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An investor purchased a bond for p dollars on Monday. For a certain nu [#permalink]
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Given: An investor purchased a bond for p dollars on Monday. For a certain number of days, the value of the bond increased by r percent per day. After this period of constant increase, the bond decreased the next day by q dollars and the investor decided to sell the bond that day for v dollars.

Asked: When did the investor sell the bond if

\(r = 100*[\sqrt{\frac{(v+q)}{p}} - 1]\)?

Monday: -
Purchase price of a bond = $p

After n number of days: -
Value of the bond = $p (1+r%)^n

Next day:-
Value of the bond = $p (1+r%)^n - q
Selling price of the bond = $v
Since the investor sold the bond for market price,
$p (1+r%)^n - q = $v
p(1+r%)^n = v+q
(1+r%)^n = (v+q)/p
(1 + r%) = ((v+q)/p)^(1/n)
r% = (v+q)/p)^(1/n) - 1
r = 100* ((v+q)/p)^(1/n) - 1)

n=2 days

Investor sold the bond after 2+1 = 3 days

IMO B
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Re: An investor purchased a bond for p dollars on Monday. For a certain nu [#permalink]
Hi Bunuel
Would you be able to provide the solution step by step for this?

Thanks
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An investor purchased a bond for p dollars on Monday. For a certain nu [#permalink]
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Expert Reply
Bunuel wrote:
An investor purchased a bond for p dollars on Monday. For a certain number of days, the value of the bond increased by r percent per day. After this period of constant increase, the bond decreased the next day by q dollars and the investor decided to sell the bond that day for v dollars. When did the investor sell the bond if

\(r = 100*[\sqrt{\frac{(v+q)}{p}} - 1]\)?


A. Two working days later.

B. Three working days later.

C. Four working days later.

D. Five working days later.

E. Six working days later.

I have given a method above that involves substitution and easier and faster method.

But if we have to do by algebra:-

Let the amount increase by r% for n days, and then it decreases by q next day to become v.
That is \(v=p(1+\frac{r}{200})^n-q\)
\(v+q=p(1+\frac{r}{100})^n\)

Substitute in the value of r given by equation
\(r = 100*[\sqrt{\frac{(v+q)}{p}} - 1]\)
so
\(r = 100*[\sqrt{(1+(\frac{r}{100})})^n - 1]\)
\(r/100=[\sqrt{(1+(r/100)})^n - 1]\)
Let r/100 be a, so \(a=(1+a)^{\frac{n}{2}}-1\) or \(a+1=(a+1)^{\frac{n}{2}}\)
Thus, 1=n/2 or n=2

So it increased for 2 days and then dropped next day
Total days = 2+1 or 3 days.­
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An investor purchased a bond for p dollars on Monday. For a certain nu [#permalink]
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