"Buy Half, Get Half KG Free!"

A seller has some quantity of rice with him. The seller offered his customer that if he/she buys half of the rice he has, he will give half kg of rice as a discount. The first customer accepts his offer and he purchased half of the rice and get half kg as extra. After selling the rice to the first customer he again makes the same offer for the second customer, and so on. The seller left with no quantity of rice after he made the fifth transaction.

The initial quantity of rice the seller had?

So the amount of rice that seller had... 

Amount of Rice The Seller Had

How he sold the rice?

We need to find the amount of the rice that seller had before each of the customer came for purchasing.

Since, no rice left after he made 5th transaction, he must had 1kg rice before this transaction. Half kg is purchased by 5th customer and half kg is given as a discount.

Before 4th customer, seller must had x kg where (x - x/2) - 1/2 = 1 i.e. x = 3 kg.

Before 3rd customer, seller must had y kg where (y - y/2) - 1/2 = 3 i.e. y = 7 kg.

Before 2nd customer, seller must had z kg where (z - z/2) - 1/2 = 7 i.e. y = 15 kg.

Before 1st customer, seller must had a kg where (a - a/2) - 1/2 = 15 i.e. y =31 kg.

Hence, the seller must had 31kg of rice initially. 

Optimize Weighing Balance

You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed.

So question again is how many minimum weights and of what denominations you need to measure all weights from 1kg to 1000kg. 

We require only......Click here to know! 

Source 

Optimisation Of Weighing Balance

What was the task given? 

Just for a moment, let's assume we have to weigh 1 to 30 Kg. Now you can weigh all those with

1,          2,          4,         6,          8,         10..........30

1           2, (2+1), 4, (4+1),6, (6+1), 8,(9+1),10..(29+1),30   .......For middle weights.


Now 6 can be weighed as 2 + 4 & 10 can be weighed as 2 + 8. We can eliminate those. That means we require only

1,        2,         4,          8,         16,

So we need weights of powers of 2.

Now if subtraction is allowed then,we require

1,        3,         6,          9,          12,          15,         18,...............30

For all weights ,

1, (3-1),3, (3+1), (6-1), 6, (6+1), (9-1),9,         12,         15,............30   

But 6 can be weighed as 9-3 , 12 as 9+3, 15 as 27-(9+3). So we can eliminate 6,12,15... This leaves only

1,         3,        9,        27,

In short, we need power of 3 only.      

For the given problem we need to weight 1 to 1000 Kg with subtraction allowed. So the maximum power of 3 that is less than 1000 is 7. To conclude, we require only 7 weights as below.

1,3,9,27,81,243,729