JEE Main 2021 (26 Aug Shift 1)

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Animated Solution for Mathematics - Sets and Relations: Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If $K %$ of them are suffering from both ailments, then $K$ can not belong to the set:

Q: Why is the intersection $K$ limited by the smaller set?

The intersection $n(H \cap L)$ is the set of patients who have both ailments. If a patient has both, they must be part of the Heart set AND the Lung set. Therefore, the number of people with both cannot exceed the number of people who have the less common ailment. If only $89\%$ have heart ailments, you cannot have $90\%$ of people with both, because that would imply more people have heart ailments than actually exist!

Q: How do I know which inequality gives the lower bound and which gives the upper bound?

The union inequality $n(H \cup L) \le 100$ forces the intersection to be large enough to 'pull' the union back down to $100\%$, giving you the lower bound ($K \ge 87$). The subset inequality $n(H \cap L) \le \min(n(H), n(L))$ prevents the intersection from being impossibly large, giving you the upper bound ($K \le 89$).

Select Answer:

Visualized Solution

Defining the Universal Set

Let the total number of patients be $100%$ .

Patients with Heart Ailment

Let $n (H)$ be the percentage of patients with heart ailment.
$n (H) = 89$

Patients with Lung Infection

Let $n (L)$ be the percentage of patients with lung infection.
$n (L) = 98$

Patients with Both Ailments

Let $n (H \cap L)$ be the percentage of patients with both ailments.
$n (H \cap L) = K$

Principle of Inclusion-Exclusion

The formula for the union of two sets is:
$n (H \cup L) = n (H) + n (L) - n (H \cap L)$

Substituting the Values

Substitute the known percentages into the formula:
$n (H \cup L) = 89 + 98 - K$

Simplifying the Union

Add the numbers:
$n (H \cup L) = 187 - K$

Maximum Possible Union

The union of $H$ and $L$ cannot exceed the total population:
$n (H \cup L) \leq 100$

Finding the Lower Bound of $K$

Substitute the union expression:
$187 - K \leq 100$
$187 - 100 \leq K$
$K \geq 87$

Maximum Possible Intersection

The intersection cannot be larger than the smallest set:
$K \leq min (n (H), n (L))$

Finding the Upper Bound of $K$

Compare the sizes of $H$ and $L$ :
$K \leq min (89, 98)$
$K \leq 89$

The Range of $K$

Combining both bounds:
$87 \leq K \leq 89$
Possible integer values for $K$ : ${87, 88, 89}$

Evaluating the Options

We need to find the set that $K$ cannot belong to.
Option 1: ${80, 83, 86, 89}$ (Contains $89$ )
Option 2: ${84, 86, 88, 90}$ (Contains $88$ )
Option 4: ${84, 87, 90, 93}$ (Contains $87$ )

Final Conclusion

Check Option 3:
Option 3: ${79, 81, 83, 85}$
None of these values are in ${87, 88, 89}$ .
Therefore, $K$ cannot belong to Option 3.

00:00 / 00:00

The Sigma Insight: Types of Sets and Set Operations

Solution Diagram

Analyzing the Universal Reality

We start with the total population, which we define as

100%

. This is our universal set. Within this, we have two distinct groups:

n (H) = 89

for heart ailments and

n (L) = 98

for lung infections.

At first glance, if you add these together, you get

187%

. However, the total cannot exceed

100%

. This is where the Principle of Inclusion-Exclusion comes into play.

The formula is defined as:

n (H \cup L) = n (H) + n (L) - n (H \cap L)

Here,

n (H \cup L)

represents the patients who have at least one ailment, and

n (H \cap L)

is the intersection

K

, representing those suffering from both.

The Constraints of Logic

We know that

n (H \cup L) = 89 + 98 - K

, which simplifies to:

n (H \cup L) = 187 - K

Now, we apply the physical constraint: the union cannot exceed the total population. Thus:

187 - K \leq 100

Solving this inequality, we find

K \geq 87

. This establishes our lower bound.

There is a second, silent constraint. The intersection

K

cannot be larger than the smallest individual set. Since

n (H) = 89

and

n (L) = 98

, the intersection

K

must satisfy:

K \leq min (89, 98) \Rightarrow K \leq 89

The Final Synthesis

Combining these two bounds, we arrive at the range:

87 \leq K \leq 89

The possible integer values for

K

are

87, 88,

and

89

We are looking for the set to which

K

cannot belong. Given the options, we evaluate the values:

Option 1 contains $89$ . Option 2 contains

88

. * Option 4 contains

87

Only Option 3, which is

{79, 81, 83, 85}

, contains none of our possible values. Thus, we have solved the mystery.

Common Traps & FAQs

Wait, why can't the union be greater than 100%?

Think of the universal set as the total number of patients in the hospital, which is

100%

. The union

n (H \cup L)

represents the percentage of patients who have at least one of the two ailments. Since you cannot have more patients than the total number of patients present, the union must be less than or equal to

100%

Why is the intersection

K

limited by the smaller set?

The intersection

n (H \cap L)

is the set of patients who have both ailments. If a patient has both, they must be part of the Heart set AND the Lung set. Therefore, the number of people with both cannot exceed the number of people who have the less common ailment. If only

89%

have heart ailments, you cannot have

90%

of people with both, because that would imply more people have heart ailments than actually exist!

How do I know which inequality gives the lower bound and which gives the upper bound?

The union inequality

n (H \cup L) \leq 100

forces the intersection to be large enough to 'pull' the union back down to

100%

, giving you the lower bound (

K \geq 87

). The subset inequality

n (H \cap L) \leq min (n (H), n (L))

prevents the intersection from being impossibly large, giving you the upper bound (

K \leq 89

Animated Solution for Mathematics - Sets and Relations: Out of all the patients in a hospital 89% are found to be suffering from heart ailment and 98% are suffering from lungs infection. If $K %$ of them are suffering from both ailments, then $K$ can not belong to the set:

Select Answer:

Visualized Solution

Defining the Universal Set

Patients with Heart Ailment

Patients with Lung Infection

Patients with Both Ailments

Principle of Inclusion-Exclusion

Substituting the Values

Simplifying the Union

Maximum Possible Union

Finding the Lower Bound of $K$

Maximum Possible Intersection

Finding the Upper Bound of $K$

The Range of $K$

Evaluating the Options

Final Conclusion

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Universal Reality

The Constraints of Logic

The Final Synthesis

Similar Questions

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Select Answer:

Visualized Solution

Defining the Universal Set

Patients with Heart Ailment

Patients with Lung Infection

Patients with Both Ailments

Principle of Inclusion-Exclusion

Substituting the Values

Simplifying the Union

Maximum Possible Union

Finding the Lower Bound of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Maximum Possible Intersection

Finding the Upper Bound of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

The Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Evaluating the Options

Final Conclusion

The Sigma Insight: Types of Sets and Set Operations

Analyzing the Universal Reality

The Constraints of Logic

The Final Synthesis

Similar Questions

In a class of 140 students numbered 1 to 140, all even numbered students opted mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is :

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In a school, there are three types of games to be played. Some of the students play two types of games, but none play all the three games. Which Venn diagrams can justify the above statement?

Finding the Lower Bound of $K$

Finding the Upper Bound of $K$

The Range of $K$

Let $A = {n \in [100, 700] \cap N : n is neither a multiple of 3 nor a multiple of 4} . T h e n t h e n u mb er o f e l e m e n t s in$ A$ is

Let $A = {n \in N ∣ n^{2} \leq n + 10, 000}$ , $B = {3 k + 1 ∣ k \in N}$ and $C = {2 k ∣ k \in N}$ , then the sum of all the elements of the set $A \cap (B - C)$ is equal to