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 How many smaller cubes are completely invisible in a n*n*n rubic's cube?7 Answersn^3- 6n^2 The number of cubes are n^3(volume). The number of cubes exposing one or more sides( 6 *( n^2) - all cubes on the sides.) Number of invisible cubes are n^3- 6n^2 .Isn't the answer none since the question says "completely invisible"? every unit is visible by at least one side so any sub cube (made of these units) has a portion exposed.N*N*N-(N*N*2+N*(N-2)*2+(N-2)*(N-2)*2)Show more responses(N-2)^3(n-2)^3 you can verify by assuming n=4. Try to visualize invisible cubes and compare it with answers.(n-2)^3 you can verify by assuming n=4. Try to visualize invisible cubes and compare it with answers.(n-2)*(n-2)*(n-2)

19 Aug 2012
 How'll you find the weight of an object attached to another object without detaching them and without using any weighing machines. Both the weights are unknown8 AnswersIf the objects are linked by a thread, I'd just put each object one side of my arm and let the heavier one go down while the lighter one goes upTheMigthyBat, no they're attached as in bolted together. You don't know the weight of either of them. And it's not about finding the heavier of them, it's about finding the exact weight of one of them without disturbing them.Cant we just see the amount of water displaced by the object and then calculate the weight? I think its easy that way, as we can control the volume immersed in water. (let it be attached in any way)Show more responsesI think we can take a perfect model of both objects and weigh them separately...by immersing them in a liquid....one by one....and checking the amount of volume displaced.......and applying principles on the liquid collectedimmersing in the liquiud is the ans to thi s ques ... bt still then they are connected together so the total volume obtained would the sum of both objects .. then how could we find the volume of each one ..Water displacement theory seems to be nice. Since there is no information regarding how they are attached we can do the following way. catch one weighted and dip the other weight in water. Some water will be displaced. Now repaeat the same in revesrese order. i.e catch the second side and dip the other weight. Now calculate the amount of water displaced in both the cases. greater displacement - lesser water displacement. Now immerse completely both the objects and find the displacement of water. by this you can arrive at the answerits simply and easy. Let us use the simple way to determine which is heavier. Let us assume the weights are connected with the thread. And use the water displacement. And we can use the six standard mathematics (A+b) whole square formulae. We can calculate the weight of the object

24 Oct 2013