Mary invested $30,000 in two accounts some at 12% per annum and the rest at 8% per annum. Her total interest for one years was $ 3200.How much was invested at each rate ?

Question

Mary invested $30,000 in two accounts some at 12% per annum and the rest at 8% per annum. Her total interest for one years was $ 3200.How much was invested at each rate ?

solution :

$3200=P1 (0.12)(1)+ (3000-P1)(0.08)(1)

P1= ?

P2= 30000-P1 ?

HOW to solve this because they have two answer:

$20,000 at 12%
$10,000 at 8%

Answer 1

well, of course they have two answers. They wanted to know the amount invested at each of two interest rates!

So, what don't you understand? They gave you the equation. If P1 is invested at 12% (which is evident from the equation), then the rest of the money (30000-P1) was invested at 8%.

Then just solve the equation for P1, and you get 20,000. The rest of the money (10,000) was invested at 8%.

Now, if it's the mechanics of the equation, consider an equivalent one with less noise.

2x + 3(10-x) = 22

If you can do that, then you can do the one above.

PS - I am not sure why they tossed in those (1) factors. Maybe because they want to explicitly note the amount of interest earned by 1 dollar.

Answer 2

To solve this problem, we can use a system of equations to represent the situation.

Let P1 represent the amount invested at 12% and P2 represent the amount invested at 8%. We know that the total amount invested is $30,000, so we can write the first equation:

P1 + P2 = 30000

Next, we know that the total interest earned is $3200. The interest on the amount invested at 12% is calculated by multiplying the amount by 0.12, and the interest on the amount invested at 8% is calculated by multiplying the amount by 0.08. Adding these two amounts together should equal $3200, so we can write the second equation:

0.12P1 + 0.08P2 = 3200

Now we have a system of equations:

P1 + P2 = 30000
0.12P1 + 0.08P2 = 3200

To solve this system, we can use the substitution method or the elimination method.

Let's solve this using the substitution method:

From the first equation, we can rewrite it as:

P2 = 30000 - P1

Now substitute this expression for P2 in the second equation:

0.12P1 + 0.08(30000 - P1) = 3200

Simplifying the equation:

0.12P1 + 2400 - 0.08P1 = 3200

Combine like terms:

0.04P1 = 800

Divide both sides by 0.04:

P1 = 800 / 0.04 = 20000

Now substitute the found value of P1 into the first equation to find P2:

P2 = 30000 - P1 = 30000 - 20000 = 10000

Therefore, Mary invested $20,000 at 12% and $10,000 at 8%.

Answer 3

To solve this problem, we can set up a system of linear equations based on the information given.

Let's assume that Mary invested P1 amount at 12% per annum. The amount invested at 8% per annum would then be (30000 - P1).

We can now set up the equation for the total interest earned:

3200 = P1 * 0.12 + (30000 - P1) * 0.08

To solve this equation, we can follow these steps:

1. Distribute and simplify the equation:
3200 = 0.12P1 + 0.08(30000 - P1)

2. Distribute further:
3200 = 0.12P1 + 2400 - 0.08P1

3. Combine like terms:
3200 = 0.04P1 + 2400

4. Move 2400 to the other side of the equation:
3200 - 2400 = 0.04P1

5. Simplify:
800 = 0.04P1

6. Divide both sides by 0.04 to solve for P1:
P1 = 800 / 0.04 = 20000

Now that we know that P1 is 20000, we can substitute this value back into the equation to find P2, the amount invested at 8% per annum:

P2 = 30000 - P1 = 30000 - 20000 = 10000

Thus, the solution to this problem is $20,000 invested at 12% and $10,000 invested at 8%.