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purchased a new mower on January 1, 20X1 for $2,500. He uses the straight-line method of...

purchased a new mower on January 1, 20X1 for $2,500. He uses the straight-line method of depreciation and estimates it to have a useful life of 4 years with no salvage value. On January 1, 20X3, Larry receives new information that the mower will continue to operate for 3 more years and the salvage value is $200. Determine the amount of depreciation recorded in 20X3.

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Answer #1

answer :$350

depreciation charge before revaluation=(cost-salvage value)/estimated life in years

salvage value before revaluation is zero

depreciation charge before revaluation=($2500-$0)/4=$625

accumulated depreciation charge for 2 year (20X1 to 20X3)=2*$625=$1250

carrying value at 1st january 20X3=cost-accumulated depreciation

carrying value at 1st january 20X3=$2500-$1250=$1250

after revaluation there is a salvage value of $200 and expectede life revised to 4 years

depreciation charge after revaluation=(carrying value-salvage value)/estimated life in years

depreciation charge after revaluation=($1250-$200)/4=$350

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