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Jack bought a new boat for £12500 The value, V, of Jack's boat at the end of n years is given by the formula V = 12 500 × (0.85^n) (a) At the end of how many years was the value of Jack's boat first less than 50% of the value of the boat when it was new? A savings account pays interest at a rate of R% per year - Edexcel - GCSE Maths - Question 9 - 2017 - Paper 3

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Jack-bought-a-new-boat-for-£12500----The-value,-V,-of-Jack's-boat-at-the-end-of-n-years-is-given-by-the-formula----V-=-12-500-×-(0.85^n)----(a)-At-the-end-of-how-many-years-was-the-value-of-Jack's-boat-first-less-than-50%-of-the-value-of-the-boat-when-it-was-new?----A-savings-account-pays-interest-at-a-rate-of-R%-per-year-Edexcel-GCSE Maths-Question 9-2017-Paper 3.png

Jack bought a new boat for £12500 The value, V, of Jack's boat at the end of n years is given by the formula V = 12 500 × (0.85^n) (a) At the end of how man... show full transcript

Worked Solution & Example Answer:Jack bought a new boat for £12500 The value, V, of Jack's boat at the end of n years is given by the formula V = 12 500 × (0.85^n) (a) At the end of how many years was the value of Jack's boat first less than 50% of the value of the boat when it was new? A savings account pays interest at a rate of R% per year - Edexcel - GCSE Maths - Question 9 - 2017 - Paper 3

Step 1

At the end of how many years was the value of Jack's boat first less than 50% of the value of the boat when it was new?

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Answer

To determine when the value of Jack's boat is less than 50% of its original price, we first calculate 50% of the original price:

50% of 12,500=0.5×12,500=6,250.50\% \text{ of } 12,500 = 0.5 \times 12,500 = 6,250.

Next, we set up the inequality based on the given formula for V:

12,500×(0.85n)<6,250. 12,500 \times (0.85^n) < 6,250.

Dividing both sides by 12,500 gives:

0.85n<0.5. 0.85^n < 0.5.

Taking the logarithm of both sides to solve for n:

log(0.85n)<log(0.5)nlog(0.85)<log(0.5). \log(0.85^n) < \log(0.5) \Rightarrow n \cdot \log(0.85) < \log(0.5).

Thus,

n>log(0.5)log(0.85).n > \frac{\log(0.5)}{\log(0.85)}.

Calculating this gives:

n>0.30100.07064.26. n > \frac{-0.3010}{-0.0706} \approx 4.26.

Therefore, Jack's boat was first worth less than 50% of its original value after 5 years.

Step 2

Work out the value of R.

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Answer

To find R, we first need to determine the interest earned on the initial investment of £5500:

Let the interest earned before tax be denoted as I. After paying 40% tax, Jack receives £79.20. The tax formula is given by:

I(0.40×I)=79.200.60×I=79.20. I - (0.40 \times I) = 79.20 \Rightarrow 0.60 \times I = 79.20.

Solving for I, we find:

I=79.200.60=132. I = \frac{79.20}{0.60} = 132.

Since interest is also given by the formula:

I=R100×5500, I = \frac{R}{100} \times 5500,

we can substitute the value of I:

132=R100×5500. 132 = \frac{R}{100} \times 5500.

Solving for R gives:

R=132×1005500=2.4. R = \frac{132 \times 100}{5500} = 2.4.

Thus, the value of R is 2.4.

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