Please provide a detailed solution for this problem

The problem below is virtually the same as -- but harder than -- PS114 in the OG13:
https://www.beatthegmat.com/og-114-t125020.html

Architj wrote:In country X, the unemployment rate among graduates dropped from 28 percent on july 1, 2000 to 19 percent on july 1, 2004. The no. of employed graduates in the country on july 1,2004 was 10 percent more than that on july 1, 2000. What was the approximate percentage change in the no. of unemployed graduates over the period july 1, 2000 to july 1, 2004?

A. 28% decrease
B. 33% decrease
C. 38% decrease
D. 43% decrease
E. 48% decrease

July 2000:
Let the total number of graduates = 1000.
Unemployed graduates = 28% of 1000 = 280.
Employed graduates = 1000-280 = 720.

July 2004:
Since the number of employed graduates increases by 10%, employed graduates = 720 + 10% of 720 = 720 + 72 = 792.

Let x = the number of unemployed graduates.
Since the unemployment rate is 19%, 19 of every 100 graduates are unemployed, while 81 are employed, implying that unemployed/employed = 19/81.
Since unemployed = x and employed = 792, we get:
x/792 = 19/81
x = (792*19)/81 = (88*19)/9 ≈ 186.

Percent decrease:
The number of unemployed graduates decreases from 280 to 186:
280-186 = 94.
Since 94/280 ≈ 1/3, the number of unemployed graduates decreases by about 33%.

The correct answer is B.