Question

What rate gives $280 as interest on a sum of $56000 in 2 years.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Rate\%=0.25\%}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Principal(p) = \ \:56000 \\ \\ \tt: \implies Simple \: Interest(S.I) =\ \: 280 \\ \\ \tt: \implies Time(t) = 2 \: years \\ \\ \red{\underline \bold{To \: Find:}}\\ \tt: \implies rate\%(r) = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies S.I= \frac{p \times r \times t}{100} \\ \\ \tt: \implies 280 = \frac{56000 \times r \times 2}{100} \\ \\ \tt: \implies 280 \times 100 = 112000 \times r \\ \\ \tt: \implies \frac{28000}{112000} = r \\ \\ \green{\tt: \implies r = 0.25\% } \\ \\ \green{\tt \therefore Rate\% \: \: is \: \: 0.25\%} \\ \\ \blue{\bold{Some \: related \: formula}} \\ \orange{\tt \circ \: A= p(1 + \frac{r}{100} )^{t} } \\ \\ \orange{\tt \circ \:A = p(1 + \frac{ \frac{r}{2} }{100} )^{2t} } \\ \\ \orange{\tt \circ \: A= p(1 + \frac{ \frac{r}{4} }{100} )^{4t} } \\ \\ \orange{ \tt \circ \: C.I = A - p} \\ \\ \orange{ \tt \circ \: A = p + S.I}$

Answer 2

Given:

• Principal,P = $56000

• Time, T = 2 Years

• Simple interest, S.I = $280

To find out:

✪Find the rate %.

Formula used:

$\boxed{ \gray{S.I} = \purple {\frac{P \times R \times T}{100} }}$

Solution:

Let the rate be x .

Then,

$S.I = \frac{P \times R \times T}{100}$

$\rightarrow \: 280 = \frac{560 \cancel{00} \times x \times 2}{ \cancel{100} }$

$\rightarrow280 = 560 \times x \times 2$

$\rightarrow280 = 1120 \times x$

$\rightarrow \: x= \frac{280}{1120}$

$\rightarrow \: x = \frac{1}{4}$

∴ Rate % p.a. = 0.25 %

Some important Terms:

• Principal, P = The money taken by borrowed or lent.

• Rate,R = The interest given by borrower to the lender every person.

• Amount, A= The total money given by borrower to the lender.

• Time, T = The time after which the money is to be returned.