Data Sufficiency

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For the first one I feel d can maximum take a value 6. So the info would not help.

2. whether r is -ve or +ve as long as p=r one can conclude. Why is the second condition needed?
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madhavi, ur first Qs -
500 is the multiple of 100 closet to x -> Can be x = 451 or 499
400 is the multiple of 100 closet to y -> Can be y = 351 or 399
So x + y = 451 + 399 = 850 = Closet to 900
or = 451 + 351 = 802 = closet to 800

So E is the best option to chose.
Ur second qs -
To me, its coming as A which I think u have also chosen. Don't getting why C is the OA(If it is ...)?

Ur third qs -
f = 30! - right?
Now by stmt - 1 10^d is a factor of f. But d can be 1, 2, 3 or so.
But in stmt - 2, it is saying that d > 6.
Here we can get the value of d and that is 7.
Now question is - how? Look, 30! = product of first 30 integers
For 10, 20 and 30 there will be 3 zeroes coming at the end. Now for every 5 at the unit digit, the product will be zeroes at the unit or ten digit. It will be like - 25 x 4, 15 x 2 and 6 x 5. In (25 x 4), we are getting 2 zeroes for the rest, only 1 zero at the end. So number of zeros that can come in 30! is (3 [For 10,20 and 30] + 2 [25 * 4] + 1 [15 * 2] + 1 [6 * 5] ) = 7.
So d = 7. Am i clear to explain?

Nice explantions camitava

Lets chk the one lft out ...

From 1 p = r

Let say p = 2 ---- case 1

1/2 > 2/6 ..Yes

Let say p = -2 ---- case 2

-1/2 > -1/3 No

hence Insufficient


From 2 r > 0 so case 2 above is eliminated ..and also p =r so only case 1 holds

hence C

Hope this helps

Thanks sujaysolanki! Again a miss from my side... I misunderstood the problem at all. God where are you? Pls help meeeeeeeeeeeee ...

Yeah... that's clear.
Thanks camitava & sujay!