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.

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 the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.

The smallest possible number of books = L.C.M of 5, 9 and 13.
Therefore, L.C.M of 5,9 and 13 is = 5 x 9 x 13 = 585.

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.

we know that m1 x d1 = m2 x d2
=> 4 x 6 = 5 x d
where d = no. of days taken by 5 men to paint
d = 24/5 days.

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are k, l, m, n & o respectively, then
k + l + m + n = 58 x 4 = 232 ---- (i) &
l + m + n + o = 60 x 4 = 240 ---- (ii)
Subtracting (i) from (ii), we get
o-k = 8 ---(iii)
Given, k/o = 8/9 ----(iv)
So from (iii) & (iv), we get
a = 64, e = 72
Therefore, number of visitors on 5th day is 72.

Average speed = 2aba+b, here a=25 b=4
= 2 x 25 x 4/(25+4) = 200/29 km/hr.

Distance covered in 5 hours 48 minutes
= Speed x time = (200/29)x (29/5) Distance covered in 5 hours 48 minutes = 40kms.

Distance of the post office from the village = (40/2) = 20 km.