Question

Last year Marta invested money in two accounts. The first account had an interest rate of 2% and the second account had an interest rate of 3%. If she invested $200 more in the first account than the second and her total interest income was $384, how much did she invest at each rate?

Answer 1

Let the money invested in second account be X .

Money invested in first acc. =X+200

Simple interest (SI)=P×R×T/100

Total interest = [ (X+200)×2×1]/100 + X×3×1/100

Therefore,

384=(2x+400+3x)/100

38400-400=5x

X=7600$ in second account and 7600+200=7800$ in first account

Answer 2

$\huge\boxed\bf \huge\mathfrak{answer}$

RS. 7800 AND RS. 7600

$\huge \red{STEP-BY-STEP} \\ \huge \red{ VERIFICATION:}$

LET, x+200 AND x BE THE FIRST AND SECOND INVESTMENT RESPECTIVELY.

THEN,

FOR FIRST INTEREST(I¹)

$= 2\% \: of \: (200 + x) \\ = 4 + 0.02x$

FOR SECOND INTEREST(I²)

$= 3\% \: of \: (x) \\ = 0.03x$

THEN,

TOTAL INTEREST(I)=I¹+I²

OR,

$384 = 4 + 0.02x + 0.03x$

OR,

$380 = 0.05x$

•°•

$x = 7600$

AGAIN,

x+200=7800

SO, THE INVESTMENTS ARE RS. 7800 AND RS. 7600 RESPECTIVELY.