Question

a sum of money was invested for 3years interest being compounded anually. the rates for successive years were 10%, 15% and 18% respectively if the compounded interest for the se ond year amounted to ₹4950 find the sum invested​

Answer 1

Answer:

Step-by-step explanation:

A=P(1+r¹/100)¹×(1+r²/100)¹

A=X(1+10/100)¹×(1+15/100)¹=>>>1.265x

Given,

1.265x-1.1x=4950

0.165x=4950

X=30000 rupees.

Hope it will help you

pls mark me as brainliest : )

Answer 2

Given :-

Rate for the 1st year = 10%

Rate for the 2nd year = 15%

Rate for the 3rd year = 18%

Amount of compound interest for the 2nd year = ₹4950

To Find :-

The sum invested​ if the compounded interest for the 2nd year amounted to ₹4950

Solution :-

We know that,

• p = Principle
• a = Amount
• r = Rate of interest
• n = Number of compound period

Given that,

Compound interest for 2nd year = Rs. 4950 and rate = 15%

According to the question,

Compound interest = $\sf P \bigg[ \bigg(1+\dfrac{r}{100} \bigg)^{n}-1 \bigg]$

$\implies \sf 4950=P \bigg[ \bigg(1+\dfrac{15}{100} \bigg)^{1}-1 \bigg]$

$\implies \sf 4950=P \bigg[ \dfrac{3}{20} \bigg]$

$\implies \sf P=\dfrac{4950 \times 20}{3}$

$\implies \sf P=Rs. \: 33000$

Hence, the amount at the end of 2nd year is Rs. 33000

For the first 2 years,

$\longrightarrow \sf A=Rs. \: 33000 \ ; \ r_1=10 \%$

$\longrightarrow \sf A=P \bigg(1+\dfrac{r_1}{100} \bigg)$

$\implies \sf 33000=P \bigg(1+ \dfrac{10}{100} \bigg)$

$\implies \sf 33000=P \bigg(\dfrac{11}{10}\bigg)$

$\implies \sf P=\dfrac{33000 \times 10}{11}$

$\implies \sf P= Rs. \: 30000$

Therefore, the sum invested is Rs. 30000

To Note :-

Compound interest is the interest calculated on the principal and the interest accumulated over the previous period.

The compound interest formula: Compound Interest = Amount – Principal