Set theory: CAT Previous Year Questions
Q. 1 In a class of 100 students, 73 like coffee, 80 like tea and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is
Check Solution
Ans: A
Let the count of individuals who favor none of the beverages be denoted by ‘n’.
Let the count of individuals who favor precisely one beverage be denoted by ‘s’.
Let the count of individuals who favor precisely two beverages be denoted by ‘d’.
Let the count of individuals who favor all three beverages be denoted by ‘t’.
The problem states that the total number of individuals surveyed is 100, leading to the equation:
n + s + d + t = 100 …… (1)
Additionally, the sum of the number of preferences for each beverage is given as 73 + 80 + 52, which can be expressed in terms of our variables. Each individual who likes exactly one drink contributes 1 to this sum. Each individual who likes exactly two drinks contributes 2 to this sum. Each individual who likes all three drinks contributes 3 to this sum. Therefore:
s + 2d + 3t = 73 + 80 + 52
s + 2d + 3t = 205 …… (2)
Subtracting equation (1) from equation (2) yields:
(s + 2d + 3t) – (n + s + d + t) = 205 – 100
d + 2t – n = 105
We need to find the difference between the maximum and minimum possible values of ‘t’.
To find the maximum value of ‘t’:
Consider the case where ‘n’ is minimized (n=0) and ‘d’ is minimized. The maximum number of people who could like all three drinks is limited by the smallest individual preference group, which is 52.
If t = 52, and to satisfy d + 2t – n = 105 with n=0, we have d + 2(52) – 0 = 105, so d + 104 = 105, which means d = 1.
Substituting into equation (1): 0 + s + 1 + 52 = 100, so s + 53 = 100, which means s = 47.
This scenario (n=0, s=47, d=1, t=52) is consistent with all given conditions. Thus, the maximum value for t is 52.
To find the minimum value of ‘t’:
We want to maximize ‘d’ to minimize ‘t’. From d + 2t – n = 105, we can express d as d = 105 + n – 2t.
Substitute this into equation (1): n + s + (105 + n – 2t) + t = 100
2n + s + 105 – t = 100
2n + s – t = -5
t = 2n + s + 5
Since ‘s’ and ‘n’ must be non-negative, the minimum value of ‘t’ will occur when ‘n’ and ‘s’ are as small as possible while maintaining consistency.
Let’s consider the total preferences sum again: s + 2d + 3t = 205.
If we set t = 5 (the smallest possible integer value of t to make progress towards minimizing), then s + 2d + 15 = 205, so s + 2d = 190.
From n + s + d + t = 100, we have n + s + d + 5 = 100, so n + s + d = 95.
If we have s + 2d = 190, and s + d = 95 – n, subtracting these gives d = 190 – (95 – n) = 95 + n.
For the smallest possible ‘t’, we want to maximize ‘d’. The maximum value ‘d’ can take is 95 (when n=0).
If d=95 and n=0, then s + 2(95) = 190, so s = 0.
Checking equation (1): n + s + d + t = 0 + 0 + 95 + 5 = 100. This satisfies equation (1).
Thus, the minimum value for t is 5.
The difference between the maximum and minimum values of t is:
Difference = 52 – 5 = 47.
The answer is option A.
Q. 2 Students in a college have to choose at least two subjects from chemistry, mathematics and physics. The number of students choosing all three subjects is 18, choosing mathematics as one of their subjects is 23 and choosing physics as one of their subjects is 25. The smallest possible number of students who could choose chemistry as one of their subjects is
Check Solution
Ans: C
Error generating problem
Q. 3 A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is
Check Solution
Ans: D
Let:
– ‘I’ represent the count of individuals playing precisely one game.
– ‘II’ represent the count of individuals playing precisely two games.
– ‘III’ represent the count of individuals playing precisely three games.
The total number of games played can be expressed as:
I + 2 * II + 3 * III = 144 + 123 + 132 = 399 …. (Equation 1)
The total number of individuals is:
I + II + III = 256 …. (Equation 2)
Subtracting Equation 1 from Equation 2 yields:
(I + II + III) – (I + 2 * II + 3 * III) = 256 – 399
-II – 2 * III = -143
II + 2 * III = 143 …. (Equation 3)
The sum of individuals playing at least two games is:
II + 3 * III = 58 + 25 + 63 = 146 …. (Equation 4)
Subtracting Equation 3 from Equation 4:
(II + 3 * III) – (II + 2 * III) = 146 – 143
III = 3
Substituting the value of III back into Equation 3:
II + 2 * (3) = 143
II + 6 = 143
II = 137
Substituting the values of II and III back into Equation 2:
I + 137 + 3 = 256
I + 140 = 256
I = 116
The count of individuals playing only tennis is calculated as:
Total playing tennis – those playing tennis and another game – those playing tennis and two other games + those playing all three games
123 – 58 – 25 + 3 = 43
Q. 4 If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be
Check Solution
Ans: D
Explanation:Let $P$ be the set of students who like pizza and $B$ be the set of students who like burger.
We are given the total number of students is 200.
Number of students who like pizza, $|P| = 105$.
Number of students who like burger, $|B| = 134$.
We want to find the number of students who like only burger, which is $|B \setminus P| = |B| – |B \cap P|$.
We know the principle of inclusion-exclusion for two sets:
$|P \cup B| = |P| + |B| – |P \cap B|$.
The total number of students is 200, so $|P \cup B| \leq 200$.
$105 + 134 – |P \cap B| \leq 200$
$239 – |P \cap B| \leq 200$
$|P \cap B| \geq 239 – 200$
$|P \cap B| \geq 39$.
Also, the number of students who like both pizza and burger cannot be more than the number of students who like pizza or the number of students who like burger.
So, $|P \cap B| \leq |P| = 105$ and $|P \cap B| \leq |B| = 134$.
Thus, $39 \leq |P \cap B| \leq 105$.
We are looking for the number of students who like only burger:
Number of students who like only burger = $|B| – |P \cap B| = 134 – |P \cap B|$.
Since $39 \leq |P \cap B| \leq 105$, we can find the possible range for the number of students who like only burger:
When $|P \cap B| = 39$ (minimum intersection), number of students who like only burger = $134 – 39 = 95$.
When $|P \cap B| = 105$ (maximum intersection), number of students who like only burger = $134 – 105 = 29$.
So, the number of students who like only burger can possibly be in the range $[29, 95]$.
Let’s check the given options:
Option A: 23. This is outside the range $[29, 95]$.
Option B: 26. This is outside the range $[29, 95]$.
Option C: 96. This is outside the range $[29, 95]$.
Option D: 93. This is within the range $[29, 95]$.
Therefore, the number of students who like only burger can possibly be 93.
Let’s verify if 93 is possible.
If the number of students who like only burger is 93, then $|B| – |P \cap B| = 93$.
$134 – |P \cap B| = 93$
$|P \cap B| = 134 – 93 = 41$.
If $|P \cap B| = 41$, then this value is within the possible range of intersection $[39, 105]$.
The number of students who like only pizza is $|P| – |P \cap B| = 105 – 41 = 64$.
The total number of students is $|P \setminus B| + |B \setminus P| + |P \cap B| = 64 + 93 + 41 = 198$.
Since $198 \leq 200$, this scenario is possible.
Correct_Option:D
Q. 5 For two sets A and B, let AΔB denote the set of elements which belong to A or B but not both. If P = {1,2,3,4}, Q = {2,3,5,6}, R = {1,3,7,8,9}, S = {2,4,9,10}, then the number of elements in (PΔQ)Δ(RΔS) is
Check Solution
Ans: 7
Explanation:AΔB is the symmetric difference of sets A and B. It can be calculated as (A union B) – (A intersection B), or equivalently, (A – B) union (B – A).
First, let’s find PΔQ:
P = {1,2,3,4}
Q = {2,3,5,6}
P ∪ Q = {1,2,3,4,5,6}
P ∩ Q = {2,3}
PΔQ = (P ∪ Q) – (P ∩ Q) = {1,2,3,4,5,6} – {2,3} = {1,4,5,6}
Next, let’s find RΔS:
R = {1,3,7,8,9}
S = {2,4,9,10}
R ∪ S = {1,2,3,4,7,8,9,10}
R ∩ S = {9}
RΔS = (R ∪ S) – (R ∩ S) = {1,2,3,4,7,8,9,10} – {9} = {1,2,3,4,7,8,10}
Now, we need to find (PΔQ)Δ(RΔS):
Let X = PΔQ = {1,4,5,6}
Let Y = RΔS = {1,2,3,4,7,8,10}
XΔY = (X ∪ Y) – (X ∩ Y)
X ∪ Y = {1,2,3,4,5,6,7,8,10}
X ∩ Y = {1,4}
XΔY = {1,2,3,4,5,6,7,8,10} – {1,4} = {2,3,5,6,7,8,10}
The number of elements in (PΔQ)Δ(RΔS) is the number of elements in {2,3,5,6,7,8,10}, which is 7.
Alternatively, we can use the property that AΔB = (A – B) ∪ (B – A).
PΔQ = (P – Q) ∪ (Q – P)
P – Q = {1,4}
Q – P = {5,6}
PΔQ = {1,4} ∪ {5,6} = {1,4,5,6}
RΔS = (R – S) ∪ (S – R)
R – S = {1,3,7,8}
S – R = {2,4,10}
RΔS = {1,3,7,8} ∪ {2,4,10} = {1,2,3,4,7,8,10}
Now, (PΔQ)Δ(RΔS) = ({1,4,5,6}) Δ ({1,2,3,4,7,8,10})
= (({1,4,5,6}) – ({1,2,3,4,7,8,10})) ∪ (({1,2,3,4,7,8,10}) – ({1,4,5,6}))
= ({5,6}) ∪ ({2,3,7,8,10})
= {2,3,5,6,7,8,10}
The number of elements is 7.
Final_Answer:7
Q. 6 If A = {$6^{2n} -35n – 1$}, where $n$ = 1,2,3,… and B = {35($n$-1)}, where $n$ = 1,2,3,… then which of the following is true?
Check Solution
Ans: A
Upon close examination of set A, it becomes evident that the expression $6^{2n} -35n – 1$ yields a result divisible by 35. Consequently, set A comprises numbers that are multiples of 35. It’s important to note, however, that set A does not encompass all multiples of 35 for varying values of $n$.
Specifically, when $n = 1$, the expression evaluates to 0. When $n = 2$, the expression results in 1225, which corresponds to the 35th multiple of 35.
Observing set B reveals that it contains all multiples of 35, including 0.
Therefore, every element present in set A will also be found in set B. Conversely, not every element in set B is guaranteed to be in set A.
Consequently, option A represents the accurate solution.