set up word problem

[soprano]

I can not figure out how to set up this problem:
Dave has $40,000 to invest. He invests part of it at 5%, one-fourth of this amount at 6%, and the rest of the money at 7%. His total annual interest income is $2,530. Find the amount invested at each rate.

 

[tutor_joel]

You need to choose what to set as a variable. So,

Dave has $40,000 to invest. This is the total dollars

He invests part of it at 5%. This is unknown.

one-fourth of this amount at 6%. This is related to the unknown amount.

and the rest of the money at 7%. This is what's left.

So, if the first part in unknown, That's your variable x. Now set it up again.

He invests part of it at 5%. This is x.

one-fourth of this amount at 6%. This is

\(\displaystyle \frac{1}{4}x\)

and the rest of the money at 7%. This is what's left.

\(\displaystyle 40,000-x-\frac{1}{4}x\)

That's the first part of the problem. Can you see how to finish it?

 

[soprano]

I am still lost. Would it be 40,000 - x - 1/4x - 0.07x + 2350

 

[masters]

I am still lost. Would it be 40,000 - x - 1/4x - 0.07x + 2350

Hi soprano,

Let's break it down.

\(\displaystyle 40000\) = total amount to be invested.

\(\displaystyle x\) = amount to be invested at 5%.

\(\displaystyle \frac{1}{4}x\) = amount to be invested at 6%.

\(\displaystyle 40000-\left(x+\frac{1}{4}x\right)\) = the remaining amount to be invested at 7%.

\(\displaystyle 2530\) = the amount of simple interest earned.

Simple interest formula tells us that \(\displaystyle I=prt\)

\(\displaystyle .05x + .06(.25x) + .07(40000-x-.25x)=2530\)

 

[Denis]

Soprano, what grade are you in?