Easy | General |

An airplane flies in a straight line from airport A to airport B, then back in a straight line from B to A. It travels with a constant engine speed and there is no wind. Will its travel time for the same round trip be greater, less or the same if, throughout both flights, at the same engine speed, a constant wind blows from A to B?

Answer

Wind will always slow down the journey!

Solution

The time during which the plane's speed is boosted is shorter than the time during which it is retarded, so the over-all effect is of retardation. Consider the example when plane's speed is same as wind. In one direction, plane's speed doubles, while in the other, it becomes zero! (i.e. it takes infinte time to finish the journey!)

Easy | General |

You are given two cords that both burn exactly one hour, not necessarily with constant speed. How should you light the cords in order to determine a time interval of exactly 15 minutes? Extra question: how to light just one cord and measure 15 minutes?

Hint

Because this lacks uniformity, we cannot break the cord in half, and expect it to burn in half the time. But we can burn both ends of a cord to finish it in half the time it would have taken otherwise!

Solution

Burn first cord at one end and the second cord at both ends. Half an hour later, second cord finishes burning. Now you can burn the other end of first cord, and it shall finish exactly in 15 minutes.

Now can you guess how to measure 15 minutes using only one cord?

Break the cord into approximately half, and burn these two cords form both their ends. If both these cords finish burning together, that means exactly 15 minutes have passed. If any one of these cords finishes first, break the other cords from approximately middle, and further burn all ends of these little cords. Continuing this way theoretically leads to exactly 15 minutes!

Now can you guess how to measure 15 minutes using only one cord?

Break the cord into approximately half, and burn these two cords form both their ends. If both these cords finish burning together, that means exactly 15 minutes have passed. If any one of these cords finishes first, break the other cords from approximately middle, and further burn all ends of these little cords. Continuing this way theoretically leads to exactly 15 minutes!

Source: Quant Interview

Enable Like and Comment Easy | Discrete Maths |

There are 10 black socks and 10 white socks (no left-right distinction) in the wardrobe. Your task is to draw the minimum number of socks at random to be sure you have a pair of a single color. How many socks should you draw?

Hint

Pigeonhole principle

Answer

3

Solution

This is the easiest example of a very powerful tool called the Pigeonhole principal, which says that if there are (N+1) pigeons to fit in N holes, atleast one hole will have 2 or more pigeons. Hence, if you pick 3 socks to come with 2 color categories, at least one category will have 2 or more socks, i.e. a pair is guaranteed with either red or black color.

Source: Common

Enable Like and Comment Easy | Discrete Maths |

Assuming that temperature varies continuously, prove that there are always two opposite points on the Earth's surface that have the same temperature.

Solution

Aritro Pathak: consider any great circle.. T(x) is the temp at the point x .. let f(x)=T(x)-T(x+pi), then f(0)=T(0)-T(pi)..f(pi)=T(pi)-T(2pi)=T(pi)-T(0) then f(0) and f(pi) have different signs, so using mean value theorem, you have that f is 0 at some point.

FunFact: There are uncountable number of such pairs.

Here is a video explanation:

https://www.youtube.com/watch?v=5Px6fajpSio

An alternate version of this puzzle is the Mountain Man

http://www.techinterview.org/post/521419748/mountain-man

FunFact: There are uncountable number of such pairs.

Here is a video explanation:

https://www.youtube.com/watch?v=5Px6fajpSio

An alternate version of this puzzle is the Mountain Man

http://www.techinterview.org/post/521419748/mountain-man

Source: Top Quant Interview

Enable Like and Comment Easy | Probability |

Probability of accidents on a road in one hour is 3/4. What is the probability of accidents in half hour?

Hint

The probability of no event in two disjoint intervals of half hours is same as probability of no event in one full hour.

Answer

1/2

Solution

Probability of no accident in 1 hour = (prob of no accident in 1/2 hour )*(no accidents in next disjoint 1/2 hour)=p^2

hence 1/4=p^2, hence p=1/2, hence probability of accident in half hour=1/2

Read Palak's detailed solution:

Assuming things to be memoryless, and thus disjoint time intervals to be independent, prob. of no accidents in [0,2t) = prob. of no accidents in [0,t) AND [t,2t) => 1-p(2t) = (1-p(t))^2, where p(t) is the prob. of accident in time interval of length t. Given that p(1h)=3/4 => p(0.5h)=1/2 and p(2h)=15/16.

The way we have defined p(t) above, it is the probability that at least 1 accident happens in time interval t. Thus, as t increases, p(t) increases.

If, instead, p(t) is defined as the fraction of vehicles that met with an accident in time interval t, then again p(2t)=p(t)+(1-p(t))*p(t)=1-(1-p(t))^2

hence 1/4=p^2, hence p=1/2, hence probability of accident in half hour=1/2

Read Palak's detailed solution:

Assuming things to be memoryless, and thus disjoint time intervals to be independent, prob. of no accidents in [0,2t) = prob. of no accidents in [0,t) AND [t,2t) => 1-p(2t) = (1-p(t))^2, where p(t) is the prob. of accident in time interval of length t. Given that p(1h)=3/4 => p(0.5h)=1/2 and p(2h)=15/16.

The way we have defined p(t) above, it is the probability that at least 1 accident happens in time interval t. Thus, as t increases, p(t) increases.

If, instead, p(t) is defined as the fraction of vehicles that met with an accident in time interval t, then again p(2t)=p(t)+(1-p(t))*p(t)=1-(1-p(t))^2

Source: Placement tests

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