Quiz – 1

1120 = P [1 + (5*12/5)/100]
P= 1000

70 = (P*4*7/2)/100
P = 500

9 = (1*40*R)/100
R = 22 1/2 %

(x*5*1)/100 + [(1500 - x)*6*1]/100 = 85
5x/100 + 90 – 6x/100 = 85
x/100 = 5
=> x = 500

4/9 P = (P*R*R)/100
R = 20/3 = 6 2/3%

Time = 2 years 4 months = 2(4/12) years = 2(1/3) years.
Amount = Rs\\\'. [8000 X (1+(15/100))^2 X (1+((1/3)*15)/100)]
=Rs. [8000 * (23/20) * (23/20) * (21/20)]
= Rs. 11109. .
:. C.I. = Rs. (11109 - 8000) = Rs. 3109.

Let the sum be Rs.P.then
P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

Amount = Rs.(30000+4347) = Rs.34347
let the time be n years
Then,30000(1+7/100)^n = 34347
(107/100)^n = 34347/30000 = 11449/10000 = (107/100)^2
n = 2years

Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. . .
So principal=RS [100*1200]/3*5=RS 8000
Amount = Rs. 8000 x [1 +5/100]^3 - = Rs. 9261.
.. C.I. = Rs. (9261 - 8000) = Rs. 1261.

difference in C.I and S.I in 2years =Rs.32
S.I for 1year =Rs.400
S.I for Rs.400 for one year =Rs.32
rate=[100*32)/(400*1)%=8%
difference between in C.I and S.I for 3rd year
=S.I on Rs.832= Rs.(832*8*1)/100=Rs.66.56