{c}}$ is:
Answer- 1
Solution:
$\frac{\text{x}^{\text{a}(\text{c}-\text{c})}}{\text{x}^{\text{b}(\text{a}-\text{c})}}\div\Big(\frac{\text{x}^\text{b}}{\text{x}^{\text{a}}}\Big)^{\text{c}}$
$\Rightarrow\frac{\text{x}^{\text{ab}-\text{ac}}}{\text{x}^{\text{ba}-\text{be}}}\div\Big(\frac{\text{x}^{\text{bc}}}{\text{x}^{\text{ac}}}\Big)$
$\Rightarrow\text{x}^{\text{ab}-\text{ac}-\text{ab}+\text{bc}}\div\text{x}^{\text{bc}-\text{ac}}$
$\Rightarrow\text{x}^{\text{bc}-\text{ac}}\div\text{x}^{\text{bc}-\text{ac}}$
$\Rightarrow1$
View full question & answer→Question 531 Mark
If $\frac{2+\sqrt{3}}{2-\sqrt{3}}=\text{a}+\text{b}\sqrt{3},$ then,
Answer- a = 7 and b = 4
Solution:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Multiplying numerator and denominator by $2\div\sqrt{3}$
So, $\frac{(2+\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}$
$=\frac{4+3+4\sqrt{3}}{4-3},$
$=7\div4\sqrt{3}$
Now equating $7\div4\sqrt{3}$ and $\text{a}\div\text{b}\sqrt{3}$
we get,
a = 7 and b = 4
View full question & answer→Question 541 Mark
If $x=\frac{2}{3+\sqrt7},$ then $(x - 3)^2=$
Answer$\text{x}=\frac{2}{3+\sqrt7}=\frac{2}{3+\sqrt7}\times\frac{3-\sqrt7}{3-\sqrt7}=\frac{2(3-\sqrt7)}{3-\sqrt7}$
$\ =\frac{6-2\sqrt7}{9-7}=\frac{6-2\sqrt7}{2}=3-2\sqrt7$
Now $(\text{x}-3)^2=(\not3-\sqrt7-\not3)^2=\big(-\sqrt7\big)^2=7$
View full question & answer→Question 551 Mark
The number $0.\overline{32}$ when expressed in the form $\frac{\text{p}}{\text{q}}$ $\big($p, q are integers and $\text{q}\neq0\big),$ is:
Answer- $\frac{29}{90}$
Solution:
Let $\text{x}=0.\overline{32}=0.32222..(1)$
Now, $10\text{x}=3.2222=3.\overline{2}...(2),$
Subtracting equation (2) and (3), we get
$90\text{x}=29$
$\Rightarrow\text{x}=\frac{29}{90}$
Hence, option (b) is correct.
View full question & answer→Question 561 Mark
There is a number $x$ such that $x^2$ is irrational but $x^4$ is rational. Then, $x$ can be :
Answer$\big(\sqrt[4]{2}\big)^2=\Big(2^{\frac{1}{4}}\Big)^2=2^{\frac{1}{4}\times2}=2^{\frac{1}{2}}=\sqrt{2},$ which is irrational.
$\big(\sqrt[4]{2}\big)^2=\Big(2^{\frac{1}{4}}\Big)^2=2^{\frac{1}{4}\times4}=2^1=2,$ which is rational.
View full question & answer→Question 571 Mark
If $3^\text{x}=64=2^6+(\sqrt{3})^8,$ then the value of x is :
Answer- 4
Solution:
$3^\text{x+64}=2^6+(\sqrt{3})^8$
But we know that,
$2^6=64$
So, $2^6+3^\text{x}=2^6+(\sqrt{3})^{2\times4}$
$\Rightarrow2^6+3\text{x}=2^6+(3)^4$
Now by equating both
We get,
$\text{x}=4$
View full question & answer→Question 581 Mark
Which of the following is the value of $\big(\sqrt{11}-\sqrt{7}\big)\big(\sqrt{11}+\sqrt{7}\big)?$
Answer- 4
Solution:
$\big(\sqrt{11}-\sqrt{7}\big)\big(\sqrt{11}+\sqrt{7}\big)$
$\big(\sqrt{11}\big)^2-\big(\sqrt{7}\big)^2$
$11-7=4$
Hence, the correct answer is option (b).
View full question & answer→Question 591 Mark
If $\text{x}=\sqrt5+2,$ then $\text{x}-\frac{1}{\text{x}}$ equals :
Answer- $4$
Solution:
$\text{x}=\sqrt{5}+2$
$\Rightarrow\frac{1}{\text{x}}=\frac{1}{\sqrt5+2}=\frac{1}{\sqrt5+2}\frac{\sqrt5-2}{\sqrt5-2}\\ \ =\frac{\sqrt5-2}{1}=\sqrt5-2$
Now, $\text{x}-\frac{1}{\text{x}}=\sqrt5+2-\big(\sqrt5-2\big)\\ \ =\sqrt5+2-\sqrt5+2=4$
Hence, correct option is (b).
View full question & answer→Question 601 Mark
If $\sqrt{2^\text{n}}=1024,$ then $3^{2\Big(\frac{\text{n}}{4}-4\Big)}=$
Answer- 9
Solution:
We have to find $3^{2\Big(\frac{\text{n}}{4}-4\Big)}$
Given $\sqrt{2^\text{n}}=1024$
$\sqrt{2^\text{n}}=2^\text{10}$
$2^{\text{n}\times\frac{1}{2}}$
Equating powers of rational exponents we get
$\text{n}\times\frac{1}{2}=10$
$\text{n}=10\times2$
$\text{n}=20$
Substituting in $3^{2\Big(\frac{\text{n}}{4}-4\Big)}$ we get
$3^{2\Big(\frac{\text{n}}{4}-4\Big)}=3^{2\Big(\frac{20}{4}-4\Big)}$
$=3^{2(5-4)}$
$=3^{2\times1}$
$=9$
Hence the correct choice is b.
View full question & answer→Question 611 Mark
The product of two irrational numbers is.
Answer- Either irrational or rational.
Solution:
$\sqrt{5}\times\sqrt{2}=\sqrt{10},$ is irrational number
$\sqrt{5}\times\sqrt{5}=5,$ is retional number
View full question & answer→Question 621 Mark
The value of $\Big\{8^{-\frac{4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}$ is:
Answer- $\frac{1}{2}$
Solution:
Find the value of $\Big\{8^{-\frac{4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}$
$\Big\{8^{-\frac{4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}=\Big\{2^{3\times\frac{-4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}$
$\big\{2^{-4}\div2^{-2}\big\}^{\frac{1}{2}}$
$\Big\{8^{-\frac{4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}=\Big\{2^{-4\times\frac{1}{2}}\div2^{-2\times\frac{1}{2}}\Big\}$
$=\Big\{2^{-2}\div2^{-1}\Big\}$
$=\Bigg\{\frac{\frac{1}{2^2}}{\frac{1}{2}}\Bigg\}$
$\Big\{8^{-\frac{4}{3}}\div2^{-2}\Big\}^{\frac{1}{2}}=\Big\{\frac{1}{2\times2}\times\frac{2}{1}\Big\}$
$=\frac{1}{2}$
Hence the correct choice is a.
View full question & answer→Question 631 Mark
A rational number between -3 and 3 is :
Answer- 0
Solution:
Since, -4.3 < -3.4 < -3 < 0 < 1.101100110001... < 3
But 1.101100110001... is an irrational number
So, the rational number between -3 and 3 is 0.
Hence, the correct option is (a).
View full question & answer→Question 641 Mark
$9^3 + (-3)^3 - 6^3 =$ ?
Answer$9^3 + (-3)^3 - 6^3 = 729 - 27 - 21 6 = 486$
View full question & answer→Question 651 Mark
$\pi$ is:
Answer- an irrational number.
Solution:
$\pi=3.14159265359...,$ which is non-terminating non-recurring.
Hence, it is an irrational number.
Hence, the correct opion is (c).
View full question & answer→Question 661 Mark
The number $1.\overline{27}$ in the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ is:
Answer- $\frac{14}{11}$
Solution:
Let $\text{x}=1.\overline{27}=1.272727...(1)$
Now, $100\text{x}=127.272727=127.\overline{27}...(2)$
Subtracting equation (1) from (2), we get
$99\text{x}=126$
$\Rightarrow\text{x}=\frac{126}{99}=\frac{14}{11}$
Hence, option (b) is correct.
View full question & answer→Question 671 Mark
Write the correct answer in the following:
The product $\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}$ equals.
Answer- $2$
Solution:
LCM of 3, 4 and 12 = 12
$\sqrt[3]{2}=\sqrt[12]{2^4}\ [\because\sqrt[\text{m}]{\text{a}}=\sqrt[\text{mn}]{\text{a}^\text{n}}]$
$\sqrt[4]{2}=\sqrt[12]{2^3}$
and $\sqrt[12]{32}=\sqrt[12]{2^5}$
$\therefore\text{product of }\sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{2^3}.\sqrt[12]{2^5}=\sqrt[12]{2^4.2^3.2^5}$
$=12\sqrt{2^{4+3+5}}=\sqrt[12]{2^{12}}=2^{12\times\frac{1}{12}}=2\ [\because\sqrt[\text{m}]{\text{a}^\text{n}}=\text{a}^{\frac{\text{n}}{\text{m}}}]$
View full question & answer→Question 681 Mark
Which of the following is a rational number?
Answer- $0$
Solution:
0 is an integer and all integers are rational numbers.
View full question & answer→Question 691 Mark
The rational number not lying between $\frac{3}{5}$ and $\frac{2}{3}$ is:
Answer- $\frac{50}{75}$
Solution:
1cm = 75
So, $\frac{3}{5}\times\frac{15}{15}$ and $\frac{2\times25}{3\times25}$
i.e. $\frac{45}{75}$ and $\frac{50}{75}$
So, $\frac{46}{75},\frac{47}{75},\frac{48}{75},\frac{49}{75}$
View full question & answer→Question 701 Mark
$(5+\sqrt{8})+(3-\sqrt{2})(\sqrt{2}-6)$ is:
Answer- Positive and irrational.
Solution:
$(5+\sqrt{8})+(3-\sqrt{2})(\sqrt{2}-6)$
$=(5+2\sqrt{2})+(3\sqrt{2}-18-2+6\sqrt{2})$
$=(5+2\sqrt{2})+(9\sqrt{2}-20)$
$=11\sqrt{2}-15$
And we know that the value of $11\sqrt{2}$ is greater than 15 so it's value will be positive,
And also sum or differences of rational and irrational is irrational.
View full question & answer→Question 711 Mark
Write the correct answer in the following:
$\frac{1}{\sqrt{9}-\sqrt{8}}$ is equal to
Answer- $3+2\sqrt{2}$
Solution:
$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{3-2\sqrt{2}}=\frac{1}{3-2\sqrt{2}}\cdot\frac{3+2\sqrt{2}}{3+2\sqrt{2}}$ $[\because\sqrt{8}=\sqrt{2\times2\times2}=2\sqrt{2}]$
$[$multiplying numerator and denominator by $3+2\sqrt{2}]$
$\frac{3+2\sqrt{2}}{9-(2-\sqrt{2})^2}$$[\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$
$=\frac{3+2\sqrt{2}}{9-8}=3+2\sqrt{2}$
View full question & answer→Question 721 Mark
If $\sqrt{3}=1.732$ and $\sqrt{2}=1.414, $ 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 731 Mark
If $\frac{3^{2\text{x}-8}}{225}=\frac{5^3}{5^{\text{x}}},$ then x =
Answer- 5
Solution:
We have to find the value of x provided $\frac{3^{2\text{x}-8}}{225}-=\frac{5^3}{5^\text{x}}$
So,
$\frac{3^{2\text{x}-8}}{225}-=\frac{5^3}{5^\text{x}}$
By cross multiplication we get
$3^{2\text{x}-8}\times5^\text{x}=3^2\times5^2\times5^3$
By equating exponents we get
$3^{2\text{x}-8}=3^2$
$2\text{x}-8=2$
$2\text{x}=2+8$
$2\text{x}=10$
$\text{x}=\frac{10}{2}$
$\text{x}=5$
And
$5^{\text{x}}=5^{3+2}$
$\text{x}=3+2$
$\text{x}=5$
Hence the correct choice is c.
View full question & answer→Question 741 Mark
Which of the following is an irrational number?
Answer- $\sqrt{23}$
Solution:
The decimal expansion of $\sqrt{23}=4.79583152...,$ which is non-terminating, nonrecurring.
Hence, it is an irrational number.
Hence, the correct opion is (a).
View full question & answer→Question 751 Mark
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is:
Answer- $6^{\frac{1}{4}}$
Solution:
An irrational number between a and b is given by $\sqrt{\text{ab}}$
An irrational number between $\sqrt{2}$ and $\sqrt{3}$ is
$\sqrt{\sqrt{2}+\sqrt{3}}=\big(\sqrt{6}\big)^{\frac{1}{2}}$
$=6^{\frac{1}{2}\times\frac{1}{2}}$
$=6^{\frac{1}{4}}$
Hence, the correct answer is option (d).
View full question & answer→Question 761 Mark
The positive square root of $7+\sqrt{48},$ is:
Answer- $2+\sqrt3$
Solution:
$\sqrt{7+\sqrt{48}}$
$=\sqrt{7+2\sqrt{12}}$
$=\sqrt{4+3+2\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4\big)^2+\big(\sqrt3\big)^2+2\times\sqrt4\times\sqrt3}$
$=\sqrt{\big(\sqrt4+\sqrt3\big)^2}$
$=\pm\big(\sqrt4+\sqrt3\big)$
Positive value is $\sqrt4+\sqrt3=2+\sqrt3$
Hence, correct option is (c).
View full question & answer→Question 771 Mark
Which of the following is not equal to $\Big[\big(\frac{5}{6}\big)^{\frac{1}{5}}\Big]^{-\frac{1}{6}}?$
Answer- $\big(\frac{6}{5}\big)^{\frac{1}{30}}$
Solution:
$\big(\frac{5}{6}\big)^{\frac{1}{5}}\frac{1}{6}$
$\Big[\big(\frac{5}{6}\big)^{\frac{1}{5}}\Big]^{-\frac{1}{6}}$
$=\big(\frac{5}{6}\big)^{\frac{1}{5}\times\big(-\frac{1}{6}\big)}$
$=\big(\frac{5}{6}\big)^{\frac{-1}{30}}$
$=\big(\frac{6}{5}\big)^{\frac{1}{30}}$
View full question & answer→Question 781 Mark
When simplified $\big(256\big)^{-\Big((4^{-\frac{3}{2}}\Big)}$ is:
Answer- $\frac{1}{2}$
Solution:
Simplify $\big(256\big)^{-\Big((4^{-\frac{3}{2}}\Big)}$
$\big(256\big)^{-\Big((4^{-\frac{3}{2}}\Big)}=\big(256\big)^{-(2^2)^{-\frac{3}{2}}}$
$=\big(256\big)^{\Big(2^{2\times-\frac{3}{2}}\Big)}$
$\big(256\big)^{-\Big((4^{-\frac{3}{2}}\Big)}=\big(256\big)^{-(2)^{(-3)}}$
$\big(256\big)^{\Big(-4^{-\frac{3}{2}}\Big)}=\big(256\big)^{\frac{1}{(-2) ^3}}$
$=\big(256\big)^{\frac{1}{-8}}$
$=\big(2^8\big)^{\frac{1}{-8}}$
$=2^{8\times\frac{1}{-8}}$
$\big(256\big)^{-\Big((4^{-\frac{3}{2}}\Big)}=2^{8\times\frac{1}{-8}}=\frac{1}{2}$
Hence the correct choice is d.
View full question & answer→Question 791 Mark
Every point on a number line represents:
Answer- a unique real number.
Solution:
On number line, we have $-\infty$ to $\infty$ numbers, consisting $-\infty...-4,-3,-2,-1,0,1,2,3,4...\infty,$ 1.12,1.14 and 1.41406532, 3.146201286295... etc. That means on number line, there are natural nmbers (1, 2, 3, 4 ...), integers, rational numbers $\frac{1}{2},\frac{1}{3},1.3333,$ irrational numbers 1.4148625385...
But if we see every number as a complete family, it becomes
Real numbers (any number which can be represent on Real axes)
So, every point on the number line reoresent a unique real number which contains every type.
Hence, option (a) is correct.
View full question & answer→Question 801 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 811 Mark
The value of $7^{\frac{1}{2}}.8^{\frac{1}{2}}$ is:
Answer- $(56)^\frac{1}{2}$
Solution:
$(7)^{\frac{1}{2}}\times(8)^{\frac{1}{2}}=(7\times8)^{\frac{1}{2}}=(56)^{\frac{1}{2}}$
Hence, the correct answer is option (b).
View full question & answer→Question 821 Mark
The rationalisation factor of $2+\sqrt3,$ is:
Answer- $2-\sqrt3$
Solution:
Rationalisation factor of any number like ${\text{a}}\pm\sqrt{\text{b}}$ is ${\text{a}}\mp\sqrt{\text{b}}.$
So. Rationalisation factor of $2+\sqrt3$ will be $2-\sqrt3$
Hence, correct option is (a).
View full question & answer→Question 831 Mark
The square root of 64 divided by the cube root of 64 is:
Answer- $2$
Solution:
We have to find the value of $\frac{\sqrt[2]{64}}{\sqrt[3]{64}}.$
So,
$\frac{\sqrt[2]{64}}{\sqrt[3]{64}}=\frac{\sqrt[2]{2\times2\times2\times2\times2\times2}}{\sqrt[2]{2\times2\times2\times2\times2\times2}}$
$=\frac{2^{6\times\frac{1}{2}}}{2^{6\times\frac{1}{3}}}$
$=\frac{\sqrt[2]{64}}{\sqrt[3]{64}}=\frac{2^3}{2^2}$
$=2^{3-2}$
$=2^1$
$=2$
The value of $\frac{\sqrt[2]{64}}{\sqrt[3]{64}}$ is 2
Hence the correct choice is b.
View full question & answer→Question 841 Mark
The number $1.\overline{3}$ in the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ is:
Answer- $\frac{1}{3}$
Solution:
Let $\text{x}=1.\overline{3}=0.3333..(1)$
Now, $10\text{x}=3.3333=3.\overline{3}...(2)$
Subtracting equation (1) from (2), we get
$9\text{x}=3$
$\Rightarrow\text{x}=\frac{3}{9}=\frac{1}{3}$
$\Rightarrow0.\overline{3}=\frac{1}{3}$
Hence, option (c) is correct.
View full question & answer→Question 851 Mark
$(-2-\sqrt{3})(-2+\sqrt{3})$ when simplified is:
Answer- Positive and rational.
Solution:
$(-2-\sqrt{3})(-2+\sqrt{3})$
$=(-2)^2-(\sqrt{3})^2$
$=4-3$
$=1$
Positive and rational
View full question & answer→Question 861 Mark
The value of $2.\overline{45}+0.\overline{36}$ is:
Answer- $\frac{31}{11}$
Solution:
Let $\text{x}=2.\overline{45}$
i.e., $\text{x}=2.4545 \ ...(\text{i})$
$\Rightarrow100\text{x}=245.4545 \ ...(\text{ii})$
On subtracting (i) and (ii), we get
$99\text{x}=243$
$\Rightarrow\text{x}=\frac{243}{99}$
Let $\text{y}=0.\overline{36}$
i.e., $\text{y}=0.3636 \ ...(\text{iii})$
$\Rightarrow100\text{y}=36.3636 \ ...(\text{iv})$
On subtracting (iii) and (iv), we get
$99\text{y}=36$
$\Rightarrow\text{y}=\frac{36}{99}$
$\therefore2.\overline{45}+0.\overline{36}=\text{x}+\text{y}=\frac{\text{243}}{\text{99}}+\frac{\text{36}}{\text{99}}=\frac{\text{279}}{\text{99}}=\frac{\text{31}}{\text{11}}$
Hence, the correct answer is option (c).
View full question & answer→Question 871 Mark
$\Big(\frac{2}{3}\Big)^\text{x}\Big(\frac{3}{2}\Big)^{2\text{x}}=\frac{81}{16}$ then x =
Answer- 4
Solution:
We have to find value of x provided $\Big(\frac{2}{3}\Big)^\text{x}\Big(\frac{3}{2}\Big)^{2\text{x}}=\frac{81}{16}$
So,
$\Big(\frac{2}{3}\Big)^\text{x}\Big(\frac{3}{2}\Big)^{2\text{x}}=\frac{81}{16}$
$\Big(\frac{2}{3}\Big)^\text{x}\Big(\frac{3}{2}\Big)^{2\text{x}}=\frac{3^4}{2^4}$
$\frac{2^\text{x}}{3^\text{x}}\frac{3^{2\text{x}}}{2^{2\text{x}}}=\frac{3^4}{2^4}$
$\frac{3^{2\text{x}-\text{x}}}{2^{2\text{x}-\text{x}}}=\frac{3^4}{2^4}$
$\frac{3^\text{x}}{2^\text{x}}=\frac{3^4}{2^4}$
Equating exponents of power we get x = 5
Hence the correct alternative is c.
View full question & answer→Question 881 Mark
Which one of the following is not equal to $\big(\sqrt[3]{8}\big)^{-\frac{1}{2}}?$
Answer- $\big(\sqrt[3]{2}\big)^{-\frac{1}{2}}$
Solution:
We have to find the value of $\big(\sqrt[3]{8}\big)^{-\frac{1}{2}}$
So,
$\big(\sqrt[3]{8}\big)^{-\frac{1}{2}}=\big(\sqrt[3]{2\times2\times2}\big)^{-\frac{1}{2}}$
$=\big(\sqrt[3]{2^3}\big)^{-\frac{1}{2}}$
$=2^{3\times\frac{1}{3}\times-\frac{1}{2}}$
$\big(\sqrt[3]{8}\big)^{-\frac{1}{2}}=2^{-\frac{1}{2}}$
$=\frac{1}{2^{\frac{1}{2}}}$
$=\frac{1}{\sqrt{2}}$
Also, $\big(\sqrt[3]{8}\big)^{-\frac{1}{2}}=2^{-\frac{2}{6}}$
Hence the correct alternative is a.
View full question & answer→Question 891 Mark
The value of $\frac{9^{\frac{1}{3}}\times27^\frac{1}{2}}{3^{\frac{-1}{6}}\times3^{\frac{1}{3}}}$ is:
Answer- 9
Solution:
$\frac{9^{\frac{1}{3}}\times27^\frac{1}{2}}{3^{\frac{-1}{6}}\times3^{\frac{1}{3}}}$
$\Rightarrow\frac{3^{\frac{2}{3}}\times3^{\frac{3}{2}}}{3^{\frac{-1}{6}}\times3^{\frac{1}{3}}}$
$\Rightarrow\frac{3^{\frac{2}{3}+\frac{3}{2}}}{3^{\frac{-1}{6}+\frac{1}{3}}}$
$\Rightarrow\frac{3\frac{4+9}{6}}{3^{\frac{-1+2}{6}}}$
$\Rightarrow\frac{3\frac{13}{6}}{3^{\frac{1}{6}}}=3^{\frac{13}{6}-\frac{1}{6}}$
$\Rightarrow3^{\frac{12}{6}}=9$
$=(\frac{1}{5})^{-1}=25$
View full question & answer→Question 901 Mark
If $\sqrt{2}=1.41$ than $\frac{1}{\sqrt{2}}=?$
Answer- 0.705
Solution:
$\frac{1}{\sqrt{2}}=\frac{1}{1.41}=0.705$
View full question & answer→Question 911 Mark
Write the correct answer in the following:
The value of 1.999... in the form $\frac{\text{p}}{\text{q}},$ where p and q are integers and $\text{q}\neq0,$ is
Answer- $2$
Solution:
Let x = 1.999...
Now, 10x = 19.999...
On subtracting Eq. (i) from Eq. (ii), we get
10x - x = (19.999...) - (1.9999...)
⇒ 9x = 18
$\therefore\text{x}=\frac{18}{9}=2$
View full question & answer→Question 921 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→Question 931 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: $\frac{0}{9}$is rational number.
Reason: $\frac{0}{9}$ is a rational number that is equal to 0.
Answer - Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
View full question & answer→Question 941 Mark
The product of a non-zero rational number with an irrational number is always a/an:
Answer- irrational number.
Solution:
The product if a non-zero rational number with an irrational number is always an irrational number.
Hence, the correct option is (a).
View full question & answer→Question 951 Mark
The value of $\sqrt{5+2\sqrt6},$ is:
Answer- $\sqrt3+\sqrt2$
Solution:
$\sqrt{5+2\sqrt6}$
$=\sqrt{3+2+2\big(\sqrt3\big)\big(\sqrt2\big)}$
$=\sqrt{\big(\sqrt3\big)^2+\big(\sqrt2\big)^2-2\big(\sqrt3\big)\big(\sqrt2\big)}$
$=\sqrt{\big(\sqrt2+\sqrt2\big)^2}$
$=\sqrt3+\sqrt2$
Hence, correct option is (b).
View full question & answer→Question 961 Mark
The value of $(243)^{\frac{1}{5}}$ is:
Answer- 3
Solution:
$(243)^{\frac{1}{5}}=(3^5)^{\frac{1}{5}}=3^{5\times\frac{1}{5}}=3$
Hence, the correct answer is option (a).
View full question & answer→Question 971 Mark
Write the correct answer in the following:
After rationalising the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}},$ we get the denominator as
Answer- 19
Solution:
$\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}$
$[$multiplying numerator and denominator by $3\sqrt{3}+2\sqrt{2}]$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{(3\sqrt{3})^2-(2\sqrt{2})^2}$ $[\text{using identity (a}-\text{b})(\text{a+b})=\text{a}^2-\text{b}^2]$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{27-8}$ $\big[\because\text{fraction}=\frac{\text{numerator}}{\text{denominator}}\big]$
$=\frac{7(3\sqrt{3}+2\sqrt{2})}{19}$
Hence, after rationalising the denominator of $=\frac{7}{(3\sqrt{3}-2\sqrt{2})}$ we get the denominator as 19.
View full question & answer→Question 981 Mark
If $9x^{+2} = 240 + 9^x$, then $x =$
AnswerWe have to find the value of $x$
Given $9x^{+2} = 240 + 9^x$
$9^x \times 9^2 = 240 + 9^x$
$9^2=\frac{240}{9^{\text{x}}}+\frac{9^{\text{x}}}{9^{\text{x}}}$
$81=\frac{240}{9^{\text{x}}}+1$
$81-1=\frac{240}{9^\text{x}}$
$80=\frac{240}{9^\text{x}}$
$9^\text{x}\times80=240$
$9^\text{x}=\frac{240}{80}$
$3^{2\text{x}}=3$
$3^{2\text{x}}=3^1$
By equating the exponents we get
$2\text{x}=1$
$\text{x}=\frac{1}{2}$
$\text{x}=0.5$
View full question & answer→Question 991 Mark
The value of $0.\bar{2}$ in the from $\frac{\text{p}}{\text{q}}$ where p and q are integers and $\text{q}\not=0$ is:
Answer- $\frac{2}{9}$
Solution:
Let x = 0.222.... ---(i)
Multiply eq. (i) by 10, we get
10x = 2.222... ----(ii)
10x - x = 2.222... -0.222...
9x = 2
$\text{x}=\frac{2}{9}$
View full question & answer→Question 1001 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: $7^1= 7$
Reason: $a^1= a$
AnswerBoth Assertion and Reason are correct and Reason is the correct explanation for Assertion.
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