how do i do this math problem, u dont have to give the answer but just tell me how to do it
The capicity of an elavator is either 20 children or 15 adults. If
12 children are currently in the elavator, how many adults can still get in?
also theres this problem
there are eight more girls then boys in high school math club. The clun has a total of 44 members . How many biys and how many girls are there?
and heres another but i would like it if someones checked to see if its right
at the conclusion of a soccer game whose two teams included
11 players each player on the winning team "gave five" to each player on the losing team. Each player on the winning team also gave five to each other player on the winning team. How many fives were given?
is it 22? if not how to do this problem?
2007-08-28 09:46:46 · 7 answers · asked by heart 2 in Science & Mathematics ➔ Mathematics
20 children=15 adult
Therefore,1 child=15/20 adult
or,12 children=15*12/20=9 adult
Hence 15-9 or 6 nore adult may get in
first of all subtract 8 from the total
It is 36,And now we have equal boys and girls
So no of boys=36/2=18
No of girls=18+8=26
2007-08-28 10:01:04 · answer #1 · answered by moona 4 · 0⤊ 0⤋
20 children = 15 adults so 4 children = 3 adults
Thus 12 children = 9 adults
Adding 6 adults would be the same as adding 8 children which would bring the total to 20 children which is the limit. So 6 adults could be added.
x= # of boys
x+ 8 = # of girls
2x+8 =44
2x= 36
x= 18 boys
x+8 = 26 girls
Each player on the winning team gives 5 to 11 opponents
So 11*11 =121 fives give to opponents
Each player on the winning team gives five to each of 10 members of his team so 10*11= 110 fives are given
Total fives is 121+110 = 231
2007-08-28 10:10:02 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
1) The idea here is that an elevator can only stand a certain amount of weight, and on average adults don't weight the same as children. Assume that children have a certain average weight, call it "c". Assume the same for adults, call it "a". This means 20c or 15a is the maximum amount of weight for the elevator. Set them equal to each other, get 12c in terms of a, and find out this difference from 15a.
2) Let b be the number of boys. Then the number of girls is b + 8. The boys plus the girls make up 44, so b + (b+8) = 44. Solve this for b, then use b to find b+8.
3) Each of the 11 players gave five to each of the 11 players on the losing team. So that's 121 fives right there. Now each of the 11 players gave high fives to each other. That means each of them have 10 more high fives. But remember that when two players give each other five, that counts as a single five, not two (if there are only two people and they give each other five, then there's only one five that took place). So take 11*10 and divide it by 2, then add it to 121.
2007-08-28 09:56:51 · answer #3 · answered by Anonymous · 0⤊ 0⤋
1) Total capacity of elevator CT = 20 C = 15 A
We need to find X adults in 12 C + X A <= CT
12/20 = 0.6 CT remains 0.4 CT. 0.4 * 15 = 6 adults
2) a) NG = NB + 8
b) NG + NB = 44
NB + 8 + NB = 44
2 NB = 36
NB = 18 boys
NG = 44 - 18 = 26 girls
3) two times the number of player in a team
2007-08-28 10:07:31 · answer #4 · answered by Joseph Binette 3 · 0⤊ 1⤋
I will answer 3 for u-
No it is not 22 sorry here is what it is.
Each kid on the winning team gave 11 high 5s so 11x11=121
Then for the loseing team you just do 121x2=242
There were 242 5s given.
2007-08-28 10:06:02 · answer #5 · answered by Anonymous · 0⤊ 0⤋
For the second problem there are 18 boys and 26 girls.
2007-08-28 10:00:41 · answer #6 · answered by Stephen M 1 · 0⤊ 0⤋
Cross multiply the ratios 20over15 * 12overX
2007-08-28 09:55:10 · answer #7 · answered by Anonymous · 0⤊ 0⤋