Class 12th

$\overrightarrow{a}*\overrightarrow{b}=\overrightarrow{c}-\left(i\right)$              

$\overrightarrow{b}*\overrightarrow{c}=\overrightarrow{a}-\left(ii\right)\left|\overrightarrow{a}\right|=2$              

Taking dot product with  $\overrightarrow{c}\&\overrightarrow{a}$ respectively in (i) & (ii) we have 

  $\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]={\left|\overrightarrow{c}\right|}^2$            

Again (i) & (ii) $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$ Þ    

Opt 1.   Projection of $\overrightarrow{a}on\overrightarrow{b}*\overrightarrow{c}=\frac{\overrightarrow{a}.\left(\overrightarrow{b}*\overrightarrow{c}\right)}{\left|\overrightarrow{b}*\overrightarrow{c}\right|}=\frac{4}{2}=2$  

 Opt 2.   $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]+\left[\overrightarrow{c}\overrightarrow{a}\overrightarrow{b}\right]=2\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=8$ (2)

(1) $\overrightarrow{b}*\left(\overrightarrow{a}*\overrightarrow{b}\right)=\overrightarrow{b}*\overrightarrow{c}$ Þ

Opt 3.   ${\left|3\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c}\right|}^2=9{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+4{\left|\overrightarrow{c}\right|}^2+2\left(3\overrightarrow{a}.\overrightarrow{b}-6\overrightarrow{a}.\overrightarrow{c}-2\overrightarrow{b}.\overrightarrow{c}\right)$  

Opt 4.  $\overrightarrow{a}*\left(\overrightarrow{c}*\overrightarrow{b}-\overrightarrow{b}*\overrightarrow{c}\right)$  

Let the total work by LCM of 20, 24 and 30 i.e. 120 units

Work done per day by A and B = a + b = $\frac{120}{20}=6$ $\frac{}{}$

Work done per day by B and C = b + c = $\frac{120}{24}=5$ $\frac{}{}$

Work done per day by A and C = a + c = $\frac{120}{30}=4$ $\frac{}{}$

2 (a + b + c) = 15

So, a + b + c = 7.5

Required number of days = $\frac{120}{7.5}=16\text{?}days$

Let the numbers be x & y

(mean proportion)2 = xy.

So, xy = 784

x = $\frac{784}{y}$ $\frac{}{}$ .… (1)

x, y and 224 are in continued proportion

y2 = 224 x .… (2)

y2 = 224 * $\frac{784}{y}$ $\frac{}{}$

y3 = 14 * 16 * 28 * 28

y = 56 and x = 14

Alternate Method

Use options and check.

Let the speeds be 2Aand A kmph respectively

$\frac{216}{3A}=6$

A = 12 and the speed of B = 3 kmph

True dip is less than the apparent dip.

d = 150 km

$d=\sqrt[]{2hR}$

$h=\frac{d^2}{2R}=\frac{{\left(150*10^3\right)}^2}{2*6.5*10^6}=1731m.$

Population covered

$\pi r^2*\sigma =3.14*150^2*2000=1413*10^5$

$\frac{\left(28-25\right)}{\left(45-28\right)}=\frac{3}{17}$

Rahul can occupy any one of the two corner positions and therefore, Rahul can occupy a position in two ways.

The other four people can seat themselves in 4! ways, that is, 24 ways.

Total ways = 2 * 24 = 48 ways.

As per the problem, we have Length of the first train = 120 m

Length of the second train = 180 m

As per the problem,

$\frac{(120+180)}{(8+s)}=5$

8 + s = 60

s = 52 m/sec