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Intern  Joined: 31 Aug 2015
Posts: 28
Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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18 00:00

Difficulty:   35% (medium)

Question Stats: 74% (02:36) correct 26% (02:40) wrong based on 155 sessions

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Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

(A) 185
(B) 180
(C) 175
(D) 190
(E) 195

source - pearson

Originally posted by excelingmat on 11 Oct 2015, 22:31.
Last edited by Bunuel on 30 Mar 2016, 11:54, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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excelingmat wrote:
Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

a) 185
b) 180
c) 175
d)190
e) 195

source - pearson

250 = n(Exactly 1 channel) + n(Exactly 2 channels) + n(Exactly 3 channels)
250 = n(Exactly 1 channel) + 50 + n(Exactly 3 channels)

Let's find the value of n(Exactly 3 channels) = x

250 = n(A) + n(B) + n(C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)
Note that each of n(A and B) is the sum of 'number of people watching exactly two channels A and B' and 'number of people watching all three channels'.
250 = 116 + 127 + 107 - n(Exactly 2 channels) - 3x + x
250 = 116 + 127 + 107 - 50 - 2x
x = 25

250 = n(Exactly 1 channel) + 50 + 25
n(Exactly 1 channel) = 175

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Intern  Joined: 19 Mar 2015
Posts: 10
Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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VeritasPrepKarishma wrote:
excelingmat wrote:
Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

a) 185
b) 180
c) 175
d)190
e) 195

source - pearson

250 = n(Exactly 1 channel) + n(Exactly 2 channels) + n(Exactly 3 channels)
250 = n(Exactly 1 channel) + 50 + n(Exactly 3 channels)

Let's find the value of n(Exactly 3 channels) = x

250 = n(A) + n(B) + n(C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)
Note that each of n(A and B) is the sum of 'number of people watching exactly two channels A and B' and 'number of people watching all three channels'.
250 = 116 + 127 + 107 - n(Exactly 2 channels) - 3x + x
250 = 116 + 127 + 107 - 50 - 2x
x = 25

250 = n(Exactly 1 channel) + 50 + 25
n(Exactly 1 channel) = 175

Hi Veritasprep,
Thanks for this great explanation. But I m little confused about this statement: 250 = n(A) + n(B) + n(C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)
I drew the venn diagram and I wasnt able to come up to this formulae. Please help!
Thanks again!
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Joined: 05 Jul 2006
Posts: 1380
Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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excelingmat wrote:
Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

(A) 185
(B) 180
(C) 175
(D) 190
(E) 195

source - pearson

Since none = 0 , therefore at least 1 = total = 250 .

and since A+B+C = 350 thus (A and B)+ (B and C)+(C and A) - (A and B and C) = 100

but (A and B)+ (B and C)+(C and A) = Exactly 2 sets + 3 (A and B and C)

Thus(A and B)+ (B and C)+(C and A) - (A and B and C) = 100 =Exactly 2 sets + 2(A and B and C) and since Exactly 2 sets = 50 thus all three (A and B and C) = 25

Exactly 1 = Total - Exactly 2 - All 3 = 250-50-25 = 175
Intern  S
Joined: 30 Jan 2016
Posts: 6
Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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VeritasPrepKarishma wrote:
excelingmat wrote:
Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

a) 185
b) 180
c) 175
d)190
e) 195

source - pearson

250 = n(Exactly 1 channel) + n(Exactly 2 channels) + n(Exactly 3 channels)
250 = n(Exactly 1 channel) + 50 + n(Exactly 3 channels)

Let's find the value of n(Exactly 3 channels) = x

250 = n(A) + n(B) + n(C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)
Note that each of n(A and B) is the sum of 'number of people watching exactly two channels A and B' and 'number of people watching all three channels'.
250 = 116 + 127 + 107 - n(Exactly 2 channels) - 3x + x
250 = 116 + 127 + 107 - 50 - 2x
x = 25

250 = n(Exactly 1 channel) + 50 + 25
n(Exactly 1 channel) = 175

Could someone please clarify the 3x+x part at the end of the equation as a substitute for what should be "neither" - n(A and B and C) in the equation ? Or am I looking at this from the wrong perspective ?
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Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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stanoevskas wrote:
VeritasPrepKarishma wrote:
excelingmat wrote:
Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

a) 185
b) 180
c) 175
d)190
e) 195

source - pearson

250 = n(Exactly 1 channel) + n(Exactly 2 channels) + n(Exactly 3 channels)
250 = n(Exactly 1 channel) + 50 + n(Exactly 3 channels)

Let's find the value of n(Exactly 3 channels) = x

250 = n(A) + n(B) + n(C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)
Note that each of n(A and B) is the sum of 'number of people watching exactly two channels A and B' and 'number of people watching all three channels'.
250 = 116 + 127 + 107 - n(Exactly 2 channels) - 3x + x
250 = 116 + 127 + 107 - 50 - 2x
x = 25

250 = n(Exactly 1 channel) + 50 + 25
n(Exactly 1 channel) = 175

Could someone please clarify the 3x+x part at the end of the equation as a substitute for what should be "neither" - n(A and B and C) in the equation ? Or am I looking at this from the wrong perspective ?

For a 3 set overlapping venn diagram,
n(A and B) = n (people who exactly watch A and B but not C) + n(A and B and C)
Similarly
n(B and C) = n (people who exactly watch B and C but not A) + n(A and B and C)
n(C and A) = n (people who exactly watch C and A but not B) + n(A and B and C)

n(A and B)+n(B and C)+n(C and A) = n(people who watch exactly 2 channels) + 3x .. equation 1

250 = n(A) + n(B) + n(C) - n(A and B) - n(B and C) - n(C and A) + n(A and B and C)

Substituting equation 1 in the above equation we get

250 = 116 + 127 + 107 - n(people who watch Exactly 2 channels) - 3x + x

Hope it is clear
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Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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Tough most GMAT takers must be familiar to the concept of 3 overlapping sets , I think the difficulty of this question's diffculty is average.
I would say Level 600-650 question.
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Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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1
My 2 cents:
I solved this question as follows:

Personally, i would to say that the only formulas i remember are
T = A + B + C - [exactly 2 sets] - 2*[all 3 sets] + None &
T = A + B + C - [among 2 sets] + [all 3 sets] + None.
If you can't solve with these 2 formulas than the best option will be dividing 3 overlapping sets into various sections as a, b, c, d, e, f and g as you might already have seen in various examples solved here.
(If not refer - https://gmatclub.com/forum/three-table- ... 05637.html)
Honestly, after 2 tries i have understood one thing that GMAT does NOT give time to anaylse question and see which formula to use. So remembering lot of formulas isn't a way to prepare for GMAT. Also, question where you can apply it directly will hardly be ever asked.
Attachments a+b+c.jpg [ 600.44 KiB | Viewed 2457 times ]

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Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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excelingmat wrote:
Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

(A) 185
(B) 180
(C) 175
(D) 190
(E) 195

We can use the equation:

Total = watched A + watched B + watched C - watched exactly two - 2(watched all 3) + watched none

250 = 116 + 107 + 127 - 50 - 2T + 0

250 = 300 - 2T

2T = 50

T = 25

So we have:

250 = watched exactly 1 + watched exactly 2 + watched all 3 + watched none

250 = n + 50 + 25 + 0

175 = n

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Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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ScottTargetTestPrep wrote:
excelingmat wrote:
Among 250 viewers interviewed who watch at least one of the three TV channels namely A, B &C. 116 watch A, 127 watch C, while 107 watch B. If 50 watch exactly two channels. How many watch exactly one channel?

(A) 185
(B) 180
(C) 175
(D) 190
(E) 195

We can use the equation:

Total = watched A + watched B + watched C - watched exactly two - 2(watched all 3) + watched none

250 = 116 + 107 + 127 - 50 - 2T + 0

250 = 300 - 2T

2T = 50

T = 25

So we have:

250 = watched exactly 1 + watched exactly 2 + watched all 3 + watched none

250 = n + 50 + 25 + 0

175 = n

Easiest and the best way from all above. Kudos as I was about to giving up on this question.

Thanks

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Re: Among 250 viewers interviewed who watch at least one of the three TV  [#permalink]

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_________________ Re: Among 250 viewers interviewed who watch at least one of the three TV   [#permalink] 30 Oct 2019, 00:31
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