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# PS Question

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PS Question [#permalink]  12 Nov 2010, 16:04
00:00

Difficulty:

5% (low)

Question Stats:

66% (01:20) correct 33% (06:15) wrong based on 6 sessions
In a survey conducted to find out the readership of three Newspapers A,B and C, it was found that the number of people who read newspaper A is at least 20 and at most 40. The number of people who read newspaper B is at least 50 and at most 70. The number of people who read newspaper C is at least 70 and at most 83. It was also found that 8 people read all the three newspapers and 85 people read at least two of the three newspapers. Find the minimum number of people who read both A and B but not C.

A)4
B)3
C)2
D)1
E)0

Plz explain how. any short cut to get the answer
[Reveal] Spoiler: OA

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Re: PS Question [#permalink]  12 Nov 2010, 18:41
Expert's post
To avoid confusion, deleting this. Given a clearer explanation at the bottom.
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Last edited by VeritasPrepKarishma on 14 Nov 2010, 18:54, edited 5 times in total.
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Re: PS Question [#permalink]  12 Nov 2010, 19:15
VeritasPrepKarishma wrote:
cleetus wrote:
In a survey conducted to find out the readership of three Newspapers A,B and C, it was found that the number of people who read newspaper A is at least 20 and at most 40. The number of people who read newspaper B is at least 50 and at most 70. The number of people who read newspaper C is at least 70 and at most 83. It was also found that 8 people read all the three newspapers and 85 people read at least two of the three newspapers. Find the minimum number of people who read both A and B but not C.

A)4
B)3
C)2
D)1
E)0

Plz explain how. any short cut to get the answer

Make the diagram of intersection of 3 sets as given below:
Attachment:
Ques.jpg

8 people read all 3 newspapers.
85 people read at least 2 so 85 - 8 = 77 read exactly 2 newspapers (the area marked by the three x)

We have to minimize those who read A and B but not C. So we have to minimize the area with 'X' (capital x).
All three x's add up to 77. To minimize the one with 'X', try and maximize the other two regions which have 'x' (small x).
Is it possible to distribute 77 between these two regions with small x? Yes, because A is at most 40 and B is at most 70. So the two small x's can be at most 32 and 62 respectively. Therefore X can easily be 0.

Note: I haven't seen intersection of 3 sets on GMAT yet (just 2 sets). Though, it is a good idea to be comfortable with the basics of how to work with three sets too.

Thanks Karishma. Thanks for the technique.
Why do we have to take maximum value for A and B.

If we give maximum values of 32 and 62 to small x regions, the value of C will be 32+64+8 = 104. According to the question, C's maximum value is 83.
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Re: PS Question [#permalink]  12 Nov 2010, 19:29
Thanks Karishma. I could not solve it myself.

It is interesting that you said that you haven't seen intersection of 3 sets on GMAT yet. However, MGMAT, Knewton and other test have them.
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Re: PS Question [#permalink]  13 Nov 2010, 04:38
Expert's post
cleetus wrote:
Why do we have to take maximum value for A and B.

If we give maximum values of 32 and 62 to small x regions, the value of C will be 32+64+8 = 104. According to the question, C's maximum value is 83.

Hey cleetus,
We are not actually giving them the maximum value. The Value of X + x + x = 77.
I want to see "Can x + x be 77 so that X is minimum, that is 0?"

But, I did overlook the fact that they have given the maximum value for C as well. Let me update the solution in that light.. and thanks for pointing it out...
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Last edited by VeritasPrepKarishma on 14 Nov 2010, 18:54, edited 1 time in total.
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Re: PS Question [#permalink]  13 Nov 2010, 05:18
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Expert's post
In fact, knowing the maximum value of C gives us a much easier solution. Since C can be at most 83 and people who read 2 or more newspapers are 85, at least 2 people will need to read A and B but not C. Look at the diagram above to see why that is so...
(Make sure you check to see that all conditions are satisfied)
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Re: PS Question [#permalink]  13 Nov 2010, 10:38
Got it.
x + x = 75. then C is having 75 + 8 = 83. Condition satisfied.
Then X = 85 - 75 - 8 = 2 rite.

Thanks Karishma
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Re: PS Question [#permalink]  13 Nov 2010, 12:15
Expert's post
Fijisurf wrote:
Thanks Karishma. I could not solve it myself.

It is interesting that you said that you haven't seen intersection of 3 sets on GMAT yet. However, MGMAT, Knewton and other test have them.

Yes because it is not a bad idea to know the concept. How to draw the circles. What the intersection means etc. And anyway, if you are very comfortable with 3 sets, 2 should be cakewalk!
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Re: PS Question [#permalink]  13 Nov 2010, 12:21
I set it up as follows:

8 = ABC
20-40 = A
50-70 = B
70-83 = C
8 = ABC
85= AB, AC, or BC

"Find the minimum number of people who read both A and B but not C."

A+B at the least is 70. Since C at the least is also 70, the minimum difference is 0.

E. +1
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Re: PS Question [#permalink]  14 Nov 2010, 08:34
Hey, I would be happy to know if my way is just lucky or its another option to solve it:

i looked only on the minimum bc we have to find the min of group AB.

I took out 8 ppl that read ABC so the new min is :
a - 12
b - 42
c - 62

Than - we know we have 85 ppl that read at least 2:
i took the min of bc and placed there all the 85 ppl that read minimum of 2 newspapers.

means - the minimum of group ab can b = 0 because we placed all the 8 of abc and the 85 ppl that read the minimum of two newspapers.

thanks.
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Re: PS Question [#permalink]  14 Nov 2010, 18:47
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Expert's post
This is the data given in the question:
Attachment:

Ques1.jpg [ 22.19 KiB | Viewed 683 times ]

The entire shaded area is given to be 85.
So x + y + z + 8 = 85 (people who read at least 2 newspapers)
or x + y + z = 77 (people who read exactly two newspapers)

The question asks us to minimize the area x (common to A and B but not to C)
There are 77 people in x + y + z so to give as few people as possible to x, we should try and give as many as possible to y and z.

But y + z + 8 + r <= 83 (because C is at most 83. We have a minimum value for C as well but let's not worry about it right now.)
So y + z + r <= 75
Now, even if r = 0, y + z can be 75 at the most. Since x + y + z = 77,
x has to be at least 2.

Let's check if it is possible to satisfy all our conditions:
If y = 25, z = 50 (We took one of the many possible values for y)
Then A could lie between 20 - 40, B could lie between 50 - 70 and C is 83. All conditions satisfied and we got the minimum value of x.

@144144 - There is a maximum limit on C. I don't think you took that into account.

Second method: You may want to do it this way:
Attachment:

Ques1.jpg [ 22.19 KiB | Viewed 683 times ]

The entire shaded area is 85 (including the red one)
Now look at this:
Attachment:

Ques.jpg [ 24.44 KiB | Viewed 669 times ]

In the diagram above, the entire shaded area is at most 83.
So the red shaded area has to be at least 85 - 83 = 2
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Re: PS Question [#permalink]  21 Nov 2010, 04:24
Nice explanation! Thanks Karishma! I found this problem very difficult. I know that every problem provides opportunities to sharpen skills but are such type of problems common to GMAT (I mean, the ones having range)?
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Re: PS Question [#permalink]  21 Nov 2010, 07:49
Expert's post
Werewolf wrote:
Nice explanation! Thanks Karishma! I found this problem very difficult. I know that every problem provides opportunities to sharpen skills but are such type of problems common to GMAT (I mean, the ones having range)?

No. GMAT questions do not have ranges of values. If I am not mistaken, this question was not from GMAT specific material. Then again, GMAT is getting a little tougher by the day. Being exposed to such questions helps in understanding the concepts. But if you wish to ignore this problem, feel free to do it.
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Re: PS Question [#permalink]  21 Nov 2010, 19:04
Expert's post
Sarang wrote:
Is IMO wrong??

The answer here is (C). Minimum is 2.
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Re: PS Question [#permalink]  21 Nov 2010, 22:17
VeritasPrepKarishma wrote:
Werewolf wrote:
Nice explanation! Thanks Karishma! I found this problem very difficult. I know that every problem provides opportunities to sharpen skills but are such type of problems common to GMAT (I mean, the ones having range)?

No. GMAT questions do not have ranges of values. If I am not mistaken, this question was not from GMAT specific material. Then again, GMAT is getting a little tougher by the day. Being exposed to such questions helps in understanding the concepts. But if you wish to ignore this problem, feel free to do it.

Thanks Karishma! That's a relief!!
Re: PS Question   [#permalink] 21 Nov 2010, 22:17
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