find the compound interest on 4000 for 2.5 years at 10%per annum
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Principal(P) = Rs 4000
Rate of Interest(r ) = 10%
Time = 2.5 years
Here first we take n = 2 years.
A = P( 1 + r /100)n Where A = Amount and n = number of years
A = 4000 (11/10)2
A = 4000 ( 121 / 100 )
A = Rs 4840
The Amount after two years = Rs 4840
Now Principal = Rs 4840
Simple interest for last ½ year is = PRT/100 = (4840 × 10 × 1 / 2) / 100 = Rs.242
Amount after 2 ½ year = Rs 4840 + Rs 242 = Rs 5082
C.I = A - P
C.I = Rs 5082 - Rs 4000 = Rs1082
∴ The compound interest is Rs 1082.
hope it'll help :)
Rate of Interest(r ) = 10%
Time = 2.5 years
Here first we take n = 2 years.
A = P( 1 + r /100)n Where A = Amount and n = number of years
A = 4000 (11/10)2
A = 4000 ( 121 / 100 )
A = Rs 4840
The Amount after two years = Rs 4840
Now Principal = Rs 4840
Simple interest for last ½ year is = PRT/100 = (4840 × 10 × 1 / 2) / 100 = Rs.242
Amount after 2 ½ year = Rs 4840 + Rs 242 = Rs 5082
C.I = A - P
C.I = Rs 5082 - Rs 4000 = Rs1082
∴ The compound interest is Rs 1082.
hope it'll help :)
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