SNAP 2007 Question Paper

It is given that

i) Current MD remembers that he earned the most(70,000)

ii) Biswajit and Dhruv did not start on 50,000

iii) Chinmay does not remember what he started on. ( It implies Chinmay was not earning 70,000). The following things are known

We know that Biswajit's profession is Lawyer and Dhruv is an Engineer. From above table we observe that Chinmay is not MD. Thus Arvind is the MD and Chinmay is Doctor

Since neither Biswajit nor Dhruv started their salary at 50,000, Chinmay starting salary was 50,000

Given Arvind's Salary$$>$$ Biswajit's Salary and Doctor's Salary $$>$$ Dhruv's Salary. We obtain the following

Chinmay's Profession is Doctor

It is given that

i) Current MD remembers that he earned the most(70,000)

ii) Biswajit and Dhruv did not start on 50,000

iii) Chinmay does not remember what he started on. ( It implies Chinmay was not earning 70,000). The following things are known

We know that Biswajit's profession is Lawyer and Dhruv is an Engineer. From above table we observe that Chinmay is not MD. Thus Arvind is the MD and Chinmay is Doctor

Since neither Biswajit nor Dhruv started their salary at 50,000, Chinmay starting salary was 50,000

Given Arvind's Salary$$>$$ Biswajit's Salary and Doctor's Salary $$>$$ Dhruv's Salary. We obtain the following

The lawyer's starting salary was Rs 60,000

It is given that 

i) Current MD remembers that he earned the most(70,000)

ii) Biswajit and Dhruv did not start on 50,000

iii) Chinmay does not remember what he started on. ( It implies Chinmay was not earning 70,000). The following things are known

We know that Biswajit's profession is Lawyer and Dhruv is an Engineer. From above table we observe that Chinmay is not MD. Thus Arvind is the MD and Chinmay is Doctor

Since neither Biswajit nor Dhruv started their salary at 50,000, Chinmay starting salary was 50,000

Given Arvind's Salary$$>$$ Biswajit's Salary and Doctor's Salary $$>$$ Dhruv's Salary. We obtain the following

Arvind received the highest salary

For the man to take $$12^{th}$$ time rest he has to work for $$\left(12\times\ 8\right)=96$$ days and have rest for 11 days

Total days lapsed will be 96+11 = 107 and hence the $$12^{th}$$ rest day is $$108^{th}$$ day.

Day 1 is monday. After this Day 8 will be Monday. In general $$\left(7k+1\right)^{th}$$ day is Monday

Therefore $$106^{th}$$ day is Monday. $$108^{th}$$ day or $$12^{th}$$ rest day will be Wednesday

If we rotate the images we find ( 1,4,5 and 6 ) to be same figure and (2,3 and 7) to be the same figure.

Odd one out is (2,3 and 7)

It is given that there are 3 racers. Hence the rank given will be 1,2 and 3

Let us assume the first statement of Babloo is true, then the second will be false(Hence Waman is not in $$2^{nd}$$ place). Based on this the ranking will be as follows

But this will imply that both of Bunty's statement is false which is not allowed. 

Therefore the first statement of Babloee is False and the second Statement is true. Ranking based on this will be 

In this, we see that one statement of Bunty is true and the other one is false. Hence this is the correct ranking

Following are the possible routes

I) P-1-4-Q

II)P-1-5-Q

III)P-2-4-Q

IV)P-2-5-Q

V)P-3-4-Q

Total 5 routes are possible

On arranging words from Option B we get "SCANTY"

On arranging words from Option C we get "ASCENT"

On arranging words from Option D we get "SACHET"

No meaningful word is obtained by rearranging word in Option A

Let the average of the number of each row and column be integer denoted by $$x$$

Then the sum of numbers in Row 1 = $$4x$$.

Similarly for row 2 and row 3 the sum will be $$4x$$ each. Hence the sum of all the numbers in the table is $$12x$$.

Let $$N$$ be the number only number that is left out of the table.

Then, (1+2+3+4+5+7+9+10+11+12+13+14+15 -N) = $$12x$$ 

$$106-N\ =\ 12x$$

Since $$x$$ is integer then $$Remainder\left(\ \frac{\ 106-N}{12}\right)=0$$

$$Remainder\left(\ \frac{\ 10-N}{12}\right)=0$$

Hence N has to be 10 which is left out. And average of each row and column will be $$\ \frac{\ 106-10}{12}=\ \frac{\ 96}{12}=8$$

Upon solving the table will look as follows

In the patter we observe $$\ \frac{product\ of\ number\ in\ traigle\ }{product\ of\ number\ in\ sqaure\ and\ circle}\ =\ \ \frac{\ 6\times\ 8}{12\times\ 8}=\ \frac{\ 5\times\ 6}{6\times\ 10}=\ \frac{\ 1}{2}$$

Let the missing number be $$N$$

$$\ \frac{\ 3\times\ 12}{4\times\ N}=\ \frac{\ 1}{2}$$

$$N=18$$