Julie put half of her savings in a savings account that pays an annual

Let the initial savings be 2x

From simple interest
xy/100 = 60
xy = 6000

From Compound interest
y(x+60)/100 = 66
xy + 60y = 6600
60y = 600 (replacing values from 1st eq)
y = 10

putting y in 1st eq
x = 600

Initial saving, 2x = 1200.

D

S.I. earned = 120
C.I. earned = 126

Time period: 2 years.
Rate= same = let us assume 10.

Difference between C.I. and S.I. for 2 years: \(\frac{PR^2}{100^2}\)

=> \(\frac{PR^2}{100^2} = 6\)

=> \(\frac{P * 10 * 10 }{ 100 * 100} = 6\)

=> P = 600 [This was an equal amount that was put in two different scheme]

Initial saving: 600 + 600 = 1200

Answer D

Given: Julie put half of her savings in a savings account that pays an annual simple interest and half in a savings account that pays an annual compound interest. After two years she earned $120 and $126 from the simple interest account and the compound interest account respectively.

Asked: If the interest rates for both accounts were the same, what was the amount of Julie's initial savings?

Let the amount of Julie's initial savings be $2x and interest rate be r%

Compound interest after 2 years = x(1+r%)^2 - x = 126
Simple interest after 2 years = 2xr% = 120
xr% = 60
r% = 60/x

x(1+60/x)^2 - x = 126
x(1+60^2/x^2 + 120/x) - x = 126
x (60^2/x^2 + 120/x) = 126
3600/x + 120 = 126
3600/x = 6
x = 3600/x = $600

Julies initial savings = 2*600 = $1200

IMO D