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Basics of Financial Mathematics Test 27

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Basics of Financial Mathematics Test 27
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  • Question 1
    1 / -0
    The time required for a sum of money to double at 55% annum compounded continuously is (in years)
    Solution
    Amount after nn years compounded annually=P(1+r100)n=P(1+\dfrac {r}{100})^n
    Where, PP is the principal amount and rr is the interest rate per annum.

    ie, 2P=P(1+r100)n2=(1+5100)n=(1+120)n2P=P(1+\dfrac {r}{100})^n\Rightarrow 2=(1+\dfrac {5}{100})^n=(1+\dfrac 1{20})^n

    ln2=n(ln2120)n=ln2ln21ln2014.3\Rightarrow \ln 2=n(\ln{\dfrac {21}{20}})\Rightarrow n=\dfrac {\ln 2}{\ln {21}-\ln {20}}\approx14.3

    Option D is the correct answer.
  • Question 2
    1 / -0
    The value of a machine depreciates every year by 5%5\%. If the present value of the machine be Rs. 100,000100,000, what will be its value after two years?
    Solution
    current value of Machine =Rs.100,00=Rs. 100,00

    After 1 year  
    value  decreased by 5%5 \%

    value of Machine  is
    Rs.95% Rs. 95 \% ofof 100,000 100,000   

    95100×100,000=Rs.95,000\Rightarrow \dfrac{95}{100} \times 100,000 = Rs. 95,000 

    After 2 years 

    again the value is  decreased by 5%5 \% from previous year' s Value 

    Now  value of Machine  is
    =95%= 95 \% ofof 95,00095,000

    = 95100× 95,000=  \dfrac{95}{100} \times  95,000 

    =0.95× 95,000=0.95\times  95,000  

    =Rs.90,250 = Rs. 90,250 is the value of machine at the of two years
  • Question 3
    1 / -0
    Which of the following asset does not depreciate?
    Solution
    Every asset is having a useful life based on its usage and nature. A predetermined rate is defined for charging a proportion of the value of the assets every year to profit & loss account which is known as depreciation.
    The land is an asset that will have an unlimited useful life and does not depreciate because of its use.
  • Question 4
    1 / -0
    Equal sums of Rs.72007200 were lent to Megha and Priya at 55 %\% per annum for a period of 44 years and 55 years respectively. Find the difference of the interest paid by Megha and Priya.
    Solution
    Simple interest on the sum PP at rate of interest RR for NN years is given as PNR100\dfrac{PNR}{100}
    Interest paid by Priya=7200×5×5100                                    =Rs 1800\text{Interest paid by Priya} = \dfrac{7200\times 5\times 5}{100}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =Rs\ 1800

    Interest paid by Megha=7200×4×5100                                     =Rs 1440\text{Interest paid by Megha} = \dfrac{7200\times 4\times 5}{100}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =Rs\ 1440

    Difference =18001440= 1800-1440
                      =Rs 360=Rs\ 360 
    Hence, D is the right option.
  • Question 5
    1 / -0
    The cost of a vehicle is Rs. 1,75,0001,75,000. If its value depreciates at the rate of 20%20\% per annum, then the total depreciation after 33 years was _________.
    Solution
    Cost of vehicle =175000= 175000
    Depreciation for first year =20100×175000=35000= \dfrac{20}{100}\times175000=35000

    Cost of vehicle after first year =17500035000=140000= 175000-35000=140000
    Depreciation for second year =20100×140000=28000= \dfrac{20}{100}\times140000 = 28000

    Cost of vehicle after second year =14000028000=112000= 140000-28000=112000
    Depreciation for third year =20100×112000=22400= \dfrac{20}{100}\times112000 = 22400

    Total depreciation after 33 years =35000+28000+22400=85400= 35000+28000+22400 = 85400
    Hence, the answer is Rs. 85,40085,400.
  • Question 6
    1 / -0
    _________ cost will still be incurred although a plant is shut down temporarily.
    Solution
    To have a true and fair view of profit & loss of the organization, depreciation of the assets has to be charged on the basis of their useful life. If a plant is shut down temporarily, there may be significant spoilage of material in the process, some significant damage to the plant. Due to some reasons like lack of demand, maintenance of plant, the plant is shut down for some time. The shutdown does not change the inherent technical nature of the plant. Hence depreciation has to be charged.
  • Question 7
    1 / -0
    The compound interest on Rs. 100000100000 at 20%20\% per annum for 22 years 33 months, compound annually is ________.
    Solution
    Given, P=100000P=100000, r=0.2r=0.2n=2+14n=2+\dfrac{1}{4} (converting 22 years 33 months to years) =n=2.25=n=2.25
    Interest is compounded anually,
    For first year =100000×0.2×1=20000=100000\times 0.2\times 1=20000
    For second year Pnew=120000P_{new }= 120000
    Interest =120000×0.2×1=24000=120000\times 0 .2\times 1=24000
    For last 0.025440.02544 year, pnew=144000p_{new}=144000
    Interest =144000×0.2×0.25=7200 = 144000\times 0.2\times 0.25=7200
    New amount =144000+7200=151200=144000+7200 = 151200
    Interest=151200+100000=51200\text{Interest} = 151200+100000=51200
  • Question 8
    1 / -0
    Depreciation is provided on _________.
    Solution
    Fixed assets are those assets that provide long term benefits to the organization and are not meant to be sold in the course of business. Hence these are recorded in the balance sheet. A portion of the cost of these assets is charged to the profit & loss account every year which is termed as depreciation.
  • Question 9
    1 / -0
    Depreciation is a charge against _______.
    Solution
    Depreciation is a reserve but is a charge against profit since it is created for the replacement of an asset. Hence, both reserves and provisions can arise as a charge against profits or as an appropriation out of profit, depending upon the nature of provision or reserve.
  • Question 10
    1 / -0
    The purpose of depreciation is to _____________.
    Solution
    The purpose of depreciation is to match the cost of a productive asset, that has a useful life of more than a year, to the revenues earned by using the asset. The asset's cost is usually spread over the years in which the asset is used.
    Depreciation is an allocation process in order to achieve the matching principle.
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