MHB Investment Problem: $12,000 Net Profit of $120

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A businessman invested $12,000 in two ventures, achieving an 8% profit on one and incurring a 4% loss on the other, resulting in a net profit of $120. The equations derived from the investment amounts and profits lead to a system that can be solved using elimination. The correct investment amounts are $5,000 at 8% and $7,000 at 4%. The net gain from the profitable investment is $400, while the loss from the other amounts to $280, confirming the overall net profit of $120. This illustrates how losses in investments can affect total returns despite profits in other areas.
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A businessman invested a total of \$12,000 in two ventures. In one he made a profit of 8% and in the other he lost 4%. If his net profit for the year was \$120, how
much did he invest in each venture?

what does it mean when we say we lost some amount in an investment? does it mean that we didn't get back the original amount that we invested?

when I solve this problem my answers are \$5000 @ 8% and \$7000 @ 4%.

but i got confused with the net income that is \$120. if I profited \$400 in a \$5000 investment I should till get a total profit of $400 in the two investments. can you please explain.
 
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I would let $x$ be the amount invested which made an 8% profit, and $y$ be the amount invested which made a 4% loss. Since the total invested is \$12,000, we may state:

(1) $$x+y=12000$$

Now, adding the profit and loss (negative profit), we get the total profit:

$$0.08x-0.04y=120$$

If we multiply this second equation by 25, we can get integral coefficients:

(2) $$2x-y=3000$$

Now, I would suggest using elimination...if you add (1) and (2), you will eliminate $y$ and get an equation solely in $x$ which you can then solve. Then using the value for $x$ which you obtain, you can find $y$ by substituting for $x$ into either (1) or (2). I suggest using (1). Your stated answers are correct.

You have a net gain of \$400 for the \$5000 investment, but a \$280 loss for the \$7000 investment, so your net gain is \$(400-280)=\$120. You now have \$12120.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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