Question

Find the Principal amount. If the difference of.S.I and C.I at 12.5% per annum for 3 years is Rs. 750.

Answer 1

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

$\green{\tt{\therefore{Principal=15360\:rupees}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies C.I- S.I= 750 \: rupees \\ \\ \tt: \implies Rate\% (r)= 12.5\% \\ \\ \tt: \implies Time(t) = 3 \: years \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Principal =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies S.I = \frac{p \times r \times t}{100} \\ \\ \tt: \implies S.I= \frac{p \times 12.5 \times 3}{100} \\ \\ \tt: \implies S.I=0.375p - - - - - (1) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies A= p(1 + \frac{r}{100} )^{t} \\ \\ \tt: \implies A= p \times (1 + \frac{12.5}{100} )^{3} \\ \\ \tt: \implies A =p \times (1 + 0.125)^{3} \\ \\ \tt: \implies A =p \times 1.423828125 - - - - - (2) \\ \\ \bold{for \: compound \: interest} \\ \tt: \implies C.I=A - p \\ \\ \tt: \implies C.I= 1.423828125p - p \\ \\ \tt: \implies C.I=0.423828125p - - - - - (3) \\ \\ \bold{For \: Difference : } \\ \tt: \implies C.I- S.I = 750 \\ \\ \tt: \implies 0.423828125p - 0.375p = 750 \\ \\ \tt: \implies 0.048828125 p = 750 \\ \\ \tt: \implies p = \frac{750}{0.048828125} \\ \\ \green{\tt: \implies p = 15360 \: rupees}$

Answer 2

Correct Question :–

• Find the Principal amount. If the difference of C.I. and S.I. at 12.5% per annum for 3 years is Rs. 750.

GIVEN :–

• C.I. - S.I. = 750 rupees

• Rate = 12.5%

• Time = 3 year

TO FIND :–

• Principal amount = ?

SOLUTION :–

• We know that –

$\\ \implies{ \bold{S.I. = \dfrac{p \times r \times t}{100} }}$

$\\ \implies{ \bold{S.I. = \dfrac{p \times 12.5 \times 3}{100} }}$

$\\ \implies{ \bold{S.I. = 0.375p }}$

• let's find C.I. –

$\\ \implies{ \bold{C.I. = A - p }}$

• We know that Amount –

$\\ \implies{ \bold{A = p \left(1 + \frac{r}{100} \right)^{t} }}$

• So that –

$\\ \implies{ \bold{C.I. = p \left(1 + \frac{r}{100} \right)^{t} - p }}$

$\\ \implies{ \bold{C.I. = p \left(1 + \frac{12.5}{100} \right)^{3} - p }}$

$\\ \implies{ \bold{C.I. = p \left( \dfrac{112.5}{100} \right)^{3} - p }}$

$\\ \implies{ \bold{C.I. = p \left( 1.125 \right)^{3} - p }}$

$\\ \implies{ \bold{C.I. = p \left(1.423828125 \right) - p }}$

$\\ \implies{ \bold{C.I. =0.423828125 p }}$

• According to the question –

$\\ \implies{ \bold{C.I. - S.I. = 750 }}$

$\\ \implies{ \bold{0.423828125 p - 0.375p = 750 }}$

$\\ \implies{ \bold{0.048828125p = 750 }}$

$\\ \implies \large{ \boxed{ \bold{p = 15,360 \: \: rupees}}}$