Question

If rate =5%pa time=2yrs ci=246 find si at the rate of 6%pa for 3 yrs

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

→ Rate = 5% P.A.

→ Time = 2 Years.

→ CI = Rs.246

→ Principal = Let P .

we know That,

→ CI = P[{1 + (R/100)}^T - 1]

Putting values we get,

→ 246 = P[{ 1 + (5/100)}² - 1]

→ 246 = P[{1 + (1/20)}² - 1]

→ 246 = P[(21/20)² - 1]

→ 246 = P[(441 - 400)/400]

→ 246 = P(41/400)

→ P = (246 * 400)/41

→ P = Rs.2400 .

_________________

Now,

→ P = 2400

→ R = 6%

→ T = 3 Years.

→ SI = (P * R * T)/100

→ SI = (2400 * 6 * 3) / 100

→ SI = 24 * 6 * 3

→ SI = Rs.432 (Ans.)

Hence, SI on the same sum will be Rs.432.

Answer 2

${ \huge{ \mathfrak{ \underline{ \underline{ \red{Solution:-}}}}}}$



${ \bold{ \orange{GIVEN- }}}$



${ \bold{ \underline{ \purple {for \: compound \: interest \: (C.I.)}}}}$



➺ Rate (R) = 5 % p.a.



➺ Time (T) = 2 years



➺ Compound Interest (C.I.) = Rs. 246



${ \underline{ \boxed{ \bold{ \pink{C.I. = P( {(1 + \frac{R }{100} )}^{T} - 1)}}}}}$



${ \bold{ 246 = P( {(1 + \frac{5}{100} )}^{2} - 1)}}$



${ \bold{ \implies{246 = P ( {(1 + \frac{1}{20} )}^{2} - 1)}}}$



${ \bold{ \implies{246 = P( {( \frac{21}{20}) }^{2} - 1)}}}$



${ \bold{ \implies{246 = P( \frac{441}{400} - 1)}}}$



${ \bold{ \implies{246 = P ( \frac{441 - 400}{400} )}}}$




${ \bold{ \implies{ \frac{41P}{400} = 246}}}$




${ \bold{ \implies{P = 6 \times 400}}}$



${ \boxed{ \bold{ \implies{ \red{ \: \: P = Rs. \: 2400 \: \: }}}}}$



Now,



${ \bold{ \underline{ \purple{for \: simple \: interest \: (S .I .)}}}}$



➺ Principal (P) = Rs. 2400



➺ Rate (R) = 6% p.a.



➺ Time (T) = 3 years



${ \underline{ \boxed{ \bold{ \pink{ \: \: S.I. = \frac{P \times R \times T}{100} \: \: }}}}}$



${ \bold{S.I. = \frac{2400 \times 6 \times 3}{100}}}$



${ \bold{ \implies{S.I. = 24 \times 6 \times 3}}}$



${ \boxed{ \boxed{ \bold{ \implies{ \red{ \: \: S.I . = Rs . \: 432 \: \: \: }}}}}}$