Read the question carefully as it can be quite tricky. There are four airplanes flying in four different directions. Each of the plane navigates a straight line at a constant speed. Please keep in mind that none of the two planes are flying parallel with each other. They began flying at the same time long time ago. Sometimes, even a collision takes place but the strong build, makes the planes continue with the journey. But even the sturdy built can only survive two collisions and after that the plane will crash.
Out of the six possible collisions, five have already taken place (do note that the collisions did not involve more than two planes at one time). Two planes have already crashed. What will be the fate of the remaining two planes?
Let us assume that the z axis denotes the time and x and y axis denote the 2D planet. The four trajectories in that case are straight lines. Since no collisions involved more than two air planes, the four lines must be lying in the same plane. Now you must be thinking that the other two planes will collide as well.
But consider the fact that they might have collided in the negative time. Thus you can't be sure with the provided facts if the two remaining planes will collide or not.
There can be both possibilities. The first assumption depicts that the four air planes fly in a manner that there is no collision between the ships. However in the second diagram, the four airplanes fly in a manner that there is collision between them.
Set the first switches on for abt 10min, and then switch on the second switch and then enter the room.
Three cases are possible
1.Bulb is on => second switch is the ans
2.Bulb is off and on touching bulb , you will find bulb to be warm
=>1st switch is the ans.
3.Bulb is off and on touching second bulb , you will find bulb to be normal(not warm)
=>3rd bulb is the ans.
A) Fill 5 ml gallon ( 5mlGallon - 5, 3mlGallon - 0)
B) Transfer to 3 ml gallon (5mlGallon - 2, 3mlGallon - 3)
C) Empty 3 ml gallon ( 5mlGallon - 2, 3mlGallon - 0)
D) Transfer 2 ml from 5 ml gallon to 3 ml gallon (5mlGallon - 0, 3mlGallon - 2)
E) Fill 5 ml gallon(5mlGallon - 5, 3mlGallon - 2)
F) Transfer 1 ml from 5 ml gallon to 3 ml gallons(5mlGallon - 4, 3mlGallon - 3)
Adam is one of the finalist in an IQ championship. As the final test, he is provided with two hourglass. One of them can measure eleven minutes while the other one can measure thirteen minutes.
He is asked to measure exactly fifteen minutes using those two hourglasses. How will he do it ?
Fifteen minutes can easily be measure using these two hour glasses.
Step 1: He will start both the hourglass.
Step 2: The moment the eleven minute hourglass is empty, he will invert it.
Step 3: When the thirteen minutes hourglass is empty, he will invert the eleven minute hourglass.
In step 3, we will have counted thirteen minutes. Since we inverted the eleven minute hourglass in step 2, it started from fresh and was inverted just for two minutes (13-11=2). In this manner when it is reversed when the thirteen minute hourglass is finished, it will have two minutes of sand left. This time when the sand finishes, he will have measured fifteen minutes. (13+2=15)
You are given with two ropes with variable width. However if we start burning both the ropes, they will burn at exactly same time i.e. an hour. The ropes are non-homogeneous in nature. You are asked to measure 45 minutes by using these two ropes.
How can you do it?
Please note that you can’t break the rope in half as it is being clearly stated that the ropes are non-homogeneous in nature.
All you have to do is burn the first rope from both the ends and the second rope from one end only simultaneously. The first rope will burn in 30 minutes (half of an hour since we burned from both sides) while the other rope would have burnt half. At this moment, light the second rope from the other end as well. Where, the second rope would have taken half an hour more to burn completely, it will take just 15 minutes as we have lit it from the other end too.
Thus you have successfully calculated 30+15 = 45 minutes with the help of the two given ropes.
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
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.
For my anniversary, I decided to surprise my wife. Since she is a voracious reader, I decided to collect a lot of books for her. On the first day of the month, I bought one book, on the second, I bought two and on the third, I bought three. This process went on till the anniversary and on the day, I had 276 books with me to gift her.
Can you calculate, on which day is our anniversary?
* They have only one torch and the river is too risky to cross without the torch.
* If all people cross simultaneously then torch light wont be sufficient.
* Speed of each person of crossing the river is different.cross time for each person is 1 min, 2 minutes, 7 minutes and 10 minutes.
What is the shortest time needed for all four of them to cross the river ?
The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 minutes. Is that it? No. That would make this question too simple even as a warm up question.
Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.
1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)
In a jar, there are some orange candies and some strawberry candies. You pick up two candies at a time randomly. If the two candies are of same flavor, you throw them away and put a strawberry candy inside. If they are of opposite flavors, you throw them away and put an orange candy inside.
In such manner, you will be reducing the candies in the jar one at a time and will eventually be left with only one candy in the jar.
If you are told about the respective number of orange and strawberry candies at the outset, will it be feasible for you to predict the flavor of the final remaining candy ?
At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.
Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.
There are a hundred statements.
1st person says: At least one of the statements is incorrect.
2nd person says: At least two of the statements is incorrect.
3rd person says: At least three of the statements are incorrect.
4th person says: At least four of the statements are incorrect.
100th person says: At least a hundred of the statements are incorrect.
Now analyze all the statements and find out how many of them are incorrect and how many are true?
The 100th statement for sure is incorrect because it says that at least 100 of the statements are incorrect.
Suppose if that is correct, then 100 statements cannot be true.
This suggests that the 100th statement is incorrect and that the first statement is true.
Similarly 99 statements cannot be true because if they were true, then two statements would become correct i.e. the 1st and the 99th.
But the 99th statement says that at least 99 are incorrect.
This suggests that the 99th statement is incorrect and that 2ndone is true.
If we keep analyzing is the same way till the end, we will find out that only the first fifty statements are true and all the remaining ones are incorrect.
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