Question

Mr.Pinto invests one-fifth of his capital at \(6\%\),one-third at \(10\%\) and the remaining at \(1\%\),each rate being simple interest per annum.Then,the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is

Answer 1

Correct Answer: 20

Let's assume Mr. Pinto's initial capital is C dollars. 
He invests one-fifth of his capital at \(6\%\),which means he invests \((\frac{1}{5})\times{C}\) dollars at \(6\%\) interest per annum.The interest earned from this investment after t years is \((\frac{1}{5})\times{C}\times0.06\times{t}\). 
He also invests one-third of his capital at \(10\%\),which means he invests \((\frac{1}{3})\times{C}\) dollars at \(10\%\) interest per annum.The interest earned from this investment after t years is \((\frac{1}{3})\times{C}\times0.10\times{t}\). 
The remaining amount,which is \((1-\frac{1}{5}-\frac{1}{3})\times{C}=(11/15)\times{C}\), is invested at \(1\%\) interest per annum.The interest earned from this investment after t years is \((\frac{11}{15})\times{C}\times0.01\times{t}\). 
Now,we want the cumulative interest income from these investments to equal or exceed his initial capital, which is C dollars.So,we can set up the following inequality: 
\((\frac{1}{5})\times{C}\times0.06\times{t}+(1/3)\times{C}\times0.10\times{t}+(\frac{11}{15})\times{C}\times0.01\times{t≥C}\) 
Now,let's solve for t: 
\((\frac{1}{5})\times0.06\times{t}+(\frac{1}{3})\times0.10\times{t}+(\frac{11}{15})\times0.01\times{t≥1}\)
Simplify: 
0.012t+0.0333t+0.0073t ≥ 1 
Combine the terms: 
0.0523t ≥ 1 
Now, divide both sides by 0.0523: 
\(t ≥ \frac{1}{0.0523}\) 
t ≥ 19.13 
Since the time (t) must be a whole number of years, the minimum number of years required for the cumulative interest income from these
investments to equal or exceed his initial capital is 20 years. 

Answer 2

Let the number of years needed be T years, and let the total investment be 15x.
\(\frac{{3x \times 6 \times T}}{{100}} + \frac{{5x \times 10 \times T}}{{100}} + \frac{{7x \times 1 \times T}}{{100}} \geq 15x\)

\(⇒\) \(\frac{75xT}{100} \geq 15x\)

\(⇒\) \(T \geq 20\)

Thus, 20 years is the minimal value of T.