Question

Calculate compound interest where P = 1000, R= 10 p.c.p.a. , N = = 2 years​

Answer 1

Step-by-step explanation:

QUESTION :- find the compound interest if Rs 1000 are invested for 2 years at this rate of 10 p.c.p.a.

ANSWER:- p=Rs.1000. r = 14% t(n)=2Years.

.°. p(1+r/100)^t. .°. 1000(1+14/100)² .°. 1000(1+7/50)² .°. 1000(1+57/7)² .°. 1000 × 57×57/ 7×7. .°. 142.85 ×3249 ×7. .°. 20.40 ×3249.

.°. Rs.66,279.6. HENCE,

Answer 2

Given :-

• Principal amount is 1000  

• Rate of interest is 10 p.c.p.a.

• Time is 2 years

To Find :-

• Compound Interest

Solution :-

~Here, we’re given the principal , rate of interest and time for which the money is invested and we need to find the Compound Interest ( CI ) after the given time. We can easily find the amount by putting the values in it’s formula and then Compound Interest.  

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Here,  

• The principal  ( P ) is Rs. 1,000

• The rate ( R ) is 10 %  

• The time ( n )  is 2 years  

• The amount will be ( A )  

• Compound Interest be ( CI )  

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As we know that ,

$\boxed{\sf{ \maltese \;\; A = P \bigg\{ 1 + \dfrac{R}{100} \bigg\}^{n} }}$

$\boxed{ \sf { \maltese \;\; CI = A -P }}$

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Finding the Amount :-

$\sf \dashrightarrow A = 1000 \bigg\{ 1 + \dfrac{10}{100} \bigg\}^{2}$

$\sf \dashrightarrow A = 1000 \bigg\{ \dfrac{11}{10} \bigg\}^{2}$

$\sf \dashrightarrow A = 1000 \times \dfrac{11}{10} \times \dfrac{11}{10}$

$\sf \dashrightarrow A = 10 \times 11 \times 11$

$\boxed{\bf{ \bigstar \;\; Amount = Rs. \; 1210 }}$

Finding the Compound Interest :-

$\sf \dashrightarrow CI = Rs.\;1210 -Rs.\;1000$

$\boxed{ \bf { \bigstar \;\; Compound\;Interest = Rs. \; 210 }}$

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Hence,  

• The compound interest is Rs. 210  

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