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# At a dinner party, 40 percent of the guests wore both jackets and ties

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Math Expert
Joined: 02 Sep 2009
Posts: 49915
At a dinner party, 40 percent of the guests wore both jackets and ties  [#permalink]

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06 Apr 2016, 01:15
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Difficulty:

35% (medium)

Question Stats:

69% (01:24) correct 31% (01:44) wrong based on 227 sessions

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At a dinner party, 40 percent of the guests wore both jackets and ties. If 50 percent of the guests who wore jackets did not wear ties, what percent of the guests wore jackets?

A. 20 percent
B. 40 percent
C. 60 percent
D. 70 percent
E. 80 percent

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Joined: 12 Aug 2013
Posts: 45
Re: At a dinner party, 40 percent of the guests wore both jackets and ties  [#permalink]

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06 Apr 2016, 02:47
lets assume there are 100 people in the party
40 have worn ties

Now out of remaining 60 , 50% have worn jackets but not ties ,which mean 30 have worn jackets but not ties .

adding all those who have worn jackets and ties + jackets only = 40 + 30 = 70 percent
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Re: At a dinner party, 40 percent of the guests wore both jackets and ties  [#permalink]

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06 Apr 2016, 03:15
2
Assume x people attended the party.

40 percent of the guests wore both jackets and ties --> 0.4x

50 percent of the guests who wore jackets did not wear ties --> It means out of the guests who wore jackets, 50% wore Jackets only and 50% wore both jackets and ties. --> Both = Jackets(only) = 0.4x

Guests who wore jackets = 0.4x + 0.4x = 0.8x --> 80%

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Joined: 22 Sep 2013
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Re: At a dinner party, 40 percent of the guests wore both jackets and ties  [#permalink]

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12 Apr 2016, 11:54
2
Assuming total as 100
Jackets + ties = 40 % => 40
50 % of who wore Jackets didnt wear ties. Let total people who wore Jackets = x
50 % of is x/2

Total who wore Jackets = Just Jackets + Jackets and ties
x = 40 + x/2
2x = 80 + x
x = 80

E
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Re: At a dinner party, 40 percent of the guests wore both jackets and ties  [#permalink]

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12 Apr 2016, 15:20
let g=total guests
let j=guests who wore jackets
.4g+.5j=j
.4g=.5j
j/g=.4/.5=80%
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Re: At a dinner party, 40 percent of the guests wore both jackets and ties  [#permalink]

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04 May 2017, 10:02
Bunuel wrote:
At a dinner party, 40 percent of the guests wore both jackets and ties. If 50 percent of the guests who wore jackets did not wear ties, what percent of the guests wore jackets?

A. 20 percent
B. 40 percent
C. 60 percent
D. 70 percent
E. 80 percent

Let total number of guests = 100
Jacket and Tie both = 40
50 percent of Guest who wore the jacket, did not wore a tie.
So the remaining 50 percent of Guest who wore jacket, wore a tie. i.e., 40

So total number of guests wearing jacket will be 40 +40 = 80
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At a dinner party, 40% of the guests wore both jackets and ties. If  [#permalink]

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17 Apr 2018, 18:01
At a dinner party, 40% of the guests wore both jackets and ties. If 50% of the guests who wore jackets did not wear ties, what percent of the guests wore jackets?

A. 20%
B. 40%
C. 60%
D. 70%
E. 80%
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Joined: 22 May 2016
Posts: 2033
At a dinner party, 40% of the guests wore both jackets and ties. If  [#permalink]

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17 Apr 2018, 20:15
Buttercup3 wrote:
At a dinner party, 40% of the guests wore both jackets and ties. If 50% of the guests who wore jackets did not wear ties, what percent of the guests wore jackets?

A. 20%
B. 40%
C. 60%
D. 70%
E. 80%

Assign a number for total guests
Let there be 20 people at this dinner party.

40%, or 8 people,
wore BOTH jacket and tie

50% of the guests who wore jackets
did not wear ties

Put those two statements together. The logic is binary:
A person wore a jacket WITH a tie or
A person wore a jacket WITHOUT a tie

40% of ALL the partygoers wore a jacket WITH a tie.
That number = 8

If 50% wore a jacket WITHOUT a tie, then
the 8 who wore a jacket WITH a tie

were the other 50% of all jacket-wearers.

8 = 50% of all jacket-wearers
8 = $$\frac{1}{2}$$ of all jacket-wearers
16 = the number of all jacket-wearers

16 people wore jackets (8 with and 8 without a tie)

What percent of the guests
wore jackets?

Total jacket wearers: 16
Total guests: 20

$$(\frac{16}{20} * 100) = (.80 * 100)$$ = 80 %

Percents

Alternatively, 40% of ALL guests wore a jacket WITH a tie,
whereas

50% of ALL WHO WORE JACKETS
wore a jacket WITHOUT a tie

The 40% category (jacket-wearers WITH ties)
constitutes the other 50% of all jacket-wearers

40% of ALL guests wore jackets WITH ties and
40% of ALL guests wore jackets WITHOUT ties

(40% + 40%) = 80% of all guests wore jackets

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Re: At a dinner party, 40% of the guests wore both jackets and ties. If  [#permalink]

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17 Apr 2018, 20:25
See the solution attached.
Attachments

IMG_20180418_085226.jpg [ 871.72 KiB | Viewed 854 times ]

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Math Expert
Joined: 02 Sep 2009
Posts: 49915
Re: At a dinner party, 40 percent of the guests wore both jackets and ties  [#permalink]

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17 Apr 2018, 22:33
Buttercup3 wrote:
At a dinner party, 40% of the guests wore both jackets and ties. If 50% of the guests who wore jackets did not wear ties, what percent of the guests wore jackets?

A. 20%
B. 40%
C. 60%
D. 70%
E. 80%

Merging topics. Please refer to the discussion above.
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Posts: 488
Re: At a dinner party, 40 percent of the guests wore both jackets and ties  [#permalink]

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30 Sep 2018, 09:29
Attachment:

VD1.jpg [ 25.02 KiB | Viewed 160 times ]

Assume 100 people attend the party. Since 40 percent of the guests wore both jackets and ties; therefore 40 people wore both of them.

It is given that 50 percent of the guests who wore jackets did not wear ties therefore the number of people who wore Jackets (only) = 40 people

Guests who wore jackets = 40 + 40 = 80 People; therefore the percent of the guests wore jackets = 80/100 = 80% or Ans is option E.
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Re: At a dinner party, 40 percent of the guests wore both jackets and ties &nbs [#permalink] 30 Sep 2018, 09:29
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