Question

Asum of rs 30000 invested in a scheme where the interest gets compounded annually and grows up to rs 51840 in 3 years.How much interest(in rs) would have got accured in six months in the same scheme had the interest been compounded quarterly?
a.3024
b.3075
c.3126
d.2975

Answer 1

Answer:

c answer not correct b answer some correct d answer not correct take tution from me then i will tell

Answer 2

$\color{lime}{ \underline{ \underline{ \color{red}{ \sf{Correct \: answer : }}}}}$

$\color{dark} \tt{{b. \: ₹3,075}} \: \color{gray}{\boxed{\checkmark}}$

$\color{lime}{ \underline{ \underline{\sf\color{blue}{ Given : }}}}$

• A sum of ₹30,000 is invested in a scheme with compounded annually.

• It extends upto ₹51,840 in duration of 3 years.

$\color{lime}{ \underline{ \underline{\sf\color{blue}{ To \: find : }}}}$

The interest in six months being compounded quarterly.

$\color{lime}{ \underline{ \underline{\sf\color{blue}{ Solution : }}}}$

We know that,

${ \underline{ \boxed{ \sf{ \blue{amount =p ( {1 + \frac{r}{100}) }^{n} }}}}} \bold \gray \dag$

Where,

p(principal) = ₹30,000

A(amount) = ₹51,840

r(rate) = ??

n(time) = 3 years.

Substituting the given values as follows :

$\sf \implies amount = p {(1 + \frac{r}{100} )}^{n}$

$\rm \implies 51840 = 30000 {(1 + \frac{r}{100}) }^{3}$

$\rm \implies \frac{51840}{30000} =( {1 + \frac{r}{100} )}^{3}$

$\rm \implies \frac{1296}{750} = {(1 + \frac{r}{100} )}^{3}$

$\rm \implies ({ \frac{6}{5} )}^{3} = {(1 + \frac{r}{100} )}^{3}$

$\sf \implies \: {(1 + \frac{1}{5}) }^{3} = {(1 + \frac{r}{100} )}^{3}$

$\gray{ \rm {( \sqrt[3]{} \: both \: sides })}$

$\rm \implies (1 + \frac{1}{5} ) = (1 + \frac{r}{100} )$

$\rm \implies (1 + \frac{20}{100} ) = (1 + \frac{r}{100} )$

$\rm \implies \frac{120}{ \cancel{100}} = \frac{100 + r}{ \cancel{100}}$

$\rm \implies r = 120 - 100$

$\bf \therefore \: r = 20\%$

After solving this problem we get :

r(rate %) = 20%

Now,

We need to find the amount after 6 months compounded quarterly.

we know that,

If any sum is compounded quarterly then we need to multiply the time(n) by 4 and rate(r) is divided by 4.

Hence,

n(time) = 6 * 4 months = 24 months

or, 2 years

And, also

r(rate) = 20/4 = 5%

$\therefore$ Formula used :

${ \underline{ \boxed{ \sf amount = p( {1 + \frac{ {r}{} }{100 \times 4} )}^{{n} {} } }}}$

Where,

p(principal) = ₹30,000

n(time) = 2 years.

r(rate %) = 5%

$☯ \begin{gathered}\underline{\boldsymbol{According\: to \:the\: question :}}\\\end{gathered}$

$\sf \longrightarrow A(amount ) = 30000( {1 + \frac{ \cancel 5}{ \cancel{100}} )}^{ 2}$

$\sf \longrightarrow A = 30000 {( \frac{21}{20}) }^{2}$

$\sf \longrightarrow A =300 \cancel{00} \times \frac{21}{2 \cancel 0} \times \frac{21}{2 \cancel 0}$

$\sf \longrightarrow A ={ { \frac{ \cancel{300} \times 21 \times 21}{ \cancel 4} }}$

$\sf \longrightarrow A =75 \times 21 \times 21$

$\sf \longrightarrow A = ₹33,075$

Amount = ₹33,075

And also,

➠ C. I. = Amount - Principal

➠ C. I. = ₹(33,075 - 30,000)

➠ C. I. = ₹3,075 ans.

Hence, proved