Question

plz answer this question!!!!!!!!

Answer 1

Solution:-
●4. answer:-
case 1. let p= x rupees , rate = r% , n = 2years

$7396 = x {(1 + \frac{r}{100}) }^{2} = ......(1)$
case 2.
$7950.70 = x {(1 + \frac{r}{100}) }^{3} ...........(2). \\ divided \: eq(2) \\ \frac{7950.4}{7396} = \frac{x( {(1 + \frac{r}{100} )}^{3} }{x( {(1 + \frac{1}{100} )}^{2} } \\ \frac{79504}{7396} = 1 + \frac{r}{100} \\ 1.075 - 1 = \frac{r}{100 } \\ r = 0.075 \times 100 = 7.5\%ans$
●5.answer:-
$p = 200000 \: \\ r = 5\% \\ t = 1year \\ by \: formula \:$
A = 200000(1+5/100) = 200000×1.05210000.
》A2 = 210000(1+8/100) = 210000×1.08= 226800
》A3 = 226800(1+12/100)= 226800× 1.12 = 254016.
》the total profit of tree years=
》 A3- P = 254016- 200000
= 54016 ans
●6. answer:-
let intrest rate = r% , time t= 2years , si = 400
now , p = (si × 100)/2×r
p = 400×100/2r = p = 20000/r .......(1).
c.i = 410 then.
410= p (1+r/100)^2 - p
410= 20000/r ( 1 + r^2/10000 + r/50) - 20000
410= 20000/r [ 1 + r^2/10000 +r/50 -1]
410 = 20000/r [ r^2/10000 +r/50]
410 = 2r + 400
2r = 410-400
2r = 10
r = 5% answer:-

rate of intrest = 5%
sum p = 20000/5 = 4000 rupees.

●7answer :-
p = 1000 rupees, s.i rate = 11% , time (t) = 3years.
s.i = 1000×11×3/100
s.i = 330
here p = 1000same amount , c.i rate = 10% , t=3years.
c.i = 1000(1+10/100)^3 - 1000
c.i = 1000×1.331 -1000
c i = 1331 -1000
c.i = 331 rupees.
So in second investment there is more profit hence it is better.
■i hope its help■