A train with 120 wagons crosses Arun who is going in the same direction, in 36 seconds. It travels for half an hour from the time it starts overtaking the Arun ( he is riding on the horse) before it starts overtaking the Sriram( who is also riding on his horse) coming from the opposite direction in 24 seconds. In how much time (in seconds) after the train has crossed the Sriram do the Arun meets to Sriram?

Let us assume the first installment is a and difference between two consecutive installments is d.
According to question,
Sum of 40 Installments = 3600
Apply the formula Sum of first n terms in an Arithmetic Progression = S_n = n/2[ 2a + (n?1)d ]
where a = the first term, d = common difference, n = number of terms.
n/2[ 2a + (n?1)d ] = 3600
Put the value of a, n and d from question,
40/2[ 2a + (40 ?1)d ] = 3600
20[ 2a + 39d ] = 3600
[ 2a + 39d ] = 3600/20 = 180
2a + 39d = 180...........................(1)

Again according to given question,
After paying the 30 installments the unpaid amount = 1/3(total unpaid amount) = 3600 x 1/3
After paying the 30 installments the unpaid amount = 1200
So After paying the 30 installments the paid amount = 3600 - 1200 = 2400
Sum of 30 installments = 2400
30/2[ 2a + (30 ?1)d ] = 2400
[ 2a + (30 ?1)d ] = 2400 x 2/30
2a + 29d = 80 x 2 = 160
2a + 29d = 160..........................(2)
Subtract the Eq. (2) from Eq. (1), we will get
2a + 39d - 2a - 29d = 180 - 160
10d = 20
d = 2
Put the value of d in Equation (1), we will get
2a + 39 x 2 = 180
2a = 180 - 78
a = 102/2
a = 51
The value of first installment = a = 51

Given, P = Rs 15000, R = 12 %
and n = 1¹/₄ = 5/4 Yr
According to the formula,
Amount = p [1 + R/(100 x 4)]⁴ⁿ
= 15000 x [1 + 12/(100 x 4)]^{4 x 5/4}
= 15000(412/400)⁵
= 15000 (103/100)⁵
= 15000 x (103/100) x (103/100) x (103/100) x (103/100) x (103/100)
= (15 x 103 x 103 x 103 x 103 x 103) / 10000000
= Rs 17389.111
= Rs 17389.12 (approx )

Let the remaining food will last for D days.
500 men had provisions for (27 ? 3) = 24 days.
( 500 + 300 ) men had provisions for D days.
More men, Less days ( Indirect Proportion )
we can write as :-
? 800 : 500 :: 24 : x ? (800 × D ) =( 500 × 24 )
? ( 500 × 24 ) /800 =15
Therefore , 15 days is correct answer .

Let us assume the first number is a and common difference is d.
According to question,
4th term of A.P = 37
a + ( n - 1 ) x d = 37
Put the value of a , n and d, we will get,
a + (4 - 1 ) x d = 37
a + 3d = 37..................(1)
sixth term is 12 more than the fourth term,
6th term = 12 + 4th term
a + ( n - 1 ) x d = 12 + 37
a + ( 6- 1 ) x d = 39
a + 5d = 39................(2)
subtract the equation (1) from (2)
a + 5d - a - 3d = 39 - 37
5d - 3d= 2
2d = 2
d = 1
Put the value of d in equation (1), we will get
a + 3 x 1 = 37
a = 37 - 3
a = 34

Second term = a + (n - 1) x d = 34 + (2 - 1) x 1 = 34 + 1 = 35
Six term = a + (n - 1) x d = 34 + (6 - 1) x 1 = 34 + 5 = 39
Sum of Second and Six term = 35 + 39 = 74
Sum of Second and Six term = 74
Answer is 74.

Principal = (P.W. of Rs. 121 due 1 year later) + (P.W. of Rs. 121 due 2 years later)
= Rs. [ 121 /(1 + 10/100) + (121 / (1 + 10/100)²]
= Rs. 210

Given, P = 185220 R = 5% (increases) and n = 3 yr
According to the formula,
Population n yr ago = P/(1 + R/100)ⁿ
= 185220/(1 + 5/100)³
= 185220/[(21/20) x (21/20) x (21/20)]
= [185220 x 20 x 20 x 20] / [21 x 21 x 21]
= 20 x 20 x 20 x 20
= 160000

S.I. for 3 years = Rs.(39 x 3) = Rs. 117.

∴ Principal = Rs. (815 - 117) = Rs. 698.

3 men ? 6 boys
? 1 man ? 2 boys
? 4 men + 4 boys ? 4 men + 2 men = 6 men
? 3 men can do a work in 18 days.
? 1 man can do a work in 18 x 3 days.
? Required number of days = (18 x 3)/6 = 9 days

1 man can finish the work in (8 x 12) = 96 days
1 woman can finish the work in (4 x 48) = 192 days
1 child can finish the work in (10 x 24) = 240 days
1 man's 1 day's work = 1/96
1 women's 1 day's work = 1/192
1 child's 1 day's work = 1/240
(10 men + 4 women + 10 children)'s 1 day's work = (10/96 + 4/192 + 10/240)
= (5/48 + 1/48 + 1/24)
= (5 + 1 + 2)/48)
= 8/48 = 1/6
Hence, they will finish the work in 6 days.