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Dahasolnce [82]
3 years ago
13

At the end of January of last year, stock A was selling at A dollars a share, and stock B was selling a B dollars a share, where

A > B. Between January and June, stock A rose x%, and stock B rose y%, where y > x. Suppose a certain investor owned N shares of A, but at the end of January last year, sold those shares and used all of that money to buy shares of B, and held those shares through the end of June. These are the only stocks in this portfolio. Do not mark more than one answer in each column.
Mathematics
1 answer:
Maslowich3 years ago
3 0

Answer: a) yNA/100

b) NA(y-x)/100

c) (NA)/B

Step-by-step explanation:

a) The total amount of dollars owned by the shares' owner = N number of shares × A dollars per share = NA dollars

This total is then transferred to buy B shares which then appreciates by y%.

The amount of increase in portfolio from January to June = y% of total dollars invested = y% of NA dollars = yNA/100

b) If the shares were left with A, the increase in portfolio from January to June would be x% and = x% of the total Dollar amount = x% of NA dollars = xNA/100

How much more money made in that time would be the difference in interest, between taking the dollars to invest in share B or keeping the dollars on investment A

That is, (yNA/100) - (xNA/100) = NA(y-x)/100

c) Total dollars available after sale of the A stock = NA

Number of B stock this dollar can buy = Total dollars available/amount of B stock per share

That is, (NA)/B

QED!

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If we shift the argument by 1 and scale by -5, we have

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