Question 1115269: Meisha has $25,000 that she wants to invest. She invests it in accounts paying 12%, 7%, and
6% simple interest. The account paying 12% is a higher-risk account, so she wants the
amount in that account to be half of the amount she has in the account paying 6% simple
interest. If her annual interest is $1945, how much is invested at each rate?
Found 5 solutions by stanbon, mananth, MathTherapy, ikleyn, josgarithmetic:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Meisha has $25,000 that she wants to invest.
She invests it in accounts paying 12%, 7%, and 6% simple interest.
The account paying 12% is a higher-risk account, so she wants the
amount in that account to be half of the amount she has in the account paying 6% simpleinterest.
If her annual interest is $1945, how much is invested at each rate?
Equations:
t + s + x = 25000
t = (1/2)x
12t + 7s + 6x = 194500
----
Modify and rearrange to get:
t + s + x = 25000
t + 0 -0.5x = 0
12t+7s+6s = 194500
--------------------------
Solve by any method you know to get:
t = 6500
s = 5500
x = 13000
============
Cheers,
Stan H.
-------------
Answer by mananth(16949)  (Show Source):
You can put this solution on YOUR website! Meisha has $25,000 that she wants to invest.
She invests x @12%, y @ 7%, and z @ 6%
The account paying 12% is a higher-risk account, so she wants the
amount in that account to be half of the amount she has in the account paying 6% simple
x=(1/2)z or z =2x
Investment
x + y + z = 25000
x+y+2x=25000
3x+y=25000......................(1)
If her annual interest is $1945,
12%x+7%y+6%z=1945
12%x+7%y+6%(2x)=1945
12x+7y+12x=194500
24x+7y=194500....................(2)
Equation for total investment
3 x + 1 y = 25000 .............1
Equation for interest
24 x + 7 y = 194500 .............2
Eliminate y
multiply (1)by 7
Multiply (2) by -1
21 x 7 y = 175000
-24 x -7 y = -194500
Add the two equations
-3 x = -19500
/ -3
x = 6500
plug value of x in (1)
3 x + 1 y = 25000
19500 + 1 y = 25000
1 y = 5500
y = 5500
Ans x = 6500
y = 5500
6500 @12%
5500 @ 7%
13000 @ 6%
Answer by MathTherapy(10791)  (Show Source):
You can put this solution on YOUR website!
Meisha has $25,000 that she wants to invest. She invests it in accounts paying 12%, 7%, and
6% simple interest. The account paying 12% is a higher-risk account, so she wants the
amount in that account to be half of the amount she has in the account paying 6% simple
interest. If her annual interest is $1945, how much is invested at each rate?
Let amount invested at 12% be S, and the amount invested at 7%, A
Then amount invested at 6% is 2S
We then get: S + A + 2S = 25,000______3S + A = 25,000 -------- eq (i)
Also, .06(2S) + .07A + .12S = 1,945______.12S + .07A + .12S = 1,945_____.24S + .07A = 1,945 ------- eq (ii)
Solve this system to get S, or amount invested @ 12%, which is $6,500
Figure out the rest!!
Answer by ikleyn(53725)  (Show Source):
You can put this solution on YOUR website! .
Meisha has $25,000 that she wants to invest. She invests it in accounts paying 12%, 7%,
and 6% simple interest. The account paying 12% is a higher-risk account, so she wants the
amount in that account to be half of the amount she has in the account paying 6% simple
interest. If her annual interest is $1945, how much is invested at each rate?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
For this problem (and for many other similar problems) there is much more simpler
and more straightforward way to solve, which I will show you right now.
Let 'x' be the amount invested at 12%.
Then the amount invested at 6% is 2x,
and the amount invested at 7% is the rest (25000 - x - 2x) = (25000-3x) dollars.
Now write an equation for the total annual interest
0.12x + 0.06(2x) + 0.07*(25000-3x) = 1945 dollars.
Simplify and find x
0.12x + 0.12x + 1750 - 0.21x = 1945,
0.12x + 0.12x - 0.21x = 1945 - 1750,
0.03x = 195
x = 195/0.03 = 6500 dollars.
Thus, $6500 was invested at 12%; 2*6500 = 13000 dollars were invested at 6% and the rest,
25000-13000-6500 = $5500 were invested at 7%.
Solved.
I neither used a system of three equation nor a system of two equations.
One equation in one unknown was enough to find all three unknowns.
It is a focus-pocus, which works successfully in many other similar problems.
Answer by josgarithmetic(39780) (Show Source):
|
|
|
