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DI - Venn Diagram-Maxima and Minima
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DI problem based on Maxima and Minima set Theory

Venkatesan S
IIM Bangalore Alumni | Cleared CAT 2015 | Cleared IIM B, C, L, K & I

U
Very good selection of sets and clarity in terms of explanation.Thank you
There is an ambiguity in the solution .if y=700,x<=3300 and z<=3000.now we sum up x,y and z .then it doesn't sum to 10700,as opposed to given in the question.please explain
Venkatesan S
a year ago
Hi Sai..! Could you pls tell me which question you are referring to?
Iam referring to the 1st question i.e..what is the minimum budget spent on at least two channels??
Venkatesan S
a year ago
the question is on minimum budget on 2 and 3 channel. to find the minimum of Y, we need to maximize X and Z.! that's why we have maximised X and Z to 5000 and the remaining 700 to Y.! could you pls tell me how X and Z becomes 3300 and 3000 respectively?
1. DATA INTERRETATION Video 5 Venkatesan S IIMB Alumni

2. An advertising agency wanted to give some ads. It has Rs. 15000 budget. The advertising channels are Newspaper, Websites and Mobile Apps. The breakup of the budget among the different channel is Newspaper and Website but not Mobile Apps Rs. 900 Mobile Apps and Website but not Newspaper RS. 1200 Newspaper or Mobile Apps but not Website Rs. 10700 The agency spent Rs. 100 for advertising in all 3 channels. The amount spent on each of the channel did not exceed one-third of the total budget. Entire budget was spent.

3. What is the minimum budget spent on at least two channels? A) 2100 B) 2200 C) 2800 D) 2900

4. If the budget spent on Mobile Apps is less than that of Newspaper, then at least how much budget was spent on Mobile Apps as well as Newspaper? A) 700 B) 900 C) 1000 D) 1100

5. If exactly half of the budget spent on Newspaper were also spent on Mobile Apps and exactly half of the budget spent on Mobile Apps were also spent on Newspaper, then how much budget spent only on Newspaper? A) 3100 B) 3300 C) 3400 D) 3800

6. If the budget spent on Newspaper is twice that of budget spent on at least two channels, then at least how much budget was spent on only Mobile Apps? A) 3700 B) 4100 C) 4300 D) 4700

7. An advertising agency wanted to give some ads. It has Rs. 15000 budget. The advertising channels are Newspaper, Websites and Mobile Apps. The breakup of the budget among the different channel is Newspaper and Website but not Mobile Apps Rs. 900 Mobile Apps and Website but not Newspaper RS. 1200 Newspaper or Mobile Apps but not Website Rs. 10700 The agency spent Rs. 100 for advertising in all 3 channels. The amount spent on each of the channel did not exceed one-third of the total budget. Entire budget was spent.

8. An advertising agency wanted to give some ads. It has Rs. 15000 budget. The advertising channels are Newspaper, Websites and Mobile Apps. The breakup of the budget among the different channel is, X 900210 100 0 Newspaper and Website but not Mobile Apps- Rs. 900 Mobile Apps and Website but not Newspaper- Rs. 1200 Newspaper or Mobile Apps but not Website -Rs. 10700 The agency spent Rs. 100 for advertising in all 3 channels. The amount spent on each of the 10700 budget. Entire budget was spent.

9. What is the minimum budget spent on at least two channels? A) 2100 B) 2200 C) 2800 D) 2900 X 900210 100 0 Max of X = 5000 Max of Z5000 X+Y+Z=10700 => Y = 700 10700

10. If the budget spent on Mobile Apps is less than that of Newspaper, then at least how much budget was spent on Mobile Apps as well as Newspaper? A) 700 B) 900 C) 1000 D) 1100 1000 X 900210 0 10 \// 120 Newspaper 900 + 100 X + Y = X + Y + Mobile 100 + 1200 + Y + Z = Y + Z + 1300 X Y 1000 YZ1300 X1000 Z1300 5000+1000 > 46001300 6000 > 5900 X + Y + Z = 5000 + Y 4600 Y + 9600 10700 10700 Y 1100

11. If the budget spent on Newspaper is twice that of budget spent on at least two channels, then at least how much budget was spent on only Mobile Apps? A) 3700 B) 4100 Thus, maximize X + Y C) 4300 D) 4700 spent on min 2 channel = 900 + 100 + Minimize Z. Max value of X = 5000. 5000 + Y + 900 + 100 = Y + 6000 X 900 2100 100 1200Y spent on min 2 channel = 2200 + Y Y + 6000 = 2 * (2200 + Y) Y + 6000 = 4400 + 2 Y = 1600 X + Y + Z = 10700 => 5000 + 1600 + Z 10700 Z 4100 10700