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Jean invests £12000 in an account paying compound interest for 2 years. In first year the rate of interest was x% .

At the end of first year the value of jean's investment is £12336.
In the second year the rate of interests is x/2 %.
What is the value of Jean's investment at the end of 2 years

User Curio
by
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2 Answers

3 votes

Answer:

£12 508.70

Explanation:

Question:

Jean invests £12 000 in an account paying compound interest for 2 years.

In the first year the rate of interest is x%.

At the end of the first year the value of Jean’s investment is £12 336.

​In the second year the rate of interest is: x/2 %

​What is the value of Jean’s investment at the end of 2 years?

Solution:

First, we find the interest rate for the first year.

Amount at beginning: £12 000

Amount after accruing interest for 1 year: £12 336

Amount of accrued interest in 1 year: £12 336 - £12 000 = £336

Interest rate:

£336 is what percent of £12 000?

percent = part/whole × 100%

percent = £336/£12 000 × 100%

percent = 2.8%

The interest rate of the first year was 2.8%.

The interest rate of the second year is 1/2 × 2.8% = 1.4%.

100% + 1.4% = 101.4% = 0.0104

Value after the second year, which is starting at £12 336 and earning 1.4% for 1 year:

1.0104 × £12 336 = £12 508.70

User Red Mak
by
7.4k points
2 votes

Answer:

Explanation:

To solve the problem, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

where:

A is the final amount

P is the principal amount (the initial investment)

r is the annual interest rate (as a decimal)

n is the number of times the interest is compounded per year

t is the time in years

We can use this formula for each of the two years and then add the results to find the final amount.

For the first year, we know that the principal (P) is £12000, the time (t) is 1 year, and the final amount (A) is £12336. We do not know the interest rate (r) or the compounding frequency (n), but we can solve for them using the given information.

Using the formula, we have:

£12336 = £12000(1 + r/n)^(n*1)

Dividing both sides by £12000, we get:

1.028 = (1 + r/n)^n

Taking the natural logarithm of both sides, we get:

ln(1.028) = n*ln(1 + r/n)

Solving for n, we get:

n = ln(1.028) / ln(1 + r/n)

For the second year, we know that the principal (P) is £12336 (the final amount from the first year), the time (t) is 1 year, the interest rate (r) is x/2%, and the compounding frequency (n) is the value we just calculated.

Using the formula, we have:

A = £12336(1 + (x/2)/n)^(n*1)

Substituting the value we calculated for n, we get:

A = £12336(1 + (x/2)/ln(1.028))^(ln(1.028)*1)

Simplifying, we get:

A = £12336(1 + (x/2)/ln(1.028))^ln(1.028)

So the value of Jean's investment at the end of 2 years is the result of compounding the interest earned in the first year and second year, which is:

Final amount = £12336(1 + (x/2)/ln(1.028))^ln(1.028) * (1 + x/200)

Note that we use the annual interest rate of x/2% in the second year, but we need to express it as a decimal by dividing by 100. Similarly, we use x/200 instead of x/100 to express the interest rate over the 2-year period.

User Steve Dennis
by
7.2k points