Problem Solving

AuBuC = 100

A + B + C = 85 => (A + B + C)' = 100 - 85 = 15

AnBnC = 5

More than one => (AuBuC) - (A + B + C)' = 100
Two prducts => (AuBuC) - [(AnB) + (BnC) + (CnA) + (AnBnC)] = (A + B + C)'

More than one = 15%
Two products = 10%

I am bad in set, so can anyone explain how to approach such questions...thanks in advance...

The following equation can always be used for triple-overlapping set :

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups)

or

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

mathewmithun wrote:I am bad in set, so can anyone explain how to approach such questions...thanks in advance...

800guy wrote:In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products?

A) 5

B) 10

C) 15

D) 20

E) 25

When a question involves overlaps among groups -- some elements in one group, some in more than one group -- use a Venn diagram so that you can see the situation more clearly.

Please refer to the attached Venn diagram when reviewing the explanation below.

The total of everything contained in the Venn Diagram is 85.
But if we add the values we've been given for each circle, we get Circle X + Circle Y + Circle Z = 50 + 30 + 20 = 100.
Why? Because we've double-counted everything contained in 2 of the circles and triple-counted everything contained in all 3 circles:

When we count all of circle X and all of circle Y, everything in both X and Y (the overlap) gets counted twice.
When we count all of circle X and all of circle Z, everything in both X and Z (the overlap) gets counted twice.
When we count all of circle Y and all of circle Z, everything in both Y and Z (the overlap) gets counted twice.
When we count all of circle X, all of circle Y, and all of circle Z, everything contained in all 3 circles (the overlap) gets counted 3 times.

So the number contained in 2 of the circles has to be subtracted from the total once, the number contained in all 3 circles has to be subtracted from the total twice.

Let B = number in 2 of the circles.
85 = 50 + 30 + 20 - B - (2*5)
85 = 90 - B
B = 5
So 5 are in 2 circles, 5 are in all 3 circles.
5 + 5 = 10 who liked more than 1 product.
Since we have 100 people, 10/100 = 10%.

The correct answer is B.

When you have groups with a double and triple overlap, remember this rule:

The number contained in 2 out of the 3 groups has to be subtracted once from the total.
The number contained in all 3 groups has to be subtracted twice from the total.

Hope this helps!

Remember below rules for these kinds of questions,

True # of items = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in at least 1/2) - (# in at least 1/3) - (# in at least 2/3) + (# in 1/2/3)

True # of items = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in only 1/2) - (# in only 1/3) - (# in only 2/3) - 2(# in 1/2/3)

Examples:
At a certain school, each of the 150 students takes between 1 and 3 classes. The 3 classes available are Math, Chemistry and English. 53 students study math, 88 study chemistry and 58 study english. If 6 students take all 3 classes, how many take exactly 2 classes?


In this case, we'd use the first formula, since we want the number who take exactly 2 classes:

150 = 53 + 88 + 58 - (doubles) - 2(triples)
150 = 199 - (doubles) - 2(6)
150 = 187 - doubles
doubles = 37

Let's just change the question a tiny bit:

At a certain school, each of the 150 students takes between 1 and 3 classes. The 3 classes available are Math, Chemistry and English. 53 students study math, 88 study chemistry and 58 study english. If 6 students take all 3 classes, how many take at least 2 classes?

In this case, we'd use the second formula, since we want the number who take at least 2 classes:

150 = 53 + 88 + 58 - (at least 2 of the 3) + (all 3)
150 = 199 - (at least 2 of 3) + 6
150 = 193 - (at least 2 of 3)
At least 2 of 3 = 43

sumanr84 wrote:Remember below rules for these kinds of questions,

True # of items = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in at least 1/2) - (# in at least 1/3) - (# in at least 2/3) + (# in 1/2/3)

True # of items = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in only 1/2) - (# in only 1/3) - (# in only 2/3) - 2(# in 1/2/3)

@ Suman,

85 =50+30+20 -X +5

X= 20

we need atleast 2.
So X+5= 20 +5 = 25

But Mich says 10..how come??

gmatmachoman wrote:

sumanr84 wrote:Remember below rules for these kinds of questions,

True # of items = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in at least 1/2) - (# in at least 1/3) - (# in at least 2/3) + (# in 1/2/3)

True # of items = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in only 1/2) - (# in only 1/3) - (# in only 2/3) - 2(# in 1/2/3)

@ Suman,

85 =50+30+20 -X +5

X= 20

we need atleast 2.
So X+5= 20 +5 = 25

But Mich says 10..how come??

Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three)

85=50+30+20-x-10

x=5

People who like more than 1 product are in 2 groups and all 3=5+5=10%

gmatmachoman wrote: @ Suman,

85 =50+30+20 -X +5

X= 20

we need atleast 2.
So X+5= 20 +5 = 25

But Mich says 10..how come??

Govi - I am watching your Maths post.

This problem is again so easy that one can solve in a minute or less.

I will use my own technique:

I + II + III = 85
I + 2 II + 3 III = 100 (= 50 + 30 + 20)

Now we have
II + 2 III = 15
II + III = 15 - III = 15 - 5 = 10 Answer

See how much time it took.

Now I would like to modify above problem slightly, Can you please try it?

Q1
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. What is the minimum possible value in percentage of the survey participants liked all three of the products.

Q2: Another similar problem:
In a class where at least one specialization is mandatory, 87% has selected Literature, 69% had selected Hotel Management, and 78% had selected Packing. What is the maximum and minimum percentage of students that have taken all three specialization? Answer: 67%, 34%

Q3: Another similar problem:
In a survey conducted among 100 men in a company, 100 men use brand A, 75 use brand B, 80 use brand C, 90 use brand D & 60 use brand E of the same product. What is the minimum possible number of men using all the 5 brands, if all the 100 men use at least one of these brands? Answer: 5

If you were able to solve Q2 and Q3, then surely, you have answer for Q1.

good luck.

It would be good for the students to use tutorteddy.com for all sorts of difficult math homework help.

Hi All,

I just need some help on this.

I tried to use the A U B U C Formula,it doesnt work.

A U B U C = A + B + C - AnB-AnC-BnC +AnBnC

we need - AnB-AnC-BnC +AnBnC

Am i correct?

Hi akhp77
,

Can u post the explanation for the other Q pls?

Thanks
karthik

Q1
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. What is the minimum possible value in percentage of the survey participants liked all three of the products.

ends up with the equation: II+2III=15 for which minimum possible value for III is 1. Correct?


Q1
Another similar problem:
In a survey conducted among 100 men in a company, 100 men use brand A, 75 use brand B, 80 use brand C, 90 use brand D & 60 use brand E of the same product. What is the minimum possible number of men using all the 5 brands, if all the 100 men use at least one of these brands?

I would request someone to try this one...

outreach wrote:The following equation can always be used for triple-overlapping set :

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in exactly 2 groups) - 2(# in all 3 groups)

or

True # of objects = (total in exactly 1 group) + (total in exactly 2 groups) + (total in exactly 3 groups)

mathewmithun wrote:I am bad in set, so can anyone explain how to approach such questions...thanks in advance...

Thanks for the equations! I have come across questions in which some objects are not in any of the groups, even though they are part of the whole set. In these questions; in such questions, I have seen people using "None" in their equations. I have seen different versions of these equations and am a bit confused. Do you equations that works for GMAT questions involving "None"?

sumanr84 wrote: Remember below rules for these kinds of questions,

True # of items = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in at least 1/2) - (# in at least 1/3) - (# in at least 2/3) + (# in 1/2/3)

True # of items = (total # in group 1) + (total # in group 2) + (total # in group 3) - (# in only 1/2) - (# in only 1/3) - (# in only 2/3) - 2(# in 1/2/3)

This formula is for only understanding and not suitable for aptitude test. If you use this formula, it will take much time.

mathewmithun wrote:Q1
In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. What is the minimum possible value in percentage of the survey participants liked all three of the products.

ends up with the equation: II+2III=15 for which minimum possible value for III is 1. Correct?

Wrong.