Question

in how many years 50000 amount to 86400 at 20% per annum, if the interest is compounded annually
plz explain fully.​

Answer 1

$\underline{ \underline{ \Large \pmb{ \sf { \pink{Answer:}} } }}$

• Number of years = 3 years

$\underline{ \underline{ \Large\pmb{ \sf { \pink{Given:}}} } }$

• Amount = Rs. 86400

• Principal = 50000

• Rate of interest = 20% p.a

$\underline{ \underline{ \Large \pmb{ \sf { \pink{To \: calculate :}} }} }$

• Time (n years)

$\underline{ \underline{ \Large \pmb{ \sf { \pink{Calculation :}} }} }$

Let us assume the number of years as "n". We'll form a suitable linear equation first and then by using the laws of exponents , we'll find the value of n.

As we know that,

$\longmapsto \underline { \boxed{\pmb{ \large\rm { A = {P \Bigg \lgroup 1 + \dfrac{R}{100} \Bigg \rgroup}^{n} }}}}$

Where,

✰ A = amount

✰ P = Principal

✰ R = Rate of interest

✰ n = time or number of years.

Clearly, this an equation. Now, substitute the values we are given in the question.

$\longmapsto \rm { 86400 = { 50000 \Bigg \lgroup 1 + \dfrac{20}{100} \Bigg \rgroup}^{n} }$

Preforming simplification in the RHS.

$\longmapsto \rm { 86400 = { 50000 \Bigg \lgroup \dfrac{100 + 20}{100} \Bigg \rgroup}^{n} }$

$\longmapsto \rm { 86400 = { 50000 \Bigg \lgroup \dfrac{120}{100} \Bigg \rgroup}^{n} }$

Transposing 50000 from RHS to LHS. Sign will be changed, as it is in the form of multiplication ; it'll become in the form of division in the RHS.

$\longmapsto \rm { \dfrac{86400}{50000} = { \Bigg \lgroup \dfrac{120}{100} \Bigg \rgroup}^{n} }$

Cancel the numerator and denominator in LHS by 5.

$\longmapsto \rm { \dfrac{17280}{10000} = { \Bigg \lgroup \dfrac{120}{100} \Bigg \rgroup}^{n} }$

Cancel the numerator and denominator in LHS by 10.

$\longmapsto \rm { \dfrac{1728}{1000} = { \Bigg \lgroup \dfrac{120}{100} \Bigg \rgroup}^{n} }$

Now, we have to write the numbers in LHS in exponential form to find the value of n. As we know that,

• 1728 = 12 × 12 × 12

That means, we can write

• 1728 = 12³

Also,

• 1000 = 10 × 10 × 10

That means, we can write

• 1000 = 10³

Now, substituting values.

$\longmapsto \rm { \dfrac{{(12)}^{3}}{{(10)}^{3}} = { \Bigg \lgroup \dfrac{120}{100} \Bigg \rgroup}^{n} }$

By using the laws of exponents,

• $\boxed{ \pmb {\rm { \dfrac{a^n}{b^n} = { \Bigg ( \dfrac{a}{b} \Bigg )}^{n} }}}$

$\longmapsto \rm { {\Bigg \lgroup \dfrac{12}{10} \Bigg \rgroup }^{3} = { \Bigg \lgroup \dfrac{120}{100} \Bigg \rgroup}^{n} }$

Cancel the denominator and numerator in RHS by 10.

$\longmapsto \rm { {\Bigg \lgroup \dfrac{12}{10} \Bigg \rgroup }^{3} = {\Bigg \lgroup \dfrac{12}{10} \Bigg \rgroup}^{n} }$

As we know that,

• If the bases are same, then their powers must be same.

Henceforth,

$\longmapsto \underline{ \boxed{\pmb{\rm \red { n = 3 \: years }}} }$

Therefore,

• Here, number of years (n) is 3 years.