SRCC GBO 2010

For the total sales, we shall add the profit to the total cost incurred.

Hence, total sales for P,Q,R, and S respectively are 30, 35, 45, 85.

Now, profit per rupee of sales for each of the companies P, Q, R, and S respectively is 5/30, 10/35, 20/45, 10/85

Clearly, profit per rupee of sales is highest in R.

The raw material cost for each of the companies P,Q,R and S respectively is 15, 15, 15, 40.

Hence, the lowest raw material cost per rupee of total cost is clearly lowest in S, which is 40/75. For the other companies, it is 15/25. 

For P, the profit/total cost is 5/20

For Q, the profit/total cost is 10/25

For R, the profit/total cost is 20/25

For S, the profit/total cost is 10/75

This ratio is clearly highest for R.

We have $$1^{-1}-\left(-1^2\right)-2=1-1-2=-2$$

Hence -2 is the answer

$$\frac{2(\sqrt{2} + \sqrt{6})}{3\sqrt{2 + \sqrt{3}}}$$

Let's rationalize the denominator.

To solve this, we shall first simplify each bracket in the expression.

Hence, the numerator becomes (18*19*20*21....36)/(1*2*3*4.....19)

Similarly, the denominator becomes (20*21*22*23.....36)/(1*2*3*4......17)

Now, on dividing the numerator by the denominator, we get the answer as 1.

Let's check this for smaller number $$(101)^2$$=10201

$$(10101)^2$$=102030201

$$(1010101)^2$$=1020304030201

The similar pattern follows for $$(101010101)^2$$. And the value will be 10203040504030201. So, there are 9 non-zero digits.

The greatest natural number that will divide 13511,13903,14589 and leave the same remainder is the HCF of (13903-13511),(14589-13903),(14589-13511) or the HCF of (392,686,1078). This comes out to be 98. Hence the answer is 98

This is a question which requires you to observe the pattern. 

For A, there are three working days and one rest day. Hence, A has a rest day on every fourth day (4, 8, 12, 16, 20, 24......1000).

For B, there are seven working days and three rest days after that. Hence, B has three rest days in the span of ten days (8/9/10, 18/19/20, 28/29/30, 38/39/40, 48/49/50..... 998/999/1000). 

From the pattern, it is clearly visible that A and B will have exactly one common day in every three-day rest period of B. This means, there will be 100 common days of rest for both A and B in the first 1000 days.