Tickle Your Grey Cells
Puzzle corner is a dedicated section for honing analytical, mathematical and reasoning skills. In this section every week you will find five Puzzles. Theses mind twisters have been carefully selected after a thorough research and provide right inputs for tickling your grey cells. Solving puzzles is considered very good exercise so as to remain mentally active. Puzzle solving enhances memory and processing speed.
Puzzle is one of the most popular category in competitive examinations. All the major Indian MBA entrance exams like CAT, MAT, XAT and SNAP have direct questions or questions based on similar logics in written test. Besides MBA exams, competitive exams like GRE, GMAT, Bank PO, SSC & other government exams include puzzle questions. Most of the companies while recruiting through campus includes Puzzles in the written tests or at interview stage. Hence those who are preparing for campus placement tests this section is a must.

Following facts are true about Ludhiana:
No two citizens have exactly the same number of coins.
No citizen has exactly 2025 coins.
There are more citizens than there are coins with any one citizen.
What is the largest possible number of the citizens of Ludhiana?


It is given that no citizens have exactly 2025 coins. Hence there are 2025 citizens with 0 to 2024 coins.
Suppose there are more than 2025 citizens. But these will violate the condition that "There are more citizens than there are coins with any one citizen." As for any number more than 2025, there will be same number of citizens as the maximum number of coins with any citizen.

There are four clustors of Oranges, Apricots and Bananas as follows:
Clustor I : 1 Orange, 1 Apricot and 1 Banana
Clustor II : 1 Orange, 5 Apricot and 7 Bananas
Clustor III : 1 Orange, 7 Apricots and 10 Bananas
Clustor IV : 9 Oranges, 23 Apricots and 30 Bananas
Clustor II costs Rs 300 and Clustor III costs Rs 390.
Can you tell how much does Clustor I and Clustor IV cost?


Assume that the values of one orange, one apricot and one banana are O, A and B respectively.
From Clustor II : O + 5A + 7B = 300
From Clustor III : O + 7A + 10B = 390
Subtracting above two equations: 2A + 3B = 90
For Clustor I :
= O + A + B
= (O + 5A + 7B)  (4A + 6B)
= (O + 5A + 7B)  2(2A + 3B)
= 300  2(90) = 300 – 180 = 120
Similarly, for Clustor IV :
= 9O + 23A + 30B
= 9(O + 5A + 7B)  (22A + 33B)
= 9(O + 5A + 7B)  11(2A + 3B)
= 9(300)  11(90) = 2700 – 990 = 1710
Thus, Clustor I costs Rs 120 and Clustor IV costs Rs 1710.

Five friends with surname Panwala, Letterwala, Talawala, Chunawala and Radhiwala have their first name and middle name as follow.
Four of them have a first and middle name of Verma.
Three of them have a first and middle name of Kanta.
Two of them have a first and middle name of Rajesh.
One of them has a first and middle name of Kiash.
Letterwala and Talawala, either both are named Kanta or neither is named Kanta. Either Panwala and Letterwala both are named Rajesh or Talawala and Chunawala both are named Rajesh. Chunawala and Radhiwala are not both named Verma. Who is named Kiash?


From (1) and (7), it is clear that Panwala, Letterwala and Talawala are named Verma.
From (6) and (5), if Letterwala or Talawala both are named Kanta, then either of them will have three names i.e. Verma, Kanta and Rajesh. Hence, Letterwala and Talawala both are not named Kanta. It means that Panwala, Chunawala and Radhiwala are named Kanta. Now it is clear that Talawala and Chunawala are named Rajesh. Also, Letterwala is named Kiash.

There are 70 employees working with Bulls Eye 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?


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


1000 cats were kept under observation to check their growth rates. After six months, there were 1000R cats. At the beginning of the 3rd year, there were roughly 2828R cats, which was 4 times what the scientists placed in there at the beginning of the 1st year. If R is a positive variable, how many cats would be there at the beginning of the 11th year?


At the beginning, there were 1000 cats. Also, there were 4000 cats at the beginning of third year which is equal to 2828R. Thus, R = 4000/2828 i.e. 1.414 (the square root of 2). Note that 2828R can be represented as 2000*R*R (R=1.414), which can be further simplified as 1000*R*R*R*R
Also, it is given that at the end of 6 months, there were 1000R cats. It is clear that the population growth is 1.414 times every six months i.e. 2 times every year. After N years, the population would be 1000*(R^(2N)) i.e. 1000*(2^N). Thus, at the beginning of the 11th year (i.e. after 10 years), there would be 1000*(2^10) i.e. 1,024,000 cats.


