Question

The simple interest on a sum of money for 2 years at 4% per annum Is rp.340 Find the:

Sum of money

The compound interest On this Sum for one year payable half -yearly at the same rate .

Answer 1

Answer:

Let sum of money be P

now, P=SI×100/R×T

=340×100/2×4

=85×50

=4250

so,sum of money is=₹4250

Answer 2

Answer:

• Sum of money = ₹4250
• Compound interest = ₹4421.7

Explanation:

Given that,

S. I. (Simple interest) on a sum of money for 2 years at 4% p.a.(per annum) is ₹340

We know,

$\rm \longrightarrow \: Simple \: interest = \dfrac{p \times r \times t}{100}$

Where, p denotes the sum of money, r is the rate, and t is the time.

According to the question,

$\rm \implies \: 340 = \dfrac{p \times 4 \times 2}{100}$

$\rm \implies \: 340 = \dfrac{8p}{100}$

$\rm \implies \: 340 = \dfrac{2p}{25}$

$\rm \implies \: \dfrac{340 \times 25}{2} = p$

$\rm \implies \: 170 \times 25 = p$

$\rm \implies \:4250 = p$

∴ Sum of money is ₹4250

Now,

Calculating compound interest for 1 year at the same rate i.e., 4% compounded half-yearly.

Since, it is compounded half-yearly,

Time = 1 year × 2 = 2 years.

Rate = 4 ÷ 2 = 2%

Amount formula :-

$\rm \longrightarrow \: p \bigg(1 + \dfrac{r}{100} \bigg)^{n}$

Where,

• p(principal) = ₹4250
• r(rate) = 2%
• n(time) = 2 years.

So,

$\rm \longrightarrow \: 4250 \bigg(1 + \dfrac{2}{100} \bigg)^{2}$

$\rm \longrightarrow \: 4250 \bigg(1 + \dfrac{1}{50} \bigg)^{2}$

$\rm \longrightarrow \: 4250 \bigg(\dfrac{51}{50} \bigg)^{2}$

$\rm \longrightarrow \: 4250 \times \dfrac{2601}{2500}$

$\rm \longrightarrow \: 4421.7$

∴ Compound interest payable is ₹4421.7