Asked by Haroon Gondal
The sum required to earn a monthly interest of Rs.1200 at 18% per annum Simple Interset is:
Answers
Answered by
Henry
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.
r = (18%/12)/100% = 0.015 = Monthly % rate expressed as a decimal.
t = 1 Mo.
Po*0.015*1 = 1200.
Po = $80,000.
Answered by
Ruchika surana
SI = P×R×T
------
100
14400(1200×12)=P(?)×18×1
--------
100
14400= 0.18 P
P = 80000
------
100
14400(1200×12)=P(?)×18×1
--------
100
14400= 0.18 P
P = 80000
Answered by
Ajay
#DKGDexterousZone
Time value of money
exercise 4A
Time value of money
exercise 4A
Answered by
p tanuja
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
R=18%
T= 1/12 year
I= P×R×T/100
1200= P×18×1/ 100×12
P= 120000×12 / 18
P =₹80000