You have ten boxes and an electronic weighing machine. In those ten boxes, you have chocolates. Each chocolate weigh 20 grams. But in one box the chocolates are defective and each weigh 19 grams exactly.
Now you can weigh in the electronic weighing machine but you can use that machine just once. How will you find out which box has the defected chocolates.
If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.
Let us begin by labelling boxes as 1, 2, 3 and so on till 10.
Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)
The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.
You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.
A petri dish kept in a lab has a colony of healthy bacteria. Every bacterium divides itself into two in exactly two minutes. Now the colony started with a single cell at 2 pm. If the petri dish was exactly half full of bacteria at 3 pm, when will the dish become full of bacteria?
Since every bacterium divides into two in two minutes and we know the fact that the dish was half filled with bacteria at 3 pm, after exactly 2 minutes, all will divide into two and will fill the petri dish. Thus, the petri dish will become full at 3:02 pm.
One man was moving to and fro in worry when a bright woman noticed him. On asking he told her that while replacing his tire, he incidentally dropped all the four nuts into a deep drain. She told him what to do and he was then able to drive back successfully till his office.
Step1. Lets say y = x
Step2. Multiply through by x xy = x2
Step3. Subtract y2 from each side xy - y2 = x2 - y2
Step4. Factor each side y(x-y) = (x+y)(x-y)
Step5. Divide both sides by (x-y) y = x+y
Step6. Divide both sides by y y/y = x/y + y/y
Step7. And so... 1 = x/y + 1
Step8. Since x=y, x/y = 1 1 = 1 + 1
Step9. And so... 1 = 2
Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules
We know that Christanio Ronaldo tells the truth on only a single day of the week. If the statement on day 1 is untrue, this means that he tells the truth on Monday or Tuesday. If the statement on day 3 is untrue, this means that he tells the truth on Wednesday or Friday. Since Christanio Ronaldo tells the truth on only one day, these statements cannot both be untrue. So, exactly one of these statements must be true, and the statement on day 2 must be untrue.
Assume that the statement on day 1 is true. Then the statement on day 3 must be untrue, from which follows that Christanio Ronaldo tells the truth on Wednesday or Friday. So, day 1 is a Wednesday or a Friday. Therefore, day 2 is a Thursday or a Saturday. However, this would imply that the statement on day 2 is true, which is impossible. From this we can conclude that the statement on day 1 must be untrue.
This means that Christanio Ronaldo told the truth on day 3 and that this day is a Monday or a Tuesday. So day 2 is a Sunday or a Monday. Because the statement on day 2 must be untrue, we can conclude that day 2 is a Monday.
So day 3 is a Tuesday. Therefore, the day on which Christanio Ronaldo tells the truth is Tuesday.
Your friends are coming over to your birthday party. Your mum has bought this delicious giant doughnut for you guys. Unfortunately she had to go somewhere. Now you have to serve all of them including you.
There are a total of nine kids including you. None of you would mind a smaller piece. Can you cut this doughnut into nine pieces in just three straight cuts?
In the figure given below, you can see that there are five squares. Supposedly if this figure is formed using different matchsticks (refer the slight gap between matchsticks), can you remove just two matchsticks so that only two squares remain ?