Question

QNETT. Calculate the amount when pa 1000, R = 10% per annum (compounded annually) Time
2 years​

Answer 1

Answer:

The amount is Rs. 1210.

Step-by-step explanation:

Solution :

Principal = Rs. 1000

Rate = 10%

Time = 2 years

$\bigstar \: {\boxed{\sf{A= P\left( 1 + \frac{R}{100}\right)^{N}}}}$

⇒ Amount = 1000(1 + 10/100)²

⇒ Amount = 1000(110/100)²

⇒ Amount = 1000(11/10)²

⇒ Amount = 1000 × 11/10 × 11/10

⇒ Amount = 10 × 11 × 11

⇒ Amount = 121 × 10

⇒ Amount = 1210

Amount = Rs. 1210

The amount is Rs. 1210.

Answer 2

$\huge\purple\star\underline\mathfrak\purple{Question :-}\purple\star$

Q. Calculate the amount when pa 1000, R = 10% per annum (compounded annually) Time 2 years

$\rm\huge\purple\star\underline\orange{Given :-}\purple\star$

• P .A = ₹1000
• 10% per annum.
• Time = 2years.

$\rm\huge\purple\star {\underline{\underline{\blue{To find :-}}}}\purple\star$

• Compound interest = ??

$\rm\huge\purple\star {\underline{\underline{\green{Solution :-}}}}\purple\star$

$\bf \purple{ \bold{ \underline{ \: here \: principal \: is \: 1000 \: { \: and } \red{\: r \: = \: 10\%per \: annum}{ \: and \: } \blue {\: n \: = 2years}}}} \\ \bf { \therefore} \: amount \: \: after \: \: 2 \: \: years = \: p ( 1 + \frac{r}{100} {)}^{n} \\ \tt \: = 1000 \times {( 1 + \frac{10}{100} )}^{2 } \\ \tt \: = \: 1000 \times {( 1 + \frac{1}{10}) }^{2} \\ \tt \: = 1000 \times {( \frac{11}{10} )}^{2} \\ \tt = 1000 \times \frac{11}{10} \times \frac{11}{10 } \\ \tt \: Amount =₹ 1210 \\ \bf \: \rm\purple{\green{ \therefore}} \green{ \underline{ \underline{compound \: interest \: = amount \: - \: principal}}} \\ \tt \: = 1210 - 1000 = ₹210$

$\rm\huge\purple\star {\boxed{\boxed{\pink{\bf{Amount \:\:= \:\: ₹1210}}}}}\purple\star$