If there are total 100 candidates,
40 candidates are really capable and
60 candidates are incapable.
Of the candidates who are really capable = 80%
40*(80/100)=32 pass the test and
of the incapable= 30%
60*(30/100)=18 pass the test.
Total candidates who passed the test = 32+18 = 50
The proportion of the really capable students who can pass the test to the total students who can pass = 32/50 = 16/25.

Total Combination of getting a card from 52 cards = C152

Because there is no replacement, so number of cards after getting first card= 51

Now, Combination of getting an another card= C151

Total combination of getting 2 cards from 52 cards without replacement= (C152×C151) 

There are total 4 Ace in stack. Combination of getting 1 Ace is = C14

Because there is no replacement, So number of cards after getting first Ace = 3

Combination of getting an another Ace = C13

Total Combination of getting 2 Ace without replacement=C14×C13

Now,Probability of getting 2 cards which are Ace =C14×C13C152×C151 = 1/221.

First girl facing North
Then she rotates 100 deg clock wise => 10 deg ES
Then again she rotates 190 deg anti clock wise => for 100 deg it is North and more 90 deg means West.
Therefore, the new direction of the girl is West.

1 2 3 4 5 6(Shankar) 7(Nitu) 8(Althaf) 9 10 11 12 13 14
Here, Althaf is 8th from front, Shankar is 9th from rear end and Nitu is between them
So minimum no. of boys standing in the queue = 14

Let first no. be "k" and multiple be "l"...start value of y with 3
then,
[(k x y) + {l x (l-1)}]
Thus, we get

15 x 3 + 3 x 2 = 51
51 x 4 + 4 x 3 = 216
216 x 5 + 5 x 4 = 1100
1100 x 6 + 6 x 5 = 6630
6630 x 7 + 7 x 6 = 46452

Therefore, the missing number is 6630.

The series follow the rule,

21 = 
77 = 92-4
165 = 132-4
285 = 172-4

5,9,13,17,21,(+4)...
In the same way next no. will be
212-4 = 441 - 4 = 437

Assume first child (the youngest) get = Rs. x

According to the question ;
each son having Rs. 30 more than the younger one

Second child will get = Rs. x + 30
Third child will get = Rs. x + 30 + 30 = x + 60
Forth child will get = Rs. x + 30 + 30 + 30 = x + 90
Fifth child will get = Rs. x + 30 + 30 + 30 + 30 = x + 120

Total amount they got = Rs. 2000

x + (x+30) + (x+60) + (x+90) + (x+120) = 2000
5x + 300 = 2000
5x = 1700
x = Rs. 340
So the youngest child will get Rs. 340.

Number of books that can be placed in the larger box is given by Volume of Box / Volume of book.
Now, n = 25 x 42 x 60 / 7 x 6 x 5
= 63000 / 210
= 300.
Therefore, 300 books can be in that box.