The compound interest accrued on an amount of ₹ 22000 at the end of two years is ₹5596.80. what would be the simple interest accrued on the same amount at the same rate in the same period?
Answers
Answer:
Let the rate of interest be R% per annum.
∴ CI = P [(1+R100)T−1]
=> 5596.8
= 22000 [(1+R100)2−1]
=> 5596.822000=(1+R100)2−1
=> ( 1 + R100)2=1+5596.822000
=> ( 1 + R100)2
= 22000+5596.822000
= 27596.822000
=> (1+R100)2=275968220000
= 1254410000
=> 1 + R100=1254410000−−−−√=112100
=> R100=112100−1=112−110100=12100
=> R = 12
∴ SI = Principal×Time×Rate100
= 22000×2×12100=Rs.5280
Step-by-step explanation:
hope it's helpful to u ;)
Let the rate of interest be R% per annum.
∴ CI = P [(1+R/100)T−1]
=> 5596.8 = 22000 [(1+R/100)²−1]
=> 5596.8/22000=(1+R/100)²−1
=> ( 1 + R/100)² =1+5596.8/22000
=> ( 1 + R/100)² = 22000+5596.8/22000
=> ( 1 + R/100)² = 27596.8/22000
=> ( 1 + R/100)² = 275968/220000
=> ( 1 + R/100)² = 12544/10000
=> 1 + R/100 =√12544/√10000 = 112/100
=> R/100 = 112/100−1 = 112−110/100 = 12/100
=> R/100 = 12/100
=> R = 12
∴ SI = Principal×Time×Rate100
= 22000×2×12/100=Rs.5280