$ 20000 will become $ 60000 at 20 % p . a . find the time ?
"si = simple interest = a - p = 60000 - 20000 = $ 40000 r = 100 si / pt = 100 * 40000 / 20000 * 20 = 10 years answer is e"
divide(subtract(60000, 20000), multiply(20000, divide(20, const_100)))