Growth Rate calculation

Question

It is given that in the current year the total Investment Capital is USD 13.8 trillion. The investment Capital contains Human capital worth of USD 3.3 trillion. The question is

If total Innovation Capital were to grow by 5% per year in the future, which of the following would be the MINIMUM required annual growth in Human Capital that would see it represent more than half of total Innovation Capital in 10 years? A. 10% B. 15% C. 20% D. 25%

I know the answer is B.

How I reached it is :

$$ \frac{1 * (x)^{10}}{4*(105)^{10}} = \frac{1}{2}\\ x^{10} = 2* (105)^{10}\\ x = (1.07 \times 105)^{10}\\ x = 112.5 $$ So I concluded the growth rate is $12.5\%$ for it to be exactly twice. For more than twice it should be $15\%$.

I used a calculator to get the value $2^{1/10}$. But in this question, I am not supposed to use a calculator. So what am I missing? How can one do this calculation or estimate an answer without a calculator for this question?

Answer 1

After Tiwa Aina's answer, let us try to approximate $\sqrt[10] 2$ without any calculator.

Let $$x=\sqrt[10] 2\implies \log(x)=\frac 1{10}\log(2)$$ Now, use the fast convergent series $$\log \left(\frac{1+t}{1-t}\right)=2 t+\frac{2 t^3}{3}+O\left(t^5\right)$$ and use $t=\frac 13$ which makes $$\log(2)\approx \frac 23+\frac 2 {81}=\frac {56}{81}\implies \log(x)\approx\frac{28}{405}$$ Now $$x=e^{\log(x)}\approx 1+\frac{28}{405}\approx 1+\frac{28}{400}=1.07$$

Answer 2

If I'm understanding you correctly, you're asking how to get $2^{1/10} \approx 1.07$ without technology. Here's how I would have done it:

First: realize that the tenth root of $2$ should be just a bit greater than $1$. The power (ha!) of exponentiation is much too great for us to multiply a number like $1.2$ by itself 10 times and only get $2$ (if you don't believe me, try it out. You'll see that a number like $1.2$ grows way too fast within 3 exponentiations).

So maybe $\sqrt[10]2 \approx 1.1$ is plausible? Let's test it:

Note that the digits of $11^n$ form the digits of the $n$th binomial expansion for $ 2 \leq n$. Indeed:

$11^2 = 121$

$11^3 = 1331$

$11^4 = 14641$

\begin{align*} 11^5 &= 161051 \\ &= 1(10^5) + 5(10^4) + 10(10^3) + 10(10^2) + 5(10^1) + 1(10^0) \\ &= \binom {5}0 (10^5) + \binom {5}1(10^4) + \binom {5}2(10^3) + \binom {5}3(10^2) + \binom {5}4(10^1) + \binom {5}5(10^0) \end{align*}

So using the fact that $1.1^n = 10^{-n} \cdot 11^n$, we can estimate the value of $1.1^{10}$.

$11^{10}$ forms the tenth binomial expansion. Its digits are reflected by the sum

$$\binom {11}0 (10^{10}) + \binom {11}1(10^9) + \binom {11}2(10^8) + \binom {11}3(10^7) +...$$ and

$$1 \cdot 10^{10} + 11 \cdot 10^9 + 55 \cdot 10^8 + 165 \cdot 10^7 = 2815...$$ which starts with $2$, so we know that $1.1^{10}$ is too big.

Let's find a lower bound now. Maybe $\sqrt[10]2 \approx 1.05$?

Well, $\frac{1.10}{1.05} \approx 1.05$ by long division, so we can say $$\frac{1.10}{1.05} \approx 1.05 \implies 1.1 \approx (1.05)^2 \implies (1.1)^5 \approx (1.05)^{10}$$ which we showed above to be $10^{-5} \cdot 11^5 = 1.61051$. So $1.05$ is too small!

So we take the middle ground and, accounting for the power of exponentiation, go a bit under the average of $1.05$ and $1.1$, giving $\boxed{1.07}$ as a good estimate.