SNAP 2006 Question Paper

Engineering, Material, Services and CIS contributes around 81% of market share.

Let us represent the sales revenue in terms of 'k'.

Total sales revenue = 100k

Sales revenue from Engineering = 31k

Sales revenue from CIS = 19k

Remaining revenue from other sectors = 50k

Profit margin = Profit/Revenue

Let the profit margin of remaining sectors be x%.

So the eqn is:

10*100k = x*50k + 12*31k + 20*19k

after cancelling 'k' from both sides, we get:

=> x = 4.94% (answer)

Let the temperature on Monday, Tuesday, Wednesday and Thursday be a,b,c and d respectively

It is given that a+b+c = 3*37 = 111...(I)

              and b+c+d = 3*34 = 102 ... (II)

Using statement (I) alone we get $$d\ =\ \ \frac{\ 4}{5}a$$. Which can be put in eqn (I), after which we can obtain the value of d

Using statement (II) alone we can add eqn(I) and eqn(II) to get value of b+c and it can be used to get value of d from eqn(II)

Using statement (III) alone we a-d = 9 which can be obtained by eqn(I) and (II) and n hence no new information is given. Statement (III) is no sufficient to get the required answer.

Case 1 : I and II

Let smaller number be  x => bigger number = x+6

$$\frac{40}{100}\cdot x=\frac{30}{100}\left(x+6\right)$$

=> x=18 , Sum of bigger and smaller number = 42

Case 2 : I and III

Let smaller number be x => bigger number = x+6

$$\frac{\left(x+6\right)}{2}\cdot\frac{3}{x}=\frac{2}{1}$$

On solving , x=18 . So sum = 42

Case 3 : II and III

Only the ratio between bigger and smaller number can be found out

Let total work to be done is 48 units

Case 1 : I and II

A takes 16 hours => B=24 hours  (since A is 50 percent more efficient than B)

A does 3 units/hour

B does 2 units/hour

Total per day work = 5 units

Total time taken = 48/5 hr

Case 2 : I and III

B takes 24 hours => A = 16 hours (since A is 50 percent more efficient than B)

Together they take = 48/5 hours

Case 3 : II and III 

B takes 24 hours

A takes 16 hours

Together they take = 48/5 hours

A=$$P\left(1+\frac{TR}{100}\right)$$

From 2 and 3 , A= 750 , T=5 and R=8

P can be found out by substituting these values in the above equation.

Average of share price for 10/8 to 12/9 is  $$\ \frac{\ 483+475+467+513}{4}$$ = 483

Price at end of September (near to beginning of october)= 535

Price at the end October = 461

$$% decrease$$ = $$\frac{535-461\ }{535}\times\ 100\%$$

= $$13.83$$%

10% is the closest option 

From the given graph, we can observe that the share price dropped 17 times in the given period.

Hence, the answer is option B.

Money spent to buy 100 shares on 31/08 is 100*461 = 46,100

Money Recieved after selling 100 shares on 10/10 = 538*100 = 53,800

Overall profit = 53,800-46,100 = 7,700