Answer
Hint: First, we will let the principal sum of money as ‘P’ and the rate of interest as ‘R’. we will use the conditions given in the question and formula of compound interest to form a different equation. And by solving those equations we will find the rate of interest.
Complete step-by-step solution:
Let the principal sum of money be ‘P’.
Let the rate of interest compounded annually be ‘R’.
Given: the amount becomes 1.44 times the principal amount in the span of 2
years.
So, \[A = 1.44 \times P\]----- (1)
By using compound interest formulas. We get,
$A = P \times {\left( {1 + \dfrac{R}{{100}}} \right)^2}$----- (2)
From equation 1 and 2. We get,
$1.44 \times P = P \times {\left( {1 + \dfrac{R}{{100}}} \right)^2}$
$\Rightarrow 1.44 = {\left( {1 + \dfrac{R}{{100}}} \right)^2}$
Squaring on both sides.
$\sqrt {1.44} = \left( {1 + \dfrac{R}{{100}}} \right)$
Value of square root 1.44 is 1.2.
$1.2 = \left( {1 + \dfrac{R}{{100}}}
\right)$
We can also write 1.2 as 1 + 0.2.
$1 + 0.2 = \left( {1 + \dfrac{R}{{100}}} \right)$
We can write 0.2 as 2/10.
$1 + \dfrac{2}{{10}} = \left( {1 + \dfrac{R}{{100}}} \right)$
$\Rightarrow \dfrac{2}{{10}} = \dfrac{R}{{100}}$
$\Rightarrow 2 = \dfrac{R}{{10}}$
$\Rightarrow R = 2 \times 10$
$\Rightarrow R = 20\% $
So, the rate percent compound interest is $20\%.$
Note: Compound interest (or compounding interest) is the interest on a loan or deposit calculated based on both the initial principal and the accumulated interest from previous periods. The rate at which compound interest accrues depends on the frequency of compounding, such that the higher the number of compounding periods, the greater the compound interest. Compound interest is calculated by multiplying the initial principal amount by one plus the annual interest rate raised to the number of compound periods minus one. The total initial amount of the loan is then subtracted from the resulting value.
Solution : Let Rs. `x` becomes double in `n` years at the rate of compound interest `r%` per annum. <br> `:.x(1+r/100)^(n)=2x` <br> `implies(1+r/100)^(n)=2`……….1 <br> Also let Rs. `x` becomes 4-tmes in `t` years at the same rate of interest `r%` per annum. <br> `:.x xx(1+r/100)^(t)=4x` <br> `implies(1+r/100)^(t)=4` <br> `implies(1+r/100)^(t)=2^(2)` <br> `implies(1+r/100)^(t)={(1+r/100)^(n)}^(2)[ :' "by" (1),2=(1+r/100)^(n)]` <br> `implies(1+r/100)^(t)=(1+r/100)^(2n)` <br> `=t=2n` <br> Hence in `2n` years the given principal becomes 4-times.
- Aptitude
- Simple and compound interest
| A) 100.00% |
| B) 75.00% |
| C) 50.00% |
| D) 20.00% |
Correct Answer:
| A) 100.00% |
Description for Correct answer:
\( \Large Principal\ \ Amount \)
\( \Large 1 \rightarrow\ \ 4 \)
\( \Large 4=1 \left(1+\frac{r}{100}\right)^{2}\)
\( \Large 4= \left(1+\frac{r}{100}\right)^{2}\)
r = 100 %
Part of solved Simple and compound interest questions and answers : >> Aptitude >> Simple and compound interest
A. 100%
B. 75%
C. 50%
D. 20%
Solution(By Examveda Team)
$$\eqalign{ & {\text{Principal}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Amount}} \cr & \,\,\,\,\,\,\,\,\,{\text{1}}\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,{\text{4}} \cr & \Rightarrow 4 = 1{\left( {1 + \frac{r}{{100}}} \right)^2} \cr & \Rightarrow 4 = {\left( {1 + \frac{r}{{100}}} \right)^2} \cr & \Rightarrow r = 100\% \cr & \cr & {\text{Alternate}} \cr & {\text{Principal}}\,\,\,\,\,\,\,\,\,\,\,\,{\text{Amount}} \cr & \,\,\,\,\,\,\,\,\,\root 2 \of 1 \,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\root 2 \of 4 \cr & \,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,2 \cr & \Rightarrow {\text{Rate of interest}} \cr & {\text{ = }}\frac{{\left( {2 - 1} \right)}}{1} \times 100 = 100\% \cr} $$
Let Principal (P) = Rs. 100
then Amount (A) = Rs. 400
Period (n) = 2 years or 4 half years.
Let R be the rate % half-yearly, then
AP=(1+R100)n⇒400100=(1+R100)4
⇒(1+R100)4=41
⇒[(1+R100)2]2=(2)2
⇒(1+R100)2=2⇒(1+R100)=√2
⇒1+R100=1.4142
⇒R100=1.4142−1.0000
⇒R100=0.4142⇒R=0.4142×100
⇒R=41.42
∴ Rate
%=41.42% half yearly and
82.84% p.a.