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Equal amounts are invested in each of three accounts paying 5% 6% and 13.5% annually. If one years combined interest income is 1200 and $.50 how much is invested in each account?

a) $1000
b) $2000
c) $3000
d) $4000

User Jim
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1 Answer

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Final answer:

To solve this problem, we can represent the amount invested in each account as variables and set up a system of equations. After solving the system, we find that $10.655 is invested in each account.

Step-by-step explanation:

To solve this problem, we can represent the amount invested in each account as variables. Let x represent the amount invested at 5%, y represent the amount invested at 6%, and z represent the amount invested at 13.5%. We can set up the following equations based on the given information:

x + y + z = 1200 (equation 1)

0.05x + 0.06y + 0.135z = 0.50 (equation 2)

Now, we can solve these equations simultaneously. From equation 1, we have x + y + z = 1200. Rearranging, we get x = 1200 - y - z. Substituting this value of x into equation 2, we get:

0.05(1200 - y - z) + 0.06y + 0.135z = 0.50

Simplifying, we have:

60 - 0.05y - 0.05z + 0.06y + 0.135z = 0.50

Combining like terms, we get:

0.01y + 0.085z = 59.5

Multiplying both sides of this equation by 100 to eliminate decimals, we get:

y + 8.5z = 5950 (equation 3)

Now, we have a system of two equations with two variables. We can solve this system using substitution or elimination method. Let's use the elimination method by multiplying equation 3 by -0.06:

-0.06y - 0.51z = -357

Next, we can add equation 1 and -0.06y - 0.51z = -357:

x + y + z - 0.06y - 0.51z = 1200 - 357

Simplifying, we get:

x + 0.94y - 0.49z = 843 (equation 4)

We now have two equations with two variables, x + 0.94y - 0.49z = 843 and -0.06y - 0.51z = -357. We can solve this system using substitution method. From equation 4, we can express x in terms of y and z:

x = 843 - 0.94y + 0.49z (equation 5)

Substituting this value of x into equation 1, we get:

(843 - 0.94y + 0.49z) + y + z = 1200

Simplifying, we have:

843 - 0.94y + y + z + 0.49z = 1200

Combining like terms, we get:

0.06y + 1.49z = 357

Multiplying both sides of this equation by 100 to eliminate decimals, we get:

6y + 149z = 35700 (equation 6)

Now, we have a system of two equations with two variables, 0.06y + 1.49z = 35700 and -0.06y - 0.51z = -357. We can solve this system using the elimination method. Let's multiply equation 6 by -0.03:

-0.18y - 4.47z = -1071

Next, we can add -0.18y - 4.47z = -1071 and -0.06y - 0.51z = -357:

-0.24y - 4.98z = -1428

Now, we can express y in terms of z:

y = (1428 - 4.98z)/0.24

Using substitution, we can substitute this value of y into equation 1:

(1428 - 4.98z)/0.24 + y + z = 1200

Simplifying, we get:

(1428 - 4.98z + 0.24y)/0.24 + y + z = 1200

Combining like terms, we have:

(1428 - 4.98z + 0.24((1428 - 4.98z)/0.24))/0.24 + ((1428 - 4.98z)/0.24) + z = 1200

Simplifying further, we get:

1414.299 + 18.111z + z = 1200

Combining like terms, we have:

19.111z + z = 1200 - 1414.299

Adding, we get:

20.111z = 214.299

Dividing by 20.111, we have:

z = 10.655

Now, we can substitute this value of z back into equation 1 to find y:

x + y + 10.655 = 1200

Substituting x = 1200 - y - 10.655, we get:

1200 - y - 10.655 + y + 10.655 = 1200

Simplifying, we have:

1200 = 1200

Since this equation is always true, we can see that y can be any value. However, we know that all the amounts invested are equal, so we can assume that y = x = z = 214.299/20.111 = 10.655.

Therefore, $10.655 is invested in each account.

User Haotian Liu
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