Question

at what rate per cent compound intrest,does a sum of money become 1.44 times of itself in 2 years

pls pls pls answer very urgent ​

Answer 1

GiVen:

Sum of the money compounded per year becomes 1.44 times of itself in 2 years.

AnSwer:

★ According to the question,

$: \implies \sf \: \: \: S = P { (\frac{ {1 + r} }{100})}^{2} \\ \\ : \implies \sf \: \: \: 1.44P = P{ (\frac{ {1 + r} }{100})}^{2} \\ \\ : \implies \sf \: \: \: 1+ \frac{r}{100} = \sqrt{1.44} \\ \\ : \implies \sf \: \: \:100+ \frac{r}{100} = 1.2 \\ \\ : \implies \sf \: \: \: 100+r = 1.2 \times 100 \\ \\ : \implies \sf \: \: \:100+r = 120 \\ \\ : \implies \sf \: \: \: r = 120-100 \\ \\ : \implies \sf \: \: \:{ \underline{ \boxed{ \sf { \purple{ \: r = 20\% \: }}}}} \: \bigstar$

Hence, rate is 20%.

Answer 2

20%

Given:-

• sum of money compunded per year become 1.44 times of itself in 2 years.

To find:-

• Rate percent

Solution:-

★ Step-by-step explanation:

$:\implies\bf S = P \bigg( \dfrac{1 + r}{100} \bigg)^2 \\ \\ \dashrightarrow\sf 1.44 \cancel{P} = \cancel{P} \bigg( \dfrac{1 + r}{100} \bigg)^2$

$\;\;\star\;\sf {\underline{Taking\;sqrt.\;both\;side}}$

$\dashrightarrow\sf \sqrt{1.44} = \sqrt{ \bigg( \dfrac{1 + r}{100} \bigg)^2} \\ \\ \dashrightarrow\sf 1.2 = \dfrac{100 + r}{100} \\ \\ \dashrightarrow\sf 1.2 \times 100 = 100 + r \\ \\ \dashrightarrow\sf 120 = 100 + r \\ \\ \dashrightarrow\sf 120 - 100 = r \\ \\ \dashrightarrow\bf {\underline{\underline{\boxed{\red{20}}}}}\bigstar$

★ Hence, rate percent is 20%.

$\rule{150}{4}$