# Simple Interest for two investments

1. Jul 8, 2017

### zak100

1. The problem statement, all variables and given/known data
Pat invested a total of 3000$. Part of the money was invested in a money market account that paid 10 percent simple annual interest, and the remainder of the money was invested in a fund that paid 8 percent simple annual interest. If the interest earned at the end of first year from these investments was 256$, how much did Pat invest at 10 percent and how much at 8 percent?

2. Relevant equations
V = P(1 + rt/100)
V= interest amount
P = Principal amount
r = rate
t = time in years

3. The attempt at a solution
P = TenP + EightP or 3000 = TenP + EightP

V= TenV + EightV or 256 = TenV + EightV
For 10%
TenV = TenP ( 1 + 0.1 *1)
For 8%
EightV = EightP(1 + 0.08 *1)
I have not developed the full solution but is it fine to work in this way or there is any other possibility? Some body please guide me.

Zulfi.

2. Jul 8, 2017

### FactChecker

So far, so good. Keep going.

EDIT: See @scottdave 's correction regarding the equations for V, TenV, and EightV.

Last edited: Jul 8, 2017
3. Jul 8, 2017

In your formulas: V = P(1 + rt/100), will give you the total Value (Principal plus Interest) at the end of the time period. If you want to find the interest earned, then you should use I = P*(rt/100). For example, if you invest 100 at 10% per year, at the end of the first year, you have V = $100(1 + (10)(1)/100) =$110. You started with $100 and end with$110, so you earn $10 (the interest). In your situation, V equals the total at the end which is 3000 that you started with plus 256 interest = 3256. 4. Jul 8, 2017 ### Ray Vickson I suggest you try to solve your two equations for the two variables TenP and EightP, to see what happens. That will show you how badly wrong is your formulation! You need to distinguish between the "interest earned" and the (total) "future value"---they are different, and you have mixed them up. Last edited: Jul 8, 2017 5. Jul 10, 2017 ### zak100 Hi, I think its solvable. There are 4 unknowns and 4 equations. Mixing is possible because question has mixed 8% Principal with 10% Principal in the amount of 3000$ and then mixed their profits also. My answer is not corect.
P = TenP = EightP
3000 = TenP + EightP
Let x1= TenP and x2 = EightP
Now V= TenV + EightV
3256 = TenV + EightV
Let y1=TenV and y2 = EightV

Thank scottdave for his advice on this.

Now 3000 = x1 + x2---- (1)
and 3256 = y1 + y2-----(2)

We have
V = P(1 + rt/100)
For 10%
y1 = x1(1+0.1)
y1 = 1.1x1---(3)

For 8%
y2 = x2(1+0.08)
y2 = 1.08x2-----(4)

Putting y1 & y2 values from (3) and (4) into (2)

3256 = 1.1x1 + 1.082x2 ----(5)
Multiplying (1) by 1.1

3300 = 1.1x1 + 1.1x2 ----(6)

Subtracting (5) from (6)

3300 = 1.1x1 + 1.1x2
-3256 = -1.1x1 -1.082
_____________________

44 = 0.0182x2
x2= 2444

Therefore EightP = 2444

Putting x2 in (1)
3000 = x1 + 2444
TenP = x1= 556

Whats wrong with my solution? Some body please guide me.

Zulfi.

6. Jul 10, 2017

### SammyS

Staff Emeritus
Mainly you have a typo.

You changed \$256 to 3256 throughout.

7. Jul 10, 2017

### Ray Vickson

Where did your figure of 1.082 come from? How did 1.08 x2 become 1.082 x2? How did 1.1 - 1.08 become 0.0182?

8. Jul 10, 2017

### FactChecker

In many problems, the hardest job is to keep the meaning of all the variables straight.
Your switch from somewhat meaningful variable names like TenP to unrecognizable names like x and y does not help to keep things straight. With your new x and y variables, I got tired of trying to understand what you are doing. I recommend using a letter that indicates interest, like 'I' rather than 'V'. Rename V to I, TenV to TenI, and EightV to EightI.

Have you corrected the equations for V, TenV, EightV as @scottdave recommended? You must correct that.

9. Jul 10, 2017

### scottdave

When you say this:
You have a typo.
It really should be:

3300 = 1.1x1 + 1.1x2
-3256 = -1.1x1 -1.08x2
----------------------------
44 = 0.02x2

x2= 2200

Then continue from there. I agree with @FactChecker that using more meaningful variable helps to keep things straight.

10. Jul 10, 2017

### zak100

Hi,
Thanks everybody.

Subtracting (5) from (6)

3300 = 1.1x1 + 1.1x2
-3256 = -1.1x1 -1.08x2
_____________________

44 = 0.02x2
x2= 2200

Therefore EightP = 2200

Putting x2 in (1)
3000 = x1 + 2200
TenP = x1= 800

Thanks scottdave for catching my error. This is the correct answer.

Zulfi.