A total of $21,000 is to be invested into 2 investments? One exceeding $3000 more.. how many in each

Question

A total of $21,000 is to be invested, some in bonds and some in certificates of deposits. If the amount invested in bonds is to exceed that in CDs by $3,000, how much will be invested in each type?

Answer 1

Hi RahSheema,

 

You can solve this problem the following way:

 

Let c = amount invested in certificates of deposit - CDs ($)

      b = amount invested in bonds ($)

 

We know that the amount invested in bonds (b) exceeds the amount invested in certificates of deposit (c), which can be represented by: b = c+3,000

We also know that the total invested in bonds and CDs is $21,000: c+b= 21,000, which can be written

 

c+ (c+3000) = 21,000

2c+3,000 = 21,000

     -3,000      -3,000 (subtract 3,000 from each side)

-------------------------

2c= 18,000

divide by 2 on both sides

c= 9,000 <-- $9,000 is invested in CDs

Now to find b, which is equal to c+3,000 thus

b= 9,000 +3,000

b= 12,000 <---$12,000 is invested in bonds.

 

To check the answer, we have to check if c+b = 21,000

$9,000 + $12,000 = $21,000 CHECK!

so we know our answer works.

 

Answer:

$9,000 is invested in certificates of deposit and $12,000 is invested in bonds.

 

 

Answer 2

Let x = the amount invested in CDs. Then x + 3000 is invested in bonds. Altogether, 21,000 is invested. So:

 

x + (x + 3000) = 21000

2x + 3000 = 21000

2x = 18000

x = 9000

 

$9,000 is invested in CDs. 9000 + 3000 = $12,000 is invested in bonds.