(i) No
(ii) Yes
(iii) No
(vi) No
(v) Yes
(vi) Yes
(vii) Yes
(viii) No
(ix) No
(x) No
(xi) Yes
(xii) Yes

Here, the perimeter of a right triangle= 60 cm
Length of the hypotenuse = 25 cm
Let the base of the right triangle be x cm
\(\therefore \)  Perpendicular of the right triangle = 60 - 25 -x
= (35 - x) cm
 By using Pythagoras Theorem, we have 
x²+(35-x)²= (25)²
\(\Rightarrow x^{ 2 }+(35-x)^{ 2 }=(25)^{ 2 }\)
\(\Rightarrow x^{ 2 }+1225+x^{ 2 }-70x=625\)
\(\Rightarrow 2x^{ 2 }-70x+600=0\)
or x²-35x+300=0
x²-15x-20x+300=0
\(\Rightarrow x(x-15)-20x(x-15)=0\)
\(\Rightarrow (x-15)(x-20)=0\)
\(\Rightarrow x=15\) or x=20
When x = 15, Base = 15 cm, Altitude = 35 - 15 = 20 cm
When x = 20, Base = 20 cm Altitude 15 cm
Now, area of the right triangle 
\(=\frac { 1 }{ 2 } \)x Base x altitude
\(=\frac { 1 }{ 2 } \) x15x20 or \(\frac { 1 }{ 2 } \) x20x15
= 150 cm²

2,5,8,11 or 11,8,5,2

1,6,11,16,.....

Yes

Ist term, a = 5.
Let common difference = d
a_n = a + ( n - 1 ) x d
\(\Rightarrow\)   a₁₀₀ = 5 + ( 100 - 1 )d
\(\Rightarrow\)     -292 = 5 + 99d
\(\Rightarrow\)  d = \(\frac{-297}{99}=-3\)
\(\therefore\)   a₅₀ = 5 + ( 50 - 1 ) x (-3)
= - 142.

Let the sum of n terms be 120.
Given AP is 3, 5, 7, ...
Here   a = 3, d = 2 and Sn = 120.
Using Sn = \(\frac{n}{2}\) { 2a + ( n - 1 )d },
We get    120 = \(\frac{n}{2}\) {6 + ( n - 1 )2}
\(\Rightarrow\)    120 = n ( n + 2 )
\(\Rightarrow\)    120 = n ( n + 2 )
\(\Rightarrow\)   n² + 2n - 120 = 0
\(\Rightarrow\)   ( n + 12 ) ( n - 10 ) = 0
\(\Rightarrow\)  Either n + 12 = 0 or n - 10 = 0
\(\Rightarrow\)       n = -12 (rejected)
\(\therefore\)      n = 10
\(\therefore\) 10 terms must be taken to make sum 120.

Here  a = 3, d = 11 - 3 = 8,
a_n = l = 803
a_n = a + ( n - 1 )d
\(\Rightarrow\) 3 + ( n - 1 )8 = 803
\(\Rightarrow\)   ( n - 1 )8 = 800
\(\Rightarrow\)   n - 1 = 100 \(\Rightarrow\) n = 101
Now   S_n = \(\frac{n}{2}(a+1)\)
\(=\frac{101}{2}(3+803)\)
= 101 x 403 = 40703

a_n = 1 - 4n
\(\Rightarrow\)  a₁ = 1 - 4 x 1 = -3
a₂ = 1 - 4 x 2 = -7
d = a₂ - a₁
= - 7 - ( - 3 ) = - 4
a₂₅ = a + 24d
- 3 = 24 x ( - 4 ) = - 99
Now,    S₂₅ = \(\frac{25}{2}({a}_{1}+{a}_{25})\)
\(=\frac{25}{2}(-3-99)\)
= 25 x ( - 51 ) = - 1275

Here a=, d=5-2=3
n=10
Now, S_n=\({n\over}[2a+(n-1)d]\)
\(\Rightarrow\ \ S_{10}={10\over2}[2\times2+9\times3]\)
=155

Here \(\sqrt { 2x+9 } +x=13\)
\(\Rightarrow \sqrt { 2x+9 } =13-x\)
On squaring both side, we get
\(\left( \sqrt { 2x+9 } \right) ^{ 2 }=(13-x)^{ 2 }\)
\(\Rightarrow 2x+9=169+x^{ 2 }-26x\)
\(\Rightarrow x^{ 2 }-28x+160=0\)
\(\Rightarrow x^{ 2 }-20x-8x+160=0\)
\(\Rightarrow (x-20)(x-8)=0\)
\(\Rightarrow \) x = 20 or x = 8

Natural numbers less than 100 and divisible by 6 are 6, 12, 18,....., 96.
  a = 6,  d = 6,  n = 16, a_n = 96
Sum = \(\frac{16}{2}\)[6 + 96] = 816

t_n  = 7 - 3n
t₁ = 7 - 3 x 1 = 4
t₂ = 7 - 3 x 2 = 1
Common difference = 1 - 4 = -3
 S_n = \(\frac{n}{2}\)[2a + (n - 1)d]
= \(\frac{25}{2}\) [2 x 4+ (25 - 1) x (-3)]
= \(\frac{25}{2}\) [8 - 72]
= - 800

(i) 2x²+  \(5\sqrt { 3\quad } +6\quad =0\)
Here a =2, b  = \(5\sqrt { 3\quad } \) ,c=6
D=b²-4ac
\(\Rightarrow D=(5\sqrt { 3 } )^{ 2 }-4\times 2\times 6\)
=75-48=27 \(\Rightarrow D>0\)
\(\therefore \) Equation has real roots given by
\(x=\frac { -b+\sqrt { D } }{ 2a } ,\frac { -b-\sqrt { D } }{ 2a } \)
\(\Rightarrow x=\frac { -5\sqrt { 3 } +\sqrt { 27 } }{ 2\times 2 } ,\frac { -5\sqrt { 3 } -\sqrt { 27 } }{ 2\times 2 } \)
\(=\frac { -\sqrt { 3 } }{ 2 } ,-2\sqrt { 3 } \)

9x²+8kx+16=0. Here a=9,,b=8k,
C=16
D=b²-4ac
=(8k)²-4X9X16=64k²-576
For equal roots,D=0
\(\Rightarrow \) (8k)²-4X9X16=64k²-576
\(\Rightarrow \) 64k²=576
\(\Rightarrow \) k²= \(\frac { 576 }{ 64 } \)
\(\Rightarrow k^{ 2 }=9\quad \therefore \quad k=\pm 3\)

mx(x-1)+49=0
\(\Rightarrow \) mx²-7mx+49=0
Here a=m,b=-7m,c=49
For equal roots ,D=0
\(\Rightarrow D=b^{ 2 }-4ac\)
\(\Rightarrow 0=(-7m)^{ 2 }-4\times m\times 49\)
\(\Rightarrow 0=49m^{ 2 }-196m\)
\(\Rightarrow 49m^{ 2 }-196m=0\)
\(\Rightarrow 7m(7m-28)=0\)
\(\Rightarrow 7m=0\) or 7m-28 =0
\(\Rightarrow m=0\) or \(m=\frac { 28 }{ 7 } =4\)
but \(m\neq 0\) [ \(\therefore \) In quadratic equation  \(a\neq 0\) ]

x² + x + 1 = 0. Here a = 1, b=1, 
C = 1
D= (1)² - 4 x 1 x 1 = -3 < 0
Equation has no real roots

Let length be x m and breadth be y m
Perimeter =2(l+b) \(\Rightarrow \) 2(x+y)=80
\(\Rightarrow \) x+y=40 \(\Rightarrow \) y=40-x
Area = lxb \(\Rightarrow \) x(40-x) =300
\(\Rightarrow \) 40-x²=300 \(\Rightarrow \) x²-40x+300=0
D=b²-4ac=(-40)²-4x1x300=1600-1200=400 > 0
\(\therefore \) Real solution is possible 
Hence it is possible to design sides of a park.

Given equation is x²-2kx+(7k-12)=0
Here a=1, b=-2k, c=7k-12
D=b²-4ac
D=(-2k)²-4X1X(7k-12)=4k²-28k+48
The equation has equal roots, then D=0
4k²-28k+48=0
k²-7k+12=0
(k-3)(k-4)=0
k=3,4

kx (x-2)+6=0
kx²-2kx+6=0
This is of the form ax²+bx+c=0
where a=k, b=-2k and c=6
Discriminant(D)=b²-4ac=(-2k)²-4 X k X 6=4k²-24k
For equal roots, D=0
4k²-24k=0
k(4k-24)=0
k=0(not possible) or 4k-24=0
4k=24
k=6

Let Ist term of the AP=a and common difference = d
A.T.Q., a₁₉ = 3 x a₆
\(\Rightarrow \) a+18d = 3(a+5d)
\(\Rightarrow a=\frac { 3 }{ 2 } d\) ........ (i)
Also, a₉ = 19 \(\Rightarrow \) a+8d = 19
\(\Rightarrow \) \(\frac { 3 }{ 2 } d\) + 8d = 19 [using eq.(i)]
\(\Rightarrow \) 19d=38 \(\Rightarrow \) d=2
When d=2, equation (i) becomes
\(a=\frac { 3 }{ 2 } \times 2=3\)
AP is 3,5,7,9,.....

Let Is term = a, common difference = d.
a₈ = 0 \(\Rightarrow\) a + 7d = 0
\(\Rightarrow\)   a = -7d
Now,   a₁₈ = a + 17
= -7d + 17d = 10d 
\(\Rightarrow\)   3 X a₁₈ = 30d          ..(i)
Also,  a₃₈ = a + 37d   
= -7d + 37d = 30d      ...(ii)
From (i) and (ii), we get
a₃₈ = 3 x a₁₈ Hence proved.

Let Ist term of AP = a and common difference = d.
\(\therefore\)      t_n = a + ( n- 1 ) d
Now    t₈ = a + 7d
\(\Rightarrow\)    37 = a + 7d           ...(i)
and   t₁₂ = a + 11d
\(\Rightarrow\)   57 = a + 11d      ..(ii)
Solving equations (i) and (ii), we get
d = 5 and a = 2
\(\therefore\)  AP is 2, 7, 12, 17, ...

Let Ist term of the AP = a and common
difference = d
Now  a5 = 1 \(\Rightarrow\) a + 4d = 1
\(\Rightarrow\)   a = 1 - 4d                   ..(i)
and  a31 = -77
\(\Rightarrow\)  a + 30d = -77
\(\Rightarrow\) ( 1 + 4d ) + 30d = -77  [ Using equation (i) ]
\(\Rightarrow\)   26d = -78 \(\Rightarrow\) d = -3
When d = -3, equation (i) becomes,
a = 1 - 4 X ( - 3 ) = 13
Let an = -17
\(\Rightarrow\) a + ( n - 1 )d = -17
\(\Rightarrow\)  13 + ( n - 1 )( - 3 ) = -17
\(\Rightarrow\)  13 - 3n + 3 = -17
\(\Rightarrow\)    -3n = -33 \(\Rightarrow\) n = 11
\(\therefore\)  11th term will be -17.

Here a=5, d=2-5=-3
Let  a_n=-22
\(\Rightarrow\) a+(n-1)d=-22
\(\Rightarrow\) 5+(n-1)(-3)=-22
\(\Rightarrow\) 5-3n+3=-22
\(\Rightarrow\) -3n=-30 \(\Rightarrow\) n=10
\(\therefore\) 10th term of the AP is -22.

Let cost price of table be Rs.x
Profit=x%, SP=x+\(\frac{x}{100}\) \(\times x\)
\(\Rightarrow \quad 96=x+\frac { { x }^{ 2 } }{ 100 } \)
\(\Rightarrow\) x²+100x-9600=0
\(\Rightarrow\) x²+160x-60x-9600=0
\(\Rightarrow\) x(x+160)-60(x+160)=0
\(\Rightarrow\) (x+160)(x-60)=0
\(\Rightarrow\) x=60, -160 [rejected]
\(\therefore\) CP=Rs.60.

Let digit at unit's place be x and digit at ten's place by y
\(\therefore\) Number 10y+x
ATQ
10y+x=7(x+y)
\(\Rightarrow\) 10y+x=7x+7y
\(\Rightarrow\) 6x=3y
\(\Rightarrow\) y=2x    ...(i)
Also 10y+x=3xy-12
\(\Rightarrow\) 10 x 2x + x=3x.2x-12
\(\Rightarrow\) 6x²-21x-12=0
\(\Rightarrow\) 2x²-7x-4=0
\(\Rightarrow\) 2x²-8x+x-4=0
\(\Rightarrow\) 2x(x-4)+1(x-4)=0
\(\Rightarrow\) (x-4)(2x+1)=0
\(\Rightarrow\) x=4 or x=\(-\frac{1}{2}\) (rejecting)
When x=4, y=2  x 4=8
\(\therefore\) Number is 10 x 8+4=84

Let denominator be x then numerator be x-1
\(\therefore\) Fraction=\(​​\frac { x-1 }{ x } \)
ATQ \( \quad \frac { x-1+3 }{ x+3 } =\frac { x-1 }{ x } +\frac { 3 }{ 28 } \)
\(\Rightarrow \quad \frac { x+2 }{ x+3 } -\frac { x-1 }{ x } =\frac { 3 }{ 28 } \)
\(\Rightarrow \quad \frac { x+2 }{ x+3 } -\frac { x-1 }{ x } =\frac { 3 }{ 28 } \)
\(\Rightarrow\) 3 x 28=3(x2+3x)
\(\Rightarrow\) x²+3x-28=0
\(\Rightarrow\) (x+7)(x-4)=0 x=-7 or x=4
Rejecting x=-7 \(\therefore\) x=4
\(\therefore\) Fraction is \(\frac { 4-1 }{ 4 }=\frac { 3 }{ 4 } \)

Let one number be x then other number be x+5
Also \(​​\frac { 1 }{ x } -\frac { 1 }{ x+5 } =\frac { 1 }{ 10 } \)
\(\Rightarrow \frac { x+5-x }{ x(x+5) } =\frac { 1 }{ 10 } \)
\(\Rightarrow\) x(x+5)=50
\(\Rightarrow\) x²+4x-50=0
\(\Rightarrow\) (x+10)(x-3)=0
\(\Rightarrow\) x=-10, x=5
When x=-10, then other number
=-10+5=-5
When x=5, then other number
=5+5=10
Hence, numbers are -10, -5 or 5, 10.

Let digit at unit's place be x and digit at ten's place by y
Number=10y+x    ....(i)
According to given condition,
10y+x=6(x+y)
\(\Rightarrow\) 10y+x=6x+6y
\(\Rightarrow\)   4y=5x y=\(\frac {3}{4}\)x   .....(ii)
Also 10y+x=3xy-6
\(\Rightarrow \quad 10\times \frac { 5 }{ 4 } x+x=3x\times \frac { 5 }{ 4 } x-6\quad (using\quad (ii))\)
\(\Rightarrow \quad \frac { 27 }{ 2 } =\frac { 15{ x }^{ 2 }-24 }{ 4 } \)
\(\Rightarrow\) 54x=15x2-24
\(\Rightarrow\) 15x²-54x-24=0
\(\Rightarrow\) 5x²-18x-8=0
\(\Rightarrow\) 5x²-20x+2x-8=0
\(\Rightarrow\) 5x(x-1)+2(x-4)=0
\(\Rightarrow\) (x-4)(5x+2)=0
\(\Rightarrow\) x=4 or x=\(-\frac{2}{5}\)
Rejecting x=\(-\frac{2}{5}\), we have x=4 when
x=4, y=\(-\frac{2}{5}\) x r=4  [Using (i)]
\(\therefore\)  Number=10 x 5+4=54

9x²-17x+8=0
9x²-9x-8x+8=0
(9x-8)(x-1)=0
x=\(8\over 9\) or x=1

5x²+11x+6=0
5x²+5x+6x+6=0
(5x+6)(x+1)=0
x=-1 or x=\(-{6\over 5}\)

x²+4x-12=0
x²+6x-2x-12=0
x(x+6)-2(x+6)=0
(x+6)(x-2)=0
x=-6 or x=2

Given equation is  4x²-2x²+3=0
-2x²+4x+3=0
Here a=-2, b=4, c=3
D=b²-4ac
=(4)²-4x(-2)x3=16+24=40

Given equation is x²+x+1=0
a=1, b=1, c=1
D=b²-4ac
D=(1)²-4x1x1
D=-3<0
given quadratic equation has no solution.