How can I solve the following problem using a set of two algebraic equations?
Mr Stein invested a total of $100,000 in two companies for a year. Company A's stock showed a 13% annual gain, while Company B showed a 3% loss for the year. Mr Stein made an 8% return on his investment over the year. How much money did he invest in each company?
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Let A represent the amount invested Company A, and B represent the amount invested in company B.
Since we know that Mr. Stein invested a total of $100,000, it follows that A and B must add up to $100,000. This gives us our first equation:
(1) `A + B = 100000`
Next, we know that he made a return of 8%. 8% of $100,000 is $8,000. So 13% of the amount invested in A minus 3% of the amount invested in B must add up to $8,000.
(2) `0.13A - 0.03B = 8000`
The above two equations represent our problem. We can now proceed to solve. From (1), we see that `A = 100000 - B` . Plugging this into (2), we get
`0.13(100000 - B) - 0.03B = 8000`
`\implies B = 31250`
`\implies A = 68750`
There are two ways to do this type of question. One way is to use one varible and use the information given to solve for the equation
The other type, which is what you wanted, uses two varibles.
Let X be the number of dollars invested in Company A and Y represent the number of dollars in Company B
Since the total investment is 100,000 dollars, we have the equation
There is another piece of information given, which is that the total return rate is 8%, so that we have another equation
money gained in Company A + money gained in Company B= total money earned
Money earned in Company A is X*13%=0.13X
Money earned in Company B = Y * (-3%)=-0.03Y (since it was a loss, there should be a negative sign)
the total money earned is : 100,000*8%=800000/100=8000 dollars
the equation is
2. 0.13 X-0.03y=8000
rewriting the first equation, we have:
X=100,000 - Y
substitute that answer into the second equation, we have:
then x = 100000-31250=68750
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