1² + 2² +...+ 19² = \(\frac { 19\times \left( 19+1 \right) \left( 2\times 19+1 \right) }{ 6 } =\frac { 19\times 20\times 39 }{ 6 } =2470\)

Let r₁, r₂ be the radii of two spheres
\(\text { Given } \frac{r_{1}}{r_{2}}=\frac{4}{7} \Rightarrow \mathrm{r}_{1}=\frac{4 r_{2}}{7}\)
Ratio of the volumes = \(\frac{V_{1}}{V_{2}}=\frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}}\)
\(=\frac{\left(\frac{4 r_{2}}{7}\right)^{3}}{r_{2}^{3}}=\frac{4^{3}}{7^{3}}\)
Ratio of volumes \(V_{1}: V_{2}=64: 343=\frac{64}{343}\)

Let the radius of the sphere be 'r' cm
Surface area = \(4 \pi r^{2}\)
when radius is increased by 25% ,then new diameter = r + 25% + r
\(=r+\frac{25 r}{100}=\frac{5 r}{4}\)
Surface area of new sphere
\(=4 \pi\left(\frac{5 r}{4}\right)^{2} \)
\(=4 \pi\left(\frac{25 r^{2}}{16}\right) \)
\(=\frac{25 \pi r^{2}}{4} \)
Increase in surface area = \(\frac{25 \pi r^{2}}{4}-4 \pi r^{2}\)
\(=\frac{25 \pi r^{2}-16 \pi r^{2}}{4} \)
\(=\frac{9 \pi r^{2}}{4} \)
Percentage increase in surface area
\(=\frac{9 \pi r^{2} / 4}{4 \pi r^{2}} \times 100 \% \)
\(=\frac{900}{16} \%=56.25 \% \)

Required Ratio = C.S.A. of the sphere: C.S.A. of the cylinder : C.S.A. of the cone
\(4\pi { r }^{ 2 }:2\pi rh:\pi rl,\ (l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { 2r }^{ 2 } } =\sqrt { 2r } units)\)
\(=4:2:\sqrt { 2 } =2\sqrt { 2 } :\sqrt { 2 } :1\).

Given that, C.S.A. of the cylinder = 88 sq. cm
2\(\pi\)rh = 88
\(2\times \frac { 22 }{ 7 } \times 14=88\) (given h = 14cm)
2r = \(\frac { 88\times 7 }{ 22\times 14 } =2\)
Therefore, diameter = 2 cm

Given \(a_{n}=\frac{5 n}{n+2}\)
To find a₆, put n = 6
\(\therefore a_{6}=\frac{5 \times 6}{6+2}=\frac{30}{8}=\frac{15}{4}\)
To find a₁₃, put n = 13
\(a_{13}=\frac{5 \times 13}{13+2}=\frac{65}{15}=\frac{13}{3} \)
\(\therefore a_{6}=\frac{15}{4} \text { and } a_{13}=\frac{13}{3} \)

All the integers 'a' must be either even or odd.
If it is even then a = 2q.
If it is odd then a = 2q + 1
Case 1:
lf a = 2q
a² = (2q)²
a² = 4q²,  remainder 0 when divided by 4.
Case 2:
If a = 2q+ 1
a² = (2q + 2)²
= 4q² + 4q + 1
= 4q ( q + 1) + 1
a² = 4m + 1 Where m = q (q + 1) is an integer
It is of the form bq + 1 where 1 is the remainder when divided by 4.
The square of any integer leaves the remainder either 0 or 1 when divided by 4.

3 + 33 + 333 + ... to n terms.
Let S_n = 3 + 33 + 333 + ...upto n terms
= 3( 1 + 11 + 111 + ... to n terms)
= \(\frac { 3 }{ 9 } \) (9 + 99 + 999 + ... to n terms)
(multiply and divide by 9)
= \(\frac { 1 }{ 3 } \)3[(10 - 1) + (100 - 1) + (1000 - 1) + to n terms]
= \(\frac { 1 }{ 3 } \) [(10 - 1) + (10² - 1) + (10³ - 1) +... + n upto n terms]
= \(\frac { 1 }{ 3 } \) {[10 + 10² + 10³ +...upto n terms] - n}
10 + 10² +... is a G.P. with a = 10, r = 10.
\(\therefore S_{n} =\frac{a\left(r^{n}-1\right)}{r-1} \)
\(S_{n} =\frac{1}{3}\left\{\left[\frac{10\left(10^{n}-1\right)}{10-1}\right]-n\right\} \)
\(=\frac{1}{3}\left[\frac{10\left(10^{n}-1\right)}{9}-n\right] \)
\(=\frac{10}{27}\left(10^{n}-1\right)-\frac{n}{3}\)
3 + 33 + 333 + ... to n terms \(=\frac{10}{27}\left(10^{n}-1\right)-\frac{n}{3}\)

Let the radius of hemispherical bowl = r
Volume of hemispherical bowl
\(=\frac{2}{3} \pi r^{3} \text { cu. units }\)
Let the height of cylindrical vessel = h
Given \(r=h+h \frac{50}{100} \Rightarrow \mathrm{r}=\mathrm{h}\left(1+\frac{50}{100}\right) \)
\(\mathrm{h}= \frac{2}{3} r \)
Now, Volume of cylindrical vessel
\(=\pi r^{2}\left(\frac{2 r}{3}\right)=\frac{2}{3} \pi r^{3}\)
Hence, Volume of juice in the cylindrical vessel
\(=\frac{\frac{2}{3} \pi r^{3}}{\frac{2}{3} \pi r^{3}} \times 100 \%=100 \%\)

Diameter of Glass = 20 cm
radius = 10 cm
water upto height = 9 cm
radius of cylindrical metal = 5 cm
height of cylindrical metal = 4 cm
Volume of water displaced = Volume of cylindrical metal
\(\pi r_{1}^{2} h_{1}=\pi r_{2}^{2} h_{2} \)
\((10)^{2} h_{1}=(5)^{2}(4) \)
\(h_{1}=\frac{100}{100}=1 \mathrm{~cm} \)
Hence, the increase in water level is 1 cm.

Let Senthil’s house number be x.
It is given that 1 + 2 + 3 +...  + (x -1) = (x + 1) + (x + 2) + .... + 49
1 + 2 + 3 + ... + (x - 1) = [1 + 2 + 3...+49] - [1 + 2 + 3+ ...+ x] 
\(\frac { x-1 }{ 2 } \left[ 1+\left( x-1 \right) \right] =\frac { 49 }{ 2 } \left[ 1+49 \right] -\frac { x }{ 2 } \left[ 1+x \right] \)
\(\frac { x\left( x-1 \right) }{ 2 } =\frac { 49\times 50 }{ 2 } -\frac { x\left( x+1 \right) }{ 2 } \)
x² - x = 2450 - x2 - x \(\Rightarrow\) 2 x 2 = 2450
x² = 1225 gives x = 35
Therefore, Senthil’s house number is 35.

The natural numbers between 300 and 600 which are divisible by 7 are 301, 308, 315, …, 595.
The sum of all natural numbers between 300 and 600 is 301 + 308 + 315 +...+ 595
The terms of the above series are in A.P.
First term a = 301; common difference d = 7; Last term l = 595.
\(n=\left( \frac { l-a }{ d } \right) +1=\left( \frac { 595-301 }{ 7 } \right) +1=43\)
Since, \({ S }_{ n }=\frac { n }{ 2 } \left[ a+l \right] \), we have \({ s }_{43 }=\frac { 43 }{ 2 } \left[ 301+595 \right] \) = 19264 

Right triangle PQR, right angled at Q and
PR = 20 cm, QR = 16 cm
PQ² = PR² - QR²
= (20)² - (16)²
= 400 - 256 = 144
PQ = 12 cm

When right triangle PQR, rotates about QR, a right circular cone is formed with PQ = 12 cm as base radius and PR = 20 cm as
slant height.
C.S.A of the Cone = \(\pi r l\) sq. units
\(=\frac{22}{7} \times 12 \times 20=754.29 \mathrm{~cm}^{2}\)
When right triangle PQR, rotates about PQR, a right circular cone is formed with
QR = 16 cm as base radius and PR = 20 cm as slant height

C.S.A of the Cone \(=\pi r l \text { sq.units } \)
\(=\frac{22}{7} \times 16 \times 20 \)
= 1005.71 cm²
Hence, C.S.A of the cone when rotates about PQ is larger.

Let x be the time taken in minutes and y be the distance travelled in km.
\(\begin{array}{|l|l|l|l|l|} \hline \text { Time taken } \mathbf{x} \text { (in minutes) } & 60 & 120 & 180 & 240 \\ \hline \text { Distance } \mathbf{y} \text { (in km) } & 50 & 100 & 150 & 200 \\ \hline \end{array}\)
(i) Observe that as time increases, the distance travelled also increases. Therefore, the variation is a direct variation. It is of the form y = kx .
Constant of variation
\(k=\frac{y}{x}=\frac{50}{60}=\frac{100}{120}=\frac{150}{180}=\frac{200}{240}=\frac{5}{6}\)
Hence, the relation may be given as
\(y=k x \ \Rightarrow \ y=\frac{5}{6} x\)
(ii) From the graph, \(y=\frac{5 x}{6},\) if x = 90 , then \(y=\frac{5}{6} \times 90=75 \mathrm{~km}\)
The distance travelled for \(1 \frac{1}{2}\)hours (i.e.,) 90 minutes is 75 km.
(iii) From the graph, \(y=\frac{5 x}{6}, \text { if } y=300\) then \(x=\frac{6 y}{5}=\frac{6}{5} \times 300=360\) minutes (or) 6 hours.
The time taken to cover 300 km is 360 minutes, that is 6 hours.

1. Table:

No. of participants ( x )	2	4	6	8	10
Amount for each participants in Rs. ( y)	180	90	60	45	36

2. Variation
Indirect Variation
3. Equation
xy = k 
xy = 2 x 180 = 4 x 90 = 6 x 60 =.... = 360
xy = 360
4.Points
(2,180), (4,190),(6,60 ) (8,45 ) ,(10,36 )
5. Solution
(i) Constant of Variation
(ii) Cash Price each participant will get if 12 participants participate = Rs. 30 /-

The distance between the point from the centre is 10 cm.

Length of the tangents PA - PB = 8.7 cm
Construction:
Steps:
(1) With O as centre, draw a circle of radius 5cm.
(2) Draw a line OP = 10 cm.
(3) Draw a perpendicular bisector of OP which cuts OP at M.
(4) With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
(5) Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA and PB = 8.7 cm

Given, diameter (d) = 6 cm, we find radius \((r)=\cfrac { 6 }{ 2 } =3cm\)

Construction
Step 1: With centre at O, draw a circle of radius 3 cm.
Step 2: Draw a line OP of length 8 cm.
Step 3: Draw a perpendicular bisector of OP, which cuts OP at M.
Step 4: With M as centre and MO as radius, draw a circle which cuts previous circle at A and B.
Step5: Join AP and BP. AP and BP are the required tangents. Thus length of the tangents are PA = PB = 7.4 cm.
Verification : In the right angle triangle OAP,PA² = OP² - OA² = 64 -9 = 55
\(PA=\sqrt { 55= } 7.4\ cm\) (approximately) .

(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47 b₁ + 54.10 = 2.47 b₂ + 54.10
2.47b₁ = 2.47b₂ ⇒ b₁ = b₂
Thus, h(b₁) = h(b₂) ⇒ b₁ = b₂, So, the function h is one – one.
(ii) If the length of the thigh bone b = 50, then the height is
h(50) = (2.47 x 50) + 54.10 = 177.6 cms
(iii) If the height of a person is 147.96 cms, then h(b) = 147.96 and so the length of the thigh bone is given by 2.47b + 54.10 = 147.96.
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cms.