For genius people,answer this?

Question

In a chess competition in school involving some boys and girls, every student had to play exactly one game with every other student, it was found in 45 games both the players were girls and in 190 games both were boys,so the number of games in which one player was a girl and other a boy is _________???
answer with proper explanation !!

Answer 1

If there were 2 girls, there would be 1 match between them.
If there were 3 girls, there would be 3 matches between them.
If there were 4 girls, there would be 6 matches between them.
If there were 5 girls, there would be 10 matches between them.
etc.
If there were 9 girls, there would be 36 matches between them.
If there were 10 girls, there would be 45 matches between them.

So we know there are 10 girls.

In general, for n players there will be n(n-1)/2 matches between them.
M(n) = n(n-1)/2
M(10) = 45

Similarly for the boys:
M(20) = 20*19/2 = 190

So there are 10 girls and 20 boys.
All together there would be M(30) matches
M(30) = 30*29/2 = 435 matches

So you want:
M(30) - M(20) - M(10)

= 435 - 190 - 45
= 200 matches between a boy and a girl

Another way to confirm this to see that the 10 girls would play each of the 20 boys, so that is 10 x 20 = 200

45 matches between two girls
190 matches between two boys
---> 200 matches between a boy and a girl

Answer 2

I can tell you there were not 90 girls, a person can not play herself.

The number of people are the combination 45 times the length of the combination 2.

Starting with 90 we have to find some number where x*(x-1) = 90

10 x 9 = 90, so we know there are 10 girls.

Do the same with boys 190 x 2 = 380 , by trial and error:
20 x 19 = 380

So there were 20 boys.

Each girl played 1 boy, so 10 girls times 20 boys = 200 games

Answer 3

Find out the number of girls: 45games let the number of girls be x
now find a serie for x this way: if 2x then 1game if 3x then 3games if 4x then 6games. That means 1/2x(x-1)=number of games. So that means 1/2x^2-1/2x=45 chance it to x^2-x-90=0 we put it in brackets and get (x-10)(x+9)=0 thus we get x=10 because x can`t be a minus number. Now do the same with the girls` games let it be y. So we get !/2Y^2-1/2Y=190 that means
(y-20)(y+19) thus we get y=20.
Now we got x boys =10 and y girls=20
Now multiply x with y because each of the 10 boys has to play against 20 girls that means 10 times 20.
So we get 200 games boy against girl.

Thanks for calling me a genius.

Answer 4

There were 10 girls (a round-robin lasting 45 games has 10 participants).

There were 20 boys (same reasoning).

All 10 girls played 20 games against a boy, so there were 10*20 = 200 games in which a girl played against a boy.

Answer 5

Everything you need in this problem is a combination. Let
B = number of boys
G =number of girls
bb = number of boy-boy games
gg = number of girl-girl games
bg = number of boy-girl games
TG = total number of games
Since you are picking two at a time,
TG = (B+G)_C_2 where C is the Combination symbol
also we know that since there are B boys
bb = 190 = B_C_2 = 20_C_2 so there are 20 boys
and
gg = 45 = G_C_2 = 10_C_2 so there are 10 girls
Now we can figure out the TG
TG = (20+10)_C_2 = 30_C_2 = 435
and from
TG = bg+bb+gg we can solve for bg to get:
bg = TG-bb-gg = 435-190-45 = 200
I hope that is clear enough to understand.

Answer 6

number of girls: x
Let's say one girl played with all the rest (x-1) games, the next one played (x-2) other games (excluding the game played with the first girl, because this game is already counted in the games played by the first girl in the example), and so on, untill last two girls played only one different game against each other. According to arithmetical progression formula:

(((x-1)+1)*(x-1))/2=45
(x*(x-1))/2=45
(x-1)*x=90
x^2-x-90=0
D=1+360=361
x=(1+19)/2=10

number of boys: y
(y-1)*y=380
y^2-y-380=0
D=1+1520=1521
y=(1+39)/2=20
so each girl played 20 games with boys or vice versa: 20*10=200
The answer is 200!

Answer 7

It would take 10 girls to play 45 games, each playing one another once.

It would take 20 boys to do the same for 190 game.

10 girls, 20 boys. Each boy plays each girl once.

20 x 10 = 200.

200

Answer 8

There are 10 girls and 20 boys. So there would be 200 games involving one boy and one girl.

Answer 9

45 x 2 = 90
190 x 2 = 380
380 + 90 = 470
190 + 45 = 235
470-235 = 235
answer 235 games

Answer 10

suppose there are m boys then each one playes m-1 games

m(m-1)/2 = 190
solving this quadratic equation m = 20

if there are n girls number of games - n(n-2)/2= 90 so n = 10

as there are 10 girls and 20 boys there are 200 games when one is a girl and one is a boy