Easy | General |

There are two beakers, one containing water, the other wine (equal volumes). A certain amount of water is transferred to the wine, then the same amount of the mixture is transferred back to the water. Is there now more water in the wine than there is wine in the water?

Answer

Equal!

Solution

The main point is that the final volumes of mixture liquids are equal. This means in jar A, volume of water + wine is same as that in jar B. Assume original volume as L. Suppose final volume in jar A is x for water and (L-x) for wine. This implies that jar B has x amount of wine and (L-x) of water. This proves that there are equal amounts of water in the wine jar and wine in the water jar.

Source: Common

Enable Like and Comment Easy | General |

Light bulbs are numbered 1 to 100, and kept off initially. First person comes and toggles all the bulbs which are multiple of 1, i.e. he switches all bulbs to on. Second person toggles all multiples of 2, i.e he turns of even bulbs. Third person comes and toggles all multiples of 3. This process continues till 100 persons pass. After this, how many bulbs are ON?

Hint

Consider the bulb number 9.

Answer

10 bulbs

Solution

We notice that for a perfect square (like 9), the number of factors are always odd, for example:

Number of factors of (16) = # [1,2,4,8,16] = 5

Note that for non-square numbers, factors are even.

As a factor toggles the state of a bulb, bulb number 9 will be toggled by 1,3 & 9. Thus bulb number 9 will switch ON, OFF, ON respectively. Note that odd number of factors cause bulb 9 to be ON at the end.

We note that for odd number of factors is the cause of bulb staying on at the end. Similarly every squared digit bulb will be switched on, and rest will remain off after all factors toggle. Thus the bulbs 1,4,9....81,100 are ON, at the end. Hence 10 bulbs are on.

Number of factors of (16) = # [1,2,4,8,16] = 5

Note that for non-square numbers, factors are even.

As a factor toggles the state of a bulb, bulb number 9 will be toggled by 1,3 & 9. Thus bulb number 9 will switch ON, OFF, ON respectively. Note that odd number of factors cause bulb 9 to be ON at the end.

We note that for odd number of factors is the cause of bulb staying on at the end. Similarly every squared digit bulb will be switched on, and rest will remain off after all factors toggle. Thus the bulbs 1,4,9....81,100 are ON, at the end. Hence 10 bulbs are on.

Easy | General |

Without explicitly calculating, find out which is bigger: e^Pi or Pi^e?

Answer

e^Pi

Solution

This is a common placement test question. Since e^x = 1 + x…., we have e^x>1+x. Now, conveneintly choosing x = (Pi / e -1) and solving, we get e^Pi > Pi^e.

Another way is to observe for f(x) = x^(1/x), differentiating: f'(x) = f(x)*(1/x^2)*(1- ln x); f is decreasing for x > e so pi^(1/pi) < e^(1/e); hence e^(pi)> pi^(e)

Another approach is to assume e^Pi > Pi^e <=> Pi*lne > e*ln Pi <=> lne/e > lnPi/Pi. Now, notice that f(x) = lnx/x is decreasing function since f'(x) is negative in range (e<x<pi), hence above assumption is true!

Another way is to observe for f(x) = x^(1/x), differentiating: f'(x) = f(x)*(1/x^2)*(1- ln x); f is decreasing for x > e so pi^(1/pi) < e^(1/e); hence e^(pi)> pi^(e)

Another approach is to assume e^Pi > Pi^e <=> Pi*lne > e*ln Pi <=> lne/e > lnPi/Pi. Now, notice that f(x) = lnx/x is decreasing function since f'(x) is negative in range (e<x<pi), hence above assumption is true!

Easy | General |

100 pirates stand in a circle. They start shooting alternately in a cycle such that 1st pirate shoots 2nd, 3rd shoots 4th and so on. They continue in circle till only one pirate is left. Which position should Captain Jack Sparrow stand to survive?

Solution

This is not a puzzle at all, this is apti/maths question, but has become quite common. You can calculate for a N=100, 73rd position is safe. For a general N, find largest m such that 2^m < N. Take t = N-2^m. Then Safe position is 2*t + 1. Please learn this formula. No need to prove it using strong induction. Proof: http://en.wikipedia.org/wiki/Josephus_problem#k.3D2

Source: Written Tests

Enable Like and Comment Easy | General |

In a dark room, there is a deck of 52 cards, with exactly 10 cards facing up, rest facing down. You need to split this into two decks with equal number of cards facing up!

Hint

The question can be solved even when up facing cards are 11. The split decks need not have equal number of up-facing cards.

Solution

Create a deck of k cards randomly from the original 52-card deck, (k=10 here) and then turn over the k-card deck. Goal is achieved!

Source: Quant Interview

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