Question

A's capital exceeds B's capital by 20.5% B invests his capital at 20% p.a. for 3 years intrest compounded annually​

Answer 1

Answer:

Given :-

A's capital exceeds B's capital by 20.5%.

B invests his capital at 20% p.a. for 3 years, interest compounded annually.

To Find :-

At what rate percentage p.a. must A invest his capital at simple interest so that at the end of 3 years both get the same amount ?

Solution :-

→ A capital = 20.5% more than B capital

→ A capital = 120.5% of B capital

→ A capital = (1205/1000) of B capital

→ A capital = (241/200) of B capital .

→ (A / B) = (241 / 200)

Now, Let us Assume that, A capital is Rs.241x and B capital is Rs.200x .

Case 1) :-

→ A capital = Principal = Rs.241x

→ Time = 3 years.

→ Rate = 20% P.A. compounded annually .

So,

→ Amount = Principal * [ 1 + (Rate/100) ]^(Time)

Putting all values we get,

→ A = (241x) * [1 + (20/100)]³

→ A = (241x) * [1 + (1/5)]³

→ A = (241x) * (6/5)³

→ A = Rs. {(241x * 216) / 125}

Case 2) :-

→ → B capital = Principal = Rs.200x

→ Time = 3 years.

→ Rate = Let R % P.A.

So,

→ Amount = Principal + (Principal * Rate * Time/100)

Putting values we get,

→ A = 200x + (200x * R * 3/100)

→ A = Rs.(200x + 6xR)

Now, we have given that, at the end of 3 years, both get the same amount .

Therefore, Comparing Both Amount Now, we get,

→ {(241x * 216) / 125} = (200x + 6xR)

→ 241x * 216 = 125(200x + 6xR)

→ 52,056x = 25000x + 750xR

→ 750xR = 52056x - 25000x

→ 750xR = 27056x

Dividing both sides by 750x now,

→ R ≈ 36.07%

Hence, to receive same amount after 3 years A must invest his capital at 36.07% P.A.