Easy | Strategy |

A traveler wants to go to the mystic. He meets a pair of twins at a fork in the road: one path leads to the jungle, the other to the mystic. One of the twins always says the truth, the other always lies. What yes/no question should he ask one of the twins to determine the path that goes to the mystic?

Hint

Ask one about the other!

Solution

The two-brother problem may be solved in two different ways:

The traveler may ask, "Would your brother agree that the road on the left leads to the mystic?" The answer is guaranteed to be 'Yes' if and only if the road on the right leads to the mystic.

The traveler may ask, "If I were to ask you whether the road on the left leads to the mystic, would you say 'Yes'?" The 'if i were to ask you' construct makes both twins tell the truth. So if the answer is 'Yes', the road on the left indeed leads to the mystic, otherwise the road on the right does so.

(solution was taken from gurmeet's blog)

The traveler may ask, "Would your brother agree that the road on the left leads to the mystic?" The answer is guaranteed to be 'Yes' if and only if the road on the right leads to the mystic.

The traveler may ask, "If I were to ask you whether the road on the left leads to the mystic, would you say 'Yes'?" The 'if i were to ask you' construct makes both twins tell the truth. So if the answer is 'Yes', the road on the left indeed leads to the mystic, otherwise the road on the right does so.

(solution was taken from gurmeet's blog)

Source: Common

Enable Like and Comment 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 | Probability |

We have a weighted coin which shows a Head with probability p, (0.5<p<1). How do we get a fair toss from this? That is, how do we toss this coin in such a way that we can have probability of winning = loosing = 50%?

Hint

Clearly we cannot have a fair toss in a single flip of this coin. But by tossing this coin 2 times, we can assign the outputs to "win" or "loose", such that we have a 50% chance for both.

Answer

Toss 2 times, mapping HT to "win", TH to "loose" and repeat the process otherwise.

Solution

Toss the coin twice. If consecutive Heads-Tails appears (HT), we "win". In case of (TH), we "loose". If (TT) or (HH) appears, repeat the process. Probability of infinite repetition (p*p*.....) + (1-p)*(1-p)*.... = 0 + 0 = 0;

Probability of H-T and T-H is equal hence it's fair now.

I proposed a faster method, "lets keep tossing the coin to form a sequence of H's & T's . I win if HT appears before TH" . Was I bluffing?

Actually yes, the probability of HT before TH is just p in second game.

Probability of H-T and T-H is equal hence it's fair now.

I proposed a faster method, "lets keep tossing the coin to form a sequence of H's & T's . I win if HT appears before TH" . Was I bluffing?

Actually yes, the probability of HT before TH is just p in second game.

Source: Common

Enable Like and Comment Easy | Probability |

Spiderman has two girlfriends, Mary Jane & Gwen Stacy. After every mission, he rushes to the central subway. Since spidey is a nice man, much impartial, he takes which-ever train arrives first. From subway, one series go towards Mary's place, and another series move towards Stacy. Trains from either series appear every 10 minutes. Also, Peter Parker sticks with the train which arrives first.

But somehow, he notices that he is spending 9 times more dates with Mary Jane than Stacy. Can you explain why?

But somehow, he notices that he is spending 9 times more dates with Mary Jane than Stacy. Can you explain why?

Hint

Do you think ratio should be 50:50? Do you think the gap between each train's arrival is 5 minutes? Can a train arrive always earlier than the other and still be at every 10 minutes gap?

Solution

Train to Mary Jane's place comes at say times 0 and 10, while the train to Stacy's place comes at times 1 and 11. So 9/10 times, train to Mary Jane's place is appearing earlier.

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