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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?
Answer & Solution
Correct Answer:
Option
C
$$A = P {\left( {1 - \frac{R}{{100}}} \right)^n}$$
Where A = Value of goods after n years
P = Initial Price
R = Rate of depriciation
$$\eqalign{ & \therefore P = \frac{{5832}}{{{{\left( {1 - \frac{{10}}{{100}}} \right)}^3}}} \cr & \Rightarrow P = \frac{{5832}}{{{{\left( {1 - \frac{1}{{10}}} \right)}^3}}} \cr & \Rightarrow P = \frac{{5832}}{{{{\left( {\frac{9}{{10}}} \right)}^3}}} \cr & \Rightarrow P = 5832 \times \frac{{10}}{9} \times \frac{{10}}{9} \times \frac{{10}}{9} \cr & \Rightarrow P = 8000 \cr} $$
Where A = Value of goods after n years
P = Initial Price
R = Rate of depriciation
$$\eqalign{ & \therefore P = \frac{{5832}}{{{{\left( {1 - \frac{{10}}{{100}}} \right)}^3}}} \cr & \Rightarrow P = \frac{{5832}}{{{{\left( {1 - \frac{1}{{10}}} \right)}^3}}} \cr & \Rightarrow P = \frac{{5832}}{{{{\left( {\frac{9}{{10}}} \right)}^3}}} \cr & \Rightarrow P = 5832 \times \frac{{10}}{9} \times \frac{{10}}{9} \times \frac{{10}}{9} \cr & \Rightarrow P = 8000 \cr} $$
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Login5832*(10/9)*(10/9)*(10/9)=8000
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.