There are three cars in a racing track. The track is made forming a perfect circle and is quite wide so that at one time, multiple cars can pass through it. The car which is leading is driving at 55 MPH and the slowest car is driving at 45 MPH. The car that is in middle of these two is driving in between the two speeds. For the time being you can say that the distance between the fastest car and the middle car is x miles and it is same between the middle car and the slowest car. Also, x is not equal to 0 or 1.
The car keeps running till the leading car catches up with the slowest car and then every car stops. Given the case, do you think that at any point, the distance between any two pairs will again become x miles?
In the question, the distance of x miles is given at a particular moment. The middle car is running at that time at 50 MPH. The distance will keep increasing with time as three are running at different speeds. It will increase by x miles every hour till a certain point of time and then it will start decreasing till the fastest car meets the slowest again.
As per question, all the cars stop then. Thus, the distance will never be x miles again.
A couple had to take shelter in a hotel for they could not proceed their journey in the rain. Having nothing to do at all, they started playing cards. Suddenly there was a short circuit and the lights went off. The husband inverted the position of 15 cards in the deck (52 cards normal deck) and shuffled the deck.
Now he asked his wife to divide the deck into two different piles which may not be equal but both of them should have equal number of cards facing up. There was no source of light in the room and the wife was unable to see the cards.
For a certain amount of time, she thought and then divided the cards in two piles. To the husband’s astonishment, both of the piles had equal number of cards facing up.
The answer is very simple. All she had to do is take the fifteen cards from the top and reverse them. This would make another pile out of that and there will be two piles - one of 15 cards and one of 37 cards. Also both of them will have the same number of inverted cards.
Just think about it and if the mathematical explanation will help you understand better, here it is.
Assume that there were p inverted cards initially in the top 15 cards. Then the remaining 37 cards will hold 15-p inverted cards.
Now when she reverses the 15 cards on the top, the number of inverted cards will become 15-p and thus the number of inverted cards in both of the piles will become same.
At first, there were 2 apples on the tree. After the wind blew, one apple fell on the ground. So there where no apples on the tree and there were no apples on the ground.
*apples => more than one apple
Another solution(non tricky one)
The wind blew so hard that the apples fell of the tree and blew along the ground.
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.
Rohit is on his way to visit your Grandma, who lives at the end of the state.It's her birthday, and he want to give her the cakes that he has made.Between his place and her grandma house, he need to cross 7 toll bridges.
Before you can cross the toll bridge, you need to give them half of the cakes you are carrying, but as they are kind trolls, they each give you back a single cake.
How many cakes do rohit have to carry with him so he can reach his grandma home with exactly 2 cakes?
Stephen died of heart attack. His soul was received in heaven. He was astonished to find himself as young as he had been in his young adult age. He looked around and there were thousands of young and naked people. His eyes searched to find someone familiar and suddenly he noticed Adam and Eve.
There are 100 doors. 100 strangers have been gathered in the adjacent room. The first one goes and opens every door. The second one goes and shuts down all the even numbered doors – second, fourth, sixth... and so on. The third one goes and reverses the current position of every third door (third, sixth, ninth… and so on.) i.e. if the door is open, he shuts it and if the door is shut, he switches opens it. All the 100 strangers progresses in the similar fashion.
After the last person has done what he wanted, which doors will be left open and which ones will be shut at the end?
Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.
The husband had booked two tickets for departure and only one for the return. The travel agent told this information to the police and they immediately came to know that he had planned the murder right from the beginning.
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
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