Set Theory - Maxima/Minima concepts - Hemant Malhotra

[hemant_malhotra]

In this note on Set theory ,will cover Maxima minima without using Venn diagram

Concept 1:

It will cover 90% of your Problems

Total= I + II + III

A + B + C = I + 2II + 3III

Following questions will revolve around these two equations

A survey shows that 41%, 35% and 60% of the people watch “A”, “B” and “C” respectively. 27% people watch exactly two of the three movies and 3% watch none. What percentage of people watch all the three movies?
a) 40%
b) 6%
c) 9%
d) 12%

I+2II+3III=41+35+60=136
II=27 (given)
So I+3III=82
I+II+III=97 ((BECAUSE 3% did not watch any movie)) (total)
I+III=70
I+3III=82
So 2III=12
So III=6

I=exactly one
II=exactly two
III=exactly three
and so on
I+II+III=exactly one +exactly two+exactly three
I+2II+3III=total means this will cover exactly two (2 times) exactly 3 (3 times )as shown in the pic

In a test in which 120 students appeared , 90 passed in History , 65 passed in Sociology and 75 passed in Political science ' 30 students passed in only one subject and 55 students in only two .5 students passed no subjects .
a) how many passed in all three ?
b) how many passed in atleast two subjects

Approach with venn diagram= 120 students appeared and 5 passed in no subjects so total passed students=115
I+II+III=115
I+2II+3III=90+65+75=230
here passed in only one subject means I=30
and II=55
so III=115-(30+55)=30
and at least 2 means II+III=115-30=85
Easy one!

In a car agency one day 120 cars were decorated with three different accessories viz, power window, AC and music system. 80 cars were decorated with power windows, 84 cars were decorated with AC and 80 cars were decorated with music systems. What is the minimum and maximum no of cars which were decorated with all of three accessories?
a) 10, 61
b) 10, 45
c) 25, 35
d) None of these

Approach
for max value of III
I+II+III=120
I+2II+3III=80+80+84=244
so we have two equations
I+II+III=120 ______________ 1)
and I+2II+3III=244 ___________________ 2)
a) max value of III
2nd equation- 1st equation
II+2III=124 for max III we have to minimize II so min II=0
so max III=62
when III=62 then I=58 that will satisfy both equations so max III=62

min III
II+2III=124 .. for min Value of III we have to maximize II
now check when i will put max value of II=124 that will not give I+II+III=120 so we have to cover this equation also
so max II=116 and II+2III=124
so 2III=8
so 4
Or method 2 - To minimize III we have to maximize II for which we have to minimize I so put I=0
so II+III=120
2II+3III=244 solve these two equations
so III=4

A survey of 200 people in a community who watched at least one of three channels BBC,CNN and DD showed that 80% of the people watched DD ,22% watched BBC and 15% watched CNN , if 5% watched DD and CNN , 10% watched DD and BBC then what percentage of people watched BBC and CNN only (CAT Question)

approach= DD=200 * 80/100=160
BBC=44
CNN=30
I+II+III=200 (given)
I+2II+3III=160+44+30=234 now 2nd -1st equation
so II+2III=34
let III=x
DD and CNN=200*5/100=10
only DD and CNN=10-x ... here x is common in all three
Only DD and BBC=20-x
10-x+20-x+(only BBC and CNN)+2x=34
so 30+(only BBC+CNN)=34
so only BBC+CNN=4
which is 2% of 200
so 2%

A group of 100 people play carom , snooker & chess.90 play chess, 80 play carom and 80 play snooker. Find the max. no. of people who play all 3 games , if each person plays at least one game.
a) 80
b) 70
c) 75
d) 65

Approach- total people==100
so I+II+III=100
I+2II+3III=250
we want maximum III
so 2nd equation-1st equation
II+2III=150
we want max III so min II =0
so 2III=150 so III=75

Question- Half of a class of 200 students enrolled for exactly one of the three activities swimming, skating and dancing. Total enrolments 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.
Q1. Find the minimum possible number of students who enrolled for at least one of the three activities.
(a) 142 (b) 143 (c) 144 (d) 141
Q2. Find the maximum possible number of students who enrolled for exactly 2 activities?
(a) 57 (b) 56 (c) 55 (d) 54

Approach= I=100
I+2II+3III=215
so 2II+3III=115
I+II+III+X=200
II+III+X=100 ( x number of students who didn't take part in any activities))
now 2II+3III=115
here II=(dancing+scating)+(scating+swimming)+(swimming+dancing)
II=a+b+c
where a=10+b
so II=10+2b+c
min value of II=10
so 2 * 10 + 3 * III=115 but that will not give III integer
so put II=11
so 2 * 11+3III=115
so 22+3 * III=115
so 3*III=93
so III=21
II+III+X=100
so 11+21+X=100
so X=58
so I+II+III=200-X=142 minimum Value
Ans2- 2II+3III=115
now we want to maximize II then we have to minimize III ... put III=0 that will give II non integer
so put III=1
so 2II=112
so II=56