Question

How much money invested at compound interest wil yield 6350.50 at the end of 3years ?for the first year the rate of interest is 5%;For second year it increases to 12% and for third year it decreases to 8%.
ANSWER CORRECTLY ​

Answer 1

Given :-

Amount = ₹6350.50

Time = 3 years

Rate for 1st year = 5%

Rate for 2nd year = 12%

Rate for 3rd year = 8%

To Find :-

The principal

Solution :-

Let the original principal be ₹x

For the first year -

P = x ; R = 5% ; T = 1 year

$Interest\: =\: {\dfrac{P \times R \times T}{100}}$

$=\: {\dfrac{x \times 5 \times 1}{100}}$

$=\: {\dfrac{x}{20}}$

Amount = Principal + Interest

$=\: x\: +\: {\dfrac{x}{20}}$

$=\: {\dfrac{20x + x}{20}}$

$=\: {\dfrac{21x}{20}}$

For the second year -

P = $=\: {\dfrac{21x}{20}}$

R = 12%

T = 1 year

$Interest\: =\: {\dfrac{P \times R \times T}{100}}$

$=\: {\dfrac{x \times 12 \times 1}{100 \times 20}}$

$=\: {\dfrac{3x}{500}}$

Amount = Principal + Interest

$=\: {\dfrac{21x}{20}}\: +\: {\dfrac{3x}{500}}$

$=\: {\dfrac{525x + 3x}{500}}$

$=\: {\dfrac{528x}{500}}$

$=\: {\dfrac{132x}{125}}$

For the third year -

P $=\: {\dfrac{132x}{125}}$

R = 8%

T = 1 year

$Interest\: =\: {\dfrac{P \times R \times T}{100}}$

$Interest\: =\: {\dfrac{132x \times \not{8}^2 \times 1}{\not{100}^{25} \times 125}}$

$=\: {\dfrac{264x}{3125}}$

Amount = Principal + Interest

$=\: {\dfrac{132x}{125}}\: +\: {\dfrac{264x}{3125}}$

$=\: {\dfrac{3564x}{3125}}$

According to question,

$=\: {\dfrac{3564x}{3125}}\: =\: 6350.50$

$x\: =\: {\dfrac{6350.50 \times 3125}{3564}}$

=> x = 5568.26

Therefore, the principal is 5568.26

Answer 2

Answer is 5568.26.                            

hope it helps..