SOLUTION: A TOTAL OF 27,000 IS INVESTED , PART OF IT AT 12% PART OF IT AT 13%. THE TOTAL INTEREST AFTER A YEAR IS 3,385 . HOW MUCH WAS INVESTED AT EACH RATE?

Question 91561: A TOTAL OF 27,000 IS INVESTED , PART OF IT AT 12% PART OF IT AT 13%. THE TOTAL INTEREST AFTER A YEAR IS 3,385 . HOW MUCH WAS INVESTED AT EACH RATE?
Found 2 solutions by stanbon, bucky:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A TOTAL OF 27,000 IS INVESTED , PART OF IT AT 12% PART OF IT AT 13%. THE TOTAL INTEREST AFTER A YEAR IS 3,385 . HOW MUCH WAS INVESTED AT EACH RATE?
--------------
Let amount invested at 12% be "x"; amount of interest is 0.12x dollars
------------
Amount invested at 13% is 27000-x; amount of int is 0.13(27000-x)=3510-0.13x dollars
-----------------------------
EQUATION:
interest + interest = 3385
0.12x + 3510-0.13x = 3385
-0.01x = -125
x = $12,500 (amount invested at 12%)
27000-12500 = $14,500 (amount invested at 13%)
============
Cheers,
Stan H.
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
There are two unknowns in this problem ... the last sentence (How much was invested at
each rate?) tells you this.
.
So let's begin by identifying the two unknowns ... call the amount invested at 12% A, and call
the amount invested at 13% B.
.
Now we can begin writing equations. For the first equation we know that the total amount
invested is $27,000. But the total amount invested is A plus B. Setting these two equal we
get a first equation:
.

.
Since A is the amount invested at 12% (or its decimal equivalent 0.12) we know that the
amount of interest made on this investment is 0.12 times A. And since B is the amount
invested at 13% (or at 0.13) we know that the amount of interest made on this investment
is 0.13 times B. The sum of these two amounts is 0.12A + 0.13B, and the problem tells you
that the total amount of interest for the year is $3385. Therefore we have another
equation based on the total interest. This equation says:
.

.
Since you can't solve a single equation with two unknowns, we need to find a way of combining
our two equations so that we end up with a single equation having only one unknown.
.
One way we can do that is to solve our first equation for one of the unknowns in terms
of the other unknown, and substitute that result into our second equation. Our first
equation said:
.

.
Subtract B from both sides of this equation and you get:
.

.
Now let's go to our second equation which said:
.

.
But we have learned that A equals 27000 - B. So in our second equation we can replace
A with 27000 - B to get:
.

.
Note that this "new" equation has only one unknown ... B. Therefore, we can solve it
for B. Begin by multiplying 0.12 times each of the terms in the parentheses. When you do
that the equation becomes:
.

.
Combining the two terms that contain a "B" reduces the equation to:
.

.
Then get rid of the 3240 on the left side by subtracting 3240 from both sides to get:
.

.
So our equation is now:
.

.
You can solve for B by dividing both sides of this equation by 0.01 to get:
.

.
Now we know that B (the amount invested at 13%) is $14,500. And since the total amount
invested was $27,000, the remaining $12,500 must be the amount invested at 12%.
.
Let's check:
.
$14,500 invested at 13% would earn 14500*0.13 = $1885
.
And $12,500 invested at 12% would earn 12500*0.12 = $1500
.
So the total earnings would be $1885 + $1500 = $3385
.
This is exactly the amount that the problem said it should be, so that checks.
.
And the total amount invested is $14,500 + $12,500 = $27,000. Again, this is exactly
the amount the problem said it was. So our answers check all the way.
.
The answers are that $14,500 is the amount invested at 13% and $12,500 at 12%.
.
Hope this helps you to understand the problem.

RELATED QUESTIONS
A total of $3600 was invested, part of it at 6% interest and the remainder at 11%. If the (answered by addingup)
A total of $5,000 was invested, part of it at 5 % interest and the remainder at 7 %. If... (answered by mananth,josgarithmetic)
a) A total of $40,000 was invested, part of it at 12% interest and the remainder at 15%. (answered by greenestamps,josgarithmetic)
$9500 is invested, part of it at 10% and part of it at 6%. for a certain year, the total... (answered by jorel1380)
$5400 is invested, part of it at 11% and part at 6% for a certain year the total yield is (answered by richwmiller)
Interest question, please help $5400 is invested, part of it at 10% and part of it at... (answered by stanbon)
Ethan invested $19,000 for 1 year, part at 11% simple interest and part at 13% simple... (answered by ikleyn,mananth,Edwin McCravy,greenestamps)
A total of $2000 was invested: part at 8% and the rest at 12% per year. If after one... (answered by macston)
6300 is invested, part of it at 12% and part of it at 9%. for a certain year the... (answered by checkley79)