In a population, patients who have high cholesterol also have high blood-pressure (BP). Some patients with high BP also have diabetes. There are no patients who have both high cholesterol and diabetes. Furthermore,
1. the total number of patients with at least one of these conditions is 75,
2. the number of patients with high cholesterol is 10,
3. the number of patients with high BP is 45, and
4. the number of patients with only high BP and no other conditions is 20.
Then the number of patients who have both diabetes and high BP is ________

Question

Consider two sets A = {2, 3, 5, 7, 11, 13} and B = {1, 8, 27}. Let f be a function from A to B such that for every element b in B, there is at least one element a in A such that f(a) = b. Then, the total number of such functions f is

Answer 1

This problem can also be solved directly with set notation instead of splitting the BP group into pieces first.

Let $C$, $P$, and $D$ denote the sets of patients with high cholesterol, high BP, and diabetes respectively. We are given $|C|=10$, $|P|=45$, $C \cap D = \varnothing$ (no patient has both cholesterol and diabetes), $C \subseteq P$ (every cholesterol patient also has high BP), and $|P \setminus (C \cup D)| = 20$ (only high BP, no other condition).

Since $C \subseteq P$, the set $P$ splits into exactly three disjoint pieces: patients in $P \cap C$, patients in $P$ with neither other condition, and patients in $P \cap D$. Because $C \subseteq P$ and $C \cap D = \varnothing$, we get $|P \cap C| = |C| = 10$.

So $|P| = |P \cap C| + |P \setminus (C \cup D)| + |P \cap D|$, which gives $45 = 10 + 20 + |P \cap D|$.

Solving, $|P \cap D| = 45 - 30 = 15$.

As a consistency check, using the inclusion-exclusion total $|C \cup P \cup D| = 75$ and substituting all known pieces confirms a non-negative value for the diabetes-only-without-BP group, so no contradiction arises.

\[\boxed{|P \cap D| = 15}\]

Answer 2

This problem can also be solved directly with set notation instead of splitting the BP group into pieces first.

Let $C$, $P$, and $D$ denote the sets of patients with high cholesterol, high BP, and diabetes respectively. We are given $|C|=10$, $|P|=45$, $C \cap D = \varnothing$ (no patient has both cholesterol and diabetes), $C \subseteq P$ (every cholesterol patient also has high BP), and $|P \setminus (C \cup D)| = 20$ (only high BP, no other condition).

Since $C \subseteq P$, the set $P$ splits into exactly three disjoint pieces: patients in $P \cap C$, patients in $P$ with neither other condition, and patients in $P \cap D$. Because $C \subseteq P$ and $C \cap D = \varnothing$, we get $|P \cap C| = |C| = 10$.

So $|P| = |P \cap C| + |P \setminus (C \cup D)| + |P \cap D|$, which gives $45 = 10 + 20 + |P \cap D|$.

Solving, $|P \cap D| = 45 - 30 = 15$.

As a consistency check, using the inclusion-exclusion total $|C \cup P \cup D| = 75$ and substituting all known pieces confirms a non-negative value for the diabetes-only-without-BP group, so no contradiction arises.

\[\boxed{|P \cap D| = 15}\]

Show Hint

The Correct Option is B

Solution and Explanation

Top Questions on Set Theory

Questions Asked in GATE ES exam