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How much will a sum of ¥4000 amount to in 1 1/2 yrs at 10% per annum compound interest,interest being paid half-yearly​

Sagot :

semsee45

Answer:

¥4630.50

Step-by-step explanation:

Compound Interest Formula

[tex]\large \text{$ \sf A=P(1+\frac{r}{n})^{nt} $}[/tex]

where:

  • A = final amount
  • P = principal amount
  • r = interest rate (in decimal form)
  • n = number of times interest applied per time period
  • t = number of time periods elapsed

Given:

  • P = 4000
  • r = 10% = 0.1
  • n = 2
  • t = 1.5

Substitute the given values into the formula and solve for A:

[tex]\implies \sf A=4000\left(1+\dfrac{0.1}{2}\right)^{2 \times 1.5}[/tex]

[tex]\implies \sf A=4000\left(1.05\right)^{3}[/tex]

[tex]\implies \sf A=4630.50[/tex]

Itz5151

Answer:

¥4630.50

Step-by-step explanation:

Here,

  • P = ¥4000
  • n = 1 1/2 = 3/2 years
  • r = 10% per annum

ATQ, as well,the interest is payable in half-yearly.

We know that,

  • [By this formula,it could easily solved]

[tex] \boxed{\rm \: A = P \bigg(1+ \cfrac{ \cfrac{r}{2} }{100} \bigg) {}^{2n} }[/tex]

Substitute the values

[tex]A = 4000 \bigg(1 + \cfrac{5}{100} \bigg) {}^{2 \times \frac{3}{2} } [/tex]

Now solve.

[tex]A = 4000 * (21/20)^3[/tex]

[tex] \implies \: A = 4000 \times \cfrac{21}{20} \times \cfrac{21}{20} \times \cfrac{21}{20} = \cfrac{21 \times 21 \times 21}{2} [/tex]

[tex] \implies \: A = \: \yen \: 4630.50[/tex]

Hence,the sum of ¥ 4000 will be amounted to ¥4630.50 in 1 1/2 years.

[tex] \rule{225pt}{2pt}[/tex]

This question arises:Why we used that formula?

Reason:

The compound interest is calculated half-yearly, the formula changes a little.In this case for r we write r/2 and for n we write 2n because a rate of r% per annum is r/2% half-yearly and n years = 2n half year So this is the reason.

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