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Foodmart and its products - Very tricky!!!

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Foodmart and its products - Very tricky!!!

by hk » Wed Jun 03, 2009 6:48 pm
Dear experts/testtakers/instructor,

I came across this problem. Can somebody give me a good explanation/method to solve this problem. Even after looking at the solution i had no idea how to solve this .. :oops: :x :evil:

Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above?

a. 5%
b. 10%
c. 15%
d. 25%
e. 30%
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by cramya » Wed Jun 03, 2009 7:27 pm
Responding to PM:

Hi HK,

There are formulas which I can find and send but sometimes I feel its too much to keep up with in addition to everything else u have to know escpecially the 3 sets one(comparitively 2 sets formula are easy to keep up with). I feel using Ven diagram and some logic almost all of these set problems can be solved within 2 -2.5 minutes or less depending on the problem given the variations in how the the questions can be worded with sets.

Hope my solution is correct and dint mess up anything silly in calculations along the way.


Draw three circles intersecting each other(don't have anything where I can draw a circle and upload so I am explaining it in detail so that u can try it yourself).

Areas unique to each circle:--

Call the parts unique to each circle a(Milk),b(Chicken),c(Eggs)

Areas common to just 2 circles:--

Milk and Chicken intersection call it x
Chicken and eggs intersection call it y
Milk and eggs intersection call it z

Area common to all three circles:--
Milk and Chicken and Eggs intersection call it k

We need to find x+y+z

a+b+c+x+y+z+k =100 (we are dealing with % and none is 0 since every one bought atleast 1 product)

i.e. a+b+c+x+y+z+10(k) =100
a+b+c+x+y+z = 90 (1)


a+x+y+10(k) = 60
a+x+y = 50 (2)

b+x+z+10(k) = 50
b+x+z= 40 (3)


c+y+10+z = 35
c+y+z = 25 (4)


Add (2),(3),(4)

a+b+c+x+y+z+x+y+z = 115

But from (1) we know a+b+c+x+y+z = 90

so 90+x+y+z= 115
x+y+z = 25


The calculation part takes less time(all we have done is addition) but with all my explanation above it may look like a long solution.

Hope this helps u to understand. Let me know if u stil have questions

D

Regards,
CR
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by hk » Wed Jun 03, 2009 7:53 pm
Dear CR,

Do you live anywhere in the US? Somewhere close to Jersey shore? If so please let me know. I would love to take you out for a Dinner!!!! You are absolutely great!!! Thanks a ton. It was actually such an easy question. What got me confused is the explanation given in the source of this question.

Check out what was the explanation given (attached)

Hail CR!!

:D
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by VP_Jim » Wed Jun 03, 2009 7:53 pm
I received a PM asking me to respond. I think that a Venn diagram approach is best, as well. If you want the formula (which is derived from Venn diagrams), it is:

Total = (Group 1) + (Group 2) + (Group 3) - 2(In all Groups) - (In two groups) + (In no groups)

So, in this question, we have:

100 = 60 + 50 + 35 - 2(10) - x + 0
x = 25

I don't recall ever seeing a problem like this on a prep test or an actual GMAT, so I'm not sure if I'd use the brain space to remember the formula.
Jim S. | GMAT Instructor | Veritas Prep
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