The sum required to earn a monthly interest of Rs.1200 at 18% per annum Simple Interset is:

Answer 1

I = Po*r*t = $1200.

r = (18%/12)/100% = 0.015 = Monthly % rate expressed as a decimal.

t = 1 Mo.

Po*0.015*1 = 1200.
Po = $80,000.

Answer 2

SI = P×R×T
------
100

14400(1200×12)=P(?)×18×1
--------
100
14400= 0.18 P

P = 80000

Answer 3

#DKGDexterousZone
Time value of money
exercise 4A

Answer 4

I =₹1200
R=18%
T= 1/12 year
I= P×R×T/100
​
1200= P×18×1/ 100×12
​
P= 120000×12 / 18
​
P =₹80000