In a period of a year an investor lost twice as much and he gained. If he originally had $10,000 and now has $8,000, what amount did he gain and loseI know he gained $2,000 and lost $4,000, but i...
In a period of a year an investor lost twice as much and he gained. If he originally had $10,000 and now has $8,000, what amount did he gain and lose
I know he gained $2,000 and lost $4,000, but i can only solve in my head. i cant put it on paper. and i need the algebraic steps to solve it. so if anyone knows, please help me. thanks
I'm assuming you were told that he started with $10,000 and ended up with $8,000.
In that case, you need to set up the equation like this:
10,000 + (x-2x) = 8,000
10,000 is, of course, what he started with. x is what he gained. 2x is what he lost. You need to do that because you know his loss was twice his gain, but you don't know the actual numbers.
So then you have 10,000 - x = 8,000
Subtract 10,000 from both sides and you have
-x = -2,000
Multiply both sides by -1 and you have
x = 2,000
When you plug in 2,000 for x, you get the right answer -- he made 2,000 (getting up to 12,000) and lost 4,000, getting back to 8,000.
Does it make more sense now?
The original investment is $10000. The invevestor has the now $8000. Therefore, in all he lost $10000-$8000 =$2000.
We have an additonal information that during the period he has loss 2 times the the profit. Therefore, the net loss of two thousand dollars is the the result of after wiping out a loss equal to profit. And now the net loss of $2000 is equal in quantity (magnitude) of profit of $2000. Therefore, he made a profit of $2000 and a loss of $4000, resulting in a net loss off $2000.
Setting in equations: If his loss is x, then his profit is x/2.
Therefore the result on the investment is $10000-x+x/2 and this is equal to $8000 .
$10000-$8000=x/2 or x/2 =$2000 is his profit.
His loss = twice (profit)= 2*(x/2)=2*$2000 =$4000.