18. All the attendees for a conference arrived by Monday morning. On Monday afternoon, 1/10 of the attendees went home. On Tuesday, 1/3 of the attendees who remained went home. On Wednesday, twice as many attendees as had gone home on Monday went home, leaving 96 attendees at the conference. If no other attendees joined of left during this time, how many attendees were at the conference on Monday morning?
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A = # of attendees on Monday morning.
On Monday afternoon there were 1/10 A went home so
On Tuesday morning there were A - 1/10A = 9/10 A attendees.
On Tuesday afternoon 1/3(9/10 A) went home so
On Wednesday morning there were 9/10 A - 1/3(9/10A) = 9/10A - (3/10A) = 6/10A
On Wednesday afternoon 2(1/10A) went home so there were 6/10A - 2/10A = 4/10A left.
We are told this left 96 attendees
So 4/10A = 96 Multiplying both sides by 10/4 we get
A = 10/4 * 96 = 10*24 = 240.
So our answer is there were 240 attendees on Monday
I agree with beckden solution. I used x instead of A for the number of attendes of the conference.
Number of attendes: x
The number of attendes went home on Monday can be symbolized: x/10
The number of attendes on Tuesday morning:
x - x/10 = 10x/10 - X/10 = 9x/10
The number of attendes went home on Tuesday afternoon:
1/3 (9x/10) = 9x/30
The number of attendes on Wednesday morning:
9x/10 - 9x/30 = 27x/30 - 9x/30 = 18x/30
The number of attendes went home on Wednesday afternoon:
2 (x/10) = 2x/10
We know that by the end left only 96 attendees. Thus:
18x/30 - 2x/10 = 96
18x/30 - 6x/30 = 96
12x/30 = 96
2x/5 = 96
x = 480/2
x = 240
I don't know if this is the best way to get to the result, but what I have done is pretty simple.
If you solve this equation you get that X = 160
Try it yourself and try to find the logic in it, sorry if I can't explain it any better.
I forgot the last day.
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