Out of 500 students,75% of boys and 72% of girls passed. Total pass percentage is 77%. How many girls passed

Let the number of Boys be B, and the number of girls be G.
That means that B + G = 500

75% of boys passed, 72% of girls passed, and 77% of students passed. So:
0.75B + 0.72G = 0.77*500 = 385

From the first equation: G = 500 - B
Substitute for G in the second equation:
0.75B + 0.72(500-B) = 385
0.75B + 0.72*500 - 0.72B = 385
0.03B + 360 = 385
Subtract 360 from both sides:
0.03B = 25
Divide both sides by 0.03
B = 833⅓

That means that there are more boys than in the entire class population! Looking closer at the question, it's not possible for the total percentage to be higher than the two partial percentages. There's something not right here...

LET B BOYS PASSED SO 75%0F B PASSED
NOW NO OF GIRLS= [500-B]

SO 72% OF [500-B] PASSED

TOTAL PASSED OUT OF 500 =77%OF 500= 385
75/100OF B+ 72/100 0F[500-B] =385

75B+ 72[500-B]=38500
75B+36000-72B= 38500
3B=2500
girls actually passed=24/49 = 72% of girls

Actually it's a silly question. The total pass percentage has to be between % of boys passed and % of girls passed.

In the question, it says 72% of girls passed. But that can't be so, if 75% of boys passed and total pass pecentage is 77%

Let A be number of boys, B be number of girls.

A+B = 500
.75A + .72B = .77(500)

The problem is that this is actually impossible- solving those two equations gives about -333 girls and about 833 boys. Someone wrote a bad question..

Let g be the total girls and b be the total boys.

b + g = 500
(0.75b + 0.72g) / 500 = 0.77

Two equations with two unknowns, solve for g.

Final answer is 0.72g.

72 percent

25% of 500 = 125 girls
72% of 125= 90

90 girls passed