Please explain how did u get 4-3 =1 for the following.and 15 had air conditioning specify all the elements in B or Only B
survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of the three popular options — air conditioning, radio and power windows — were already installed. The survey found:
15 had air conditioning
2 had air conditioning and power windows but no radios
12 had radio
6 had air conditioning and radio but no power windows
11 had power windows
4 had radio and power windows
3 had all three options.
2 had air conditioning and power windows but no radios
12 had radio
6 had air conditioning and radio but no power windows
11 had power windows
4 had radio and power windows
3 had all three options.
What is the number of cars that had none of the options? (CAT 2003)
1. 4 2. 3 3. 1 4. 2
1. 4 2. 3 3. 1 4. 2
Answer: We make the Venn diagram and start filling the areas as shown:
Total Number of cars according to the diagram = 2 + 6 + 3 + 1 + 5 + 2 + 4 = 23.
Therefore, number of cars having none of the given options = 25 - 23 = 2.
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Therefore, number of cars having none of the given options = 25 - 23 = 2.
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A survey about Tv viewership was conducted on 100 respondents. The results are,
93 liked sony,89 liked Zee, 81 liked Stratv, 75 liked zee cinema, 78 like Mtv.
1 did not like any of the above.
Find the minimum number of people liking all the the 5 channels.
93 liked sony,89 liked Zee, 81 liked Stratv, 75 liked zee cinema, 78 like Mtv.
1 did not like any of the above.
Find the minimum number of people liking all the the 5 channels.
The total number of people watching any of these channels = (100-1)=99(I don't watch any).
Now the surplus = [(93+89+81+75+78) - 99] = 317. Now for every increase in the number of viewers watching two channels, there is actually a 20% increase from the initial (I am not sure whether we have to consider all the 5 combinations or not). Then for every increase in the number of people watching 3 channels, there is a resultant increase of [8-5]/5 = 60% (Here also are we to consider the total number of combinations possible). Again for every increase in the number of people watching 4 channels, the resultant increase = [9-5]/5 = 80%.
Now here I consider the combinations of groups possible for each of these groups:
2 channel groups = 5.
3 channel groups = 3.
4 channel groups = 2.
So the surplus we have been able to cover till now = (100+180+160) = 340. So there need not be any people watching all the five channels.
Now the surplus = [(93+89+81+75+78) - 99] = 317. Now for every increase in the number of viewers watching two channels, there is actually a 20% increase from the initial (I am not sure whether we have to consider all the 5 combinations or not). Then for every increase in the number of people watching 3 channels, there is a resultant increase of [8-5]/5 = 60% (Here also are we to consider the total number of combinations possible). Again for every increase in the number of people watching 4 channels, the resultant increase = [9-5]/5 = 80%.
Now here I consider the combinations of groups possible for each of these groups:
2 channel groups = 5.
3 channel groups = 3.
4 channel groups = 2.
So the surplus we have been able to cover till now = (100+180+160) = 340. So there need not be any people watching all the five channels.
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A:0 B:13 c:27 D:55
ppl having south indian food = 920 ppl having north indian food = 910 ppl having American food = 820 ppl having Chinese food = 780 ppl having Italian food = 790 ppl having Continental food = 800 hence, total= 920 + 910 + 820 + 780 + 790 + 800 = 5020 Of the 1000 ppl, 7 dont like any of the 6 types. Remaining = 1000 -7 = 993 Surplus = 5020 - 993 (Becoz 7 ppl dont like any of the 6 types of food) = 4027 Accomodating the surplus in the portions of 5 intersecting groups, we can have at most 993 * 4 = 3972 ppl surplus. Remaining surplus to be accomodated in the portion where all 6 groups intersect = 4027 -3972 = 55 Hence, answer (D)55.......... ---------------------------------------------------------------------------------
"In a college, where every student follows at least one of the three activities- drama, sports, or arts- 65% follow drama, 86% follow sports, and 57% follow arts. What can be the maximum and minimum percentage of students who follow
· all three activities
· exactly two activities"
· all three activities
· exactly two activities"
(X + 2Y + 3Z) = 208%
(X + Y + Z) = 100%
(X + Y + Z) = 100%
[If Z has to be fixed (be min or max) either X or Y has to be zero (they can not be negative, remember). Now who will correspond to max and who to min - only question. If X = 0, Y + Z will 'eat away' maximum Z, leaving small Z. So it will be used to calculate min Z. Vice-versa.]
In this case,
for min Z, X = 0%
=> Z = 8%, X = 0%, Y = 92%
for max Z, Y = 0%
=> Z = 54%, X = 46% and Y = 0%
for min Z, X = 0%
=> Z = 8%, X = 0%, Y = 92%
for max Z, Y = 0%
=> Z = 54%, X = 46% and Y = 0%
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1.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 minimum percentage of people who like all three?
Ans: for min of all three= 100-((100-70)+(100-75)+(100-80))
=>100-(30+25+20)=100-75=25
2.In a college, where every student follows at least one of the three activities- drama, sports, or arts-
65% follow drama, 86% follow sports, and 57% follow arts. What can be the maximum and minimum percentage of students who follow
· all three activities
· exactly two activities
Ans:min For all three activities= 100-((100-65)+(100-86)+(100-57))
=>100-(35+14+43)=100-92=8%
max For exactly two=100-8=92%(since x+2y=100 and min of all three can be used)
What is the minimum percentage of people who like all three?
Ans: for min of all three= 100-((100-70)+(100-75)+(100-80))
=>100-(30+25+20)=100-75=25
2.In a college, where every student follows at least one of the three activities- drama, sports, or arts-
65% follow drama, 86% follow sports, and 57% follow arts. What can be the maximum and minimum percentage of students who follow
· all three activities
· exactly two activities
Ans:min For all three activities= 100-((100-65)+(100-86)+(100-57))
=>100-(35+14+43)=100-92=8%
max For exactly two=100-8=92%(since x+2y=100 and min of all three can be used)
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If I solve the question by drawing the venn diagram the max value I am getting as 26.
While if I solve this question in this way then I m getting rt ans.
total students - 120
surplus = 59+67+73-120=79
Total number of students study more than one sub = 34+33+26=93
b'coz surplus is 79 but being accomodated is 93, that's why max no. of students study more than 3 subjects is 93-79=14
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Half of a class of 200 students enrolled for exactly one of the three activities swimming, skating and dancing. Total enrollments were 80 in swimming, 75 in skating and 60 in dancing from the class. Number of students who enrolled for skating and swimming only was 10 more than the number of students who enrolled for skating and dancing only.
Q 1 Find the maximum possible number of students who enrolled for exactly 2 activities?
Q 2 Find the minimum possible number of students who enrolled for at least one of the three activities.
Soln :
I think you have misunderstood the problem. The total number of people attending the classes is NOT 100. The total number of people attending exactly one of the classes is 100. Hence what you are doing is wrong -- since you are assuming a surplus over the value of 100 but the total value is NOT 100 in the first place.
This is how I think u do this -
Assume a, b, c to be the three sets. Assume m, n, p to be the number of people attending exactly one of the 3 classes. Assume x, y, z to be the intersection of a and b, b and c, c and a. Let l be the intersection of all three.
then we have 2 equations -
m + n + p + 2(x + y + z) + 3l = 80 + 75 + 60 = 215
but m + n + p = 100
so, 2(x + y + z) + 3l = 115 -----eqn 1.
also
m + n + p + x + y + z + l = total number of students.
100 + x + y + z + l cannot be more that 157. ( from eqn 1)
so x + y + z + l <= 57 ----- eqn 2. (and required value of x + y + z is max when value of eqn 57.)
solving 1 & 2 we get l = 1; x + y + z = 56
hence the maximum value of students attending exactly two classes is 56.
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In this case we assign the variables to every area of the Venn diagram and form the conditions keeping two things in mind:
· try to express the areas in the Venn diagram through least number of variables.
· all the numbers will be zero or positive. No number can be negative.
Out of 210 interviews of IIM- Ahmedabad, 105 CAT crackers were offered tea by the interview panel, 50 were offered biscuits, and 56 were offered toffees. 32 CAT crackers were offered tea and biscuits, 30 were offered biscuits and toffees, and 45 were offered toffees and tea. What is the
· maximum and minimum number of CAT crackers who were offered all three snacks?
· maximum and minimum number of CAT crackers who were offered at least one snack?
Answer: Let’s make the Venn diagram for this question. Since we want to assume least number of variables, we can see that assuming a variable for the number of students who were offered all three snacks will help us express all the other areas. Let the number of students who were offered all three snacks = x.
In the above diagram, we have expressed all the areas in terms of x. To decide maximum value of x, we note that 32 - x, 45 - x and 30 - x will be zero or positive. Therefore, the maximum value of x will be 30. (30 is the lowest among 30, 32 and 45). To decide minimum value of x, we note than x - 19 and x – 12 will be zero or positive. Therefore, x cannot be less than 19 (19 is the higher number between 19 and 12).
Therefore, maximum and minimum number of CAT crackers who were offered all three snacks = 30 and 19.
The number of CAT crackers who were offered at least one snack = Total number of CAT crackers in the Venn diagram = x + 28 + 32 - x + x + 45 - x + x - 19 + 30 - x + x - 12 = 104 + x.
As the maximum and minimum values of x are 30 and 19, respectively, the maximum and minimum value of 104 + x will be 134 and 123, respectively.
Maximum and minimum number of CAT crackers who were offered at least one snack = 134 and 123.
ANY WAY TO SOLVE FOR THE VALUE OF X?
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Q.A survey was conducted among 500 ppl each of whom likes at least one of apple, orange , banana.The number of ppl who like apple is 240, those who like orange are 250 and those who like banana are 290.
Q1. If 60 ppl like only apple and banana, then what is the maximum possible number of ppl who like only orange?
1)120 2) 130 3)140(ans)
Q2. If 120 ppl like only apple, then what is the maximum possible number of ppl who like only orange and banana?
1)170 2)160(ans) 3)180
Q3. What is the maximum possible number of ppl who like all the 3 fruits?
1)110 2)120 3)150 4)130 5)140(ans)
A+B+C+D+E+F+G = 500 A+B+C+2(D+E+F)+3G=780 D+E+F+2G=280 Apples =A+D+G+E=240 Banana=C+E+F+G=250 Orange=B+D+F+G=290 1) Given e =60, From equation D+E+F+2G=280, D+F+2G=220 In order to maximize B in equation B+D+F+G=290 put D and F = 60 and hence B will be 140 3) In equation D+E+F+2G=280 put D, E, F = 0 to maximize G which will be 140   |
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