In the first year of a couple's marriage, the wife’s earnings were 40

If the Salary of Husband in the first year = 60, amount, P, that he invested is 60*0.3 = 18

Annual Interest = \(\frac{P*r*t}{100}\) = \(\frac{18*10*1}{100.}\)

Since we are finding percentage of increase in the interest, common terms (i.e is 1/100) gonna cancelled out in the last. So, i just substitute it with a random variable. Also, I personally don't like to involve fractions or decimals in my calculations (helps me to do most of the calculations mentally).

Solution:

Let the combined income of the couple in the first year be 1000. Then, the wife earned 0.4 * 1000 = 400 and the husband earned 1000 - 400 = 600. The wife invested 400 * 0.4 = 160 at 5% for a year; thus, the interest she earned was 160 * 0.05 = 8. The husband invested 600 * 0.3 = 180 at 10% for a year; thus, the interest he earned was 180 * 0.1 = 18. In total, the couple earned an interest of 8 + 18 = 26 in the first year.

In the second year, the couple earned 1000 * 1.1 = 1100 in total. Let h be the earnings of the husband. Then, the wife earned 5h/6 and in total, they earned h + 5h/6 = 11h/6. Setting this equal to 1100, we obtain:

11h/6 = 1100

h/6 = 100

h = 600

So, the husband earned 600 in the second year and the wife earned 1100 - 600 = 500. The wife invested 500 * 0.48 = 240 at 5% for a year; thus, she earned an interest of 240 * 0.05 = 12 in the second year. The husband invested 600 * 0.5 = 300 at 10% for a year; thus, he earned an interest of 300 * 0.1 = 30 in the second year. In total, the couple earned an interest of 30 + 12 = 42 in the second year.

We can use the percent greater than formula to obtain that the interest earned in the second year was 100 * (42 - 26)/26 = 100 * 16/26 ≈ 61.53 percent, which is approximately 60%.

Answer: D

Thanks for the solution, just 1 query why are you not considering the interest earned in 2nd year from the investment made in the 1st year.
Thanks in Advance.

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