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by Prajakt » Wed Sep 05, 2012 7:13 pm
Q. 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?

A. 10
B. 15
C. 20
D. 5
E. 0

I'll post the answer tomorrow.
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by neelgandham » Wed Sep 05, 2012 10:38 pm
Before I answer the question, Can I ask you to cite the source of the question ? I am pretty sure that this isn't a GMAT question. But thanks a lot for the mental exercise :D

Solution :

Total number of men surveyed = 100
Let A, B, C, D and E be the number of men who use A, B, C, D and E respectively. I wil try and explain by introducing one brand at a time. Here it goes.

Step 1: Brand E
Number of men who use both A and E = A + E = 60
Number of men who use just A = 100 - 60 = 40.

Step 2: Brand B
To minimise the value of A + E + B, let us maximise the value of A+B and A+E
Number of men who use both A and B = A + B = 40.
Number of men who use both A and E = A + E = 25.
Number of men who use A, E, B = 100 - 40 - 25 = A + E + B = 100 - (Sum of all the above) = 35.

Step 3: Brand C
To minimise the value of A + E + B + C, let us maximise the value of E + B + C, A + E + C and A + E + B
Number of men who use both E,B and C = E + B + C = 40
Number of men who use both E,A and C = A + E + C = 25
Number of men who use both E,B and A = A + E + B = 20
Number of men who use both E,B,A and C = A + E + B + D = 100 - (Sum of all the above) = 15

Step 4: Brand D
To minimise the value of A + E + B + C + D, let us maximise the value of E + B + C + D, A + E + C + D, A + E + B + D and A + E + C + D
Number of men who use both E,B,D and C = E + B + C + D = 40
Number of men who use both E,D,A and C = A + E + C + D = 25
Number of men who use both E,B,A and D = A + E + B + D = 20
Number of men who use both E,D,A and C = A + E + C + D = 10
Number of men who use both E,B,A,D and C = A + E + B + C + D = 100 - (Sum of all the above) = 5

So, the minimum possible number of men using all the 5 brands is 5
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by das.ashmita » Thu Sep 06, 2012 12:30 am
Use Don't Use
A 100 0
B 75 25
C 80 20
D 90 10
E 60 40

total no. of men who do not use at least 1 brand = 0+25+20+10+40 = 95
Hence, minimum no. of men using all the 5 brands = 100 - 95 = 5

Ans : D

You can refer
https://www.beatthegmat.com/least-value- ... 64-45.html
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by GMATGuruNY » Thu Sep 06, 2012 5:40 am
Prajakt wrote:Q. 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?

A. 10
B. 15
C. 20
D. 5
E. 0

I'll post the answer tomorrow.
To MINIMIZE the number who use all 5 brands, we must MAXIMIZE the number who use 4 of the brands.

Since A=100, the maximum value of BCDE = 0.
Since B=75, the maximum value of ACDE = 100-75 = 25.
Since C=80, the maximum value of ABDE = 100-80 = 20.
Since D=90, the maximum value of ABCE = 100-90 = 10.
Since E=60, the maximum value of ABCD = 100-60 = 40.

Thus:
The MAXIMUM number who use 4 of the brands = 0+25+20+10+40 = 95.
Thus:
The MINIMUM number who use all 5 brands = 100-95 = 5.

The correct answer is D.
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by Prajakt » Thu Sep 06, 2012 8:36 pm
neelgandham wrote:Before I answer the question, Can I ask you to cite the source of the question ? I am pretty sure that this isn't a GMAT question. But thanks a lot for the mental exercise :D

Solution :

Total number of men surveyed = 100
Let A, B, C, D and E be the number of men who use A, B, C, D and E respectively. I wil try and explain by introducing one brand at a time. Here it goes.

Step 1: Brand E
Number of men who use both A and E = A + E = 60
Number of men who use just A = 100 - 60 = 40.

Step 2: Brand B
To minimise the value of A + E + B, let us maximise the value of A+B and A+E
Number of men who use both A and B = A + B = 40.
Number of men who use both A and E = A + E = 25.
Number of men who use A, E, B = 100 - 40 - 25 = A + E + B = 100 - (Sum of all the above) = 35.

Step 3: Brand C
To minimise the value of A + E + B + C, let us maximise the value of E + B + C, A + E + C and A + E + B
Number of men who use both E,B and C = E + B + C = 40
Number of men who use both E,A and C = A + E + C = 25
Number of men who use both E,B and A = A + E + B = 20
Number of men who use both E,B,A and C = A + E + B + D = 100 - (Sum of all the above) = 15

Step 4: Brand D
To minimise the value of A + E + B + C + D, let us maximise the value of E + B + C + D, A + E + C + D, A + E + B + D and A + E + C + D
Number of men who use both E,B,D and C = E + B + C + D = 40
Number of men who use both E,D,A and C = A + E + C + D = 25
Number of men who use both E,B,A and D = A + E + B + D = 20
Number of men who use both E,D,A and C = A + E + C + D = 10
Number of men who use both E,B,A,D and C = A + E + B + C + D = 100 - (Sum of all the above) = 5

So, the minimum possible number of men using all the 5 brands is 5

My friend gave me this one! I told him the same that it's not a GMAT problem. But after spending some time on similar problems, I realized that there is a trick which can allow you to answer in less than a minute. :)
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by neelgandham » Thu Sep 06, 2012 10:55 pm
Prajakt - Do you mind sharing the TRICK ?
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by Prajakt » Thu Sep 06, 2012 11:31 pm
neelgandham wrote:Prajakt - Do you mind sharing the TRICK ?
Sure! I'll be more than happy to do that! It is already shared by das.ashmita and GMATGuruNY.

Sum of the difference from 100 = (100-100)+(100-75)+(100-80)+ (100-90)+ (100-60)=0+25+20+10+40=95
Again, take the difference from 100 =>(100-95)= 5 ====> That's the ANSWER!
You can use it for all "At least" type Venn Diagram Questions.
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by manihar.sidharth » Fri Sep 07, 2012 9:44 am
Why not just take the highest Common factor which should imply whats asked.
So the HCF in this case is 5 and hence the answer.
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by KishoreReddy » Thu Dec 14, 2017 6:42 am
manihar.sidharth wrote:Why not just take the highest Common factor which should imply whats asked.
So the HCF in this case is 5 and hence the answer.
HCF method doesn't always get you the answer. If you want to try, just Try the following question with HCF method.

In a town, 90% of the people read The Hindu, 74% read Indian Express, 70% read Times of India. All the people read at least 1 newspaper. What is the minimum and maximum no of people who read all the newspapers.

A. 34%, 67%
B. 33%, 68%
C. 30%, 70%
D. None
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