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%? $2,000 $3,000 $4,000

Answer 1

Answer:

$2000

Step-by-step explanation:

6,000 dollars is invested in two different accounts earning 3% and 5% interest.

Let x be the amount invested in account earning 3% interest

So, 6000-x is the amount invested in account earning 5% interest

$S.I. =\frac{P\times R\times T}{100}$

So, Simple interest for account earning 3% interest

$S.I. =\frac{x \times 3 \times 1}{100}$

$S.I. =0.03x$

Simple interest for account earning 5% interest

$S.I. =\frac{(6000-x) \times 5 \times 1}{100}$

$S.I. =0.05(6000-x)$

$S.I. =300-0.05x$

Since we are given that the two accounts earned $220 in interest

So, $0.03x+300-0.05x=220$

$−0.02x+300=220$

$80=0.02x$

$\frac{80}{0.02}=x$

$4000=x$

So, 4000 was invested at 3%

(6000-x)=6000-4000=2000 was invested at 5%

Hence 2000 was invested at 5%