# How many people are there in room A after the change of rooms (below)?There were 1200 people in Room A and Room B. 30% of 800 people in Room A were women. 40% of the people in Room B were men....

How many people are there in room A after the change of rooms (below)?

There were 1200 people in Room A and Room B. 30% of 800 people in Room A were women. 40% of the people in Room B were men. After some people in both rooms had changed rooms, 25% of the people in Room A and 75% in room B were women.

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

Number of people in Room A initially = 800

Number of people in room B initially = 1200-800 = 400

In room A initially there were 30% of 800 are women.

Number of women in room A initially = 30/100*800 = 240

Number of men in room A initially = 800-240 = 560

In room B initially there were 40% 400 are of men.

Number of men in room B initially = 40/100*400 = 160

Number of women in room B initially = 400-160 = 240

So;

Total number of men = 160+560 = 720

Total number of women = 240+240 = 480

AFTER CHANGE

Number of people in room A = x

Number of people in room B = 1200-x

So after change 25% of people in room A and 75% in room B are women. But total women ill be 480.

0.25x+0.75(1200-x) = 480

x = 840

So there are 840 people in room A after change.

lemjay | High School Teacher | (Level 2) Senior Educator

Posted on

Let,    A - # of people in Room A

B - # of people in  Room B

The total number of people in A and B is 1200. So we have,

`A+B=1200`        (EQ.1)

> Then, let's consider the number of people in  Room A.

`A_f + A_m = A`

where `A_f` and `A_m` represents the number of females and males in Room A.

The given values for Room A are:

`A=800 `                          `A_f = 0.30(800)=240`

Then, let's solve for `A_m` .

`240+A_m=800`

`A_m=800-240`

`A_m=560`

> Next, let's consider the number of people in Room B.

Susbtituting value of A to EQ.1, then Room B has:

`800+B=1200`                 `B=400`

The given number of males in Room B  is:

`B_m=0.40(400)=160`

Then, solve for number of females in Room B. `B_f` is:

`B_f + B_m=400`

`B_f+160=400`

`B_f=240`

So we have,

Room A        Room B          Total

#of females:         240              240               480  (total # of females)

# of males  :         560              160               720 (total # of males)

Total # of People:  800             400              1200

> When some of the people changed rooms,  we have

`A'_f=0.25A'`                     `B'_f=0.75B'`

Note that the total number of females in both room is still the same which is 480. So,

`A'_f + B'_f=480`

`0.25A'+0.75B'=480`     (EQ.2)

To solve for A' (new # of people in Room A), express EQ.2 in one variable. To do so, we need to take note that the total number of people is both rooms is unchanged.

`A'+B'=1200`

`B'=1200-A'`

Subtitute this to EQ.2 .

`0.25A'+0.75(1200-A')=480`

`0.25A'+900-0.75A'=480`

`900-0.5A'=480`

`-0.5A'=-420`

`A'=840`

Hence, after some of the people changed rooms, Room A has a total of 840 people.