Question

Mathematics, 17.03.2021 23:50 burners

Answer 1

9514 1404 393

Answer:

1. 10 study all three
2. 8 study only math and chem

Step-by-step explanation:

1. Adding the numbers of those who study a given subject counts ...

• once -- those who study only one subject
• twice -- those who study exactly two subjects
• three times -- those who study all three subject.

That total is 40 +48 +44 = 132. When we subtract from this the total number of students, we get 132 -80 = 52, the number who study exactly 2 subjects plus twice the number who study all three.

From this, we can subtract the number who study exactly 2 subjects, leaving twice the number who study all three. 52 -32 = 20. So, 10 students study all three subjects.

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2. Adding the numbers of those who study physics and another subject, we have 20+24 = 44. Subtracting twice the number who study all three gives the total of those how study only two subjects, one of which is physics. That total is 44 -20 = 24. So, of those 32 who study exactly two subjects, the remaining 32 -24 = 8 must be students who study only math and chemistry.

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It may be helpful to see the matrix of equations the problem statement describes. If we use the variables m, p, c, mp, mc, pc, mpc in that order to represent numbers of 1-subject, 2-subject, and 3-subject students, the augmented matrix looks like this.

  $\left[\begin{array}{ccccccc|c}1&1&1&1&1&1&1&80\\0&1&0&1&0&1&1&40\\1&0&0&1&1&0&1&48\\0&0&1&0&1&1&1&44\\0&0&0&1&0&0&1&20\\0&0&0&0&0&1&1&24\\0&0&0&1&1&1&0&32\end{array}\right]$

The solution is m=20, p=6, c=12, mp=10, mc=8, pc=14, mpc=10. The bold values are the ones this question is asking about.