Aclassroom has some students in it. the ratio of boys to girls is 3: 2. after a minute, a boy enters the classroom and two girls leave. the ratio of boys to girls is now 2: 1. how many students are in the classroom at this point?

24

Step-by-step explanation:

I'm not sure what formulas apply here, so I just used multiples and elimination to find the answer.

If the starting ratio is b:g--> 3:2 and 1 boy enters and 2 girs leave, the ratio, without figuring the multiples of 3 and 2, would leave you with 4:0 and we know that's not the case, because in the end we still have half the number of girls as we do boys (so we can't have 0 girls).  I took multiples of 3 and 2.  First I multiplied both by 2 to get a ratio equivalent to 3:2 which is 6:4.  Now if 1 boy enters and 2 girls leave, the ratio is 7:2 which does not reduce to 2:1.

Next, I multiplied both by 3 to get a ratio equivalent to 3:2 which is 9:6.  Now if 1 boy enters and 2 girls leave, the ratio is 10:4 which does not reduce to 2:1.

Next, I multiplied both by 4 to get a ratio of 12:8.  If 1 boy enters and 2 girls leave, the ratio is 13:6 which does not reduce to 2:1.

Next, I multipled both by 5 to get a ratio of 15:10.  If 1 boy enters and 2 girls leave, the ratio is 16:8 which does reduce to 2:1.

That means that originally, there were 25 students in the room and at the end of the comings and goings, there were 24.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

the answer is -1 1/5, also a

hope i

-lavira : )