Question 25721: On the first day of school, the percentage of boys in a particular class is 60%. During the school year, six girls move away, and are replaced in the class by six boys; this makes the class roster 75% boys. Find the number of boys and girls in the class on the first day of school.
Found 3 solutions by stanbon, mewtwo1, josmiceli:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Let total number in the class be "A".
EQUATION:
60%A+6=75%A
15%A=6
A=6/0.15= 40 Number in the class
60%(40)=24 boys in class on day one.
40%(40)=16 girls in class on day one.
Cheers,
Stan H.
Answer by mewtwo1(3) (Show Source):
You can put this solution on YOUR website! 24Boys and 16Girls
Find total of Boy and Girls in class.
15% = 6, 15/100 = 6/X, X x 15/100 = 6/X x X, which is 15X/100 = 6
100 x 15X/100 = 6 x 100, which is15X = 600,
1/15 x 15X = 600 x 1/15, which is X = 40
(Number of Boys and Girls in class is 40)
Find numerical value for 60% of 40
60/100 = X/40, 40 x 60/100 = X/40 x 40, which is 2400/100 = X
X = 2400/100 IS X = 24. 24 Boys – 40 = 16 Girls
24 Boys and 16 Girls
Explation:
If 6 girls were taken away and the percentage of boys when from 60% to 75%, then 6 students equal 15%. Now, from here there are a few ways to do this… unfortunately I always seem to find the most difficult way possible to solve things… sorry. So, if 15 % (15/100), is (=) 6, then 100% of number of students must be… X (unknown), now to solve for X. 15 over 100 equal 6 over X.
15/100=6/X, so X is currently a denominator (1/X), it must first be a numerator (X/1 or X … it’s the same thing). Cross-multiply by multiplying Both Sides of the equation by X/1. This will cause the left side to have 15X/100 and the right side to have 6X/X (which can now be looked at as just 6x1 = 6.
15X/100 = 6, is the current stage of our problem. We must get X ALONE on one side, so next move the 1/100 part of 15X/100 to the right side by multiplying Both Sides by 100. So 15X/100 x 100 (or 100/1) because 100/100 is 1 this side can now be read as 15X x 1 = 15X. The right side is 6 x 100 = 600.
15X = 600, is the current stage. Because 15X can also be read as 15 x X, we can divide Both Sides by 15 ( the same as multiplying by 1/15). So the left side is 15X/15, which is like 15/15 x X, WHICH is like 1 x X which is just X. The right side is 600/15 = 40. So X = 40 …….good…but, we still didn’t answer the original question.
So, X (100% of the number of students) is 40. The question told us that 60% of the students were boys, so 60/100 = X/40 (X being the numerical value of 60%), Multiply Both Side by 100 and then 40, which should end up as 100X= 2400, divied Both Side by 100, so now X =24…. 24 boys and the rest of the 40 must be 16 girls.
24Boys and 16Girls
Answer by josmiceli(19441)  (Show Source):
You can put this solution on YOUR website! Let b = number of boys on the 1st day of school
let g = number of girls on the 1st day of school
so, b + g equal number of boys + number of girls
60% of (b + g) is boys on the 1st day
later on, 6 girls leave and 6 boys are added
That means (b + g) does not change, but boys are now 75% of (b + g)
so, on the 1st day,
(a).....b/(b + g) = .60
and later on,
(b).....(b + 6)/(b + g) = .75
solve the first equation for b
b = .6b + .6g
.4b = .6g
b = (3/2)g
now solve the second equation for g
.75(b + g) = b + 6
.75b + .75g = b + 6
.75g = .25b + 6
express as fractions
(3/4)g = (1/4)b + 6
3g = b + 24
g = (1/3)b + 8
substitute b = (3/2)g (from before)
g = (1/3)(3/2)g + 8
g = (1/2)g + 8
2g = g + 16
g = 16
now substitute this value in either of the two equations (a) or (b)
b/(b + 16) = 3/5
b = (3/5)(b + 16)
(5/3)b = b + 16
(2/3)b = 16
b = (3/2)16
b = 24
so, there are 16 girls and 24 boys on the first day of school
If I add 6 boys and subtract 6 girls like the problems says
are the boys 75% of (boys + girls)?
(24 + 6) /((24 + 6) +(16 - 6)) = .75
30/(30 + 10) = 30/40 = .75
answer checks
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