If an investment principal p decreased d percent in each of the first

In the second year why are we multiplying \(P(1- \frac{d}{100}) with \frac{d}{100}\)
Could you please explain the 2nd year working in detail? I'm a bit lost

Because the new value in 2nd year will be the principal from 1st year which is \(P*(1-\frac{d}{100})\) and this value will be decreased by d percent in 2nd year.

Question:
If an investment principal \(p\) decreased \(d\) percent in each of the first two years and recovered \(i\) percent in the third year, which of the following represents the value of the investment at the end of the third year?

Initial value = \(p\)
After 1 year, the value decreased by \(d\)%; thus, the value becomes \((100 - d)\)% of the initial value

=> \(V1\) (value after 1 year) = \((100-d)\)% of \(p\)

After 1 more year, the value decreased by \(d\)%; thus, the value becomes \((100 - d)\)% of the value \(V1\)

=> \(V2\) (value after 2 years) = \((100-d)\)% of \((100-d)\)% of \(p\)


After 1 more year, the value increased by \(i\)%; thus, the value becomes \((100 + i)\)% of the value \(V2\)

=> \(V3\) (value after 3 years) = \((100+i)\)% of \((100-d)\)% of \((100-d)\)% of \(p\)

\(= [(100+i)/100] * [(100-d)/100] * [(100-d)/100] * p\)

\(= [1+(i/100)] * [1-(d/100)] * [(1-(d/100)] * p\)

\(= p*[1-(d/100)]^2 * [1+(i/100)]\)

Answer D


Hope this helped

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