Question

A sum of money is placed at simple interest for 3 years at 10% per annum and then the amount is invested for 2 years at the same rate at compund interest. If the total amount of 5 years became Rs. 471900,what was tha sum?Find it.

Answer 1

Answer:

solution for compound interest :

A = 471900 i = 10/100 =0.1 n = 2

A = P(1+i)ñ

471900 = P(1+0.1)²

P = 471900/1.21

P = 390000

solution for first 3 year simple interest :

P = A-i = 390000-i

i = ??

R = 10 N = 3

i = PRN / 100

i = (390000-i) (10) (3) /100

10i = 11,70,000-3i

13i = 11,70,000

i = 11,70,000/13

i = 90000

Money invested by A is = 390000-90000 = 300,000

HOPE IT'S HELPING

Answer 2

The principal amount was ₹3,00,000.

Given:

• A sum of money is placed at simple interest for 3 years at 10% per annum and,
• The amount then invested for 2 years at the same rate at compound interest.
• If the total amount of 5 years became Rs. 471900.

To find:

• What was the sum? Find it.

Solution:

Formula to be used:

Simple interest: $\bf SI= \frac{PRT}{100}$

here,

P: Principal amount

R: Rate of interest

T:Time in years.

Amount after adding compound Interest:$A=P\left(1 + \frac{R}{100} \right) ^{T}$

here,

A: Total sum(Principal amount+Compound interest)

Step 1:

Write equation from the given statement.

ATQ,

Let the principal amount be ₹P.

R= 10%

T: 3 years

So,

$SI = \frac{P \times 10 \times 3}{100}$

$\bf SI = 0.3P$

Total amount: P+SI

$T_1 = P+ 0.3P$

$\bf T_1 = 1.3P$

Thus,

1.3P will be invested on compound interest for next two year at 10% per annum.

Step 2:

Write an equation for compound interest.

Here,

P= 1.3P

R= 10%

T: 2 years

So,

$A = 1.3P\left( {1 + \frac{10}{100} }\right)^{2}$

$A = 1.3P\times \frac{121}{100}$

ATQ,

Total amount after 5 years will be ₹471900, so put this value.

$471900 = \frac{1.3 \times 121P}{100}$

$P= \frac{47190000}{121 \times 1.3}$

$\bf P = 3,00,000$

Thus,

The principal amount was ₹3,00,000.

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