If a sum of money becomes \(\frac{11}{8}\) times in 7.5 years at simple rate of interest, then at the same rate of interest it will becomes \(\frac{19}{12}\) times in:
Answer & explanation
Correct answer: option 4
Let principal be P and rate be R,
Amount = P(1 + \(\frac{7.5R}{100}\))
⇒ \(\frac{11}{8}\)P = P(1 + \(\frac{7.5R}{100}\))
⇒ \(\frac{11}{8}\) = 1 + \(\frac{7.5R}{100}\)
⇒ \(\frac{7.5R}{100}\) = \(\frac{3}{8}\)
⇒ R = 5
Now,
⇒ \(\frac{19}{12}\) = P(1 + \(\frac{5T}{100}\))
⇒ \(\frac{19}{12}\) = 1 + \(\frac{5T}{100}\)
⇒ T = \(\frac{35}{3}\)
= 11 years 8 months