In a survey conducted to find out the readership of three Ne

Question

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

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

KarishmaB · Accepted Answer

This is the data given in the question:

Attachment:

Ques1.jpg [ 22.19 KiB | Viewed 13271 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.
Answer (C).

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 13271 times ]

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

Attachment:

Ques.jpg [ 24.44 KiB | Viewed 13125 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

KarishmaB · Answer

To avoid confusion, deleting this. Given a clearer explanation at the bottom.

cleetus · Answer

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.

Fijisurf · Answer

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.

KarishmaB · Answer

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...

KarishmaB · Answer

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)

cleetus · Answer

Got it.
x + x = 75. then C is having 75 + 8 = 83. Condition satisfied.
Then X = 85 - 75 - 8 = 2 rite.

Thanks Karishma

KarishmaB · Answer

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!

rockzom · Answer

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

144144 · Answer

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.

Werewolf · Answer

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)?

KarishmaB · Answer

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.

KarishmaB · Answer

The answer here is (C). Minimum is 2.

Werewolf · Answer

Thanks Karishma! That's a relief!!

jlgdr · Answer

Is the answer C or E, I thought it was C but the answer bar shows E. Would someone please confirm and/or change the answer bar? Thanks! J

KarishmaB · Answer

Answer is (C). I have updated the OA.

m3equals333 · Answer

thanks for the explanation/Venn diagram Karishma, very helpful.

Sarunyu · Answer

My method is quite simple.
Let's translate the question as follows:
A: 20-40
B: 50-70
C: 70-83
At least two: 85
All: 8

The number of people who read  only  two of the three newspapers: AB+AC+BC = 85-8 = 77 ---- (1)

The maximum number of people who read BC and AC: AC+BC = 83-8 = 75 ---- (2)
(Assuming the maximum value of C equally assumes the minimum value of AB.)

(1)-(2) AB = 2

The answer is "C".

MathRevolution · Answer

A: 20 ≤ A ≤ 40 
B: 50 ≤ B ≤ 70
C: 70 ≤ C ≤ 83

At least two: 85

For C, there should be 2 newspaper other than C that has been read

Answer C

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