Business analyst Interview Questions

Personally, I would say that this question isn't about giving presentations; it's about how you handle uncomfortable situations that can arise in business with little warning. How do you give a presentation on a subject you know little about? Don't give a speech, instead, utilize a roundtable discussion method to bring any knowledge the audience might have into the discussion in the hopes that the group's knowledge will help to make the presentation more fruitful and beneficial for everyone. Often, interview questions are not about specific job responsibilities, but rather about skills they are looking for in potential candidates.

A basic template for presentation is responding to this set of questions: What, Why, Where, When, Who, and How? Some people add How much?

To be honest , I would just play it off . I would use all my knowledge and understanding of the topic and say it . If I don’t know nothing , then I don’t say nothing . Easy you shouldn’t be forced to give a presentation with a lack of knowledge, that is plain dumb . Don’t play yourself . It’s better not to say nothing and just excuse your self :

Not very clear formulation of problem . I move V (in thousand) bananas on L feets., what I get in point L 1.V-L/V Or 2.(V-L) In first case I spend 1/V moving on 1 feet, but in second energy do not depend of load.

Sorry my Error not 1/v but V and not (V-L/V) but (V-V*L)

how do you mathematically formulate it? I mean I see how it's 500 from intuition. But, if I were to approach the problem from a purely mathematical way, how would I formulate it?

@Shilpa, Could you share what you found? Thanks!

It's not 500. Not sure if correct but I calc 666.

It should be around 666 bananas.

How did you arrive at 666?

let us say you move x feet before you drop your bananas and then go back to point A, and let's say you pick 1000 bananas first time. You eat x bananas when you get to point x, and since you need go back to Point A, you will keep another x bananas, then you drop 1000-2x bananas at point X, next time suppose you pick y bananas, when you get to point X, you will eat y-x bananas, and pick former dropped 1000-2x bananas, all these bananas add up to 1000, so we got y-3x+1000=1000, that's y-3x=0. Since we can pick max 1000 bananas, so max y is 1000, x is 333, that is say, we can go as far as 333 feet and then we must go back. since after we get to point b and don't have any more bananas for us to go back to point A, seems 333 is the answer

1000-2x+1000-2x+1000-x=2000 x=200 (first stop point @ 200 feet) 1000-2y+1000-y=1000 y=333 (second stop point @ 533 feet) so you can move 1000-(1000-533)=533 to Point B.

Hey ltss587ATgmailDOTcom - I understand how you got to 1000-2x+1000-2x+1000-x (3000-5x) which is effectively number of banas left at Point X. Why did you equate it to 2000? what is the rationale to it? Also how did you get the second equation? 1000-2y+1000-y=1000 Wei - In your equation, though it makes sense, but u didnot go back the third time to pick up the 1000 bananas. At this time, y-3x=0., you still have a 1000 bananas left? Could you please explan. I really appreciate it.

Hey G It goes like this 1000-2x+1000-2x+1000-x=2000 ( equate it to 2000 because after this u will wont need to go back third time as the maximum u can carry will be over in two time) x=200 (first stop point @ 200 feet) 1000-2y+1000-y=1000 ( equate it to 1000 because after this u will wont need to go back second time as the maximum u can carry will be over once y=333 (second stop point @ 533 feet) so you can move the remaining 1000 distance in (1000-533)=533 to Point B. So 533 bananas left

sorry guys its 833 bananas You have to do it a foot at at time to max out You'll have to carry 3 loads until you eat 1000 bananas so 3x = 1000 or x = 333 ft. next you have to carry 2 loads until you eat the next 1000 bananas so 2x = 1000 or x = 500 ft. Now you have exactly 1000 bananas left so only one load fo the remaining 167 feet this gives you 833 bananas left.

It is 333 bananas!!!

833 is the maximum left and best explanation. Do it on a number line. Maximize each load and trip. The second step must be 2 trips closest to the goal ending with 1000 bananas because there be one final trip at maximum load the shortest distance. So stops at 333+500+167 gets the most product moved at least cost. Logistics.

Technically this isn't really asking anything.

Technically you don't have to eat any bananas. If you move them less than a foot each time, you don't have to eat any bananas.

3000, pickup 1000 bananas and walk 1/2 feet, go back get another 1000 and put them @ 1/2 ft location and then get the rest bunch.

833 is correct. Starting at 1000 feet mark. Carry 1000 bananas drop 999 at 999 foot mark. Go back and bring 2*1000 similar way and drop at 999 ft. Now you have 3*999 bananas at 999 feet. Keep doing this until 3*x exceeds 1000. You are now at 666 foot mark with 1998 (3000-334*3) bananas. Now do two trips and drop bananas at each foot traveled. At 177 (833) foot mark you will have 1000 bananas. Now you will loose 177 bananas to end

Incomplete question but this is making the presumption that the answer required is 'what is the largest number of bananas I get get to point B'? Te answer is 2 You must eat one banana per foot traveled. Whether forward or backwards, a foot is still a foot, so a banana must be eaten. If I take 1000 bananas and drop 334 of them off after 333 feet, I will have just enough to get back to point A. I grab another 1000. I walk 333 feet, arriving with 667 bananas. I pick up 333 of those I left the first time, leaving one, and walk another 333 feet. I reach the 666 feet mark with 667 in my possession. I leave two there. One my way back, I must pick up the remaining banana from the 333 foot mark in order to get back to point A. (665+1) I pick up the remaking thousand. At 666 feet, I have 334 left. I pick up the 2 that I left there, then eat the 334 necessary for me to reach point B. I have 2 left I grab the remaining 1000. I reach the 666 mark with 334 bananas. I take the one I left there. I walk 334 feet, arriving at point B with a single banana

Sorry for the fat fingers in my answers! I ate too many bananas and was feeling ill, lol

Second apology. Somehow seems as if my response sequence got messed up. I'll try this again!

I must eat a banana whether traveling to or from point A. I take 1000 and walk 333 feet I arrive with 667 and leave 334 of them I get back to point A empty handed and grab another 1000. At 333 feet, I take 333 of those I left the first time, resuming my journey with 1000 bananas. I walk to the 666 feet mark arriving with 667. I leave 2 and head back with 665 in my possession. I pick up the one I left at the 333 point mark and make it back to point A empty handed. The last thousand is enough fuel to get me from A to B, so I arrive at point B with the two bananas I had left at 666 feet. The answer is 2

The answer can be googled, they offer a generalized form. For this particular example it is 533

3000. Since the direction of the distance between point A and point B is not specified, we can assume point A is at a higher elevation than point B. We can then either 1) slide the bananas down the slope from point A to point B if we assume the their is a slope with a low enough coefficient of friction, or 2) assume that Point A is directly above point B and simply drop the bananas. Since in both scenarios we do not have to move, there is no need to eat any bananas. (Note: for the second scenario, the problem does not specify the condition the bananas should be in when they arrive at point B.)

Its a silly question and it deserve a silly answer as well. Answer: both have flat bottom .....

Both have expiration date. If they are not sold/used by the expiration date/flight time, they are useless

Both are designed for efficient transportation

They are both designed to be lightweight, which translates into saving on transportation costs and energy.

They both are designed to hold things safely. They both are designed to fit the maximum 'product' in a given area.(milk cartons are square so you can fit more in a fridge than round bottles) They both have tons of federal regulation regarding their design and how they hold their respective contents. The contents of both have extremely low margins and often serve as loss leaders. ie milk is sold cheap to get customers in to buy other goods and seats are sold cheap to get customers to buy other goods like ryan air. The cost for both are a small proportion of the total price. They both are designed in a very utilitarian manner, but also have luxury versions. paper cartons vs glass milk bottles and coach vs first class seats

Indirect complimentary objects that came with the purchase of a product! Those objects are both necessary for the sale of milk, and a flight. The customer is not considering those two objects when they are making their purchase. *Admittedly a seat on a flight has more importance than the carton

Packaging is the first line of attack to product segmentation, first vs economy, carton vs glass.

They both can both be filled or empty

Both are in an aisle.

You should have a basic knowledge of credit card concepts, e.g. credit limits, balances, interest rates. The interviewer will be happy to explain any uncommon terms that they use during the case (e.g. I wasn't familiar with the term "utilization ratio" and my interviewer was happy to tell me what it meant). You don't need to know about the crazy derivatives or anything like that. Put it this way: The interviewers won't penalize you if you ask them what a particular term means during a case. However, if they ask you what consumers might care about in choosing a credit card or bank, you should be able to name basic concepts such as the ones above.

this problem is conditional probability because previous events effect probability of current. If the chambers are loaded with 3 bullets at random: % of survival = 3/6 * 2/5 *1/4 = 5% chance of shooting 3 blanks consecutively. If the bullets are in 3 consecutive chambers, then your % drastically changes. 50% chance on first shot. If blank, then you know that 3 permutations are no longer available: chamber 123,612, 561. Therefore chamber 5 has a 66% chance to contain a bullet chamber 6 has a 33% chance to contain a bullet chamber 2 has a 33% chance of containig the bullet chamber 3 has a 66% chance to contain a bullet chamber 4 must contain a bullet Chance of dying with 3 consecutive shots if the bullets have to be consecutive is 0.5*0.666*0.333 = 11%

--Spin-shoot-spin-shoot-spin-shoot: We are selecting a random barrel each time (50% chance of containing a bullet) Probability of surviving = (1/2)*(1/2)*(1/2) = 1/8 = .125 --Spin-shoot-shoot First chamber has 50% chance of containing bullet. If we survive the first shot, there are still 3 bullets (2 blanks) left, but only five remaining chambers Probability of surviving = (1/2)*(2/5) = 2/10 = .2 .2 > .125, so the second option is safer.

Martin is right except for the last part. It would be Probability of surviving = (1/2)*(2/3) = .33 This is because if you survived the first shot you are in 3 of the possible blank chambers that are all next to each other. If you first shot blank #1, you will move and shoot blank #2 next. If you shot blank #2 first, you will move and shoot blank #3 next. If you shot blank #3 first, you will move and shoot the first bullet (1/3 chance of getting bullet conditional on surviving the first shot). Given then you survived the first shot with 50% chance, you started in one of those blank chambers. And since the blanks are adjacent to each other (since the bullets are all adjacent), your chances will be 2/3 for survival. And 1/2*2/3 = 1/3

None of them, I want to live.

How about we just not play Russian roulette.

rather not play

I choose option C, never point a firearm at anything you are not willing to destroy.