During a high school drama rehearsal, 15 girls left. Twice as many boys as girls remained.?

Question

Later, 45 boys departed, leaving five times as many girls as boys. Determine the initial number of boys and girls at the rehearsal.

plz show ur work =) thx

Answer 1

Let's assume there were 'g' girls and 'b' boys at the beginning.

With the given assumptions we have,

g-15 = 2b ---- (i)

and

g-15 = 5 (b-45) ---- (ii)

from (i) & (ii) we get,

2b=5(b-45)

or, 2b=5b-225
or 2b-5b = -225
or -3b = -225
or b = 75

from (i) we get

g-15 = 2b = 2X75=150

so, g = 150+15 = 165

So there were 165 girls and 75 boys initially

cool

Answer 2

if you can't solve an easy question like this on your homework, maybe you should go ask your teacher for extra help.

You want to find the initial number of boys and girls, so:

Let b = initial # of boys
g = initial # of girls

15 girls leave:
g - 15
leaving twice the number of boys as girls:
2(g - 15) = b

then 45 boys leave:
b - 45
leaving five times as many girls as boys; remember that the new number of girls is g - 15, instead of just g:
5(b - 45) = g - 15
5b - 225 = g - 15

you know from the first part that
b = 2(g - 15)
b = 2g - 30
so you can substitute that into your second equation:
5(2g - 30) - 225 = g - 15
10g - 150 - 225 = g - 15
10g - 375 = g - 15

now bring the G's to one side and the values to the other:
10g - g = -15 +375
9g = 360
divide, and you get:
g = 40
so the initial number of girls was 40.
you still need to find the initial number of boys, so substitute this value into your first equation:
2(g - 15) = b
2(40 - 15) = b
2(25) = b
50 = b

so the initial number of boys was 60.

check your answers by substituting them into the problem

you started with 40 girls and 15 left, leaving 25. double that and you get the initial number of boys, which was 50, which is what your answer said. so now you've got 25 girls and 50 boys. then 45 boys leave, and 5 remain. there are now five times as many boys as girls, just as the question said. so the answers are correct: 40 girls and 50 boys.

sorry if that was long, but like i said, talk to your teacher if you cannot solve problems as simple as this.

Answer 3

Let G be the number of girls initially
and B be the number of boys initially

15 girls leave therefore remaining number of girls is G- 15

The number of boys is twice as much as the number of girls therefore 2 x the number of girls left = the number of boys initially (no boys have left yet)
Therefore we get this equation:
2(G-15) = B
B= 2G -30

Later 45 boys leave so the number of boys drop to B- 45

Now there are 5x as many girls as boys so 5x the number of boys = the number of girls left:
5(B - 45 )= (G - 15)
5B - 225 = G - 15

Now substitute in the 1st eqn:
5(2G - 30) - 225 = G - 15
10 G -150 - 225 = G - 15
9G= -15 + 225 + 150 = 360
G= 360/9 = 40

Now B= 2G -30 = 2*40 - 30 = 50

Therefore initially there are 40 girls and 50 boys

Answer 4

Let boys = x
girls = y -15
x = 2(y - 15).....Eq 1
y - 15 = 5(x - 45).....Eq 2
Substitute the value of 'y - 15' of Eq 2 in Eq 1
x = 2(y - 15)
x = 2[5(x - 45)]
x = 2(5x - 225)
x = 10x - 450
x - 10x = -450
-9x = -450
9x = 450
x = 450/9
x = 50
Substitute the value of 'x' in Eq 2
y - 15 = 5(x - 45)
y - 15 = 5(50 - 45)
y - 15 = 5*5*
y - 15 = 25
y = 25 + 15
y = 40
The initial number of boys was 50.
The initial number of girls was 40.

Answer 5

2(g-15)=b ...(1)
5(b-45)=g-15 ...(2)
By solving (1) and (2), you will get 40 girls and 50 boys initially.

Answer 6

Let B=initial boys, G=initial girls.

G-15=2B
5(G-15)=B-45

G=2B+15

5(2B+15-15)=B-45
10B=B-45
9B=45
B=5

G=2B+15=(2*5)+15=25