Question

A total of $7000 is invested: part at 8% and the remainder at 14%. How much is invested at each rate if the annual interest is $710?

Answer 1

let's say the amounts are "a" and "b"

we know the total is 7000, so.. .whatever "a" and "b" are, they add  up to 7000
thus
    a + b = 7000

how much is 8% of "a"? well, 8% of anything is (8/100) * anything
so 8/100 * a or 0.08a

how much is 14% of "b"? well, 14/100 * b or 0.14b

whatever those percentage amounts are, we know, they yielded $710
thus
   0.08a + 0.14b = 710

thus    $\bf \begin{cases} a+b=7000\implies \boxed{b}=7000-a\\ 0.08a+0.14b=710\\ ----------\\ 0.08a+0.14\left( \boxed{7000-a} \right)=710 \end{cases}$

solve for "a", to see how much was invested at 8%

what about "b"?  well, b = 7000 - a

Answer 2

$4500 is invested at 8% and $2500 is invested at 14% if the annual interest is $710.

What is annual interest rate?

An annual percentage rate is expressed as an interest rate. It calculates what percentage of the principal you'll pay each year by taking things such as monthly payments into account.

According to the question

A total of $7000 is invested: part at 8% and the remainder at 14%.

Let x be the amount invested in 8% and (7000 - x) amount invested in

14 %.

$\frac{x(8)(1)}{100}+\frac{(700-x)(14)(1)}{100} =710$

$8x+ 98000 - 14x = 71000$

$6x = 27000$

x = $4500

$7000 - x$

= $7000 - 4500$

= $2500

Thus, $4500 is invested at 8% and $2500 is invested at 14% if the annual interest is $710.

Find out more information about annual interest rate here

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