Let the original number of persons be x
Then, share of each person \(=Rs \frac{6500}{x}\)
If the number of persons is increased by 15.
The, new share of each person \(=Rs \frac{6500}{x+15}\)
\(\therefore \quad \frac{6500}{x+15}=\frac{6500}{x}-30\)
50.

Let length of the shortest side = x m.
Then, hypotenuse = (2x + 6)m and third side
= (2x + 6 - 2)m = (2x + 4)m
By Pythagoras theorem,
(2x + 6)² = x² + (2x + 4)² [∵ H² = p2 + B²]
⇒ 4x² + 24x + 36 = x² + 4x² + 16x + 16 [∵ (a + b)² = a² + 2ab + b²)
⇒ x² + 4x² + 16x + 16 - 4x² - 24x - 36 = 0
⇒ x² - 8x - 20 = 0
By quadratic formula,
\(x=\frac{-(-8) \pm \sqrt{(-8)^{2}-4 \times 1 \times(-20)}}{2 \times 1}\)
\(\left[\because x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} ; \text { here } a=1, b=-8 \text { and } c=-20\right]\)
\(\Rightarrow \quad x=\frac{8 \pm \sqrt{64+80}}{2}\)
\(\Rightarrow \quad x=\frac{8 \pm \sqrt{144}}{2} \)
\(\Rightarrow \quad x=\frac{8 \pm 12}{2}\)
\(\Rightarrow \quad x=\frac{8+12}{2} \text { or } x=\frac{8-12}{2}\)
\(\Rightarrow \quad x=\frac{20}{2} \text { or } x=\frac{-4}{2}\)
⇒ x = 10 or x = - 2
But length of side cannot be negative.
∴ x = 10
Hence, shortest side is 10 m, hypotenuse is 2 x 10 + 6 = 26 m
and third side = 2 x 10 + 4 = 24 m.

\(\therefore \) x = -2 is a root of 3x² +7x+p=0
\(\Rightarrow 3(-2)^{ 2 }+7\times (-2)+p=0\)
\(\Rightarrow p=2\)
\(\therefore \quad x^{ 2 }+k(4x+k-1)+p=0\) becomes
\(x^{ 2 }+4kx+k^{ 2 }-k+2=0\)
D = (4k)²-4X 1(k²-k+2)
=16k²-4k²+4k-8
= 12k²+4k-8
\(\therefore \) Roots are equal
\(\therefore \) 12k²+4k-8=0
\(\Rightarrow 3k^{ 2 }+k-2=0\)
\(\Rightarrow 3k^{ 2 }+3k-2k-2=0\)
\(\Rightarrow 3k(k+1)-2(k+1)=0\)
\((k+1)(3k-2)=0\)
\(k=-1,k=\frac { 2 }{ 3 } \)

Let x be length of piece and y be rate per metre
\(\Rightarrow\) xy=200   ...(i)
In new condition, the length of piece be (x+5) m and the price per metre be Rs.(y-2)
\(\Rightarrow\) (x+5)(y-2)=200   ...(ii)
\(\Rightarrow\) xy-2x+5y-10=200
\(\Rightarrow\) 2x-5y+10=0   ...(iii)
[Putting xy=200 in (ii)]
Now putting ya in (iii), we get
\(\Rightarrow\) 2x-5+10=0
\(\Rightarrow\) 2x²-1000+10x=0
\(\Rightarrow\) x²+5x-500=0
\(\Rightarrow\) x²+25x-20x-500=0
\(\Rightarrow\) x(x+25)-2x(x+25)=0
\(\Rightarrow\)  (x-20)(x+25)=0
\(\therefore\) x=20
or x=-25 (rejecting) [Length is not negative]
\(\therefore\) When x=20, y=\(\frac{200}{20}\)=10
\(\therefore\)  Length of the piece=20m and price or the piece per metre=Rs.10

I.et the distance covered bv peacock (AD) be x m l
Distance covered by snake (DC) be x m 
In right angled \(\triangle ABD\) AD² =AB²+BD²
\(\Rightarrow \) x²=(9)²+(27-x)²
\(\Rightarrow \) x²=81+729+x²-54x
\(\Rightarrow 810=54x\Rightarrow x=\frac { 810 }{ 54 } \)
x=15 and 27-x = 27-15=12
The snake is caught at 12m from the hole.