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likoan [24]
2 years ago
4

Neil bought a house for £235 000

Mathematics
2 answers:
Mnenie [13.5K]2 years ago
6 0

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

NNADVOKAT [17]2 years ago
3 0

Answer:

£ 253,484.16

Step-by-step explanation:

Cost of house for first year = £235,000

<u>Year 1</u>

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

<u>Year 2</u>

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

<u>Year 3</u>

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

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