a = assumed mean = 150
\(\overset { - }{ =a+ } \quad \frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times h\)
\(=150+\frac { -12 }{ 50 } \times 20\)
= 150 - 4.8 = 145.2
\(\frac { N }{ 2 } =\frac { 50 }{ 2 } =25\)
Median class = 120 -140
I = 120,/= 14, cf. = 12
Median = l + \(\left( \frac { N }{ 2 } -c.f \right) \times h\)
\(=120+\frac { 25-12 }{ 14 } \times 20\)
= 120 + 18.57 138.57
Mode = 3 Medain - 2 Mean
= 3 x 138.57 - 2 x 145.2
= 415.71 - 290.4 = 125.31
Hence, mean = 145.2, median = 138.57, mode = 125.31

In \(\Delta AOP\), tan 30°=\(\frac{H-120}{OP}\)
\(\frac { 1 }{ \sqrt { 3 } } =\)\(\frac{H-120}{OP}\) 

OP = (H -120)\(\sqrt{3}\) ...(i)
In \(\Delta OPA'\), tan 60°=\(\frac{H+120}{OP}\)
OP =\(\frac { H+120 }{ \sqrt { 3 } } \)...(ii)
From (i) and (ii), we get
\(\frac { H+120 }{ \sqrt { 3 } } \)=\(\sqrt{3}\)(H-120)
So height of cloud H = 240 m.

Given
AB + AD = BC + CD
AD = BC + CD-AB
AD= h + d - x 
In rt \(\triangle\)ACD,
AD² = AC² + DC²
\(\Rightarrow\) (h + d - x)² = (x + h)²+ d²
\(\Rightarrow\) (h + d - x)² - (x + h)² = d²
\(\Rightarrow\) (h + d - x - x - h) (h + d - x + x + h) = d²   [\(\therefore\) (a² - b²) = (a - b)(a + b)]
\(\Rightarrow\) (d - 2x)(2h + d) = d²
\(\Rightarrow\)2hd + d²-4hx-2xd = d²
\(\Rightarrow\) 2hd = 4hx + 2xd
=2 (2h+d)x
\(\Rightarrow x=\frac { hd }{ 2h+d } \)

2x + 3y = 12
\(\Rightarrow \quad y=\frac { 12-2x }{ 3 } \)

x - y = 1
\(\Rightarrow\)  y = x -1

Plotting the above points and drawing a line joining them, we get the graph of the equations 2x + 3y = 12 and x - y = 1. Clearly, the two lines intersect at point P(3, 2).
Hence, x = 3 and y = 2 is the solution of the system.
Area of shaded region = Area of \(\Delta\)PAB
= \(\frac { 1 }{ 2 } \) \(\times\) base \(\times\) height
=\(\frac { 1 }{ 2 } \) \(\times\) AB \(\times\) PM
=\(\frac { 1 }{ 2 } \)\(\times\) 5 \(\times\) 3
= 7.5 square unit.

LHS = \(2(\sin ^{ 6 }{ \theta } +\cos ^{ 6 }{ \theta } )-3(\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } )+1\)
\(=2[{ (\sin ^{ 2 }{ \theta } ) }^{ 3 }+{ (\cos ^{ 2 }{ \theta } ) }^{ 3 }]-3(\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } )+1\)
\(=[(\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } )(\sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } -\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } )]-3\sin ^{ 4 }{ \theta } -3\cos ^{ 4 }{ \theta } +1\)\(=-\sin ^{ 4 }{ \theta } -\cos ^{ 4 }{ \theta } -2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } +1\)
\(=1-(\sin ^{ 4 }{ \theta } -\cos ^{ 4 }{ \theta } -2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } )\)
\(=1-[{ (\sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } ) }^{ 2 }]=1-1=0=RHS\)
Hence proved. 

Use Pythagoras theorem, to find the distance of the foot of the ladder from the building.

= 8 m

The graph of given linear equations intersect each other at origin. So, the given pair of linear equations is consistent. x=0, y=0

T_p = a + ( p - 1 )d
\(\Rightarrow\)     \({1\over q}=a+(p-1)d\)                    ...(i)
T_q = a + ( p - 1 )d
\(\Rightarrow\)     \({1\over q}a+(q-1)d\)                         ...(ii)
Subtracting (ii) from (i), we get
\({p-q\over pq}=(p-q)d\Rightarrow d={1\over pq}\)
Putting d = \({1 \over pq}\) in (i), we have
\({1\over q}=a+{(p-q)\over pq}\Rightarrow a={1\over pq}\)
\(\therefore\)     S_pq = \({pq\over2}[2a+(pq-1)d]\)
or,      S_pq = \({pq\over2}\left[ {{2\over pq}+{(pq-1)\over pq}} \right]\)
\(={1\over 2}(pq+1)\)

Speed of motor boat in still water is 24 km/h
 Let speed of the stream be x km/h
Speed of the boat in downstream direction 
=(24+x)km/h
Distance covered in downstream direction
= 32 km
Time taken in downstream direction 
\(=\frac { 32 }{ 24+x } \) hours
Speed of the boat in upstream direction (24 +x) km/h
Distance covered in upstream direction  = 32 km 
ATQ \(\frac { 32 }{ 24-x } -\frac { 32 }{ 24+x } =1\)
\(\Rightarrow \frac { 32(24+x)-32(24-x) }{ (24-x)(24+x) } =1\)
\(\Rightarrow 64x=576-x^{ 2 }\)
\(\Rightarrow x^{ 2 }+64x-576=0\)
\(\Rightarrow x^{ 2 }+72x-8x-576=0\)
\(\Rightarrow (x+72)(x-8)=0\)
x=-72 (rejected) or x=8
\(\therefore \) x=8 speed of stream = 8 km/hr