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Solution

Venn diagram

Well, it has been stated that no teacher teaches both guitar and flute. This obviously implies that there are no teachers who teach all the three musical instruments. The reason: Suppose there are a few teachers who teach all the three musical instruments - guitar, vioin and flute, then this means these teachers are teaching guitar and flute. But we have been told no teacher teaches both guitar and flute. Thus, the number of teachers teaching all the three instruments is zero (I hope the reasoning is right!).

Now I can draw the Venn diagram as above. The set of guitar teachers is represented by the circle labelled "Guitar", the set of violin teachers is represented by the circle labelled "Violin", and the set of flute teachers is represented by the circle labelled "Flute".

The region coloured green is common to all the three circles and, hence, represents the set of teachers who teach all the three musical instruments. But this is zero as we have reasoned.

The region coloured yellow is common to the "Guitar" and "Violin" circles and, therefore, is the set of teachers who teach both guitar and violin but not flute. The number of teachers who teach only guitar and violin but not flute = (the number of teachers who teach guitar and violin) - (number of teachers who teach guitar, violin and flute). So, the number of teachers who teach only guitar and violin but not flute = 4 - 0 = 4. (Note: The number of teachers who teach guitar and violin could include the teachers teaching all the three instruments, hence the substraction.)

The region coloured light blue is common to the "Flute" and "Violin" circles and, therefore, represents the set of teachers who teach both flute and violin, but not guitar. We have to find this number. Let it be "x".

What remains is the pink region. It is common to the "Guitar" and "Flute" circles and, therefore, represents the set of teachers teaching both guitar and flute but not violin. But, we have been told that there are no such teachers. This set, therefore, has zero elements.

What remains to be determined is how many teachers teach only guitar, how many teach only violin and how many teach only flute. We have been told that ten teachers teach guitar. Therefore, these ten teachers include those who teach only guitar, those who teach guitar and violin, those who teach guitar and flute, and those who teach all the three instruments.

Therefore, number of teachers teaching only guitar + 4 + 0 + 0 = 10.

Therefore, the number of teachers who teach only guitar = 10 - 4 = 6.

Similarly we work out that the number of teachers who teach only violin = (5 - x) and the number of teachers who teach only flute = (7 - x).

The total number of teachers is 20. These 20 teachers include those who teach only guitar, those who teach guitar and violin, those who teach guitar and flute, those who teach all the three instruments, those who teach only violin, those who teach both violin and flute but not guitar, and those who teach only flute.

Therefore, 6 + 4 + 0 + 0 + (5-x) + x + (7 - x) = 20.

Thus, 22 - x = 20, or x = 2. So, the number of teachers who teach both flute and violin (but not guitar) is 2. We have been asked to find the number of teachers who teach both flute and violin. This number includes the number of teachers who teach all the three instruments. But, in the present case, the number of teachers who teach all the three instruments is zero, and, hence, the number of teachers teaching both flute and violin is the same as the number of teachers teaching both flute and violin but not guitar: 2.

Okay, for those who like to use formulae we have:

n(AUnion symbolBUnion symbolC) = n(A) + n(B) + n(C) - n(AIntersection symbolB) - n(AIntersection symbolC) - n(BIntersection symbolC) + n(AIntersection symbolBIntersection symbolC).

Here "A" is the set of guitar teachers, "B" is the set of violin teachers and "C" is the set of flute teachers.

n(AUnion symbolBUnion symbolC) = 20.

n(A) = 10, n(B) = 9, n(C) = 7, n(AIntersection symbolB) = 4, n(AIntersection symbolC) = 0, let n(BIntersection symbolC) = x (this is the number of teachers who teach both flute and violin but not guitar), and n(AIntersection symbolBIntersection symbolC) = 0.

Therefore, 20 = 10 + 9 + 7 - 4 - 0 - x + 0.

Thus 22-x = 20. So, x = 2.

So, the number of teachers who teach only violin and flute but not guitar = 2. We have been asked to find out the number of teachers who teach both violin and flute. This number includes the number of teachers who teach all the three instruments.

Therefore, the number of teachers who teach both violin and flute = number of teachers who teach only violin and flute but not guitar + the number of teachers who teach all the three musical instruments.

So, the number of teachers who teach both violin and flute = 2 + 0 = 2.

Did I get that right? Or did I goof up? Do let me know.

Some useful links for
your career:


  • Union Public Service Commission - www.upsc.gov.in
  • IIT-Kharagpur - www.iitkgp.ac.in
  • Indian Statistical Institute - www.isical.ac.in
  • Indian Institute of Technology Madras - www.iitm.ac.in
  • Indian Institute of Management, Ahmedabad - www.iimahd.ernet.in
  • Indian Institute of Mass Commission - www.iimc.nic.in
  • IIT Bombay - www.iitb.ac.in
  • Indian School of Mines, Dhanbad - www.ismdhanbad.ac.in
  • Birla Institute of Technology, Ranchi - www.bitmesra.ac.in
  • Central Institute of Fisheries Nautical and Engineering Training - www.cifnet.nic.in
  • Indian Institute of Information Technology, Allahabad (Deemed University) - www.iiita.ac.in
  • Central Marine Fisheries Research Institute, Kochi - www.cmfri.com
  • Tata Institute of Social Sciences, Mumbai - www.tiss.edu