Question

A total of $11,000 is invested in two accounts. Part was invested at 4% and the rest was invested at 7%. If the investments earn $680 peer year, how much was invested at each rate? I=prt

Answer 1

Answer:

The amount invested at 4% was $3,000 and the amount invested at 7% was $8,000

Step-by-step explanation:

we know that

The simple interest formula is equal to

$I=P(rt)$

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

$t=1\ year\\ P_1=x\\P_2=11,000-x\\I=\ 680\\r_1=4\%=4/100=0.04\\r_2=7\%=7/100=0.07$

substitute in the formula above

$I=P_1(r_1t)+P_2(r_2t)$

$680=0.04x+(11,000-x)0.07$      

solve for x

$680=0.04x+770-0.07x$

$0.07x-0.04x=770-680$

$0.03x=90$

$x=\ 3,000$

so

$\ 11,000-x=\ 11,000-\ 3,000=\ 8,000$

therefore

The amount invested at 4% was $3,000 and the amount invested at 7% was $8,000