A sum of money becomes 2200 after three years and 4400 after six years on compound interest is
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Let the sum be P and rate of interest be R% per annum.
Amount after 3 years = 2200
P(1 + R/100)T = 2200
P(1 + R/100)3 = 2200 --- ( 1)
Amount after 6 years = 4400
P(1+R/100)T = 4400
P(1 + R/100)6 = 4400 --- (2)
(2) ÷ (1) => [P(1+R/100)6]/[P(1+R/100)3] = 4400/2200 = 2
=> (1+R/100)3 = 2 (Substituting this value in equation 1)
=> P × 2 = 2200
P = 22002 = 1100
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