Let the number of rounds done by SUV be $x$, car be $y$, jeep be $z$ and auto be p.
Thus the number of passengers transported will be $7x+5y+8z+2p$ ....   (1)
Checking options respectively:
1$\to x=1,z=2,p=1$
       Substituting these in (1)
       7+8 $\times$ 2+2=25.
       So option A is correct.
2$\to x=2,y=2$
       Substituting these in (1)
       7$\times$2+85$\times$2=24
       So option B is incorrect.
3$\to x=1,y=2,z=1$
       Substituting these in (1)
       7+5$\times$2+8=25.
       So option C is correct.
4$\to$ Similarly substituting the values of option D. We will find number of people transported=26
 

Total number of work days required =10 x 20 = 200
Since 10 men worked for 10 days, 10 x 10 = 100 days work is already done.
So, 200 - 100 = 100 days of work is left. 
Once, 10 more men joined, total number of workers=20
So, number of days required more $=\frac{100}{20}=5$
Thus, total number of days = 10 + 5 = 15.

Remaining number of men = 250-50 =200 men
             Remaining number of days = (350-25) days = 325 days.
            250 men had provisions for 325 days
           1 man had provisions for (250 × 325) days  
          200 men had provisions for $\frac{(250\times 325)}{200}$ days  
                                                                                       
                                          = 406.25days =406 days
        hence, the remaining food will last for 406 days.
 

Let the required distance be y.

$\begin{array}{ccc}Distance\left(inkm\right)& 363& y\\ Amountofpetrol\left(inl\right)& 24.2& 7\end{array}$

The distance travelled and the amount of petrol consumed are directly proportional
$\frac{A}{B}$= Constant
$\frac{A_1}{B_1}$ = $\frac{A_2}{B_2}$

$\frac{24.2}{363}$ = $\frac{7}{y}$
$24.2\times y=363\times 7$

$y=\frac{7\times 363}{24.2}=105$ km

$$\begin{array}{ccc}Numberofgifts& 3& y\\ Price\left(InRupees\right)& 550& 1100\end{array}$$
$\frac{3}{550}=\frac{y}{1100}\Rightarrow y=\frac{3\times 1100}{550}=6$ gifts

Let the number of pipes be y.

$\begin{array}{ccc}\text{Number of pipes}& 7& y\\ \text{Time (in minutes)}& 220& 81\end{array}$

The number of pipes and the time are inversely proportional.
AB = constant
$A_1\times B_1$ = $A_2\times B_2$
$\frac{7}{y}$ = $\frac{81}{220}$

$\Rightarrow y=\frac{7\times 220}{81}=19.01\approx 19$

Shane has to use 19 pipes to fill the tank in 1 hour 21 minutes.

4 oxen = 12 cows
$\Rightarrow$ $1$ ox = $\frac{12}{4}$ cows
$\Rightarrow 7\text{oxen}\equiv \frac{12}{4}\times 7\text{cows}=21\text{cows}$
$\Rightarrow$  (7 oxen + 3 cows) $\equiv$ (21 cows + 3 cows) = 24 cows
Now, 12 cows can graze the field in 32 days.
So, 1 cow can graze the field in $(32\times 12)$ days.
[less cows, more days]
24 cows can graze the field in $\frac{32\times 12}{24}=16days$
[more cows, less days]   
Hence, 9 oxen and 3 cows can graze the field in 16 days.

$$\begin{array}{ccc}Numberofbottles& 238& y\\ Time\left(Inhours\right)& 2& 7\end{array}$$
The time taken and the number of bottles filled are directly proportional.
$\frac{A}{B}$= Constant
$\frac{A_1}{B_1}$ = $\frac{A_2}{B_2}$
$\frac{2}{238}=\frac{7}{y}$
$\Rightarrow y=\frac{7\times 238}{2}$
$\Rightarrow y=833$ bottles

Let the required number of days with 30 men working for 6 hours be x.
$\begin{array}{ccc}\text{Number of men}& 39& 30\\ \text{Number of hours}& 5& 6\\ \text{Number of days}& 12& x\end{array}$
Number of men, number of hours and number of days are inversely proportional and work done is constant in both the cases, hence
$\Rightarrow 39\times 12\times 5=30\times x\times 6x=\frac{39\times 12\times 5}{30\times 6}x=13\text{days}$

Given distance s is directly proportional to square of time t.
That is:
 $s=k.t^{2}20=4kk=5\text{So,}s=5t^{2}605=5t^2$

t = 11 seconds