Sets problems

[Zealot]

Please advise me solve this question

Q. In Net's hair salon 42% of the customers dye their hair blond, 36% dye their eyebrows blond and 35% straighten their hair. 20% of the customers have neither of these three procedures, while 12% take all of these three procedures. Which portion of the customers come for exactly 2 of these 3 procedures?

[EdYo]

IMO 21%

You can use 3 circles Venn diagram or the formula. But usually the diagram will make sure that you won't make any mistake in thinking process. Since I can't draw the diagram here i use the formula
Call A, B, C are 3 procedures

--> P(AB) + P(AC) + P(BC) = P(A) + P(B) + P(C) - [ 100 - P(neither ABC)] - P(ABC) = 42 +36 + 35 - (100 -20) - 12 = 21

[clock60]

21 for me too

[arindambanerji]

Yea 21 for me as well...

42 +36 + 35 - 80 - 12 = 21

[Zealot]

Well the formula says:

p(A U B U C)= p(A) + p(B) + p(C) - p(A and B) - p(B and C) - p(A and C) + p(A and B and C)

And as per this

p(A U B U C)= 100 - neither (ABC) = 100-20 = 80

80 = 42 + 36 + 35 - p(A and B) - p(B and C) - p(A and C) + 12

p(A and B) + p(B and C) + p(A and C) = 113-80+12 = 45

Is that right?

[sh_vivek]

yea, IMO 45 too. it should be +12 and not -12.

[Fiver]

I do not know the formulae, but here's my approach:
A+B+C is 42+36+35=113, while the total absolute value of A+B+C should have been 80. Hence the extra (113-80) constitutes of the ones that are counted more than once.
Counted 3 times are the ones present in all 3 sets; hence we need to reduce 2 times this value.
Counted 2 times are the ones present in A&B or A&C or B&C; hence we need to reduce 1 time this value.
Hence 33 = 2(values in all 3 sets) + 1(values in either of 2 sets)
33 = 2*12 + (value common to exactly 2 of the 3 sets)
9 = value common to exactly 2 of the 3 sets.
My answer happens to be different from all the rest.
I'm I missing something here?
Zealot what's the OA?

[Zealot]

Fiver, I am afraid I don't have the answer to this Q. as I picked this from a GMAT practice software. Yes, your appraoch makes more sense as (A' and B' and C') is 20% we all missed that part.

[mx+b]

Originally Posted by Zealot

Well the formula says:

p(A U B U C)= p(A) + p(B) + p(C) - p(A and B) - p(B and C) - p(A and C) + p(A and B and C)

And as per this

p(A U B U C)= 100 - neither (ABC) = 100-20 = 80

80 = 42 + 36 + 35 - p(A and B) - p(B and C) - p(A and C) + 12

p(A and B) + p(B and C) + p(A and C) = 113-80+12 = 45

Is that right?

Your answer is absolutely correct. Unfortunately, it's not what the question is asking.

The question is asking for the people who get exactly two, but not three, treatments. So you need to find p(A and B minus (A and B and C)) plus p(B and C minus (A and B and C)) plus p(A and C minus (A and B and C)). This works out to 9.

This is a tricky one. I got 21 on my first go-round, and it took me a while to figure out what I did wrong. The 21 answer is double-counting p(A and B and C).

[mx+b]

Fiver, that's a beautifully simple way of looking at it. I wish I'd thought about it that way the first time around!