### Brain Teasers

At what time immediately prior to Six O'clock the hands of the clock are exactly opposite to each other.

Give the exact time in hours, minutes and seconds.

**Answer**

**4 hrs. 54 min. 32.74 seconds**

It is obvious that between 5 O'clock and 6 O'clock the hands will not be exactly opposite to each other. It is also obvious that

the hands will be opposite to each other just before 5 O'clock. Now to find exact time:

The hour hand moves 1 degree for every 12 degrees that the minute hand moves. Let the hour hand be X degree away from 5

O'clock. Therefore the minute hand is 12X degree away from 12 O'clock.

Therefore solving for X

Angle between minute hand and 12 O'clock + Angle between 12 O'clock and 4 O'clock + Angle between 4 O'clock and hour

hand = 180

12X + 120 + (30-X) = 180

11X = 30

Hence X = 30/11 degrees

(hour hand is X degree away from 5 O'clock)

Now each degree the hour hand moves is 2 minutes.

Therefore minutes are

= 2 * 30/11

= 60/11

= 5.45 (means 5 minutes 27.16 seconds)

Therefore the exact time at which the hands are opposite to each other is

= 4 hrs. 54 min. 32.74 seconds

Brain Teaser No : 00002

Ali Baba had four sons, to whom he bequeathed his 39 camels, with the proviso that the legacy be divided in the following way :

The oldest son was to receive one half the property, the next a quarter, the third an eighth and the youngest one tenth. The four

brothers were at a loss as how to divide the inheritance among themselves without cutting up a camel, until a stranger appeared

upon the scene.

Dismounting from his camel, he asked if he might help, for he knew just what to do. The brothers gratefully accepted his offer.

Adding his own camel to Ali Baba's 39, he divided the 40 as per the will. The oldest son received 20, the next 10, the third 5 and

the youngest 4. One camel remained : this was his, which he mounted and rode away.

Scratching their heads in amazement, they started calculating. The oldest thought : is not 20 greater than the half of 39? Someone

must have received less than his proper share ! But each brother discovered that he had received more than his due. How is it

possible?

**Answer**

They took their percentages from 40 and not from 39, so they got more than their share.

The oldest son got 1/2 of 40 = 20 which is 0.5 more

The second son got 1/4 of 40 = 10 which is 0.25 more

The third son got 1/8 of 40 = 5 which is 0.125 more

The youngest son got 1/10 of 40 = 4 which is 0.1 more

And the stranger got 1/40 of 40 = 1 which is 0.025 more (As he is not supposed to get anything)

All these fractions add to = 0.5 + 0.25 + 0.125 + 0.1 + 0.025 = 1 which stranger took away.

Brain Teaser No : 00003

In a hotel, rooms are numbered from 101 to 550. A room is chosen at random. What is the probability that room number starts with

1, 2 or 3 and ends with 4, 5 or 6?

**Answer**

There are total 450 rooms.

Out of which 299 room number starts with either 1, 2 or 3. (as room number 100 is not there) Now out of those 299 rooms only

90 room numbers end with 4, 5 or 6

So the probability is 90/450 i.e. 1/5 or 0.20

Brain Teaser No : 00005

A man went into a fast food restaurant and ate a meal costing Rs. 105, giving the accountant a Rs. 500 note. He kept the change,

came back a few minutes later and had some food packed for his girl friend. He gave the accountant a Rs. 100 note and received

Rs. 20 in change. Later the bank told the accountant that both the Rs. 500 and the Rs. 100 notes were counterfeit.

How much money did the restaurant lose? Ignore the profit of the food restaurant.

**Answer**

**He lost Rs.600**

First time restaurant has given food worth Rs.105 and Rs. 395 change. Similarly second time, food worth Rs.80 and Rs.20

change. Here, we are not considering food restaurant profits.

Brain Teaser No : 00006

You are working in a store that stocks bangles. Three boxes of bangles have been incorrectly labeled. The labels say Red Bangles,

Green Bangles and Red & Green Bangles.

How can you re-label the boxes correctly, by taking only one bangle from one box?

**Answer**

Keep in mind that boxes are incorrectly labeled.

Take out one bangle from the box labeled "Red & Green Bangles". There are 2 possibilities:

If that bangle is Red, it means that box contains Red Bangles. The box labeled as "Green Bangles" contains Red & Green

Bangles. And box labeled "Red Bangles" contains Green bangles.

RED GREEN BOX --- Red Bangles

RED BOX --- Green Bangles

GREEN BOX --- Red & Green Bangles

If that bangle is Green, it means that box contains Green

Bangles. The box labeled as "Green Bangles" contains Red

Bangles. And box labeled "Red Bangles" contains Red &

Green Bangles.

RED GREEN BOX --- Green Bangles

RED BOX --- Red & Green Bangles

GREEN BOX --- Red Bangles

Brain Teaser No : 00007

On one side of a card is written :

"THE SENTENCE ON THE OTHERSIDE OF THIS CARD IS TRUE."

On turning the card over you find:

"THE SENTENCE ON THE OTHERSIDE OF THIS CARD IS FALSE."

Which sentence is true?

**Answer**

It's a Paradox. Both the sentences are contradictory to each other. If you say that the first sentence is true, then the second will

contradict it and vice versa.

Brain Teaser No : 00008

3 blocks are chosen randomly on a chessboard. What is the probability that they are in the same diagonal?

**Answer**

There are total of 64 blocks on a chessboard. So 3 blocks can be chosen out of 64 in 64C3 ways.

So the sample space is = 41664

There are 2 diagonal on chessboard each one having 8 blocks. Consider one of them.

3 blocks out of 8 blocks in diagonal can be chosen in 8C3 ways.

But there are 2 such diagonals, hence favourables = 2 * 8C3 = 2 * 56 = 112

The require probability is

= 112 / 41664

= 1 / 372

= 0.002688

Brain Teaser No : 00010

In a contest of intelligence, three problems A, B and C were posed.

Among the contestants there were 25 who solved at least one problem each.

Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C.

The number of participants who solved only problem A was one more than the number who solved problem A and at least

one other problem.

Of all students who solved just one problem, half did not solve problem A.

How many students solved only problem B?

**Answer**

**6 students solved only problem B**

X => Students who solved only problem A

Y => Students who solved only problem B

Z => Students who solved only problem C

P => Students who solved both problem B and problem C

From 4 :

Students who solved only problem A = Students who solved only problem B + Students who solved only problem C

X = Y + Z

From 3 :

Students who solved problem A and at least one other = X - 1

From 2 :

(Y + P) = 2 * (Z + P)

Y + P = 2 * Z + 2 * P

Z = (Y - P) / 2

From 1 and Figure:

X + X - 1 + Y + Z + P = 25

2*X + Y + Z + P = 26

2*(Y + Z) + Y + Z + P = 26 (from 4)

3*Y + 3*Z + P = 26

3*Y + 3* (Y - P) / 2 + P = 26 (from 2)

6*Y + 3*Y - 3*P + 2*P = 52

9*Y - P = 52

Y = (52 + P) / 9

Now, it is obvious that all values are integer. Hence, P must be 2 and Y must be 6.

So 6 students solved only problem B.

Brain Teaser No : 00011

When Alexander the Great attacked the forces of Porus, an Indian soldier was captured by the Greeks. He had displayed such

bravery in battle, however, that the enemy offered to let him choose how he wanted to be killed. They told him, "If you tell a lie, you

will put to the sword, and if you tell the truth you will be hanged."

The soldier could make only one statement. He made that statement and went free. What did he say?

**Answer**

The soldier said, "You will put me to the sword."

The soldier has to say a Paradox to save himself. If his statement is true, he will be hanged, which is not the sword and hence

false. If his statement is false, he will be put to the sword, which will make it true. A Paradox !!!

Brain Teaser No : 00014

Five horses ran in the race.

There were no ties.

Sikandar did not come first.

Star was neither first nor last.

Mughal Glory came in one place after Sikandar.

Zozo was not second.

Rangila was two place below Zozo.

In what order did the horses finish?

**Answer**

It's simple.

Let's find the possible places horses can finish. Possibilities are:

Sikandar - 2,3,4 (not 5th as Mughal Glory came one place after him)

Star - 2,3,4

Mughal Glory - 3,4,5

Zozo - 1,3 (not 4th & 5th as Rangila is two place after him)

Rangila - 3,5

So the result is:

1 Zozo

2 Star

3 Rangila

4 Sikandar

5 Mughal Glory

Brain Teaser No : 00015

In the town called Alibaug, the following facts are true:

No two inhabitants have exactly the same number of hairs.

No inhabitants has exactly 2025 hairs.

There are more inhabitants than there are hairs on the head of any one inhabitants.

What is the largest possible number of the inhabitants of Alibaug?

**Answer**

**2025**

It is given that no inhabitants have exactly 2025 hairs. Hence there are 2025 inhabitants with 0 to 2024 hairs in the head.

Suppose there are more than 2025 inhabitants. But these will violate the condition that "There are more inhabitants than there

are hairs on the head of any one inhabitants." As for any number more than 2025, there will be same number of inhabitants as

the maximum number of hairs on the head of any inhabitant.

Brain Teaser No : 00016

At what time after 4.00 p.m. is the minutes hand of a clock exactly aligned with the hour hand?

**Answer**

**4:21:49.5**

Assume that X minutes after 4.00 PM minute hand exactly aligns with and hour hand.

For every minute, minute hand travels 6 degrees.

Hence, for X minutes it will travel 6 * X degrees.

For every minute, hour hand travels 1/2 degrees.

Hence, for X minutes it will travel X/2 degrees.

At 4.00 PM, the angle between minute hand and hour hand is 120 degrees. Also, after X minutes, minute hand and hour hand

are exactly aligned. So the angle with respect to 12 i.e. Vertical Plane will be same. Therefore,

6 * X = 120 + X/2

12 * X = 240 + X

11 * X = 240

X = 21.8182

X = 21 minutes 49.5 seconds

Hence, at 4:21:49.5 minute hand is exactly aligned with the hour hand.

Brain Teaser No : 00017

A card contains following three sentences:

a. THIS SENTENCE CONTAINS FIVE WORDS.

b. THIS SENTENCE CONTAINS TWO VERBS.

c. EXACTLY ONE SENTENCE ON THIS CARD IS TRUE.

Is sentence C true or false?

**Answer**

It's a paradox.

You cant say it true or false, as your answer will contradict itself.

Brain Teaser No : 00018

A barber in a certain small town shaves all the men who do not shave themselves, but never shaves any who do shave

themselves.

Does the barber shave himself? Note that the barber is a man.

**Answer**

It's a paradox.

You can't say it true or false, as your answer will contradict itself.

Brain Teaser No : 00021

A rich man died. In his will, he has divided his gold coins among his 5 sons, 5 daughters and a manager.

According to his will: First give one coin to manager. 1/5th of the remaining to the elder son. Now give one coin to the manager and

1/5th of the remaining to second son and so on..... After giving coins to 5th son, divided the remaining coins among five daughters

equally.

All should get full coins. Find the minimum number of coins he has?

**Answer**

We tried to find out some simple mathematical method and finally we wrote small C program to find out the answer. The

answer is 3121 coins.

Here is the breakup:

First son = 624 coins

Second son = 499 coins

Third son = 399 coins

Forth son = 319 coins

Fifth son = 255 coins

Daughters = 204 each

Manager = 5 coins

Brain Teaser No : 00022

There were N stations on a railroad. After adding X stations 46 additional tickets have to be printed.

Find N and X.

**Answer**

Let before adding X stations, total number of tickets

t = N(N-1)

After adding X stations total number of tickets are

t + 46 = (N+X)(N+X-1)

Subtracting 1st from 2nd

46 = (N+X)(N+X-1) - N(N-1)

46 = N2 + NX - N + NX + X2 - X - N2 + N

46 = 2NX + X2 - X

46 = (2N - 1)X + X2

X2 + (2N - 1)X - 46 = 0

Now there are only two possible factors of 46. They are (46,1) and (23,2)

Case I: (46,1)

2N - 1 = 45

2N = 46

N = 23

And X = 1

Case II: (23,2)

2N - 1 = 21

2N = 22

N = 11

And X = 2

Hence, there are 2 possible answers.

Brain Teaser No : 00023

There is a grid of 20 squares by 10 squares. How many different rectangles are possible?

Note that square is a rectangle.

**Answer**

**11550**

The Generic solution to this is:

Total number of rectangles = (Summation of row numbers) * (Summation of column numbers)

Here there are 20 rows and 10 columns or vice versa. Hence, total possible rectangles

= ( 20 + 19 + 18 + 17 + 16 + .... + 3 + 2 + 1 ) * ( 10 + 9 +8 + 7 + .... + 3 + 2 + 1)

= ( 210 ) * (55)

= 11550

Hence, total 11,550 different rectangles are possible.

If you don't believe it, try formula on some smaller grids like 4x2, 3x2, 3x3 etc...

Brain Teaser No : 00024

A person wanted to withdraw X rupees and Y paise from the bank. But cashier made a mistake and gave him Y rupees and X

paise. Neither the person nor the cashier noticed that.

After spending 20 paise, the person counts the money. And to his surprise, he has double the amount he wanted to withdraw.

Find X and Y. (1 Rupee = 100 Paise)

**Answer**

As given, the person wanted to withdraw 100X + Y paise.

But he got 100Y + X paise.

After spending 20 paise, he has double the amount he wanted to withdraw. Hence, the equation is

2 * (100X + Y) = 100Y + X - 20

200X + 2Y = 100Y +X - 20

199X - 98Y = -20

98Y - 199X = 20

Now, we got one equation; but there are 2 variables. We have to apply little bit of logic over here. We know that if we

interchange X & Y, amount gets double. So Y should be twice of X or one more than twice of X i.e. Y = 2X or Y = 2X+1

Case I : **Y=2X**

Solving two equations simultaneously

98Y - 199X = 20

Y - 2X = 0

We get X = - 20/3 & Y = - 40/2

Case II : **Y=2X+1**

Solving two equations simultaneously

98Y - 199X = 20

Y - 2X = 1

We get X = 26 & Y = 53

Now, its obvious that he wanted to withdraw Rs. 26.53

Brain Teaser No : 00025

What is the remainder left after dividing 1! + 2! + 3! +
+ 100! By 7?

Think carefully !!!

**Answer**

A tricky one.

7! onwards all terms are divisible by 7 as 7 is one of the factor. So there is no remainder left for those terms i.e. remainder left

after dividing 7! + 8! + 9! + ... + 100! is 0.

The only part to be consider is

= 1! + 2! + 3! + 4! + 5! + 6!

= 1 + 2 + 6 + 24 + 120 + 720

= 873

The remainder left after dividing 873 by 7 is 5

Hence, the remainder is 5.

Brain Teaser No : 00026

Find the last digit of summation of the series:

199 + 299 + 399 + 499 +
+ 9899 + 9999

**Answer**

**The last digit of the series is 0.**

We group the sum as follow:

(199 + 1199 + ... + 9199) + (299 + 2299 + ... 9299) + ...... + (999 + 1999 + ... + 9999) + (1099 + 2099 + 3099 + ... 9099)

All the terms in a single group have the same last digit (i.e. last digits of 199 + 1199 + ... + 9199 are same, is 1, & similarly for the

other groups).

Also, there are 10 terms in each group except for the last one. Therefore the last digit of the sum of terms in first 9 groups is 0.

(as whatever be the last digit, we have to multiply it by 10) And the last digit of the sum of the terms in the group 10 is obviously

0.

Hence, the last digit of the series is 0.

Brain Teaser No : 00027

Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger

than the original number.

**Answer**

The answer is **285714**.

If its rightmost digit is placed at its left end, then new number is 428571 which is 50% larger than the original number 285714.

The simplest way is to write a small program. And the other way is trial and error !!!

Brain Teaser No : 00028

There are 10 boxes containing 10 balls each. 9 boxes contain 10 balls of 10 kg each and one box contains 10 balls of 9 kg each.

Tool is available for proper weighing. How can you find out the box containing 9 kg balls?

You are allowed to weigh only once. You can remove balls from the boxes. All balls are of same size and color.

**Answer**

1. Mark the boxes with numbers 1, 2, 3, 4, ... upto 10

2. Take 1 ball from box 1, take 2 balls from box 2, take 3 balls from box 3, take 4 balls from box 4 and so on

3. Put all of them on the scale at once and take the measurement.

4. Now, subtract the measurement from 550 ( 1*10 + 2*10 + 3*10 + 4*10 + 5*10 + 6*10 + 7*10 + 8*10 + 9*10 + 10*10)

5. The result will give you the box number which has a ball of 9 Kg

Brain Teaser No : 00029

A fly is flying between two trains, each travelling towards each other on the same track at 60 km/h. The fly reaches one engine,

reverses itself immediately, and flies back to the other engine, repeating the process each time.

The fly is flying at 90 km/h. If the fly flies 180 km before the trains meet, how far apart were the trains initially?

**Answer**

**Initially, the trains were 240 km apart.**

The fly is flying at the speed of 90 km/h and covers 180 km. Hence, the fly flies for 2 hours after trains started.

It's obvious that trains met 2 hours after they started travelling towards each other. Also, trains were travelling at the speed of

60 km/h. So, each train traveled 120 km before they met.

Hence, the trains were 240 km apart initially.

Brain Teaser No : 00030

A certain street has 1000 buildings. A sign-maker is contracted to number the houses from 1 to 1000. How many zeroes will he

need?

**Answer**

**The sign-maker will need 192 zeroes.**

Divide 1000 building numbers into groups of 100 each as follow:

(1..100), (101..200), (201..300), ....... (901..1000)

For the first group, sign-maker will need 11 zeroes.

For group numbers 2 to 9, he will require 20 zeroes each.

And for group number 10, he will require 21 zeroes.

The total numbers of zeroes required are

= 11 + 8*20 + 21

= 11 + 160 + 21

= 192

Brain Teaser No : 00031

Find sum of digits of D.

Let

A= 19991999

B = sum of digits of A

C = sum of digits of B

D = sum of digits of C

(HINT : A = B = C = D (mod 9))

**Answer**

**The sum of the digits od D is 1.**

Let E = sum of digits of D.

It follows from the hint that A = E (mod 9)

Consider,

A = 19991999

< 20002000

= 22000 * 10002000

= 1024200 * 106000

< 10800 * 106000

= 106800

i.e. A < 106800

i.e. B < 6800 * 9 = 61200

i.e. C < 5 * 9 = 45

i.e. D < 2 * 9 = 18

i.e. E <= 9

i.e. E is a single digit number.

Also,

1999 = 1 (mod 9)

so 19991999 = 1 (mod 9)

Therefore we conclude that E=1.

Brain Teaser No : 00032

Find the smallest number N which has the following properties:

1. its decimal representation has 6 as the last digit.

2. If the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as great as the

original number N.

**Answer**

**The smallest such number is 153846.**

Assume that the number N is

N = BnBn-1Bn-2 ... B3B26

as its given that 6 is the last digit.

Now after erasing 6 and putting it in front of the remaining digits, we get

Nnew = 6BnBn-1Bn-2 ... B3B2

Also given that Nnew is 4 times the N. Also note that the last digit Nnew is second last digit of N and so on. The required result is

BnBn-1Bn-2 ... B3B26

X 4

--------------------

6BnBn-1Bn-2 ... B3B2

So start multiplying and put nth digit of Nnew to (n + 1)th digit of N and you will get result as

1 5 3 8 4 6

X 4

---------------

6 1 5 3 8 4

Hence, **the number is 153846**

Brain Teaser No : 00033

There are 9 coins. Out of which one is odd one i.e weight is less or more. How many iterations of weighing are required to find odd

coin?

**Answer**

It is always possible to find odd coin in 3 weighings and to tell whether the odd coin is heavier or lighter.

1. Take 8 coins and weigh 4 against 4.

o If both are not equal, goto step 2

o If both are equal, goto step 3

2. One of these 8 coins is the odd one. Name the coins on heavier side of the scale as H1, H2, H3 and H4. Similarly,

name the coins on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter.

Weigh (H1, H2, L1) against (H3, H4, X) where X is one coin remaining in intial weighing.

o If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.

If both are equal, L4 is the odd coin and is lighter.

If L2 is light, L2 is the odd coin and is lighter.

If L3 is light, L3 is the odd coin and is lighter.

o If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2

If both are equal, there is some error.

If H1 is heavy, H1 is the odd coin and is heavier.

If H2 is heavy, H2 is the odd coin and is heavier.

o If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4

If both are equal, L1 is the odd coin and is lighter.

If H3 is heavy, H3 is the odd coin and is heavier.

If H4 is heavy, H4 is the odd coin and is heavier.

3. The remaining coin X is the odd one. Weigh X against the anyone coin used in initial weighing.

o If both are equal, there is some error.

o If X is heavy, X is the odd coin and is heavier.

o If X is light, X is the odd coin and is lighter.

Brain Teaser No : 00034

My friend collects antique stamps. She purchased two, but found that she needed to raise money urgently. So she sold them for

Rs. 8000 each. On one she made 20% and on the other she lost 20%.

How much did she gain or lose in the entire transaction?

**Answer**

**She lost Rs 666.67**

Consider the first stamp. She mades 20% on it after selling it for Rs 8000.

So the original price of first stamp is

= (8000 * 100) / 80

= Rs 6666.67

Similarly, consider second stamp. She lost 20% on it after selling it for Rs 8000

So the original price of second stamp is

= (8000 * 100) / 80

= Rs 10000

Total buying price of two stamps

= Rs 6666.67 + Rs 10000

= Rs 16666.67

Total selling price of two stamps

= Rs 8000 + Rs 8000

= Rs 16000

Hence, she lost Rs 666.67

Brain Teaser No : 00035

In a sports contest there were m medals awarded on n successive days (n > 1).

1. On the first day 1 medal and 1/7 of the remaining m - 1 medals were awarded.

2. On the second day 2 medals and 1/7 of the now remaining medals was awarded; and so on.

3. On the nth and last day, the remaining n medals were awarded.

How many days did the contest last, and how many medals were awarded altogether?

**Answer**

**Total 36 medals were awarded and the contest was for 6 days.**

On day 1: Medals awarded = (1 + 35/7) = 6 : Remaining 30 medals

On day 2: Medals awarded = (2 + 28/7) = 6 : Remaining 24 medals

On day 3: Medals awarded = (3 + 21/7) = 6 : Remaining 18 medals

On day 4: Medals awarded = (4 + 14/7) = 6 : Remaining 12 medals

On day 5: Medals awarded = (5 +7/7) = 6 : Remaining 6 medals

On day 6: Medals awarded 6

I got this answer by writing small program. If anyone know any other simpler method, do submit it.

Brain Teaser No : 00036

A number of 9 digits has the following properties:

The number comprising the leftmost two digits is divisible by 2, that comprising the leftmost three digits is divisible by 3,

the leftmost four by 4, the leftmost five by 5, and so on for the nine digits of the number i.e. the number formed from the

first n digits is divisible by n, 2<=n<=9.

Each digit in the number is different i.e. no digits are repeated.

The digit 0 does not occur in the number i.e. it is comprised only of the digits 1-9 in some order.

Find the number.

**Answer**

The answer is **381654729**

One way to solve it is Trial-&-Error. You can make it bit easier as odd positions will always occupy ODD numbers and even

positions will always occupy EVEN numbers. Further 5th position will contain 5 as 0 does not occur.

The other way to solve this problem is by writing a computer program that systematically tries all possibilities.

Brain Teaser No : 00038

The population of an island consists of two and only two types of people : the knights, who invariably tell the truth and the knaves

who always lie.

three of the inhabitants called X, Y and Z were standing together. A newcomer to the island asked, "Are you a knight or a

knave?" X mumbled his answer rather indistinctly, so the stranger could not quite make out what he had said. The

stranger than asked Y, "What did X say?" Y replied, "X said that he was a knave." Whereupon Z said, "Don't believe Y,

he's lying." What are Y and Z?

Suppose that the stranger asked X, instead, "How many knights among you?" Again X replies indistinctly. So the stranger

asks Y, "What did X say?" Y replies, "X said that there is one knight among us." Then Z says, "Don't believe Y, he is lying!"

Now what are Y and Z?

There are only two inhabitants, X and Y. X says, "At least one of us is a knave." What are X and Y?

Suppose X says, "Either I am a knave, or Y is a knight?" What are X and Y?

Consider once more X, Y and Z each of who is either a knight or a knave. X says, "All of us are knaves." Y says, "Exactly

one of us is a knight." What are X, Y and Z?

**Answer**

**Teaser 1 : **A Simple one. The statement made by Y is false - "X said that he was a knave".

**Case 1 Case 2 Case 3 Case 4**

**X **Knight Knight Knave Knave

**Y **Knight Knave Knight Knave

Analyse the above 4 cases. In all the cases statement made by Y is contradicory and therefore false. Hence, Y is Knave and Z

is Knight.

**Teaser 2 : **Again the statement made by Y is false - "X said that there is one knight among us". Analyse these statement with 4

possible cases as above. In all the cases statement made by Y is false. Hence, Y is Knave and Z is Knight.

**Teaser 3 : **X is Knight and Y is Knave.

**Teaser 4 : **Both are Knight.

**Teaser 5 : **X and Z are Knaves, Y is Knight.

Brain Teaser No : 00039

Find next number in the series :

3, 7, 31, 211, ?

**Answer**

**1831**

All the numbers in the series are Prime Numbers. So the next number will also be a prime number.

**Two consecutive numbers in the**

**series**

**(A and B)**

**Total Prime numbers bewtween A**

**and B**

**(C)**

**Prime number just before**

**B**

**(D)**

**E = (C +**

**D)**

3 and 7 1 5 6

7 and 31 6 29 35

31 and 211 35 199 234

Number after 7 is the prime number on skipping 6 prime numbers after 7 i.e. 31

Number after 31 is the prime number on skipping 35 prime numbers after 31 i.e. 211

Hence, number after 211 is the prime number on skipping 234 prime numbers after 211 i.e. **1831**

**The other possible answer is 1891.**

Subtract 1 from each number in the given series: 2, 6, 30, 210

First number = 2*1 = 2

Second number = 2*3 = 6

Third number = 6*5 = 30

Fourth number = 30*7 = 210

Fifth number = 210*9 = 1890

Sixth number = 1890*11 = 20790

Thus, the pattern is : multiply previous number by next odd number and add one to the multiplication. Thus, the series is 3, 7,

31, 211, 1891, 20791, ...

Thanks to N. Anand for this much more simpler answer.

Brain Teaser No : 00040

Assume for a moment that the earth is a perfectly uniform sphere of radius 6400 km. Suppose a thread equal to the length of the

circumference of the earth was placed along the equator, and drawn to a tight fit.

Now suppose that the length of the thread is increased by 12 cm, and that it is pulled away uniformly in all directions.

By how many cm. will the thread be separated from the earth's surface?

**Answer**

The cicumference of the earth is

= 2 * PI * r

= 2 * PI * 6400 km

= 2 * PI * 6400 * 1000 m

= 2 * PI * 6400 * 1000 * 100 cm

= 1280000000 * PI cm

where r = radius of the earth, PI = 3.141592654

Hence, the length of the thread is = 1280000000 * PI cm

Now length of the thread is increasd by 12 cm. So the new length is = (1280000000 * PI) + 12 cm

This thread will make one concentric circle with the earth which is slightly away from the earth. The circumfernce of that circle is

nothing but (1280000000 * PI) + 12 cm

Assume that radius of the outer circle is R cm

Therefore,

2 * PI * R = (1280000000 * PI) + 12 cm

Solving above equation, R = 640000001.908 cm

Radius of the earth is r = 640000000 cm

Hence, the thread will be separatedfrom the earth by

= R - r cm

= 640000001.908 - 640000000

= 1.908 cm

Brain Teaser No : 00041

A polygon has 1325 diagonals. How many vertices does it have? taken from www.johnsjm.blogspot.com

**Answer**

The formula to find number of diagonals (D) given total number of vertices or sides (N) is

N * (N - 3)

D = -----------

2

Using the formula, we get

1325 * 2 = N * (N - 3)

N2 - 3N - 2650 = 0

Solving the quadratic equation, we get N = 53 or -50

It is obvious that answer is **53 **as number of vertices can not be negative.

Alternatively, you can derive the formula as triange has 0 diagonals, quadrangel has 2, pentagon has 5, hexagon has 9 and so

on......

Hence the series is 0, 0, 0, 2, 5, 9, 14, ........ (as diagram with 1,2 or 3 vertices will have 0 diagonals).

Using the series one can arrive to the formula given above.

Brain Teaser No : 00042

An emergency vehicle travels 10 miles at a speed of 50 miles per hour.

How fast must the vehicle travel on the return trip if the round-trip travel time is to be 20 minutes?

**Answer**

**75 miles per hour**

While going to the destination, the vehicle travels 10 mils at the speed of 50 miles per hour. So the time taken to travel 10 miles

is

= (60 * 10) / 50

= 12 minutes

Now it's given that round-trip travel time is 20 minutes. So the vehicle should complete its return trip of 10 miles in 8 minutes.

So the speed of the vehicle must

= (60 * 10) / 8

= 75 miles per hour