interest story problem

Question

the annual interest on a $19000 investment exceeds the interest earned on a $13000 investment by $534. The $19000 is invested at a 0.6% higher rate of interest than the $13000. What is the interest rate of each investment?

Answer 1

Hi Anjali,

 

You've got yourself a tough problem there.  Don't lose hope, I think we can sort things out. 

 

Let's give ourselves some numbers.  We'll let R_1 be the interest rate on the $19,000, expressed as a number between 0 and 1, and I_1 the amount of interest earned on that $19,000.  We'll let r_2 be the interest rate on the $13,000, expressed as a number between 0 and 1, and i_2 the interest. Also, we'll give ourselves the definition of annual interest:

I_1 = $19000 x R_1

i_2 = $13000 x r_2

 

So what do we know?  The first sentence tells us that

I_1 = i_2 + $534. 

which is the same as

$19000 x R_1 = $13000 x r_2 + $534

 

The second sentence tells us that

R_1 = r_2 + 0.006

 

But if that's so, we can substitute "(r_2 + 0.006)" for R_1:

$19000 x R_1 = $13000 x r_2 + $534

$19000 x (r_2 + 0.006) = $13000 x r_2 + $534

 

distribute the 19000 across:

$19000 x r_2 + $19000 x 0.006 = $13000 x r_2 + $534

$19000 x r_2 + $114                 = $13000 x r_2 + $534

 

subtract $13000 x r_2 from both sides, and subtract $114 from both sides while you're at it. 

$6000 x r_2  =  $420

 

Solve for r_2:

r_2 = 420 / 6000.

 

I'll let you handle that, eh?

 

And then we already know that

R_1 = r_2 + 0.006

 

Since you now know r_2, you can add 0.006 to get R_1.

 

Fair enough?

 

Cheers,

Mike N.