Question 1203977: Joseph invested a total of $10,000$10,000 in two accounts. After a year, one account lost 7.8%, while the other account gained 6.7%. In total, Joseph lost \$562.50. How much money did Joseph invest in each account?Write and solve a system of equations to answer the following questions.Joseph invested in the account with 7.8% loss.Joseph investedin the account with 6.7% gain.
Found 4 solutions by josgarithmetic, greenestamps, ikleyn, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source):
You can put this solution on YOUR website! The account which gained 6.7%, y quantity invested
The account which lost 7.8%, 10000-y quantity invested
Starting total quantity was 10000 dollars. (Assuming you made typing mistake)
Ending quantity was  dollars, or 9437.5 dollars.
Forming the equation by description:
 --------first, think how this makes sense; and then when you are satisfied, solve for y.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The overall percentage change was -562.50/10000 = -.05625 = -5.625%.
Before doing any calculations, note that this percentage is much closer to -7.8% than it is to +6.7%, so by far the larger amount was invested in the account that lost 7.8%.
A solution using the typical formal algebraic method...
x = amount invested in account that gained 6.7%
10000-x = amount invested in account that lost 7.8%
The overall change was a loss of $562.50:
ANSWER: $1500 was invested in the account that gained 6.7%; the other $8500 in the account that lost 7.8%
CHECK: .067(1500)-.078(8500) = 100.50-663 = -562.50
You can also solve any 2-part "mixture" problem like this informally, performing virtually the same calculations but without the formal algebra.
The three percentages in the problem are -7.8, -5.625, and +6.7.
Look at those percentages on a number line and determine that -5.625 is (2.175)/(14.5) = .15 of the way from -7.8 to +6.7.
That means .15 or 15% of the total was invested in the account that gained 6.7%.
ANSWER: 15% of the $10,000, or $1500, was invested in the account that gained 6.7%; the other $8500 was invested in the account that lost 7.8%.
Note that in this problem, since the numbers were "ugly", the informal method wasn't much easier than the formal algebraic method. But with "nice" numbers that are easy to work with mentally, the informal method can get you to the solution with far less effort.
Answer by ikleyn(52778)  (Show Source):
You can put this solution on YOUR website! .
Joseph invested a total of $10,000 in two accounts. After a year, one account lost 7.8%,
while the other account gained 6.7%. In total, Joseph lost $562.50.
How much money did Joseph invest in each account?
Write and solve a system of equations to answer the following questions.
Joseph invested in the account with 7.8% loss.
Joseph invested in the account with 6.7% gain.
~~~~~~~~~~~~~~~~~~
Let x be the amount which was invested and gained 6.7%.
Let y be the amount which was invested and lost 7.8%.
Then you can write two equations.
First equation is OBVIOUS: it is
x + y = 10000 dollars. (1)
It says that the total of the two accounts is $10,000, exactly as the problem states.
Second equation is about the net loss.
It is also obvious:
0.078y - 0.067x = 562.50 dollars. (2)
Indeed, 0.078y is the lost amount, while 0.067x is the gained amount.
We take 0.078y with the sign "+", since in equation (2) we consider the lost amount as positive.
You may consider the loss as negative value, but then you should be consistent and write all the terms of equation (2) with opposite signs.
So, you either write the "loss" equation in the form (2),
or you change all the terms in this equation to opposite.
It is your choice to write (2) as is or with opposite signs - the meaning of the equation (2) will be the same.
So, for what follows, let's write these two equations as
x + y = 10000 (1)
-0.067x + 0.078y = 562.50 (2)
So, we consider the loss in (2) with the sign "+". As soon as you wrote these equations, the setup is completed.
To solve the system, let's multiply equation (1) by 0.067 (all the terms).
Keep equation (2) as is. You will get then
0.067x + 0.067y = 670.00 (1')
-0.067x + 0.078y = 562.50 (2')
Now add equations (1') and (2'). The terms 0.067x and -0.067x will kill each other,
and you will get a single equation with unknown y
0.067y + 0.078y = 670 + 562.50,
or 0.145y = 1232.5, which gives y = 1232.5/0.145 = 8500.
Then from equation (1), x = 10000 - 8500 = 1500.
So, the invested amount to the account which gained 6.7% was 1500 dollars.
the invested amount to the account which lost 7.8% was 8500 dollars.
CHECK. -0.067*1500 + 0.078*8500 = (I use my calculator) = 562.5 dollars, the net loss. ! correct !
Solved.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
Joseph invested a total of $10,000$10,000 in two accounts. After a year, one account lost 7.8%, while the other account gained 6.7%. In total, Joseph lost \$562.50. How much money did Joseph invest in each account?Write and solve a system of equations to answer the following questions.Joseph invested in the account with 7.8% loss.Joseph investedin the account with 6.7% gain.
Let amounts invested in the 7.8% and 6.7% accounts be H, and L, respectively
We then get the following INVESTMENT equation: H + L = 10,000____L = 10,000 - H ---- eq (i)
Loss "suffered" on 7.8% investment: - .078H
Gain on 6.7% account: .067L
Therefore, we get the following RETURN-ON-INVESTMENT (ROI) equation:
- .078H + .067L = - 562.5 ---- eq (ii)
- .078H + .067(10,000 - H) = - 562.5 ---- Substituting 10,000 - H for L in eq (ii)
- .078H + 670 - .067H = - 562.5
- .078H - .067H = - 562.5 - 670
- .145H = - 1,232.5
Amount invested in 7.8% account, or 
Obviously, amount invested in the 6.7% account, or L = $10,000 - $8,500 = $1,500.
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