There are 345 students at a college who have taken a course in calculus, 212 who have taken a course in discrete mathematics, an

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There are 345 students at a college who have taken a course in calculus, 212 who have taken a course in discrete mathematics, an

d 188 who have taken courses in both calculus and discrete mathematics. How many students have taken a course in either calculus or discrete mathematics

Mathematics

1 answer:

AnnZ [12.3K] · Answer 1 · 2026-02-23T17:34:37+03:00

Answer:

There are 369 students who enrolled in either calculus or discrete mathematics.

Step-by-step explanation:

I'll create a Venn diagram to illustrate these figures.

Let’s define:

A represents the count of students who completed calculus.

B stands for those who took discrete mathematics.

We have the following:

$A = a + (A \cap B)$

Here, a denotes the students who studied calculus exclusively, while $A \cap B$ represents those who took both subjects.

Using the same reasoning:

$B = b + (A \cap B)$

There are 188 students taking both calculus and discrete mathematics.

This implies that $A \cap B = 188$

A total of 212 students studied discrete mathematics.

<pBased on this, $B = 212$

345 students have attended a calculus course.

<pSo, from this information, $A = 345$

We find the total number of students who took either calculus or discrete mathematics.

$(A \cup B) = A + B - (A \cap B) = 345 + 212 - 188 = 369$

A total of 369 students participated in either calculus or discrete mathematics.