**If you have a 5-liter jug and a 3-litre jug, how would you measure exactly 4 litres?
**Step 1: Fill 5 litre of water in Jug 1

Step 2: Transfer the water from 5 liters Jug 1 to 3 litres Jug 2.

2 litres remaining in Jug 1.

Step 3: Empty Jug 2.

Step 4: Transfer 2 litres from Jug 1 to Jug 2.

Now, Jug 1 = Empty, Jug 2 = 2 litres.

Step 5: Refill Jug 1 with 5 litres of water.

Step 6: Transfer 1 litre from Jug 1 to Jug 2 (it already has 2 litres from Step 4)

Now, 5 litres – 1 litre = 4 litres in Jug 1

**If you have 15L data, what will you use Excel or R? And why?
**It is preferred to use R for a very large data set as excel could only take data upto 2^20 (1048576).

**If we throw two dice, what will be the probability of getting 4 in both the dice?
**Overall outcomes of throwing 2 dice are:

{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Out of these, getting a 4 in both the dice is represented by just 1 outcome, ie, (4,4).

So, the probability of getting 4 in both the dice = Case of getting 4 in both the dice/Total no. of cases=1/36.

**If there are 730 accidents in a year on a particular road, what is the probability of getting 3 accidents in a day and at least 3 accidents in a day?
**If there are 730 accidents in a year on a particular road, there are 730/365=2 accidents in a day on that road. So, this follows Poi(2). Thus, the probability of getting 3 accidents in a day =

P(X = 3) = (e^-2*2^3)/3! = 0.18044.

And, probability of getting at least 3 accidents in a day =

P( X > = 3) = 1 – p(X < = 2) = 1- (P(x = 0) + P(x = 1) + P(x = 2)) = 0.3233.

**Suppose in a triangle there are 3 ants sitting at each corner . What will be the probability that ants will collide?
**Let’s start by reversing the way the riddle is asked and find the probability that there will

Well, what if all of the ants are walking in the same direction? Then there will never be a collision between any of the ants because they are all walking in the same direction. And, the only time they will be walking in the same direction is if they are all walking either counter-clockwise around the triangle, or clockwise around the triangle. So, there are only two scenarios in which a collision will not happen between the ants.

Now that we know that there are only two scenarios where the ants will not collide, we have to ask ourselves how many different ways are there for the ants to move on the sides of the triangle? Well, each ant can move in 2 different directions. Because there are 3 ants, this means that there are 23 (which equals eight) possible ways that the ants can move. And since we already know that there are only 2 ways in which the ants can avoid collision entirely, this means that there are 6 scenarios where the ants will collide. And 6 out of 8 possible scenarios, means that the probability of collision is 6/8, which equals 3/4 or .75. Thus, the probability of the ants colliding is 0.75.

**Guesstimate of the weight of an aircraft.
**The total weight W of the aircraft is simply the sum of the weight of all of the individual components.

W = w(fuselage) + w(wing) + w(engines) + w(payload) + w(fuel) + …

To generalize, if we have a total of “n” discrete components, the weight of the aircraft is the sum of the individual i component weights with the index i going from 1 to n.

W=summation of w(i) (i=1 to n).

**Suppose every day when you travel to the office you take the same route while going and coming back. While going you travel at 40km/hr and while coming back you travel at 60km/hr. What is the average speed in which you travel?
**Let us consider distance D. Then, we know, speed = distance/time.

So, 40=(D/T1) and 60=(D/T2).

T1=(D/40) and T2=(D/60).

Total time T=T1+T2.

(D/40)+(D/60)=(D/24).

Total distance = D+D = 2D.

We know, Average speed = Total Distance/Total Time = 2D/(D/24) = 48Km/hour.

**Suppose in a hole 22metre deep there is an ant and it wants to come out. It climbs up by 5metre before again slipping by 2metre in the hole, how many attempts the ant needed to take before coming out of the well, assuming that it will not get down (slip back)?
**The ant would need to make 7 attempts. Everytime the ant climbs 5 meters and slips by 2 meters, in total it would have climbed by 3 metres. In the 6th attempt it would have climbed by (5*3)+5 metres = 20 metres and then slip by 2 metres coming back down to 18 metres. In the 7th attempt, it would have climbed 5 more metres bringing him to 23 metres and thus, out of the well.

**Suppose a number x, first increases by 25% and then decreases by 25%. After all this adjustment, will the answer be less than x, equal to x or greater than x?
**If a number x, first increases by 25% and then decreases by 25%, the resulting answer would be less than x.

**How many 4 letter words can you make from the word ‘GIFT’?
**The number of 4 letter words we can make from the word ‘GIFT’ = 4! = 4*3*2*1 = 24.

**If you are going to receive Rs 100 after 10 years, what is the present value of it at 3% rate of interest?
**If I am going to receive Rs. 100 after 10 years, the present value of it at 3% rate of interest = Rs.100*(1.03)^-10 = Rs.74.41.

**Suppose you have Rs 200 and you have invested Rs.100 in 2 different banks. One bank offers CI of 12% pa and other bank offers SI of 12%pa. Which bank will give you a higher interest rate by the sixth month?
**The bank that offers 12% pa SI will have an amount after 6 months = 100(1+0.5*0.12)=106. This gives us 6% interest at the sixth month. The bank that offers 12% pa CI will have an amount after 6 months = 100(1.12)6/12=105.83. This gives us 5.83% interest at the sixth month. Thus, the bank offering 12% pa SI will give us a higher interest rate.

**Give a practical example of Binomial Distribution.
**Suppose that X is the number of sixes obtained when a fair die is thrown 10 times. Then,

P(X=x)=10Cx*(1/6)^x*(5/6)^(10-x) and the probability of exactly one six in 10 throws is 10C1(1/6)^1*(5/6)^9=0.3230.

**Suppose you are measuring the rainfall everyday for 365 days. What will be the mode?
**If we are measuring rainfall for 365 days, we would have the data for 365 days. The highest rainfall recorded in those 365 days would be the mode of the data.