Question

the simple interest on a certain sum for 3 years at 10% per annum is rupees 829.50 find the sum ​

Answer 1

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

$\green{\tt{\therefore{Principal=2765\:rupees}}}$

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

$\green{\underline \bold{Given:}} \\ \tt: \implies Simple\:Interest(S.I) = 829.5\: rupees \\ \\ \tt: \implies Rate\%(r) = 5\% \\ \\ \tt: \implies Time(t) = 3 \: years \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Principal(p)= ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies S.I = \frac{p \times r \times t}{100} \\ \\ \tt: \implies 829.5= \frac{p \times 10 \times 3}{100} \\ \\ \tt: \implies 829.5\times 100=p \times 30 \\ \\ \green{\tt: \implies p= 2765\: rupees} \\ \\ \bold{for \: amount : } \\ \tt: \implies A= p + S.I\\ \\ \tt: \implies A= 2765 + 829.5\\ \\ \green{\tt: \implies A= 3594.5 \: rupees} \\ \\ \blue{ \bold{Some \: related \: formula}} \\ \orange{\tt \circ \: A = p(1 + \frac{r}{100})^{t}} \\ \\ \orange{\tt \circ \: A = p(1 + \frac{ \frac{r}{2} }{100})^{2t}} \\ \\ \orange{\tt \circ \: A = p(1 + \frac{ \frac{r}{4} }{100})^{4t}}$

Answer 2

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$\tt{\huge{\purple{ .............. }}}$

QUESTION :

the simple interest on a certain sum for 3 years at 10% per annum is rupees 829.50 find the sum .

SOLUTION :

Let the Principle be Rs. X

SI = X × 3 × 10 / 100 = 0.3 X

0.3 X = 829.5

=> X = 8295 / 3 = Rs. 2765

Answer : The required sum is Rs. 2765.