Let x litres of 30% acid solution is required to be added. Then
Total mixture = (x + 600) litres
Therefore 30% x + 12% of 600 > 15% of (x + 600)
and 30% x + 12% of 600 < 18% of (x + 600)
or \(\frac{30 x}{100}+\frac{12}{100}(600)>\frac{15}{100}(x+600) \)
and \(\frac{30 x}{100}+\frac{12}{100}(600)<\frac{18}{100}(x+600)\)
or 30x + 7200 > 15x + 9000
and 30x + 7200 < 18x + 10800
or 15x > 1800 and 12x < 3600
or x > 120 and x < 300,
i.e. 120 < x < 300
Thus, the number of litres of the 30% solution of acid will have to be more than 120 litres but less than 300 litres.

We have,
\(\tan 2 x=-\cot \left(x+\frac{\pi}{3}\right)=\tan \left(\frac{\pi}{2}+x+\frac{\pi}{3}\right)\)
or \(\tan 2 x=\tan \left(x+\frac{5 \pi}{6}\right) \)
Therefore \(2 x=n \pi+x+\frac{5 \pi}{6}, \text { where } n \in \mathbf{Z} \)
or \(x=n \pi+\frac{5 \pi}{6}, \text { where } n \in \mathbf{Z}\)

In 60 minutes, the minute hand of a watch completes one revolution. Therefore, in 40 minutes, the minute hand turns through \(\frac{2} {3}\) of a revolution. Therefore \(\theta=\frac{2}{3} \times 360^{\circ}\) or \(\frac{4\pi}{3}\) radian. Hence, the required distance travelled is given by
\(l=r \theta=1.5 \times \frac{4 \pi}{3} \mathrm{~cm}=2 \pi \mathrm{cm}=2 \times 3.14 \mathrm{~cm}=6.28 \mathrm{~cm}\)

Let a \(\in\) A. Then a \(\in\) A \(\cup\) B. Since A \(\cup\) B = A \(\cap\) B , a \(\in\) A \(\cap\) B. So a \(\in\) B. Therefore, A \(\subset\) B. Similarly, if b \(\in\) B, then b \(\in\) A \(\cup\) B.
Since A \(\cup\) B = A \(\cap\) B, b \(\in\) A Ç B. So, b \(\in\)A. Therefore, B \(\subset\) A. Thus, A = B

Let A = { –1, 0, 1 }. The subset of A having no element is the empty set \(\phi\). The subsets of A having one element are { –1 }, { 0 }, { 1 }. The subsets of A having two elements are {–1, 0}, {–1, 1} ,{0, 1}. The subset of A having three elements of A is A itself. So, all the subsets of A are \(\phi\), {–1}, {0}, {1}, {–1, 0}, {–1, 1}, {0, 1} and {–1, 0, 1}.

∴ \(\left| \frac { { z }_{ 1 }+{ z }_{ 2 }+1 }{ { z }_{ 1 }-{ z }_{ 2 }+1 } \right| \)
= \(\left| \frac { 2-i+1+i+1 }{ 2-i-1-i+1 } \right| =\left| \frac { 4 }{ 2-2i } \right| \)
= \(\frac { 4 }{ |2-2i| } =\frac { 4 }{ \sqrt { (2)^{ 2 }+(-2)^{ 2 } } } \)
= \(\frac { 4 }{ \sqrt { 4+4 } } =\frac { 4 }{ \sqrt { 8 } } =\frac { 4 }{ 2\sqrt { 2 } } =\sqrt { 2 } \)

We have
L.H.S.=\({tan({\pi\over4}+x)\over tan({\pi\over4}-x)}\)
\(={{tan{\pi\over4}+tan x\over 1-tan{\pi\over4}tanx}\over{tan{\pi\over4}-tan x\over 1+tan{\pi\over4}tanx}}\)
\(\left[\begin{array}{r} \because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B} \\ \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B} \end{array}\right]\)
\(={{1+tanx\over1-tanx}\over{ 1-tanx\over 1+tan x} }={(1+tanx)^2\over (1-tan x)^2}\)
\(={sin({x-y\over 2})\over cos({x-y\over2})}=tan({x-y\over2})=R.H.S.\)

We have 
L.H.S. =\(sin^2{\pi\over6}+cos^2{\pi\over3}-tan^2{\pi\over 4}\)
\(=({1\over2})^2+({1\over2})^2-(1)^2={1\over4}+{1\over4}-1\)
\(={1+1-4\over4}={-2\over4}={-1\over2}=R.H.S\)

cot \(({-15\pi\over4})\)=cot \(({-15\times 180^o\over4})\)
= cot (- 675°)
cot (- 675°) =cot (- 2 x 360° + 45°)
= cot 45° = 1

sin 765° = sin (2 x 360° + 45°)
= sin 45° =\(1\over \sqrt{2}\)

Let x litre of 2% boric acid solution be added to 640litres of 8%boric acid solution. Then
Total quantity of mixture = (640 + x) litres
Total boric acid in (640 + x) litres of mixture
= \(\frac { 2x }{ 100 } +\frac { 8 }{ 100 } \times 640=\frac { x }{ 50 } +\frac { 256 }{ 5 } \)
It is given that the resulting mixture must be more than 4% but less than 6% boric acid
\(\therefore\) \(\frac { 4 }{ 100 } (640+x)<\frac { x }{ 50 } +\frac { 256 }{ 5 } <\frac { 6 }{ 100 } (640+x)\)
\(\Rightarrow\)  \(\frac { 640+x }{ 25 } <\frac { x+2560 }{ 50 } <\frac { 1920+3x }{ 50 } \)
\(\Rightarrow\) 1280 + 2x \(\Rightarrow\) 1280 + 2x \(\Rightarrow\) x < 1280 and -2x < -640
\(\Rightarrow\) x < 1280 and x > 320
\(\Rightarrow\) 320 Thus 2% boric acid solution must be more than 320 litres but less than 1280 litres.

M.I of  \(\sqrt { 5 } \)+3i = \(\frac { 1 }{ \sqrt { 5 } +3i } \)
= \(\frac { 1 }{ \sqrt { 5 } +3i } \times \frac { \sqrt { 5 } -3i }{ \sqrt { 5 } -3i } \)
= \(\frac { \sqrt { 5 } -3i }{ (\sqrt { 5 } )^{ 2 }+(3i)^{ 2 } } =\frac { \sqrt { 5 } -3i }{ 5-9i^{ 2 } } \)
= \(\frac { \sqrt { 5 } -3i }{ 5+9 } \)
\(=\frac{\sqrt{5}}{14}-\frac{3 i}{14}\)

Let r₁ and r₂ be radii of two circles in which arcs of same length I subtend angles \(\theta _1\) = 60° and \(\theta _2\) = 75° respectively.
\(\therefore \theta _1 ={1\over r_1}\Rightarrow ({60\times {\pi\over 180}})^C={l\over r_1}\)
\(\Rightarrow r_1={3l\over \pi}\)
Also, \(\theta_2 ={l\over r_2}\Rightarrow 75\times {\pi\over 180}={l\over r_2}\)
\(\Rightarrow r_2={12l\over 5\pi} ....(ii)\)
From (i) and (ii)
\({r_1\over r_2}={{3l\over \pi}\over{12l\over 5\pi}}\Rightarrow {r_1\over r_2}={3l\over \pi}\times {5\pi\over12l}={5\over4}\)
Hence r₁ : r₂ = 5 : 4.

Here G = {7,8} and H = {5, 4, 2}
We know that the Cartesian product P × Q of two non-empty sets P and Q is defined as
P × Q = {(p, q): p∈ P, q ∈ Q}
\(\therefore\) G x H={(7,5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)} and H x G = {(5,7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}

Number of elements in set A = 3
Number of elements in set B = 3
Number of elements in A x B = 3 x 3 = 9.

Let the length of the shortest side of the triangle be x cm.
Then length of the longest side =3x cm.
Thus the length of the third side =(3x−2) cm.
Since the perimeter of the triangle is at least 61 cm,
x+3x+(3x−2)  ≥ 61
⇒7x−2≥61
⇒7x≥61+2
⇒7x≥63⇒x≥9 
Thus the minimum length of the shortest side is 9 cm.

Also, IQ = \(\frac { MA }{ CA } \) \(\times \) 100
\(\therefore \) 80 \(\le \) \(\frac { MA }{ CA } \)\(\times \) 100 \(\le \)140
\(\Rightarrow \) 80 \(\le \)\(\frac { MA }{ 12 } \) \(\times \) 100 \(\le \) 140
9.6 \(\le \) MA \(\le \) 16.8
range of mental age is[9.6,16.8]

Let x be the marks obtained by Ravi in the third unit test.
Since the student should have an average of at least 60 marks.
\( \frac{70+75+x}{3} \geq 60 \)
\( \Rightarrow 145+x \geq 180\)
\( \Rightarrow x \geq 180-145 \)
\( \Rightarrow x \geq 35 \)
Thus the student must obtain a minimum of 35 marks to have an average of at least 60 marks.

Let x be the marks obtained by student in the annual examination. Then
\(\frac{62+48+x}{3} \geq 60\)
or 110 + x  \(\geq\) 180
or x \(\geq\)70
Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.

2(2x+3)−10<6(x−2)
⇒4x+6−10<6x−12
⇒4x−4<6x−12
⇒−4+12<6x−4x
⇒8<2x
⇒4 Thus, all real numbers x, which are greater than 4, are the solution of the given inequality.
Hence, the solution set of the given inequality is (4, \(\infty \))

\({ \left[ { i }^{ 18 }+{ \left( \frac { 1 }{ i } \right) }^{ 25 } \right] }^{ 3 }\) =\(\left[ ({ i }^{ 2 })^{ 9 }+\frac { 1 }{ ({ i }^{ 2 })^{ 12 }.i } \right] ^{ 3 }\)
= \(\left[ (-1)^{ 9 }+\frac { 1 }{ (-1)^{ 12 }.i } \right] ^{ 3 }\)
=\(\left[ -1+\frac { 1 }{ i } \right] ^{ 3 }\) =(-1-i)³   \(\left( \because \frac { 1 }{ i } =-i \right) \)
= -(1+i)³=[1+i³+3i+3i²]
= -[1+3i-3-1]
= -[ -2+ 2i]
= 2-2i
=2(1-i)

We know that, degree measure \(=\frac { 180 }{ \pi } \times \) Radian measure
\(\therefore \)  Required degree measure \(=\left( \frac { 180 }{ \frac { 22 }{ 7 } } \times 6 \right) \quad \left[ \therefore \pi =\frac { 22 }{ 7 } \right] \)
\(=\left( \frac { 1080\times 7 }{ 22 } \right) ^{ o }=\left( 343\frac { 7 }{ 11 } \right) ^{ 0 }\)
\(={ 343 }^{ o }+\frac { 7 }{ 11 } \times 60min\)  \(\left[ \because { 1 }^{ o }=60' \right] \)
\(={ 343 }^{ o }+38\frac { 2 }{ 11 } min={ 343 }^{ o }+38'+\frac { 2 }{ 11 } min\)
\(={ 343 }^{ o }+38'+\frac { 2 }{ 11 } \times 60''\)
\(={ 343 }^{ o }+38'+10.9''={ 343 }^{ o }38'11''(approx.)\)
Hence, the degree measure of 6 rad is 
\({ 343 }^{ o }38'11''(approx.).\)

Here, given degree measure have seconds and minutes. So, firstly convert seconds into minutes.
\(\therefore \) \(1"=\left( \frac { 1 }{ 60 } \right) ^{ ' }\) 
\(\therefore \) \(30''=\left( \frac { 30 }{ 60 } \right) ^{ ' }=\left( \frac { 1 }{ 2 } \right) ^{ ' }\)
\(\therefore \) Total minutes to convert
\(=37'+\frac { { 1 }^{ ' } }{ 2 } =\left( \frac { 74+1 }{ 2 } \right) ^{ ' }=\left( \frac { 75 }{ 2 } \right) ^{ ' }\)
\(\therefore \) \(1'=\left( \frac { 1 }{ 60 } \right) ^{ o }\)  
\(\therefore \) \(\left( \frac { 75 }{ 2 } \right) ^{ ' }=\left( \frac { 75 }{ 2\times 60 } \right) ^{ o }=\left( \frac { 5 }{ 8 } \right) ^{ o }\)
\(\therefore \) Total degree to convert
\(={ 48 }^{ o }+\left( \frac { 5 }{ 8 } \right) ^{ o }\)
\(=\left( \frac { 384+5 }{ 8 } \right) ^{ o }=\left( \frac { 389 }{ 8 } \right) ^{ o }\)
\(\because \) Radian measure \(=\frac { \pi }{ 180 } \times \) Degree measure
\(\therefore \)  \(\left( \frac { 389 }{ 8 } \right) ^{ o }=\frac { \pi }{ 180 } \times \frac { 389 }{ 8 } \)
\(=\frac { 389\pi }{ 1440 } \) rad
Hence, 48^o 37' 30'' is \(\frac { 389\pi }{ 1440 } \) rad.

(i). It is false, because if P = {m, n} and Q = {n, m}, then P x Q = {m,n} x {n,m} = {(m,n), (m,m),(n,n),(n,m)}
(ii). It is false, correct statement is "If A and B are non-empty sets, then AxB is a non-empty set of ordered pairs (x,y), such that \(x\in B\) and \(y \in B^{\prime \prime}\)
(iii). It is true, because \(B\cap \phi =\phi \)
\(\therefore \quad A\times (B\cap \phi )=A\times \phi =\phi \)

\((f+g)(x)=x^{2}+2 x+1,(f-g)(x)=x^{2}-2 x-1,\)
\((\text { fg })(x)=x^{2}(2 x+1)=2 x^{3}+x^{2},\left(\frac{f}{g}\right)(x)=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \)

We have,
\((f+g)(x)=x^{2}+2 x+1,(f-g)(x)=x^{2}-2 x-1\)
\((\text { fg })(x)=x^{2}(2 x+1)=2 x^{3}+x^{2},\left(\frac{f}{g}\right)(x)=\frac{x^{2}}{2 x+1}, x \neq-\frac{1}{2} \)

Given,
\(f\left( x \right) ={ x }^{ 2 }-x-2\)
\(A=\left\{ -1,0,2,5,6,11 \right\}\)
\(B=\left\{ -2,-1,0,18,28,108 \right\} \)
Clearly,
\(f(-1)={ \left( -1 \right) }^{ 2 }-(-1)-2=0\)
\(f(0)={ \left( 0 \right) }^{ 2 }-0-2=-2\)
\(f(2)={ \left( 2 \right) }^{ 2 }-2-2=0\)
\(f(5)={ \left( 5 \right) }^{ 2 }-5-2=18\)
\( f(6)={ \left( 6 \right) }^{ 2 }-6-2=28\)
\(f(11)={ \left( 11 \right) }^{ 2 }-11-2=108\)
Now, \(f(A)=\left\{ f\left( x \right) :x\in A \right\}\)
\(=\{ 0,-2,18,28,108\}\)
Here, we see that \( -1\in B\quad but-1\notin f(A)\)
\( \therefore \quad f(A)\neq B\)

Let M be the set of students in Mathematics class and P be the set of students in Physics class.
Given that, n(M)=60 and n(P)=90
Two classes meet at same hour
\(M\cap P=\phi \Rightarrow n(M\cap P)=0\)
Now, \(n(M\cup P)=n(M)+n(P)-n(M\cap P)\)
=60+90-=150

Given, A = {1,3,5}   n(A) = 3
Number of elements in P(A) = 2³ = 8
\(\therefore\)   Number of elements in P(P(A)) = 2⁸ = 256

Let A, B and C denote the set of students who study English, Economics and Mathematics, respectively.
Then, we have,
Total number of students, n (A \(\cup \) B \(\cup \) C) = 40
Number of students who study English, n (A) = 16
Number of students who study Eoconomics, n (B) = 22
Number of students who study Mathematics, n (C) = 26
Number of students who study English and Economics, n (A  \(\cap \) B) = 5
Number of students who study Mathematics and Economics, n( B \(\cap \) C) = 14
and number students who study all subjects,
n (A \(\cup \) B \(\cup \) C) = 2
Clearly, n (A \(\cup \) B \(\cup \) C)  = n(A) + n(B) - n(C) - n (A \(\cap \) B) - n( B \(\cap \) C)  - n( A \(\cap \) C) -  n (A \(\cap \) B \(\cap \) C)                       
40 = 16 +22+ 26 - 5- 14 n ( C \(\cap \) A) + 2
40 = 66 - 19 - n ( C \(\cap \) A)
n ( C \(\cap \) A) = 47 - 70
=7
Hence the number of students who study English and Mathematics are 7.

Let C and T respectively donote the set of students who play Cricket and Tennis, and set U donotes the set of all students in a class.
Then n(C) = 25, n(T) = 20, n(C\(​​\)\(\cap \)T) = 10 and n (U)= 60
We know that, n(C\(​​\)\(\cup \)T) = n(C) + n(T) - n(C\(​​\)\(\cap \)T)
\(         \Rightarrow \) n(C\(​​\)\(\cup \) T) = 25 + 20 - 10 = 45 - 10 = 35
Now the Number of students who play neither game
n(C'\(\cap \)T' )   = n(C'\(\cup \)T' ) 
= n (U) -  n(C\(\cup \)T) = 60 - 35= 25
Hence, 25 students play neither games.