Question

$6,000 dollars is invested in two different accounts earning 3% and 5% interest. at the end of one year, the two accounts earned $220 in interest. how much money was invested at 5%?

Answer 1

Answer-

$2000 were invested at 5%

Solution-

Let x amount of money was invested at 5% and (6000-x) amount was invested as 3%

We know that,

$i=\dfrac{P\cdot r\cdot t}{100}$

Putting the values,

$i_1=\dfrac{P_1\cdot r_1\cdot t}{100}$

$i_1=\dfrac{x\cdot 5\cdot 1}{100}=\dfrac{5x}{100}$

And

$i_2=\dfrac{P_2\cdot r_2\cdot t}{100}$

$i_2=\dfrac{(6000-x)\cdot 3\cdot 1}{100}=\dfrac{3(6000-x)}{100}$

According to the question,

$\Rightarrow \dfrac{5x}{100}+\dfrac{3(6000-x)}{100}=220$

$\Rightarrow \dfrac{5x+3(6000-x)}{100}=220$

$\Rightarrow 5x+3(6000-x)=22000$

$\Rightarrow 5x+18000-3x=22000$

$\Rightarrow 5x-3x=22000-18000$

$\Rightarrow 2x=4000$

$\Rightarrow x=2000$

Therefore, $2000 were invested at 5%

Answer 2

The 2 values invested are x and 6000-x

I=PRT
interest=principal times rate times time

time is 1 year so t=1
r=0.03 ad 0.05
we want to find 5% so x is wha we solve for

the interests are
0.03(6000-x) and 0.05x
they add to 220
220=0.03(6000-x)+0.05x
220=180-0.3x+0.05x
220=180+0.02x
minus 180 both sides
40=0.02x
divide both sides by 0.02
2000=x
$2000 was invested at 5%