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You have 101 coins, and you know that 50 of them are counterfeit. Every true coin has the same weight, an unknown integer, and every false coin has the same weight,which differs from that of a true coin by 1 gram. You also have a two-pan pointer scale that will show you the difference in weight between the contents of each pan. You choose one coin.
Can you tell in a single weighing whether it’s true or false?
Well, this trick will help you to identify that coin!
What was the task given?
Yes, you can tell that whether the coin is true or false with single weighting.Just divide 100 coins into 2 groups of 50 coins each & put into 2 pans of weighing balance.
Let's assume true coin weighs 1 gram (or 2 gram) & fake coin weighs 2 gram (or 1 gram). Remember, if the sum of 2 integers is even then difference between two is bound to be even. And if the sum of those is odd then difference between them has to be odd.
CASE 1 :
If the coin that you are holding is true then the total weight on the balance will be
50 + (50x2) = 50 + 100 = 150 (or 50x2 + 50 = 150). So, the total sum of weights in 2 pans is even, hence difference between them has to be even. For example, if those 150 grams are distributed as 80 vs 70 then difference between them is 10 which is even.
CASE 2 :
If the coin you are holding is fake then the total weight on the balance will be
51 + (49x2) = 51 + 98 = 149 (or 51x2 + 49x1 = 153).
Here, total is odd hence the difference must be odd too. For example, if above 149 grams are distributed as 90 vs 59 then pointer of balance will point at 31 which is odd.
Conclusion:
In short, you have to notice the difference between 2 weights on the pans. If it's even then the coin you are holding is true and if difference is odd then you are holding a fake coin.
The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels.
What is the most bananas you can bring over to your destination?
As many as 'these' numbers of bananas can be saved!
What was the puzzle?
Let A be the starting point and B be the destination in this transportation. If the camel is taken with 1000 bananas at start, to reach the point B which is 1000 km away from A, it needs 1000 bananas. So there will be no bananas left to return back to point A.
That's why we need to break down the journey into 3 parts.
Part 1 :
For every 1 km the camel needs to -
1. Move ahead 1 km with 1000 bananas but eat 1 banana in a way.
2. Leave 998 bananas at the point and take 1 banana to return back to previous point.
3. Pick up another 1000 bananas and move forward while eating 1 banana.
4. Drop 998 bananas at the same point. Return back to previous point by consuming 1 banana.
5. Pick left over 1000 bananas and move 1km forward while consuming 1 banana to same point where 998 + 998 bananas are dropped. Now, the camel doesn't need to return back to previous point. So, 998 + 998 + 999 are carried to the point.
That is for every 1km, the camel needs 5 bananas.
After 200 km from point A, the camel eats of 200x5 = 1000 bananas and at this point the part 1 ends.
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PART 2 :
1. Move ahead 1 km with 1000 bananas but eat 1 banana in a way.
2. Leave 998 bananas at the point and take 1 banana to return back to previous point.
3. Pick up another 1000 bananas and move forward to the point where 998 bananas left while eating 1 banana.
Now, the camel needs only 3 bananas per km.
So for next 333 km, the camel eats up 333x3 = 999 bananas.
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PART 3 :
So far, the camel has travelled 200 + 333 = 533 km from point A and needs to cover 1000 - 533 = 467 km more to reach at B.
Number of bananas left are 3000 - 1000 - 999 = 1001.
Now, instead of wasting another 3 bananas for next 1 km here, better drop 1 banana at the point P2 and move ahead to B with 1000 bananas. This time the camel doesn't need to go back at previous points & can move ahead straightaway.
For the remaining distance of 467 km, the camel eats up another 467 bananas and in the end 1000 - 467 = 533 bananas will be left.
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There are two empty bowls in a room. You have 50 white balls and 50
black balls. After you place the balls in the bowls, a random ball will
be picked from a random bowl. Distribute the balls (all of them) into
the bowls to maximize the chance of picking a white ball.
This is the way to maximize the chances!
What was the task given?
Let's distribute 50 black ball in one bowl & other 50 white ball in another bowl.
Then,
Probability (White Ball) = (1/2)(0/50) + (1/2)(50/50) = 0.5.
Now, if 1 white ball is kept in 1 bowl and other 49 white + 50 black = 99 balls in other bowl, then
Probability (White Ball) = (1/2)(1/1) + (1/2)(49/99) = 0.747.
That's nearly equal to 3/4 which is certainly higher than the previous case. And that's the way of maximizing the probability of white ball.
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...
