{tex}12576 = 4052 \times 3 + 420{/tex}
{tex}4052 = 420 \times 9 + 272{/tex}
{tex}420 = 272 \times 1 + 148{/tex}
{tex}272 = 148 \times 1 + 124{/tex}
{tex}148 = 124 \times 1 + 24{/tex}
{tex}124 = 24 \times 5 + 4{/tex}
{tex}24 = 4 \times 6 + 0{/tex}
HCF of 12576 and 4052 is '4'.

You can check the marks distribution here : <a href="https://mycbseguide.com/cbse-syllabus.html">https://mycbseguide.com/cbse-syllabus.html</a>

Given,
{tex}\frac { 1 } { ( a + b + c ) } = \frac { 1 } { a } + \frac { 1 } { b } + \frac { 1 } { c }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { ( a + b + c ) } - \frac { 1 } { c } = \frac { 1 } { a } + \frac { 1 } { b } \Rightarrow \frac { c - ( a + b + c ) } { c ( a + b + c ) } = \frac { b + a } { a b }{/tex}
{tex}\Rightarrow \quad \frac { - ( a + b ) } { c ( a + b + c ) } = \frac { ( a + b ) } { a b }{/tex}

On dividing both sides by (a+b)
{tex}\Rightarrow \quad \frac { - 1 } { c ( a + b + c ) } = \frac { 1 } { a b }{/tex}

Now cross multiply
{tex}\Rightarrow{/tex} c(a + b + c) = -ab 
{tex}\Rightarrow{/tex} c² + ac + bc + ab = 0

{tex}\Rightarrow{/tex} c(c +a) + b(c +a) = 0
{tex}\Rightarrow{/tex} (c + a) (c + b) = 0

{tex}\Rightarrow{/tex} c + a = 0 or c + b = 0
{tex}\Rightarrow{/tex} c = -a or c = -b.
Therefore, -a and -b are the roots of the  equation.

HCF of 234 and 111
234 = 2 x 3 x 3 x 13
111 = 3 x 37
Therefore, HCF = 3
So we can write it
3 = 111 - (12 x 9)
3 = 111 - {(234 – 111 x 2) x 9}[where 12 = 234 - 111 x 2]
3 = 111 - {234 x 9 – 222 x 2 x 9}
3 = 111 - (234 x 9) + (111 x 18)
3 = 111 + (111 x 18) - (234 x 9)
3 = 111[1 + 18] – 234 x 9
3 = 111 x 19 – 234 x 9
3 = 234 x (-9) + 222 x 19

Hence, HCF of 234 and 111 in the form of
234x + 111y is 234 x (-9) + 111 x 19

Given, 

ax + by = a - b   .......1  

bx - ay = a + b   .......2

Multiply by a in equation 1 and b in equation 2, we get

a² x + aby = a² - ab   .......3  

b² x - aby = ab + b²   .......4

Add equation 3 and 4, we get

      a² x + aby + b² x - aby = a² - ab + ab + b²

=> a² x + b² x = a² + b² 

=> x(a² + b² ) = a² + b²

=> x = (a² + b² )/(a² + b² ) 

=> x = 1

Put value of x in equation 1, we get

=> a + by = a - b

=> by = a - b - a

=> by = -b

=> y = -b/b

=> y = -1

So, x = 1, y = -1

We have,
tan 5° tan 25° tan 30° tan 65° tan 85°
= (tan 5° tan 85°) (tan 25° tan 65°) tan 30°  {tex}\left[ \begin{array} { c } { \because \tan 85 ^ { \circ } = \tan \left( 90 ^ { \circ } - 5 ^ { \circ } \right) = \cot 5 ^ { \circ } } \\ { \tan 65 ^ { \circ } = \tan \left( 90 ^ { \circ } - 25 ^ { \circ } \right) = \cot 25 ^ { \circ } } \end{array} \right] {/tex}
= (tan 5° cot 5°) (tan 25° cot 25°) tan 30°

={tex}\left(tan5^\circ\times\frac1{\tan5^\circ}\right)\left(tan25^\circ\times\frac1{\tan25^\circ}\right){/tex}   {tex}\left[cot\theta=\frac1{\tan\theta}\right]{/tex}
{tex}= 1 \times 1 \times \frac { 1 } { \sqrt { 3 } } = \frac { 1 } { \sqrt { 3 } } {/tex}

We have,
f(x) = abx² + (b² - ac)x - bc
= abx² + b²x - acx - bc
= bx(ax + b ) - c(ax + b)
= (ax + b) (bx - c)
Now r(x)=0 if 

ax+b=0 or bx-c=0

{tex}\style{font-family:Arial}{\begin{array}{l}\style{font-size:12px}{\mathrm i}\style{font-size:12px}.\style{font-size:12px}{\mathrm e}\style{font-size:12px}.\style{font-size:12px}\;\style{font-size:12px}{\mathrm x}\style{font-size:12px}=\style{font-size:12px}-\frac{\style{font-size:12px}{\mathrm b}}{\style{font-size:12px}{\mathrm a}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm{or}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm x}\style{font-size:12px}=\frac{\style{font-size:12px}{\mathrm c}}{\style{font-size:12px}{\mathrm b}}\\\end{array}}{/tex}

Thus, the zeroes of f(x) are : 

{tex}\style{font-family:Arial}{\style{font-size:12px}{\mathrm\alpha}\style{font-size:12px}=\style{font-size:12px}-\frac{\style{font-size:12px}{\mathrm b}}{\style{font-size:12px}{\mathrm a}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm{and}}\style{font-size:12px}\;\style{font-size:12px}{\mathrm\beta}\style{font-size:12px}=\frac{\style{font-size:12px}{\mathrm c}}{\style{font-size:12px}{\mathrm b}}}{/tex}

{tex}\text{α+β=-}\frac{\mathrm b}{\mathrm a}\;+\frac{\mathrm c}{\mathrm b}=\frac{-\mathrm b^2+ac}{ab}=-\frac{\mathrm( b^2-ac)}{ab}--(1){/tex}

{tex}\text{αβ=}\frac{\mathrm b}{\mathrm a}\;\times-\frac{\mathrm c}{\mathrm b}=-\frac{\mathrm c}{\mathrm a}----(2){/tex}

Now for f(x) = abx² + (b² - ac)x - bc

A = ab , B= b² - ac, C = -b

{tex}-\frac BA=-\frac{b^2-ac}{ab}--(3){/tex}

{tex}\frac CA=\frac{-bc}{ab}=-\frac ca----(4){/tex}

From (1) & (3) and (2)  & (4)  we conclude:

{tex}\begin{array}{l}\text{α+β=-}\frac{\mathrm B}{\mathrm A}\\\end{array}{/tex}

{tex}\begin{array}{l}\text{αβ=}\frac C{\mathrm A}\\\end{array}{/tex}

Let {tex}\alpha,\beta{/tex} be the zeros of the polynomial {tex}f(x)=x^2-8x+k{/tex}.
Sum of zeroes = {tex} \alpha + \beta = - \left( \frac { - 8 } { 1 } \right) = 8{/tex} and, Product of zeroes = {tex} \alpha \beta = \frac { k } { 1 } = k{/tex}
Now, 
{tex} \alpha ^ { 2 } + \beta ^ { 2 } = 40{/tex}

{tex} \Rightarrow \alpha ^ { 2 } + \beta ^ { 2 }+2 \alpha\beta-2 \alpha\beta= 40{/tex}
{tex} \Rightarrow \quad ( \alpha + \beta ) ^ { 2 } - 2 \alpha \beta = 40{/tex}
{tex} \Rightarrow \quad 8 ^ { 2 } - 2 k = 40{/tex}
{tex} \Rightarrow \quad 2 k = 64 - 40 {/tex}

{tex}\Rightarrow 2 k = 24 {/tex}

{tex}\Rightarrow k = 12{/tex}

First find  the HCF of 575 and 225 by Using Euclid's division algorithm,
575 = 225{tex}\times{/tex} 2 + 125
225 = 125{tex}\times{/tex} 1 + 100
125 = 100{tex}\times{/tex} 1 + 25
100 = 25 {tex}\times{/tex}4 + 0
So, HCF of 575 and 225 = 25

Let speed of the train be x km/hr and speed of the car be y km/hr

Total distance = 600 Km

In first case, he travelled 320 km by train and rest i.e. 600 -320 =280 Km by Car.

In 2nd case, he travelled 200 Km by train and rest i.e. 400 Km by Car.
Using formula ; {tex}time=\frac{distance}{speed}{/tex} & according to question,

{tex}\frac{320}{x}+\frac{280}{y}=\frac{26}{3}{/tex} ..(i) (first case, here time = 8hr 40 minutes= 26/3 hrs)
{tex}\frac{200}{x}+\frac{400}{y}=\frac{55}{6}{/tex} ..(ii) (2nd case, here time = 30 minutes i.e. 1/2 hrs greater than that of 1st case)

multiplying eq. (i) by 5 and eq. (ii) by 8, we get
{tex}\frac{1600}{x}+\frac{1400}{y}=\frac{130}{3}{/tex} ..(iii)
{tex}\frac{1600}{x}+\frac{3200}{y}=\frac{220}{3}{/tex} ..(iv)
subtracting (iii) by (iv), we get
{tex}\frac{-1800}{y}=\frac{-90}{3}{/tex}
y = {tex}\frac{1800 \times 3}{90}{/tex}
y = 60 km/hr
Putting y = 60 in eq. (i), we get
{tex}\frac{320}{x}+\frac{280}{60}=\frac{26}{3}{/tex}
{tex}\frac{320}{x}=\frac{26}{3}-\frac{14}{3}{/tex}
{tex}\frac{320}{x}=\frac{12}{3}{/tex}
x = {tex}\frac{320 \times 3}{12}{/tex}
x = 80 km/hr

Hence, speed of the train = 80 Km/hr
speed of the car = 60 Km/hr

It is conventional.

{tex}2^{\mathrm n}\times5^{\mathrm n}{/tex}

{tex}\text{=10}^n{/tex}

{tex}\text{If n=0 then 10}^0\text{=1}{/tex}

{tex}\text{If n>0 then 10}^n\text{ will end with 0 }{/tex}

{tex}\mathrm{If}\;\mathrm n<0\;\mathrm{then}\;10^{\mathrm n}\;\mathrm{ends}\;\mathrm{with}1\;(\mathrm e.\mathrm g.\;0.1,0.01,0.001){/tex}

Hence  for all values of n,   {tex}2^n\times 5^n{/tex} can never end with 5.

Now,
9350 = 2730{tex}\times {/tex}3 + 1160
2730 = 1160 {tex}\times {/tex}2 + 410
1160 = 410{tex}\times {/tex}2 + 340
410 = 340 {tex}\times {/tex}1 + 70
340 = 70 {tex}\times {/tex}4 + 60
70 = 60 {tex}\times {/tex}1 + 10
60 = 10 {tex}\times {/tex}6 + 0
Therefore, HCF = 10

Move opposite to tree and measure the distance from bank of river to where angle of elevation is 30,

Then breadth of river and height of tree can be calculated

4u² - 8u = 0
4u(u - 2) = 0
Therefore, u = 0 and u = 2

{tex}A = \frac { h _ { b } b } { 2 }{/tex}

Let x students planned the picnic.
Then, (x - 8) students attended the picnic.
Total bus charges = Rs. 1440
{tex}\therefore \quad \frac { 1440 } { ( x - 8 ) } - \frac { 1440 } { x } = 30{/tex}
{tex}\Rightarrow \frac { 1 } { ( x - 8 ) } - \frac { 1 } { x } = \frac { 30 } { 1440 } \Rightarrow \frac { x - ( x - 8 ) } { ( x - 8 ) x } = \frac { 1 } { 48 }{/tex}
{tex}\Rightarrow \quad \frac { 8 } { \left( x ^ { 2 } - 8 x \right) } = \frac { 1 } { 48 } \Rightarrow{/tex} x² - 8x = 384 [ by cross multiplication]
{tex}\Rightarrow{/tex} x² - 8x - 384 = 0 {tex}\Rightarrow{/tex} x² - 24x + 16x - 384 = 0
{tex}\Rightarrow{/tex} x(x - 24) + 16(x - 24) = 0 {tex}\Rightarrow{/tex} (x - 24)(x + 16) = 0
{tex}\Rightarrow{/tex} x - 24 = 0 or x + 16 = 0 {tex}\Rightarrow{/tex} x = 24 or x = -16
{tex}\Rightarrow{/tex} x = 24 [{tex}\because{/tex} number of students cannot be negative]
Thus, 24 students planned the picnic.

1. Number of students who attended the picnic = (24 - 8) = 16
2. Share of 24 students = Rs. 1440

Share of 8 students = Rs. {tex}\left( \frac { 1440 } { 24 } \times 8 \right){/tex} = Rs. 480
{tex}\therefore{/tex} money paid towards the fee = Rs. 480