Maths

When S1 runs 1900 metres, S3 would run 1862 metres. Hence, the start given is  38/1900 * 2000 = 40 metres

$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}.......\left(i\right)$

$\frac{x-2}{1}=\frac{y-\lambda }{2}=\frac{z-3}{4}......\left(ii\right)$

$\overrightarrow{OA}=\left(i+2j+k\right)\overrightarrow{OB}=\left(2\widehat{i}+\lambda \widehat{j}+3\widehat{k}\right)$

$\overrightarrow{a}=2i+j+3k\&\overrightarrow{b}=i+2j+4k$

$\therefore$ shortest distance = $\left|\frac{\left(\overrightarrow{OB}-\overrightarrow{OA}\right).\left(\overrightarrow{a}*\overrightarrow{b}\right)}{\overrightarrow{a}*\overrightarrow{b}}\right|=\frac{1}{\sqrt[]{38}}$

$\Rightarrow \left|14-5\lambda \right|=1$           

$\Rightarrow \lambda =3\&\frac{13}{5}$          

Integer value of $\lambda$ = 3

$36x^2+36y^2-108x+120y+c=0$

$\Rightarrow x^2+y^2-3x+\frac{10}{3}y+\frac{c}{36}=0$              

The circle not interesting either axes

$\therefore g^2<c\&f^2<c$               

  $\Rightarrow \frac{9}{4}<\frac{c}{36}\Rightarrow c>81-\left(i\right)\&\frac{25}{9}<\frac{c}{36}\Rightarrow c>100......\left(ii\right)$            

(i)  $\cap$ (ii) -> c > 100 .(iii)

Point of intersection between x – 2y – 4 = 0 & 2x – y – 5 = 0 is (2 - 1) lies inside the circle.

$\therefore 4+1-6-\frac{10}{3}+\frac{c}{36}<0\Rightarrow c<156.......\left(iv\right)$         

-> $\left(iii\right)\cap \left(iv\right)$ 100 < c < 156

$\left(\frac{A}{B}\right)*\left(\frac{B}{C}\right)*\left(\frac{C}{D}\right)=\frac{A}{D}$

$\Rightarrow \frac{5}{6}*\frac{13}{14}*\frac{9}{4}=\frac{195}{112}$

The new time is $\frac{1}{3}$  of the old time. If time becomes $\frac{1}{3}$ , then new speed should have been   $\frac{3}{1}$ of the initial speed. Therefore

$\frac{3S-S}{S}*100=200\mathrm{\%}$

Using the successive percentage change rule, we have

$-25+x-\frac{25x}{100}=20$

$\frac{3x}{4}=45$

x =60%

We have to find last non zero digit by 4 so we will eliminate 0 in question so now it becomes 104925 = 4 as 4odd = 4