Analytical Question #9

You have a bucket of jelly beans in three colors – red, green and blue. With your eyes closed, reach in the bucket and take out two jelly beans of the same color. How many jelly beans do you have to take to be certain of getting two of the same color?

Four. Pick just three jelly beans, and it's possible you'd have one of each color and therefore no match. With four jelly beans, at least two have to be the same color.

Analytical Question #8

If you have a 5 quart pail, a 3 quart pail, and infinite supply of water, how would you measure exactly 4 quarts?

A. Fill the 3Q pail with water ==> 5Q[0], 3Q[3]
B. Fill the 5Q pail with the water from the 3Q pail ==> 5Q[3], 3Q[0]
C. Fill the 3Q pail with water ==> 5Q[3], 3Q[3]
D. Fill the 5Q pail with the water from the 3Q pail ==> 5Q[5], 3Q[1]
E. Empty the 5Q pail ==> 5Q[0], 3Q[1]
F. Fill the 5Q pail with the water from the 3Q pail ==> 5Q[1], 3Q[0]
G. Fill the 3Q pail with water ==> 5Q[1], 3Q[3]
H. Fill the 5Q pail with the water from the 3Q pail ==> 5Q[4], 3Q[0]

Analytical Question #7

There are two fuses which burn at different rates among themselves and between themselves. They only burn for one hour. For example, one could burn the entire way except one inch in a minute, then spend the last 59 minutes on that last inch. How can you exactly time 45 minutes using them?

• Light both ends of Fuse 1 and one end of Fuse 2 at the same time.
• Fuse 1 will exactly take 30 minutes to burn out since it started burning on both the ends at the same time.
• Light the other end of Fuse 2 the instant after Fuse 1 burns out.
• At that point Fuse 2 has burned for 30 minutes and has 30 minutes remaining. From this point Fuse 2 will exactly take 15 minutes to burn out as the remaining burning time will be halved.

So the total time is 30 + 15 = 45 minutes.

Analytical Question #6

You have twelve coins which look alike and are of equal weight except for one, which may be heavier or lighter. How can you find the odd coin and if it is heavier or lighter by using a balance and only three weighings?



W = Weighing
Create 3 groups with 3 coins each: G1(ABCD), G2(EFGH), G3(IJKL)


I. W1 ==> G1[ABCD] <> G1 (lighter) or G2 (heavier) has the odd coin
Move G2[EFG] to G4, G1[ABC] to G2, and G3[IJK] to G1
Groups: G1[IJKD], G2[ABCH], G3[L], G4[EFG]
I.A. W2 ==> G1[IJKD] <> G1[D] or G2[H] is the odd coin
I.A.1 W3 ==> G1[D] <> G1[D] is the odd coin and is lighter
I.A.2 W3 ==> G1[D] = G3[L] ==> G2[H] is the odd coin and is heavier
I.A.3 W3 ==> G1[D] > G3[L] ==> Not Possible
I.B. W2 ==> G1[IJKD] = G2[ABCH] ==> G4[EFG] has the odd coin and is heavier
I.B.1 W3 ==> G4[E] <> G4[F] is the odd coin
I.B.2 W3 ==> G4[E] = G4[F] ==> G4[G] is the odd coin
I.B.3 W3 ==> G4[E] > G4[F] ==> G4[E] is the odd coin
I.C. W2 ==> G1[IJKD] > G2[ABCH] ==> G2[ABC] has the odd coin and is lighter
I.C.1 W3 ==> G2[A] <> G2[A] is the odd coin
I.C.2 W3 ==> G2[A] = G2[B] ==> G2[C] is the odd coin
I.C.3 W3 ==> G2[A] > G2[B] ==> G2[B] is the odd coin
----------------------------------------------------------------------------------
II. W1 ==> G1[ABCD] = G2[EFGH] ==> G3[IJKL] has the odd coin
II.A. W2 ==> G1[ABC] <> G3[IJK] has the odd coin and is heavier
II.A.1 W3 ==> G3[I] <> G3[J] is the odd coin
II.A.2 W3 ==> G3[I] = G3[J] ==> G3[K] is the odd coin
II.A.3 W3 ==> G3[I] > G3[J] ==> G3[I] is the odd coin
II.A. W2 ==> G1[ABC] = G3[IJK] ==> G3[L] is the odd coin
II.A.1 W3 ==> G3[I] <> G3[L] is heavier
II.A.2 W3 ==> G3[I] = G3[L] ==> Not Possible
II.A.3 W3 ==> G3[I] > G3[L] ==> G3[L] is lighter
II.A. W2 ==> G1[ABC] > G3[IJK] ==> G3[IJK] has the odd coin and is lighter
II.A.1 W3 ==> G3[I] <> G3[I] is the odd coin
II.A.2 W3 ==> G3[I] = G3[J] ==> G3[K] is the odd coin
II.A.3 W3 ==> G3[I] > G3[J] ==> G3[J] is the odd coin
----------------------------------------------------------------------------------
III. W1 ==> G1[ABCD] > G2[EFGH] ==> G1 (heavier) or G2 (lighter) has the odd coin
Move G2[EFG] to G4, G1[ABC] to G2, and G3[IJK] to G1
Groups: G1[IJKD], G2[ABCH], G3[L], G4[EFG]
III.A. W2 ==> G1[IJKD] <> G2[ABC] has the odd coin and is heavier
III.A.1 W3 ==> G2[A] <> G2[B] is the odd coin
III.A.2 W3 ==> G2[A] = G2[B] ==> G2[C] is the odd coin
III.A.3 W3 ==> G2[A] > G2[B] ==> G2[A] is the odd coin
III.B. W2 ==> G1[IJKD] = G2[ABCH] ==> G4[EFG] has the odd coin and is lighter
III.B.1 W3 ==> G4[E] <> G4[E] is the odd coin
III.B.2 W3 ==> G4[E] = G4[F] ==> G4[G] is the odd coin
III.B.3 W3 ==> G4[E] > G4[F] ==> G4[F] is the odd coin
III.C. W2 ==> G1[IJKD] > G2[ABCH] ==> G1[D] or G2[H] is the odd coin
III.C.1 W3 ==> G1[D] <> Not Possible
III.C.2 W3 ==> G1[D] = G3[L] ==> G2[H] is the odd coin and lighter
III.C.3 W3 ==> G1[D] > G3[L] ==> G1[D] is the odd coin and heavier

Analytical Question #5

There is a game in which 3 players play each other and only one wins. If 81 players have participated in a tournament, how many matches will be played to decide the winner?

Round 1: 81/3 = 27 Games ==> 27 Winners
Round 2: 27/3 = 9 Games ==> 3 Winners
Round 3: 9/3 = 3 Games ==> 3 Winners
Round 4: 3/3 = 1 Game ==> 1 Winner
Total Games Played to decide the Winner = 27 + 9 + 3 + 1 = 40.

OR

Since each game eliminates 2 players and there are 80 players to be eliminated, the total number of games palyed to decide the winner is 80/2 = 40.

Analytical Question #4

You have eight balls of the same size. 7 of them weigh the same, and one of them weighs slightly more. How can you find the ball that is heavier by using a balance and only two weighings?

Choose any 6 balls and weigh 3 against 3

• if they weigh the same, you have another weighing for the remaining 2 balls and you can find the heavier one
• if they don’t weigh the same, choose any 2 balls from the group of 3 which was heavier and weigh them
  ==> if they weigh the same, the remaining ball is the heavier one
  ==> if they don't weigh the same, you just found the heavier one

Analytical Question #3

There is the gun with six chambers and two bullets placed in adjecent chambers. Close the barrel, spin it, and pull the trigger. The slot was found empty. Now if I point the gun against you and want to pull the trigger, which one do you prefer , that I spin the barrel first or that I just pull the trigger?

X X _ _ _ _
1 2 3 4 5 6

• When fired initially, the position should have been any of 3, 4, 5, 6.Had it been 6, I will get hit next time , otherwise not. i.e. out of next 4 new positions 4, 5, 6, and 1, the chances that I get hit is 1/4.
• If spun again the chance I get hit is 2/6 = 1/3.

So I would go with out spinning

Analytical Question #2

A train leaves Los Angeles at 15mph heading for New York. Another train leaves New York at 20mph heading for Los Angeles on the same track. If a bird, flying at 25mph, leaves from Los Angeles at the same time as the train and flies back and forth between the two trains until they collide. How far will the bird have traveled?

Distance between NY and LA = d miles
Time taken for the trains to collide = d/(15+20) hours
Distance travalled by the bird in that time = (d/35)*25 = 5d/7

Analytical Question #1

You've got someone working for you for seven days and a gold bar to pay them. The gold bar is segmented into seven connected pieces. You must give them a piece of gold at the end of every day. If you are only allowed to make two breaks in the gold bar, how do you pay your worker?

• Day 1 : Break (1) a single piece from the bar and pay it to the worker.
• Day 2 : Break (2) a two piece from the bar and pay the worker in exchange for the single piece.
• Day 3 : Pay the single piece to the worker.
• Day 4 : Pay the four piece to the worker inexchange for the 3 pieces (2 piece + 1 piece)
• Day 5: Pay the single piece to the worker.
• Day 6: Pay the two piece to the worker in exchange for the single piece.
• Day 7: Pay the single piece to the worker.