`P(1 + R/100)^5 = 2P `
`rArr (1+ R/100)^5 = 2` .........................................(i)
Let `P(1 + R/100)^n = 8P` `rArr (1 + R/100)^n = 8 = 2^3 = {(1 + R/100)^5}^3 ` [using(i)]
`rArr (1+ R/100)^n = (1 + R/100)^15`
`rArr n = 15`1
`:.` Required time = 15 years .
`12000 xx (1 + R/100)^5 = 24000`
`rArr (1 + R/100)^5 = 2`
`:.` `[(1 + R/100)^5]^4 = 2^4 = 16`
`rArr (1 + R/100)^20 = 16`
`rArr P(1 + R/100)^20 = 16P`
`rArr 12000(1 + R/100)^20 = 16 xx 12000 ` = 192000
S.I. on Rs. 4624 for 1 year = Rs. (4913 - 4624 ) = Rs. 289.
`:.` Rate = `((100 xx 289)/(4624 xx 1))`% = `25/4`% = `6 1/4`%
Now, `x(1 + (25)/(4 xx 100))^2 = 4624`
or ` x xx 17/16 xx 17/16 = 4624`
`:.` `x = (4624 xx 16/17 xx 16/17)` = Rs. 4096.
S.I. on Rs. 800 for 1 year = Rs. (840 - 800) = Rs. 40
`:.` Rate = `((100 xx 40)/(800 xx 1))`% = 5%
Mr. Dua invested money in two schemes A and B offering compound interest @ 8 p.c.p.a. and 9 p.c.p.a. respectively . if the total amount of interest accrued through two schemes together in two years was Rs. 4818.30 and the total amount invested was Rs. 27,000 what was the amount invested in Scheme A ?
Let he investment in scheme A be Rs. `x`.
Then , investment in scheme B = Rs. (27000 - `x`).
`:.` ` [x xx {(1 + 8/100)^2 - 1} + (27000 x){(1 + 9/100)^2 - 1}]` = 4818 .30
`hArr (x xx 104/625) + (1881(27000 -x))/(10000) = 481830/100`
`hArr 1664x + 1881(27000 - x) = 48183000`
`hArr (1881x - 1664x = (50787000 - 48183000)`
`hArr 217x = 2604000`
`hArr 2604000/217` = 12000
`x` = 12000.
A person lent out a certain sum on simple interest and the same sum on compound interest at a certain rate of interest per annum . he noticed that the ratio between he difference of compound interest and simple interest of 3 years and that of 2 years is 25 : 8. The rate of interest per annum is :
Let the principal be Rs. P and rate of interest be R% per annum.
Difference of C.I. and S.I. for 2 years
=`[P xx (1 + R/100)^2 - P] - ((P xx R xx 2)/(100)) = (PR^2)/(104)`
Difference of C.I. and S.I. for 3 years.
= `[P xx (1 + R/100)^3 - P] - ((P xx R xx 2)/(100)) ` = `(PR^2)/(10^4)((300 + R)/(100))`
`:.` `((PR^2)/(10^4)((300 +R)/(100)))/((PR^2)/(10^4)) = 25/8`
`rArr ((300 + R)/(100)) = 25/8`
`rArr R = 100/8` = `12 1/2`%
Amount of Rs. 100 for 1 year when compounded half -yearly.
= Rs. `[100 xx (1 + 3/100)^2]` = Rs. 106.09
`:.` Effective rate = (106.09 - 100)% = 6.09%
Difference in C.I and S,I, for 2 years = Rs. (696.30 - 660 ) = Rs. 36.30
S.I. for one year = Rs. 330.
`:.` S.I. on Rs. 330 for 1 year = Rs. 36.30
`:.` Rate `((100 xx 36.30)/(330 xx 1))`% = 11%
Let he sum be Rs. `x` , Then
C.I. when compounded half -yearly
= `[x xx (1 + 10/100)^4 - x] = 4641/10000 x`
C.I. when compounded annually = `[x xx (1 + 20/100)^2 - x] = 11/25 x`
`:.` ` 4641/10000 x - 11/25 x = 482`
or `x = (482 xx 10000)/(241) ` = 20000.
`x` = 20000.
For first year, S.I. = C.I.
Now, Rs. 16 is the S.I. on S.I. for 1 year.
Rs. 10 id S.I. on Rs. 100.
`:.` Rs. 16 is S.I. on Rs.` (100/10 xx 16)` = Rs. 160
So, S,I, on principal for 1 year at 10% is Rs. 160
`:.` principal = Rs. `((100 xx 160)/(10 xx 1))` = Rs. 1600
Amount for 2 years compounded half yearly = Rs. `[1600 xx (1 + 5/100)^4]` = Rs. 1944.81
`:.` C.I. = Rs. (1944.81 - 1600) = Rs. 344.81
S.I. = Rs. `((1600 xx 10 xx 2)/(100))` = Rs. 320.
`:.` (C.I.) - (S.I.) = Rs. (344.81 - 320) = Rs. 24.81.
