Que 16: There are 100 water bottles. One of the bottles contains poisoned water. If a rat drinks poisoned water he will die after one hour of it. You have 1 hour to find out which bottle has the poison. How many minimum rats are needed to figure out which bottle contains poison?
Ans 16:
We can use digital logic to find out the poisoned bottle. Also, the rat will die in 1 hour and we have only 1 hour to figure it out we can give that water to rat only once. If we use the digital logic 100 numbers can be represented in 8 bits. If we convert each number into a digital number and give the water to those rats only if that position contains 1. Suppose the poison is in the 10th (00001010)bottle we will give the water to the 2nd and 4th rats. After 1 hour, we check which rats have died(represent logic 1) and which have not died(represent logic 0) and we can form back the number and we can figure it out. So, we need 8 rats to figure out the poisoned water bottle.

Que 17: You have a box with 10 pairs of blue socks, 10 pairs of black socks, and 10 pairs of gray socks of the same size. You can pick one sock at a time. How many times do you have to pick the socks from the box to ensure that you have at least one pair of socks with the same color?
Ans 17:
You have a total of 30 pairs of socks. If you pick 30 socks there might be a chance that all the socks are of the same leg. So, if we pick 1 more sock It will ensure that we will have a pair of socks. So we need to pick 31 socks from the box.

Que 18: You have a rectangular sheet with a circular hole anywhere in the sheet, you know the center of the sheet and the center of the hole, How will you divide the sheet into two parts such that their areas are equal?
Ans 18:
If we draw a straight line from the center of the rectangle in any direction it will divide the rectangle into 2 equal halves and the same in the circle. So we can draw a line that passes through the center of both rectangle and circle. This line will divide the sheet into two parts with equal area.

Que 19: You have a balance scale, You need to measure integer weights from 1kg. to 40kgs. only. How many minimum numbers of weights do you need if you can put weights on one side of the scale?
Ans 19:
We have weights that add up to 40 kg. we have 2 choices either we put it on balance or won’t put it. Hence, we can use binary weights which can count from 1 to 40. So the weights will be 20, 21, 22, 23, 24, 25 which is 1, 2, 4, 8, 16, 32. A total of 6 weights are needed.

Que 20: In the previous question, what is the minimum number of weights needed if you can put your weights on both sides of the scale.
Ans 20:
In this question, we can put the weights on either side of the balance scale or don’t put them on the scale. So, you will have 3 options +1, 0, or -1. We can use base 3 combinations of the weights 30, 31, 32, 33 which is 1, 3, 9, 27. A total of 4 weights are needed.

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Que 11: Calculate the number of squares in an 8×8 chessboard.
Ans 11: To calculate the number of squares in an 8×8 chessboard, we will calculate first with smaller numbers like 2×2 and 3×3.
In 2×2 there will be 4+1 = 5 squares.
In 3×3 there will be 9+4+1 = 14 squares.
In 4×4 there will be 16+9+4+1 = 30 squares.
Following the pattern we can calculate for 8×8 chessboard will be 64+49+36+25+16+9+4+1 = 204

Que 12: you have a 3 liter and a 5 liters buckets (Assume unlimited water source available). How will you measure 4 liters of water.
Ans 12:
We can follow the below sequence to get 4 liters of water. We can make the combination of 5 + 5 – 3 – 3 = 4 ltr. Refer to Table 6 for the sequence of activity.

Que 13: Find the angle between the minute and hour hands of the clock at 4:20 PM.
Ans 13:
First calculate the position of the minute hand w.r.t. 12 o’clock at 20 min. = (20*360o/60) = 120o. Then we can calculate the position of hour hand w.r.t 12 o’clock at 4:20 PM, we can write it as 420/60 = 41/3 = 13/3. To convert into the angle (13/3×360o / 12 ) = 130o. The Angle formed by the hour and minute hand will be 10o.

Que 14: You have two sand timers, which can show 4 minutes and 7 minutes respectively. Use both the sand timers(at one time or one after another or any other combination) and measure a time of 9 minutes.
Ans 14: Start the 7-minute sand timer and the 4-minute sand timer.
Once the 4-minute sand timer ends turn it upside down instantly.
Once the 7-minute sand timer ends turn it upside down instantly.
After the 4-minute sand timer ends turn the 7-minute sand timer upside down(it has now a minute of sand in it) So effectively 8 + 1 = 9.

Que 15: You have 9 balls, 8 of which have the same weight. The remaining one is defective and heavier than the rest. You can use a balance scale to compare weights to find which is the defective ball. How many measurements do you need so that you will be surely able to do it?
Ans 15:
We can split the balls into 3 groups with 3 balls in each group. Then we can compare any 2 groups if they are equal then defective ball is in the 3rd group or it will be in the heavier group. Then we will get 3 balls in which 1 is heavier, again we can compare 2 balls if they are equal 3rd ball is defective else heavier ball. So, in total, we need to measure 2 times to find out heavier balls.

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Que 6: You have two non-uniform threads, each takes 30 mins to burn if we burn it from one side. How will you measure 45 minutes with those two threads without cutting or breaking them?
Ans  6: If we burn it from one side, it will take 30 mins. to burn. So if we burn it from both the side it will burn in 15 mins. So, to measure 45 mins. 1st we will burn one thread from one side for 30 mins and immediately after it we will burn 2nd thread from both sides to measure 15 mins. and thus we can measure 45 mins.  

Que 7: You have 10 packets of chocolate, each packet consists of 10 chocolates. Each chocolate is 10gm except in one packet which has 11gms chocolate each, all the packets look similar. Use a digital weighing machine once to find out which packet has 11 gm of chocolates.
Ans  7: As we can use the weighing machine once, we have to make some combination of the chocolates in such a way that one weigh will tell the packet. One such possible combination is to take 1 chocolate from the 1st packet, 2 chocolate from the 2nd packet, and so on.
So, the total chocolates become 550gm. If 1st packet has the 11gm chocolate the measured weight will be 560gm. Likewise, if the nth packet has 11 gm chocolate measured weight will be (550+10n)gm.

Que 8: If a Pen and a pencil cost Rs 10, if the price of a pen is Rs 9 more than that of a pencil. What is the cost of a pen and pencil? 
Ans  8:
Suppose the price of the Pen is x and the price of the pencil is y.
             By the question  x + y = 10 and x – y = 9
             Solving both equations x(Pen) will be Rs 9.5 and y(Pencil) will be Rs 0.5. 

Que 9: There are 25 horses. You can conduct a race among at most 5 to find out their relative speed. At no point, you can find out the actual speed of the horse in a race. Find out how many races are required to get the top 3 fastest horses.
Ans  9: Make 5 groups of 5 horses and run 5 races. Suppose five groups are a,b,c,d, and e and the next alphabet represents its rank in the group(of 5 horses) eg. d3 means horse in group d and has ranked 3rd in his group. [ 5 RACES DONE ] 
a1 b1 c1 d1 e1 
a2 b2 c2 d2 e2 
a3 b3 c3 d3 e3 
a4 b4 c4 d4 e4 
a5 b5 c5 d5 e5 

Now make a race of (a1,b1,c1,d1,e1).[RACE 6 DONE] suppose result is a1>b1>c1>d1>e1 
which implies a1 must be FIRST. 
b1 and c1 MAY BE(but not must be) 2nd and 3rd. 
For the II positions, the horse will be either b1 or a2.
(we have to find the top 3 horses therefore we choose horses b1,b2,a2,a3, and c1 to race among them [RACE 7 DONE]. 
The only possibilities are : 
c1 may be third 
b1 may be the second or third 
b2 may be third 
a2 may be the second or third 
a3 may be third 
The final result will give ANSWER. suppose result is a2>a3>b1>c1>b2 
then answer is a1,a2,a3,b1,c1. 

Que 10: Prove that the number between 2 consecutive odd prime numbers(eg 17 19), is always divisible by 6.
Ans  10: To prove this, we need to prove that the number between two consecutive odd prime numbers is divisible by 2 and 3.
The number between two odd numbers will be even so it will be divisible by 2.
In each consecutive 3 odd number 1 number is always divisible by 3,  let’s say two consecutive odd prime numbers are n and n+2, then n-2 and n+4 will be divisible by 3 and hence the number between them n+1 will also be divisible by 3.

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Que 1: There are few bacteria in a bottle which gets double every day. On the 10th day, that bottle gets full. On which day the bottle was one-fourth.
Ans  1:  If the bacteria is double each day and on the 10th day, the bottle is getting full it means, It was half on the 9th day and one-fourth on the 8th day.

Que 2: 100 coins are lying on a table. 10 of them are heads up and 90 are tails up. You can’t see and feel the coins. How can we split the coins into two piles such that there is the same number of heads up in each pile?
Ans  2: We can make 2 piles of 10 and 90. Suppose if x coins are heads and 10-x tails on the 1st pile then it means there are 10-x heads are there in the 2nd pile. If we flip all the coins in 1st pile, then there will be 10-x heads up which will be equal to heads in 2nd pile.

Que 3: Four persons want to cross a bridge. They take 1min., 2min., 5min. and 10min. to cross the bridge respectively. If only two people can cross the bridge at a time and they need a torch. They have only one torch, so each time they cross they have to bring back that torch. Both the person crossing the bridge should have to walk together. In how much minimum time they all will cross the bridge?
Ans  3: Suppose we have persons a, b, c, and d which take 1min., 2min, 5min. and 10min. to cross the bridge respectively. We will follow the following sequence :

a and b cross the bridge:  2min.
a will return back:              1min.
c and d cross the bridge:   10 min.
b will return back:              2min.
a and b cross the bridge:    2min.
So in total 17 min. they all will cross the bridge.

Que 4: There are three closed boxes. There are apples in one box, oranges in 2nd box, and mixed fruit (apple and orange) in 3rd box. By mistakenly the labels of the items are incorrectly tagged on all the boxes. You can draw any one fruit from any box and label them correctly. 
Ans  4: Pick a fruit from the basket labeled ‘Mix fruit’. We know from the question that this basket does not contain ‘Mix fruit’ in that box.

If this fruit is an apple, then label this Basket as ‘Apple’. Now we’ve determined that the basket labeled as ‘Mix fruit’ only contains Apples.

The basket labeled as ‘Oranges’, but we know that since the label is incorrect, this basket either has only apples in it or has Mix fruit. Since we already know which basket contains only apples, we know that the basket labeled as ‘Oranges’ contains ‘Mix fruit’. So label it as ‘Mix fruit’. The 3rd basket will be labeled as ‘Oranges’.

Que 5: There are three persons at different places on a circular track, They start running in a random direction with the same speed, what is the probability that they will not cross each other?
Ans  5:
Each person can run in two directions clockwise or anticlockwise. So, all the people will have 2 options. So there will be a total of 8 possibilities in which they can run. They won’t cross each other if they all run in the same direction clockwise or anticlockwise. So, In 2 cases out of 8, they will not cross, Probability will be 2/8 = ¼.

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