A man invests a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year, he earned $92 in interest. How much money was invested in each account?

$1,500 at 2%; $2,300 at 4%
$1,400 at 2%; $2,200 at 4%
$1,000 at 2%; $1,800 at 4%

Question

A man invests a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year, he earned $92 in interest. How much money was invested in each account?

$1,500 at 2%; $2,300 at 4%
$1,400 at 2%; $2,200 at 4%
$1,000 at 2%; $1,800 at 4%

tkhunny · Answer

Why not calculate the amounts and see?

$1,500 at 2%; $2,300 at 4%
----- 2300 - 1500 = 800 Good
----- 1500*0.02 + 2300*0.04 = ??
$1,400 at 2%; $2,200 at 4%
----- 2200 - 1400 = 800 Good
----- 1400*0.02 + 2200*0.04 = ??
$1,000 at 2%; $1,800 at 4%
----- 1800 - 1000 = 800 Good
----- 1000/0.02 + 1800*0.04 = ??

whpalmer4 · Answer

Not hard to write the equations to solve it:

$$I = Prt$$First account:$$I_1 = Prt = P(0.02)(1)$$Second account:$$I_2 = (P+800)(0.04)(1)$$$$I_1+I_2 = 92$$$$0.02P + 0.04(P+800) = 92$$Solve for $P$ for the amount in the first account, $P+800$ is the amount in the second account.