(Puzzles) Challenging Mathematical Puzzles (29)
Puzzles : Challenging Mathematical Puzzles
89. There are four groups of Mangoes, Apples and Bananas
as follows:
Group I : 1 Mango, 1 Apples and 1 Banana
Group II : 1 Mango, 5 Apples and 7 Bananas
Group III : 1 Mango, 7 Apples and 10 Bananas
Group IV : 9 Mango, 23 Apples and 30 Bananas
Group II costs Rs 300 and Group III costs Rs 390.
Can you tell how much does Group I and Group IV cost?
Answer: Group I costs Rs 120 and Group IV costs Rs 1710
Assume that the values of one mango, one apple and one banana are M, A and B
respectively.
From Group II : M + 5A + 7B = 300
From Group III : M + 7A + 10B = 390
Subtracting above to equations : 2A + 3B = 90
For Group I :
= M + A + B
= (M + 5A + 7B)  (4A + 6B)
= (M + 5A + 7B)  2(2A + 3B)
= 300  2(90)
= 300  180
= 120
Similarly, for Group IV :
= 9M + 23A + 30B
= 9(M + 5A + 7B)  (22A + 33B)
= 9(M + 5A + 7B)  11(2A + 3B)
= 9(300)  11(90)
= 2700  990
= 1710
Thus, Group I costs Rs 120 and Group IV costs Rs 1710
90. TicTacToe is being played. One 'X' has been placed
in one of the corners. No 'O' has been placed yet.
Where does the player that is playing 'O' has to put his first 'O' so that 'X'
doesn't win?
Assume that both players are very intelligent. Explain your answer
Answer : 
"O" should be placed in the center.
Let's number the positions as:
1  2  3

4  5  6

7  8  9
It is given that "X" is placed in one of the corner position. Let's
assume that its at position 1.
Now, let's take each position one by one.
* If "O" is placed in position 2, "X" can always win by
choosing position 4, 5 or 7.
* If "O" is placed in position 3, "X" can always win by
choosing position 4, 7 or 9.
* If "O" is placed in position 4, "X" can always win by
choosing position 2, 3 or 5.
* If "O" is placed in position 6, "X" can always win by
choosing position 3, 5 or 7.
* If "O" is placed in position 7, "X" can always win by
choosing position 2, 3 or 9.
* If "O" is placed in position 8, "X" can always win by
choosing position 3, 5 or 7.
* If "O" is placed in position 9, "X" can always win by
choosing position 3, or 7.
If "O" is placed in position 5 i.e. center position, "X"
can't win unless "O" does something foolish ;))
Hence, "O" should be placed in the center.
91. Amit, Bhavin, Himanshu and Rakesh are sitting around
a table.
 The Electronics Engineer is sitting to the left of the Mechanical
Engineer.
 Amit is sitting opposite to Computer Engineer.
 Himanshu likes to play Computer Games.
 Bhavin is sitting to the right of the Chemical Engineer.
Can you figure out everyone's profession?
Answer : 
Amit is the Mechanical Engineer. Bhavin is the Computer Engineer. Himanshu and
Rakesh are either Chemical Engineer or Electronics Engineer.
Amit and Bhavin are sitting opposite to each other. Whereas Chemical Engineer
and Electronics Engineer are sitting opposite to each other.
We cannot find out who is Chemical Engineer and Electronics Engineer as data
provided is not sufficient
92. Five friends with surname Batliwala, Pocketwala,
Talawala, Chunawala and Natakwala have their first name and middle name as
follow.
1. Four of them have a first and middle name of Paresh.
2. Three of them have a first and middle name of Kamlesh.
3. Two of them have a first and middle name of Naresh.
4. One of them have a first and middle name of Elesh.
5. Pocketwala and Talawala, either both are named Kamlesh or neither is named
Kamlesh.
6. Either Batliwala and Pocketwala both are named Naresh or Talawala and
Chunawala both are named Naresh.
7. Chunawala and Natakwala are not both named Paresh.
Who is named Elesh?
Answer : 
Pocketwala is named Elesh.
From (1) and (7), it is clear that Batliwala, Pocketwala and Talawala are named
Paresh.
From (6) and (5), if Pocketwala or Talawala both are named Kamlesh, then either
of them will have three names i.e. Paresh, Kamlesh and Naresh. Hence, Pocketwala
and Talawala both are not named Kamlesh. It means that Batliwala, Chunawala and
Natakwala are named Kamlesh.
Now it is clear that Talawala and Chunawala are named Naresh. Also, Pocketwala
is named Elesh.
93. Mr. Wagle goes to work by a bus. One day he falls
asleep when the bus still has twice as far to go as it has already gone.
Halfway through the trip he wakes up as the bus bounces over some bad potholes.
When he finally falls asleep again, the bus still has half the distance to go
that it has already traveled. Fortunately, Mr. Wagle wakes up at the end of his
trip.
What portion of the total trip did Mr. Wagle sleep?
Answer : 
Mr. wagle slept through half his trip.
Let's draw a timeline. Picture the bus route on a line shown below:
 ________  ________________
Start 1/3 1/2 2/3 End
 shows time for which Mr. Wagle was not sleeping
_____ shows time for which Mr. Wagle was sleeping
When Mr. Wagle fell asleep the first time, the bus sill had twice as far to go
as it had already gone, that marks the first third of his trip.
He wake up halfway through the trip i.e slept from 1/3 mark to the 1/2 mark. He
fell sleep again when the bus still had half the distance to go that it had
already traveled i.e 2/3 mark.
Adding up, all sleeping times,
= (1/2  1/3) + (1  2/3)
= 1/6 + 1/3
= 1/2
Hence, Mr. wagle slept through half his trip.
94. In your sock drawer, you have a ratio of 5 pairs of
blue socks, 4 pairs of brown socks, and 6 pairs of black socks.
In complete darkness, how many socks would you need to pull out to get a
matching pair of the same color?
Answer :  4
You have a bucket of jelly beans. Some are red, some are
blue, and some green. With your eyes closed, pick out 2 of a like color.
How many do you have to grab to be sure you have 2 of the same?
Answer : 
You have a bucket of jelly beans. Some are red, some are blue, and some green.
With your eyes closed, pick out 2 of a like color.
How many do you have to grab to be sure you have 2 of the same?
If you select 4 Jelly beans you are guaranteed that you will have 2 that are the
same color.
95. There are 70 employees working with BrainVista of which 30 are females. Also,
 30 employees are married
 24 employees are above 25 years of age
 19 married employees are above 25 years, of which 7 are males
 12 males are above 25 years of age
 15 males are married.
How many unmarried females are there and how many of them are above 25?
Answer : 
15 unmarried females & none are above 25 years of age.
Simply put all given information into the table structure and you will get the
answer.

Married 
Unmarried 

Below 25 
Above 25 
Below 25 
Above 25 

Female 
3 
12 
15 
0 
Male 
8 
7 
20 
5 
There is a safe with a 5 digit number as
the key. The 4th digit is 4 greater than the second digit, while the 3rd digit
is 3 less than the 2nd digit. The 1st digit is thrice the last digit. There are
3 pairs whose sum is 11.
Find the number.
Answer :  65292
As per given conditions, there are three possible combinations for 2nd, 3rd and
4th digits. They are (3, 0, 7) or (4, 1, 8) or (5, 2, 9)
It is given that there are 3 pairs whose sum is 11. All possible pairs are (2,
9), (3, 8), (4, 7), (5, 6). Now required number is 5 digit number and it
contains 3 pairs of 11. So it must not be having 0 and 1 in it. Hence, the only
possible combination for 2nd, 3rd and 4th digits is (5, 2, 9)
Also, 1st digit is thrice the last digit. The possible combinations are (3, 1),
(6, 2) and (9, 3), out of which only (6, 2) with (5, 2, 9) gives 3 pairs of 11.
Hence, the answer is 65292.
96. 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