The second, seventh, ninth and eleventh letters of the word ORGANISATION are R, S, T and O respectively. Clearly, only one word can be formed using these letters,
which is SORT. The third letter in SORT is R.

We know that   Speed=distancetime 

distance= speed * time

⇒ d=40×518×9

   d= 100 mts.

Let the sum be Rs. p, rate be R% p.a. and time be T years.

Then,

 P×T×(R+2)100-P×T×R100=108 ⇒2PT= 10800 ...... (1)

 And, 

 P×R×(T+2)100-P×R×T100=180 ⇒2PR = 18000 ......(2)

Clearly, from (1) and (2), we cannot find the value of p
So, the data is not sufficient.

Let the total capacity of tank be 90 liters.
Capacity of tank filled in 1 minute by K = 3 liters.
Capacity of tank filled in 1 minute by L = 15 liters.
Therefore, capacity of the tank filled by both K and L in 1 minute = 18 liters.
Hence, time taken by both the pipes to overflow the tank = 90/18 = 5 minutes.

Let the cost price be Rs. 100
Then, market price is Rs. 140
Now, the first discount is of 28.57% Rs.140 =~ Rs. 40
Hence, its selling price = 140 - 40 = Rs. 100
Now since Rana is selling at cost price, any further discount will be equal to loss percentage.
The next dicount of 12.5% will be the loss percentage to Rana.

As we know that a week has 7 days,

k weeks k days = (7k + k) days = 8k days.

Class can have a public method for specific data type conversions.
for example:
class B
{
double value;
public  B(int i )
operator double()
{
return value;
}
};
B BObject;
double i = BObject; // assigning object to variable i of type double.
now conversion operator gets called to assign the value.

From the given data,

12 children 16 days work,
One child’s one day work = 1/192.

8 adults 12 days work,
One adult’s one day’s work = 1/96.

Work done in 3 days = ((1/96) x 16 x 3) = 1/2

Remaining work = 1 – 1/2 = 1/2

(6 adults+ 4 children)’s 1 day’s work = 6/96 + 4/192 = 1/12

1/12 work is done by them in 1 day.

1/2 work is done by them in 12 x (1/2) = 6 days.