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Neil bought a house for £235 000

In the first year the value of the house depreciated by 4%
In each of years 2 and 3 the value of the house increased by 6%
Work out the value of the house at the end of year 3

2 Answers

1 vote

Answer:

Based on the given conditions, formulate: 235000×(1+1-4%)

Calculate the sum or difference:235000×(2-0.04)

Calculate the sum or difference:235000×1.96

Calculate the product or quotient:460600

get the result:460600

So the answer is 460600

User Mobility
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4.5k points
4 votes

Answer:

£ 253,484.16

Explanation:

Cost of house for first year = £235,000

Year 1

Depreciation for first year is 4% = 4/100

Depreciation in £ = 235,000 x 4/100 = £9,400

Effective value of house at end of first year
= 235000 - 9400 = £225,600

Year 2

In the second year the house value increased by 6% = 6/100

So total increase in £ at the end of year 2 with an initial value of £225,600 = £225,600 x 6/100 = £13,536

So actual value of house at end of year 2
= £225,600 + £13,536
= £239,136

Year 3

The value increased by 6% for year 3.

Value at beginning of year 3 = £239,136

Increase in value at 6% = £239,136 x 6/100 = £14,348.16

Total Value at end of year 3

= £239,136 + £14,348.16

= £253,484.16

Answer: £253,484.16

Note
Another way of quickly calculating this is noting that 4% depreciation means the new value is (1 - 4%) of initial value = 1 - 4/100 = 0.96

An increase of 6% means the new value will be 1.06 times the initial value

So the value of the house after 4% depreciation for the first year and 6% appreciation for the next 2 years

= 235000 x (0.96)(1.06)(1.06) = 253,484.16

User Roger Steinberg
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