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There is a decrease of 10% yearly on an article. If this article was bought 3 years ago and present cost is Rs. 5,832 then what was the cost of article at buying time?

A. Rs. 7,200

B. Rs. 7,862

C. Rs. 8,000

D. Rs. 8,500

Answer: Option C

Solution(By Examveda Team)

$$A = P {\left( {1 - \frac{R}{{100}}} \right)^n}$$
Where A = Value of goods after n years
P = Initial Price
R = Rate of depriciation
$$\eqalign{ & \therefore P = \frac{{5832}}{{{{\left( {1 - \frac{{10}}{{100}}} \right)}^3}}} \cr & \Rightarrow P = \frac{{5832}}{{{{\left( {1 - \frac{1}{{10}}} \right)}^3}}} \cr & \Rightarrow P = \frac{{5832}}{{{{\left( {\frac{9}{{10}}} \right)}^3}}} \cr & \Rightarrow P = 5832 \times \frac{{10}}{9} \times \frac{{10}}{9} \times \frac{{10}}{9} \cr & \Rightarrow P = 8000 \cr} $$

This Question Belongs to Arithmetic Ability >> Interest

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Comments ( 4 )

  1. Kuna Sahoo
    Kuna Sahoo :
    5 years ago

    Plz solution the question

  2. Bharti Kumari
    Bharti Kumari :
    6 years ago

    5832*100*100*100/90*90*90= 8000

  3. Khanda Bipin
    Khanda Bipin :
    6 years ago

    Ans:C
    5832*(10/9)*(10/9)*(10/9)=8000

  4. White House
    White House :
    7 years ago

    Answer option is C.
    use effective percent formula for 10% for three years
    we got 27.1% decreased worth of article
    100-27.1=72.9=present worth of article
    72.9=Rs.5832
    100=Rs.x
    by cross multiplication
    x=(5832*100)/72.9
    we get,
    x=8000=worth of article three years ago.

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