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
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