A sum of money amounts to Rs. 702 in 2 years and Rs. 783 in 3 years. The rate percent is?

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.