Question

elation between A, P, r, n, CI
a) If Rs 4,800 is invested at a compound interest rate of 10% per annum then find
(i) the amount after 2 years (ii) the interest earned in 2 years​

Answer 1

✯ GIVEN :-

• The investment is ₹4,800.
• At a rate of 10%.

✯ TO FIND :-

• (i) The amount after 2 years.
• (ii)The Interest earned after 2 years.

✯ SOLUTION :-

• We know that,

➜ $\sf {A = P(1+r/100)^{n}}$.

Where,

• P(principal) = ₹4,800.
• R(rate) = 10%.
• T(time) = 2 years.

$\bf \:a \: (amount )= P(1 + \frac{r}{100} )^n \\ \bf \implies a\: = 4800 \times(1 + { \frac{ \cancel{10}}{ \cancel{100} })}^{2} \\ \implies \bf \: a \: = 4800 \times (1 + \frac{1}{10} )^{2} \\ \bf \implies \: a = 48 \cancel{00} \times \frac{11}{1 \cancel0} \times \frac{11}{1 \cancel0} \\ \implies \bf a \: = 48 \times 11 \times 11 \\ \bf \implies {\fcolorbox{orange}{pink} {a = ₹5808\:\:\: ans.}}$

$\bf \therefore \: c.i. = a(amount) + p(principal) \\ \bf \dashrightarrow \: c.i. = 5808 + 4800 ={ \fcolorbox{black}{lightgreen} {₹10608\:\:\: ans. }}$

Hence,

• (i) Amount after 2 years is ₹5,808.
• (ii) Interest earned after 2 years is ₹10,608.

✯ ADDITIONAL INFO ✯

• A(Amount) = $\sf (1+frac{r}{100})^n$.
• C. I. (compound interest)= Amount + principal.
• S. I. (simple interest)= $\sf \frac{r \times t \times p}{100}$

Answer 2

Answer:

✯ GIVEN :-

The investment is ₹4,800.

At a rate of 10%.

✯ TO FIND :-

(i) The amount after 2 years.

(ii)The Interest earned after 2 years.

✯ SOLUTION :-

We know that,

➜ \sf {A = P(1+r/100)^{n}}A=P(1+r/100)n .

Where,

P(principal) = ₹4,800.

R(rate) = 10%.

T(time) = 2 years.

\begin{gathered} \bf \:a \: (amount )= P(1 + \frac{r}{100} )^n \\ \bf \implies a\: = 4800 \times(1 + { \frac{ \cancel{10}}{ \cancel{100} })}^{2} \\ \implies \bf \: a \: = 4800 \times (1 + \frac{1}{10} )^{2} \\ \bf \implies \: a = 48 \cancel{00} \times \frac{11}{1 \cancel0} \times \frac{11}{1 \cancel0} \\ \implies \bf a \: = 48 \times 11 \times 11 \\ \bf \implies {\fcolorbox{orange}{pink} {a = ₹5808\:\:\: ans.}}\end{gathered}a(amount)=P(1+100r)n⟹a=4800×(1+10010)2⟹a=4800×(1+101)2⟹a=4800×1011×1011⟹a=48×11×11⟹a = ₹5808ans.

\begin{gathered} \bf \therefore \: c.i. = a(amount) + p(principal) \\ \bf \dashrightarrow \: c.i. = 5808 + 4800 ={ \fcolorbox{black}{lightgreen} {₹10608\:\:\: ans. }}\end{gathered}∴c.i.=a(amount)+p(principal)⇢c.i.=5808+4800=₹10608ans. 

Hence,

(i) Amount after 2 years is ₹5,808.

(ii) Interest earned after 2 years is ₹10,608.

✯ ADDITIONAL INFO ✯

A(Amount) = \sf (1+frac{r}{100})^n(1+fracr100)n .

C. I. (compound interest)= Amount + principal.

S. I. (simple interest)= \sf \frac{r \times t \times p}{100}100r×t×p