Math, asked by kdm0006, 9 months ago

A certain sum of money is invested at the rate of 5% per annum compound interest, the interest compounded annually.If the difference between the interests of third year and first year is Rs.102.50 , find the sum.

Answers

Answered by RvChaudharY50
364

Solution :-

→ Let Principal = P

→ Rate = 5% per annum compounded Annually .

→ Diff. b/w CI of 3rd year - CI of first year = Rs.102.5

First Year :-

Amount = P(1+R/100)^T

→ A = P(1 + 5/100)¹

→ A = P(1 + 0.05)

→ A = 1.05P

So,

CI of first year = Amount - Principal = 1.05P - P = 0.05P.

__________________

Second Year :-

→ Amount = P(1+R/100)^T

→ A = P(1 + 5/100)²

→ A = P(1 + 0.05)²

→ A = (1.05)²P

→ A = 1.1025P

So,

→ CI of first year = Amount - Principal = 1.1025P - P = 0.1025P.

__________________

Third Year :-

→ Amount = P(1+R/100)^T

→ A = P(1 + 5/100)³

→ A = P(1 + 0.05)³

→ A = (1.05)³P

→ A = 1.157625P

So,

→ CI of first year = Amount - Principal = 1.157625P - P = 0.157625P

__________________

Now,

CI for third year = CI till third year-CI till second year

→ CI for third year = 0.157625P - 0.1025P = 0.055125P

__________________

Therefore,

→ Diff. b/w 3rd year CI & 1st year CI = 0.055125P - 0.05P = 0.005125P.

__________________

Hence,

0.005125P = 102.5

→ P = (102.5) ÷ (0.005125)

→ P = Rs.20,000 (Ans.)

Hence, The Required sum is Rs.20,000 .

Answered by rupalysharma5
7

Step-by-step explanation:

ANSWER

Let the sum (Principle) = Rs. 100

C.I. of 1st year = Rs. \frac{100\times 5\times 1}{100} = Rs. 5$$

And, amount of 1st year = Rs. 100 + Rs. 5=Rs. 105$$

⇒ The principke for 2nd year = Rs. 105

C.I. of 2nd year =Rs.

100

105×5×1

=Rs.5.25

And, amount of 2nd year =Rs.105+Rs.5.25=Rs.110.25

⇒ The principle for 3rd year =Rs.110.25

C.I. of 3rd year =Rs.

100

110×5×1

=Rs.5.5125

Difference between C.L of 1st year and C.L of 3rd year =Rs.5.5125−Rs.5=Rs.0.5125

Now, when the difference of interest =Rs.0.5125,sum=Rs.100

And, when the difference of interest =Rs.61.50,sum=Rs.

0.5125

100

×61.50=Rs.12,000

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