Salary of Husband in the first year = 60
Interest he gets = 60*0.3*10*x= 180x, where \(x=\frac{1}{100 }\)
Salary of Wife in the first year = 40
Interest she gets = 40*0.4*5*x= 80x
Total interest earned by the couple in the
first year =
260xSalary of Husband in the second year = 60
Interest he gets = 60*0.5*10*x= 300x
Salary of Wife in the second year = 50
Interest she gets = 40*0.48*5*x= 120x
Total interest earned by the couple in the
second year =
420xIncrease in the interest = \(\frac{420-260}{260} *100\) \(= \frac{800}{13} ≈ \frac{780}{13} = 60\) %
Bunuel wrote:
In the first year of a couple's marriage, the wife’s earnings were 40 percent of the combined earnings of the couple. The wife invested 40 percent of her earnings at an annual return of 5 percent and the husband invested 30 percent of his earnings at an annual return of 10 percent. In the second year of their marriage, the combined earnings of the couple increased by 10 percent and the wife’s earnings were five-sixths of her husband’s earnings. The wife invested 48 percent of her earnings and the husband invested 50 percent of his earnings in their respective investment instruments of the previous year. If the couple made no other investments and took out the interest earned in the first year at the beginning of the second year, by approximately what percent was the interest earned by the couple in the second year greater than the interest earned by the couple in the first year of their marriage? The interest income from the couple’s investments is not considered in their earnings.
A. 30%
B. 40%
C. 50%
D. 60%
E. 70%