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by yellowho » Mon Mar 14, 2011 9:35 pm
According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the max percentage of people who like two items?


80 people like cherries pair with 75% people who like bananas=> 75% people like bananas and cherries
5% of people left over who likes only cherries; Now pair them up with the people who likes apples. so 5%.


Total Double = 75%(bananas + cherries) + 5%(cherries+apples)= 80%;

Answer if 75% .


It seems that OA paired 70 apples with 75 to produce 70 with 5 left-over and then paired the 5 with the 80 to give 75%.
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by srcc25anu » Mon Mar 14, 2011 10:09 pm
A&B = 70 + 75 - 100 = 45
A&C = 70 + 80 - 100 = 50
B&C = 75 + 80 - 100 = 55

A&B + A&C + B&C = 150

total = A + B+c - (A&B + B&C + A&C) - ALL 3
100 = 70 + 75 + 80 - 150 - ALL 3
ALL 3 = 25

A&B ONLY = 45 - 25 = 20
A&C ONLY = 50-25 = 25
B&C ONLY = 55-25 = 30

THEREFORE {ALL 3 + A&B + A&C + B&C} = 100 and sum of 2 items only = 20+25+30 = 75
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by HSPA » Mon Mar 14, 2011 10:10 pm
I see word "atleast" and question is for MAX value

out of 100 atleast 70 like apple... so there can be 100 people who like apple
out of 100 atleast 75 like banana... so there can be 100 people who like banana

So MAX people who like two shall be 100

IMO : 100

Minimum number of people who like two items shall be 75....
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by yellowho » Mon Mar 14, 2011 11:01 pm
Can someone comment on how I did this problem? I know its wrong but it sounds so logical =). I really need to fix the way I'm thinking.
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by ankur.agrawal » Mon Mar 14, 2011 11:59 pm
yellowho wrote:Can someone comment on how I did this problem? I know its wrong but it sounds so logical =). I really need to fix the way I'm thinking.
I have sent a pm to GmatGuru ( Mitch) for giving a strategic solution for these kind of problems. I also get stuck with these.
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by GMATGuruNY » Tue Mar 15, 2011 3:08 am
yellowho wrote:According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the max percentage of people who like two items?


80 people like cherries pair with 75% people who like bananas=> 75% people like bananas and cherries
5% of people left over who likes only cherries; Now pair them up with the people who likes apples. so 5%.


Total Double = 75%(bananas + cherries) + 5%(cherries+apples)= 80%;

Answer if 75% .


It seems that OA paired 70 apples with 75 to produce 70 with 5 left-over and then paired the 5 with the 80 to give 75%.
100 people total.
Let A = apple lovers, B = banana lovers, C = cherry lovers.
To minimize the overlap among the 3 groups, we need to minimize the number within each group.

Let A = 70, B = 75, C = 80.
A + B = 70+75 = 145, so the overlap -- the number who like both A and B -- is 45.
Some of these 45 people also like C.
45 + C = 45 + 80 = 125, so the overlap -- the number who like all 3 fruits -- is 25.
Thus, at least 25 people like all 3 fruits.

To determine the maximum number who like exactly 2 of the fruits, we can use the overlapping groups formula:

Total = G1 + G2 + G3 - (those in exactly 2 groups) - 2*(those in exactly 3 groups)

100 = 70 + 75 + 80 - x - 2*25
100 = 175 - x
x = 75.
Thus, the maximum number who like exactly 2 of the fruits is 75.
Last edited by GMATGuruNY on Tue Mar 15, 2011 3:36 am, edited 2 times in total.
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by HSPA » Tue Mar 15, 2011 3:14 am
Hi Mitch,

How did you take A= apple lovers = 70... does the word atleast have no siginificane here... atleast 70 means >=70... or

I havent understood what I am missing here???
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by GMATGuruNY » Tue Mar 15, 2011 3:21 am
HSPA wrote:Hi Mitch,

How did you take A= apple lovers = 70... does the word atleast have no siginificane here... atleast 70 means >=70... or

I havent understood what I am missing here???
Since we're trying to minimize the number in all 3 groups, we need to minimize the number within each group.

Let's see what happens to the overlap if we increase the value of A:
Let A = 71.
A + B = 71 + 75 = 146, so the overlap -- the number who like both A and B -- is 46.
Some of these 46 people also like C.
46 + C = 46 + 80 = 126, so the overlap -- the number who like all 3 fruits -- is 26.

Notice that increasing the value of A increases the number in all 3 groups.
Thus, to minimize the overlap among the 3 groups, we need to minimize the number within each group.
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by HSPA » Tue Mar 15, 2011 3:54 am
What could be the minimum count of people who like two varities?
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by GMATGuruNY » Tue Mar 15, 2011 6:37 am
HSPA wrote:What could be the minimum count of people who like two varities?
0.

If 100% like all 3 fruits, then none like exactly 2 of the fruits.
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by yellowho » Tue Mar 15, 2011 7:56 pm
Mitch,

What's wrong with my logic? What logic mistake am I making?

Also, It seems that you calculated the doubles by calculating the triples first. Is there a way to do the other way?
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by nafiul9090 » Thu Jun 23, 2011 5:44 pm
GMATGuruNY wrote:
HSPA wrote:Hi Mitch,

How did you take A= apple lovers = 70... does the word atleast have no siginificane here... atleast 70 means >=70... or

I havent understood what I am missing here???
Since we're trying to minimize the number in all 3 groups, we need to minimize the number within each group.

Let's see what happens to the overlap if we increase the value of A:
Let A = 71.
A + B = 71 + 75 = 146, so the overlap -- the number who like both A and B -- is 46.
Some of these 46 people also like C.
46 + C = 46 + 80 = 126, so the overlap -- the number who like all 3 fruits -- is 26.

Notice that increasing the value of A increases the number in all 3 groups.
Thus, to minimize the overlap among the 3 groups, we need to minimize the number within each group.
hello mitch

could you please shed some light on the following same type of problem...i cant get this

problem.
In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football.7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football.If 18 students does not play any of these given sports, ho many students play exactly two of these sports?

regards nafi
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by GMATGuruNY » Thu Jun 23, 2011 7:58 pm
nafiul9090 wrote: hello mitch

could you please shed some light on the following same type of problem...i cant get this

problem.
In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football.7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football.If 18 students does not play any of these given sports, ho many students play exactly two of these sports?

regards nafi
1. Draw a Venn Diagram
2. Plug in for the number of students who play all 3 sports.
3. Determine the other values in the Venn Diagram, working from the center out.
3. Check whether the values in the Venn Diagram satisfy the conditions given.

Since only 4 people play Cricket and Football, the number who play all 3 sports is likely to be 3 or less.
Start with the middle possible value: let the number of people who play all 3 sports = 2.
Here's the Venn Diagram:

Image

Number who play exactly 2 sports:
Since 7 in total play Hockey and Cricket, 7-2 = 5 play only Hockey and Cricket but not Football.
Since 4 in total play Cricket and Football, 4-2 = 2 play only Cricket and Football but not Hockey.
Since 5 in total play Hockey and Football, 5-2 = 3 play only Hockey And Football but not Cricket.

Working our way from the center out, we can determine the number who play exactly 1 sport:
Number who play only Hockey = 20-5-2-3 = 10.
Number who play only Cricket = 15-5-2-2 = 6.
Number who play only Football = 11-3-2-2 = 4.

Adding all the values in the circles to the 18 who do not play any sports, we get:
(10+3+2+5+6+2+4) + 18 = 50.
Success!

The number who play exactly 2 sports = 5+2+3 = 10.
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by nafiul9090 » Thu Jun 23, 2011 8:40 pm
Since 4 people play cricket, the number who play all 3 sports is likely to be 3 or less.
Start with the middle value: let the number of people who play all 3 sports = 2.
hello mitch

do we always take middle number as the number of people who play all 3 groups???

regards nafi
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by GMATGuruNY » Fri Jun 24, 2011 7:11 am
nafiul9090 wrote:
Since only 4 people play Cricket and Football, the number who play all 3 sports is likely to be 3 or less.
Start with the middle value: let the number of people who play all 3 sports = 2.
hello mitch

do we always take middle number as the number of people who play all 3 groups???

regards nafi
I imagined that the question asked the following:

How many students play all three sports?

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

Given this situation, I would start by plugging in the middle answer choice, which says that 2 students play all 3 sports.
If answer choice C didn't work, I could determine whether I needed a bigger or smaller answer choice.
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