Here, Number=12ⁿ where n stand for any natural number .
Now 12ⁿ= (2² x3)ⁿ
Now , For 12ⁿ to end with 0, it should have 2 as well as 5 in its Prime factors to end with 0, Also to end with 5 , it requires at least a single multiple of 5 in its Prime Factors, So 12ⁿ cannot end with the digit 0 or 5.

63

1

It is a quadratic equation.

The total fare (in Rs.) after each kilometre is 15, 23, 31, 39,.... 
No

20

x³ + 6x² + 3x + 10 thus we know that,
a + a + b + a + 2b = 6
\(\Rightarrow\) 3(a + b) = 6 \(\Rightarrow\) a + b = 2
a(a + b) + (a + b)(a + 2b) + a(a + 2b) = 3
\(\Rightarrow\) (a + b)(2a + 2b) + a(a + 2b) = 3
\(\Rightarrow\) 2 x 2 x 2 + a² + 2ba = 3
 \(\Rightarrow\)  a² +2ba=-5 \(\Rightarrow\) ab= \(\frac{-5-a^{2}}{2}\)
a(a+b)(a+2b) = -1
a(2)(a+b) = -1
\( a^{2}+a b =-\frac{1}{2} \) 
\(\Rightarrow \quad a^{2}+\left(\frac{-5-a^{2}}{2}\right) =-\frac{1}{2}\)
\(\Rightarrow\) 2a² - 5 - a² = - 1 ~ a2 = 4
 \(\Rightarrow\) a= ± 2
 a+b=2
\(\Rightarrow\)  b=0 or 4

123.4 kg

Given \(\triangle NSQ\cong \triangle MTR\) and \(\angle1=\angle2\)
To prove \(\triangle PTS\sim \triangle PRQ\)
Proof Since, \(\triangle NSQ\cong \triangle MTR\)
\(\therefore\) SQ = TR  ... (i)
Also, \(\angle1=\angle2\)
\(\Rightarrow\) PT = PS .... (ii)
[since, sides opposite to equal angles are also equal]
From Eqs.(i) and (ii), \(\frac{PS}{SQ}=\frac{PT}{TR}\)
\(\Rightarrow ST\parallel QR\)
[by converse of basic proportionality theorem]
\(\therefore \angle 1=\angle PQR\)   and \( \angle 2=\angle PRQ\)
[\(\therefore\)  atternate exterior angle]
In \(\triangle PTS\) and  \(\triangle PRQ\),
\(\angle P=\angle P\)    [common angle]
\(\angle 1=\angle PQR\)  [proved above]
and \(\angle 2=\angle PRQ\)
\(\therefore \triangle PTS\sim \triangle PRQ\)
[by AAA similarity criterion]

A trapezium ABCD in which \(PQ\parallel AB\), draw a line AC, which intersects PQ at O.
Join AC, which intersects PQ at O.
Given, \(AB\parallel DC\) and \(PQ\parallel AB\)
Then, \(AB\parallel PQ\parallel DC\)

In \(\triangle ADC\), \(PO\parallel DC\)
By basic proportionality theorem,
\(\frac{AP}{PD}=\frac{AO}{OC}\)  ... (i)
In \(\triangle CAB\), \(OQ\parallel AB\)
By basic proportionality theorem,
\(\frac{AO}{OC}=\frac{BQ}{QC}\) ... (ii)
On comparing Eqs. (i) and (ii), we get
\(\frac{AP}{PD}=\frac{BQ}{QC} \Rightarrow \frac{AP}{18}=\frac{35}{15}\)
[\(\because\)  PD = 18 cm, BQ = 35 cm and QC = 15 cm]
\(\Rightarrow AP=\frac { 18\times 35 }{ 15 } \) = 42 cm
\(\therefore\)  AD = AP + PO
= 42 + 18 = 60 cm

By property of rectangle, we know that its opposite sides are of equal lengths.
i.e. DC=AB \(\Rightarrow\) x+3y=13   ...(i)
and AD=BC \(\Rightarrow\) 3x+y=7   .....(ii)
On multiplying Eq. (ii) by 3 and then subtracting Eq. (i), we get

\(\Rightarrow\) x=1
On putting x=1 in Eq. (i), we get
3y=12 \(\Rightarrow\) y=4
Hence, x=1 and y=4.

Given that, x, y ane 40⁰ are the angles of a triangle.
\(\therefore\) x+y+40⁰=180⁰  [\(\because\) sum of all the angles of a triangle is 180⁰ ]
\(\Rightarrow\)x+y=140⁰     ...(i)
Also, x-y=30⁰    ...(ii)
On adding Eqs. (i) and (ii), we get
2x=170⁰ \(\Rightarrow\)  x=85⁰
On putting  x=85⁰ in Eqs. (i), we get
85⁰+y=140⁰  \(\Rightarrow\)   y=55⁰
Hence, the required values of x and y are 85⁰ and 55⁰ respectively.

Here, 327.7081 is terminating. So, it represents a rational number.
Thus, \(327.7081=\frac { 3277081 }{ 10000 } =\frac { p }{ q } \)
Here, q = 10⁴ = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
= 2⁴ x 5⁴ = (2 x 5)⁴
So, the prime factors of q are 2 and 5.

Given AB and CD are two common tangents of circles C₁, and C₂ (say).
To prove AB = CD 
Proof Let F and G be the centres of the circles C, and C2'
respectively.
Join AF, FC, FB and FD.

Now, tangents FB and FD are drawn from an external point F to the circle C₂.
FD = FB ... (i)
Since, radius is perpendicular to the tangent at the point of contact.
AF⊥AD and FC丄BC
⇒ ㄥFAD=ㄥFCB=90⁰
In right angled ΔAFD and ΔCFB,
AF=FC
ㄥA=ㄥC=90⁰
and FD=FB
∴   ΔAFD=ΔCFB
Then, AD=BC

Since, the total number of terms (n) = 11 odd
\(\because\)     Middle most term \(= \frac {\left(n+1\right)}{2}\) th term
\(\Rightarrow\) \(30 = \left (\frac {11+1}{2}\right)\) th term
\(\Rightarrow\)  6th term = 30
and \({a}_{6}=30\)  
\(\Rightarrow\)  a + (6-1)d = 30       
\(\Rightarrow\) a + 5d = 30
\(\because\)Sum of n terms of an AP,  ...(i)
\({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)  
\(\therefore\)    \({ S }_{ 11 }=\frac { n }{ 2 } \left[ 2a+\left( 11-1 \right) d \right] \) 
\(\therefore\)   \({ S }_{ 11 }=\frac { n }{ 2 } \left( 2a+10d \right) \)   
= 11(a+5d)
\(\Rightarrow\) \({ S }_{ 11 }=11\times 30\)   
= 330 [from Eq.(i)]

Let the three parts of 207 be a -d, and a  + d
\(\therefore\)     a - d + a + a + d = 207
3a  = 207
a = \(\frac{207}{3}\)
a = 69
Also,   ( a - d ) a = 4623
\(\Rightarrow\)    ( 69 - d ) 69 = 4623
\(\Rightarrow\)      69 - d = 67
\(\Rightarrow\)  d = 69 - 67 = 2
Hence, the three parts of 207are 69 - 2, 69 + 2
i.e., 67, 69 and 71.

Water displaced by a person = 0.04 m³ 
Water displaced by 500 persons = 500 x 0.04 = 20 m³
Volume of cuboid = Volume of water displaced
Ibh=20
80 x 50 x h=20 
h=\(\frac{20}{80\times{50}}\)=\(\frac{1}{200}\)m
⇒ h=0.5 cm

scalene triangle

\(100 (\sqrt {3} + 1) \ m\)

n=5, 11 here both answers are correct be cause sum of 6th term to 11trm is zero

(ii) The fee charged every month by a school from classes I to XII, when the monthly fee for class I is Rs.250, and it increased by Rs.50 for the next higher class

Let,diameter of inner most circle = x
Diameter of middle circle = 2x
Diameter of outer most circle = 3x
∴ Area of inner most circle = π(x)²
Area of middle circle =π(4x² - x²)
=π(3x²)
Area of outer most circle =π(3x)²-π(2x)²
=π(9x² - 4x²)
=π X 5x²
Ratio of the areas of three regions.
πx²: π(3x²):π(5x²)=1:3:5

Yes

Here,   t₈ = \(\frac{{t}_{2}}{2}\Rightarrow\) a + 7d = \(\frac{a+d}{2}\)
\(\Rightarrow\)  2a + 14d = a + d \(\Rightarrow\) a = - 13d (i)
and  t₁₁-\(\frac{{t}_{4}}{3}=1\)
\(\Rightarrow\)  a + 10d - \(\frac{a+3d}{3}=1\)
\(\Rightarrow\frac{3a+30d-a-3d}{3}=1\)
\(\Rightarrow\)  2a + 27d = 3
\(\Rightarrow\)   2 ( - 13d ) + 27d = 3  [ using (i) ]
\(\Rightarrow\)     -26d + 27d = 3
\(\Rightarrow\)   d = 3,
\(\therefore\)  from (i)  a = -13 x 3 = - 39
Therefore,,  t₁₅ = a + 14d
= - 39 + 14 x 3
= - 39 + 42 = 3