Amount = Rs. `[200(1 + 5/100)^3 + 200(1 + 5/100)^2 + 200(1 + 5/100)]`
= Rs.` [200 xx 21/20 xx 21/20 xx 21/20 + 200 xx 21/20 xx 21/20 + 200 xx 21/20]`
= Rs.` [200 xx 21/20 (21/20 xx 21/20 + 21/20 + 1)]` = Rs. 662.02.
Time = `2 73/365` years = `2 1/5` years
`:.` Amount = Rs.` [20480 xx (1 + (25)/(4 xx 100))^2 (1 + (1/8 xx 25/4)/(100))]`
= Rs.` (20480 xx 17/16 xx 17/16 xx 81/80)`
= Rs. 23409.
`:.` C.I. = Rs. (23409 - 20480) = Rs. 2929
Amount = Rs.` [25000 xx(1 + 12/100)^3]`
=Rs.`(25000 xx 28/25 xx 28/25 xx 28/25)`
= Rs. 35123.20.
`:.` C.I. = Rs. (35123.20 - 25000) = 10123.20
Amount = Rs.` [8000 xx (1 + 5/100)^2]`
= Rs.`(8000 xx 21/20 xx 21/20)`
= Rs. 8820.
Let each instalment be Rs. `x` Then ,
(P.W. of Rs. `x` due 1 year hence ) + (P.W. of Rs. `x` due 2 years hence ) + (P.W. of Rs. `x` due 3 years hence ) = 7620.
`:.` ` (x)/((1 + (50)/(3xx 100))` + ` (x)/((1 + (50)/(3xx 100))^2 `+` (x)/((1 + (50)/(3xx 100))^3` = 7620
`hArr (6x)/(7) + (36x)/(49) + (216x)/(343) = 7620`
`hArr 294x + 252x + 216x = 7620 xx 343`
`hArr x = ((7620 xx 343)/(762)) = 3430`
`:.` Amount of each instalment = Rs. 3430.
`P(1 + R/100)^15 = 2P rArr (1 + R/100)^15 = (2P)/(P) = 2` ..........................(i)
Let` P(1 + R/100)^n = 8P rArr (1 + R/100)^n = 8 = 2^3 = {(1 + R/100)^15}^3 ` [using (i)]
`rArr (1 + R/100)^n = (1 + R/100)^45`
`rArr n = 45`
Thus, the required time = 45 years
Let the sum be Rs. P, Then,
`P(1 + R/100)^3 = 6690`................................(i)
and ` P(1 + R/100)^6 ` ....................................(ii)
On dividing , we get` (1 + r/100)^3 = 10035/6690 = 3/2`
Substituting this value in (i) , we get :
`P xx 3/2 = 6690`
or` P = (6690 xx 2/3) = 4460`
Hence , the sum is Rs. 4460.
S.I. on Rs. 7350 for 1 year = (8575 - 7350 = Rs. 1225.
`:.` Rate = `((100 xx 1225)/(7350 xx 1))% = 50/3 `% = `16 2/3`%
Let the sum be Rs. `x` Then ,
`x (1 + (50)/(3 xx 100))^2 = 7350`
`hArr x xx 7/6 xx 7/6 = 7350`
`hArr x = (7350 xx 36/49)`
`hArr x = 5400`
`:.` Sum = Rs. 5400.
Let the rate be R% p.a. Then,
`[18000(1 + R/100)^2 - 18000] - ((18000 xx R xx 2)/(100)) = 405`
`hArr 18000[((100 +R)^2)/(10000) - 1 - (2R)/(100)] = 405`
`hArr 18000[((100 +R)^2 - 10000 - 200R)/(10000)] = 405`
`hArr 9/5 R^2 = 405`
`hArr R^2 = ((405 xx 5)/(9))`
`hArr R^2 = 225`
`hArr R= 15`
`:.` Rate = 15%
Let the two parts be Rs. `x` and Rs. `(1301 - x)`
` x (1 + 4/100)^7 = (1301 - x ) (1 + 4/100)^9`
`hArr (x)/((1301 - x)) = (1 + 4/100)^2 = (26/25 xx 26/25)`
`hArr 625x = 676(1301 - x)`
`hArr 1301x = 676 xx 1301`
`hArr x = 676`
So, the two parts are Rs. 676 and Rs. (1301 - 676) i.e. 676 and Rs. 625
