Question

Three halls contained 9,876 chairs altogether. One-fifth of the chairs were transferred from the first hall to the second hall. Then, one-third of the chairs were transferred from the second hall to the third hall and the number of chairs in the third hall doubled. In the end, the number of chairs in the three halls became the same. How many chairs were in the second hall at first?'

Answer 1

interesting

Answer 2

hard?

Answer 3

not really, too many equations but i'm working on it ;)

Answer 4

kk, ty

Answer 5

this is very simple.
in the end there were equal number of chairs in the three halls.
that means each hall has 9876/3 = 3292 chairs in the end
suppose there were x chairs in the first hall in the beginning.
one fifth of the chairs were transferred to hall 2
that means x - x/5 = 3292
find x. that is the total number of chairs in hall 1 in the beginning

suppose there were y number of chairs in hall 2 in the beginning
x/5 chairs were transfered to hall 2 from hall 1
that means the chairs in hall 2 after the transfer was y+(x/5)
one third of these chairs were transfered to hall 3, leaving hall 2 with 3292 chairs
that means
(y+(x/5)) - (y+x/5) / 3 = 3292
find y. that is your answer.

Answer 6

I think i have it just give me a second to type it out.

Answer 7

i see you started working from the last clue in the question which is much simpler than what i was working out, had a couple of interruptions but i guess it was worth it ;) I'm arriving at the same answer so i might as well not show it lol :P the answer is right there ;)

Answer 8

there! the answer is not there anymore!

Answer 9

lol solve for y and you have the answer :P what i'm getting to looks like i'll arrive at the same answer so yh i stopped working it out lol

Answer 10

good

Answer 11

Assume A=Hall 1 B=Hall 2 and C=Hall 3

A=4115 B=4115 C= 1646

Since all rooms have the same number at the end, we know that after transferring a fifth of it's chairs, A=3292. so, x-(x/5)=3292
5x-x=16460
4x=16460
x=4115
So, Abegins with 4115. Now we know that at the beginning B+C=9876-4115
B+C=5761
Now we add 1/5 of A So that B+C=5761+823
B+C=6584 after the first move

We also know that after the first move, 1/3 of the chairs from B Doubles the amount of C so, B/3=C
since we now have 2 equations with the same variables after step one we can solve:
B/3=C
B=3C


B+C=6584
3C+C=6584
4C=6584

B+1646=6584
B=4938


C=1646

So, after step 2 A=3292 B=4938 C=1646

Since we are looking for room two after step one, we just reverse step one, so
At Step 1: B=4938-(4115-3292)
B=4938-(823)
B=4115

So, at the beginning Hall 2 has 4115 Chairs.

Answer 12

sorry that took so long, just wanted to double check the answer :)