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

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

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

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.

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

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.

Prajakt - Do you mind sharing the TRICK ?

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.

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.

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