Let the number of red balls be x and white balls be y.
According to the question,
\(\frac { 1 }{ 2 } y=\frac { 1 }{ 3 } x\) or \(2x-3y=0\quad \)...(i)
and 3(x + y) - 7y = 6
or 3x - 4y = 6                        ...(ii)
Multiplying eqn. (i) by 3 and eqn. (ii) by 2 and then subtracting, we get
6x - 9y = 0
6x - 8y = 12
-   +        -
\(\underline { \overline { \quad -y=-12 } } \)
\(\therefore\) y =12
2x - 36 = 0
Substituting 12 in eqn. (i),
\(\therefore\) x = 18
\(\therefore\) x = 18,y = 12
Hence, number of red balls = 18
and number of white balls = 12

Let the actual speed of the train be x km/hr and actual time taken be y hr.
Distance = Speed \(\times\) Time
\(\because\) =xy km
According to the given condition,
xy = (x + 10)(y - 2)
\(\Rightarrow\) xy = xy - 2x + 10y-20
\(\Rightarrow\) 2x - 10y + 20 = 0
\(\Rightarrow\) x - 5y + 10 = 0 [divide by 2] ... (i)
and xy = (x - 10)(y + 3)
\(\Rightarrow\) xy = xy + 3x - 10y - 30
\(\Rightarrow\) 3x-10y- 30 = 0 ... (ii)
On multiplying eqn. (i)by 3 and subtracting eqn. (ii) from eqn. (i),
3 \(\times\) (x - 5y + 10) - (3x - 10y - 30) = 0
\(\Rightarrow\) - 5y = - 60
\(\therefore\) y = 12
On substituting y = 12 in eqn. (i),
x - 5 x 12 + 10 = 0
\(\Rightarrow\) x - 60 + 10 = 0
\(\Rightarrow\) x = 50
Hence, the distance covered by the train
= 50 \(\times\) 12
= 600 km.

Let number of right answers be x.
Let number of wrong answers be y.
As per question
4x - Y = 70    .....(i)
5x - 2y = 80   .....(ii)
2 \(\times\) eq. (i) - eq. (ii)
8x - 2y = 140
5x - 2y = 80
\(\underline { -\quad +\quad \quad - } \)
3x = 60
\(\Rightarrow\) x = 20
Substituting the value of x in eq (i) to get value of y,
4(20) - Y = 70
\(\Rightarrow\) 80-y = 70
\(\therefore\) Y = 10
Hence total number of questions are = 20 + 10 = 30 
Value: Honesty.

Given, a pair of linear equations:
8x + 5y = 9
\(\Rightarrow\) 8x + 5y - 9 = 0   ... (i)
and 3x + 2y = 4
\(\Rightarrow\) 3x + 2y-4 = 0    ... (ii)
On comparing eqn. (i) and (ii) with ax + by + c = 0,
a₁ = 8, b₁ = 5, c₁ = - 9
and a₂ = 3, b₂ = 2, c₂ = - 4
By cross-multiplication method,
\(\frac { x }{ \left| \begin{matrix} 5 & -9 \\ 2 & -4 \end{matrix} \right| } =\frac { y }{ \left| \begin{matrix} -9 & 8 \\ -4 & 3 \end{matrix} \right| } =\frac { 1 }{ \left| \begin{matrix} 8 & 5 \\ 3 & 2 \end{matrix} \right| } \)
\(\because \frac { x }{ \left| \begin{matrix} { b }_{ 1 } & { c }_{ 1 } \\ { b }_{ 2 } & { c }_{ 2 } \end{matrix} \right| } =\frac { y }{ \left| \begin{matrix} { c }_{ 1 } & { a }_{ 1 } \\ { c }_{ 2 } & { a }_{ 2 } \end{matrix} \right| } =\frac { 1 }{ \left| \begin{matrix} { a }_{ 1 } & { b }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{matrix} \right| } \)
\(\Rightarrow \quad \frac { x }{ \{ (5)(-4)-(2)(-9)\} } =\frac { y }{ \{ (-9)(3)-)-4)(8)\} } =\frac { 1 }{ \{ 8\times 2-\times 5\} } \)
\(\Rightarrow \quad \frac { x }{ -2 } =\frac { y }{ 5 } =\frac { 1 }{ 1 } \)
\(\Rightarrow \quad \frac { x }{ -2 } =\frac { 1 }{ 1 }\) and \(\frac { y }{ 5 } =\frac { 1 }{ 1 } \)
\(\Rightarrow\)  x = -2 and y = 5
By substitution method:
From (ii),  3x = 4 - 2y
\(\Rightarrow \quad x=\frac { 4-2y }{ 3 } \)       ....(iii)
Substitute the value of x in (iii) in (i),
\(\therefore \quad 8\left( \frac { 4-2y }{ 3 } \right) +15y=9\)
\(\Rightarrow\) 32 - 16y + 15y = 27
\(\Rightarrow\)   -y = 27 - 32
\(\Rightarrow\)   y = 5
\(\therefore\) from (iii) \( x=\frac { 4-2(5) }{ 3 } \)
\(=\frac { 4-10 }{ 3 } \)
x = -2
\(\Rightarrow \quad x=-2\) and \( y=5\)

LHS = \((\sin { \theta } +cosec\theta )[{ (\sin { \theta } +\cos { \theta } ) }^{ 2 }-1]\)
\(=(\sin { \theta } +cosec\theta ){ (\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } +2\sin { \theta } .\cos { \theta } }-1)\)
\(=2\sin { \theta } +2\cos { \theta } =2p\) 

\(\cot ^{ 2 }{ \theta } \tan ^{ 2 }{ \theta } =\frac { 1 }{ \tan ^{ 2 }{ \theta } } .\tan ^{ 2 }{ \theta } =1\)

\(\frac { 3(1-\sqrt { 3 } ) }{ 2 } \)

Let the thief was caught in x minutes after committing a theft. 
Since thief runs at a uniform speed of 50 m/minute
\(\therefore\)  Distance covered in x minutes = 50x m
Policeman runs to catch the thief and he runs 60m in first minute and increases his speed by 5 m/min every succeeding minute
\(\therefore\)  60, 65, 70, 75, ....
Which is an A.P., with first term 60 and common difference 5.
Now, according to the statement of the question,
We have
\(={{x-2}\over{2}}[2(60)+(x-3)6]=50x\)
\(\Rightarrow\)  ( x - 2 ) [ 120 + 5x - 5 ] = 100x
\(\Rightarrow\) 120x - 240 + 5x² - 15x - 10x + 30 = 100x
\(\Rightarrow\)    5x² - 5x - 210 = 0
\(\Rightarrow\)     x² - x - 42 = 0
\(\Rightarrow\)   ( x - 7 )( x + 6 ) = 0
\(\Rightarrow\) x = 7 or x = - 6
Rejecting x  = -6, which is not possble
Thus, x = 7
Hence, after 7 minutees, teh policeman will catch the thief.

n = 27
Middle most term = \(\frac { n+1 }{ 2 } =\frac { 28 }{ 2 } =14\)
t₁₃ + t₁₅ = 2 + 2 = 4
t₁₂ + t₁₆ = 4 + 4 = 8
t₁₁ + t₁₇ = 6 + 6 = 12
t₁₀ + t₁₈ = 8 + 8 = 16
t₁ + t₂₇ = 26 + 26 = 52
Hence AP becomes 4,8,12,16,......,52
a = 4, d = 4, a_n = 52, n = 13
\({ S }_{ 13 }=\frac { 13 }{ 2 } \left[ 4+52 \right] =364m\)
and distance covered to collect the books = 364 m
Total distance covered = 364 + 364 = 728 m
Maximum distance she travelled carrying a flag is 26 m

The multiples of 4 between 10 and 250 be 12, 16, 20, ..., 248.
Here, a = 12, d = 16 - 12 = 4 and a_n = 248.
From formula, a_n = a + ( n - 1 )d, we get
12 + ( n - 1 )4 = 248
\(\Rightarrow\)  4 ( n - 1) = 248 - 12   \(\Rightarrow\)  4 ( n - 1 ) = 236
\(\Rightarrow\)  n - 1 = \({236 \over 4}\)  =  59   \(\Rightarrow\)  n = 59 + 1 = 60
\(\because\)         S_n = \({n\over2}\) ( a + l )    \(\Rightarrow\)   S₆₀ = \({60\over2}\) ( 12 + 248 ) = 30 x 260 = 7800