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Stuck on an algebra problem?

A pension fund has a total of $1 million invested in stock of the ABC Company and bonds of the DEF Corporation. The ABC stock yields 12% in cash each year, and the DEF bonds yields 10% in cash each year. The pension fund received a total of $115,000 in cash from ABC stock and DEF bonds last year. How much money was invested in ABC Stock?

A) $750,000.00

B) $600,000.00

C) $500,000.00

D) $333,333.33

E) $250,000.00

I'm doing something wrong since my answer isn't any of these answer choices. Help in solving this?

Update:

This is an SAT question so I don't have anything to study off of specifically and I don't understand it. I don't need to know the right answer, I could just choose any random letter and be done with it, but I want to know the steps to solve it. And currently Y!A is the most available resource.

4 Answers

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  • 1 decade ago
    Best Answer

    Ok, this initially looks like a compound interest problem. It isn't. It's just multiplication. You don't have to figure out multiple years, so the interest never "compounds." It's just rate times principal.

    Let's use the following variables:

    a = amount invested with ABC

    d = amount invested with DEF

    We know from the first sentence that:

    a + d = 1,000,000

    We know the rate of return for each from the second sentence.

    r(a) = 0.12

    r(d) = 0.10

    Finally, we know the total return is 115,000. This has to be the sum of the returns from each company. So...

    a * r(a) + d * r(d) = 115,000

    0.12a + 0.10d = 115,000

    Now we have our equations:

    {Eq. 1} a + d = 1,000,000

    {Eq. 2} 0.12a + 0.10d = 115,000

    First thing I'm going to do is multiply Eq. 2 by 100 to clear the decimals

    {Eq. 2} 0.12a + 0.10d = 115,000

    {Eq. 2} 12a + 10d = 11,500,000

    Now let's see if we can find a way to subtract them to eliminate a variable. What if we multiply Eq. 1 by 10:

    {Eq. 1} a + d = 1,000,000

    {Eq. 1} 10a + 10d = 10,000,000

    Now we can subtract Eq. 1 from Eq. 2:

    {Eq. 2} 12a + 10d = 11,500,000

    {Eq. 1} 10a + 10d = 10,000,000 <--- subtract, eliminating d and forming a 3rd equation

    {Eq. 3} 2a = 1,500,000 <-- divide by 2

    {Eq. 3} a = 750,000 <--- Answer!

    So, 750,000 was invested with ABC.

    Let's check

    If 750k was invested with ABC, 250k was invested with DEF. So let's plug that into Eq. 2

    0.12(750,000) + 0.10(250,000) = 115,000 <--- simplify using calculator

    90,000 + 25,000 = 115,000 <--- add

    115,000 = 115,000 <-- Correct!

  • 1 decade ago

    Let x = investment in ABC in $1000, y = investment in DEF in $1000.

    Then x + y = 1,000

    0.12x + 0.10y = 115

    Subtract 0.1 times the first equation from the second equation to get

    0.02x = 15. Multiply both sides by 50 to get 750.

    So the correct answer is A.

  • 1 decade ago

    First of all not to be mean all. Yahoo answers is not for homework answers, that's why is called homework to help you but obviously your not trying hard and giving up. Look at your book, notes, ask a parent to give you hints but never expect answers that just will make you slow

    Source(s): READ!
  • 1 decade ago

    B

    Source(s): internet and brain
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