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I encountered this kind of question in an employment exam. I am just curious how to compute this. I already forgot the actual numbers so I will change them to variables.

There is a survey for a product for A number of people regarding its quality and price. B number of people are satisfied both with the quality and price. Find the number of people who are both dissatisfied with quality and price.

Here’s the result of survey: Price:

– Satisfied: C number of people

– Dissatisfied: D number of people

Quality:

– Satisfied: E number of people

– Dissatisfied: F number of people

To be clear, C + D = A and E + F = A.

A, B, C, D, E and F are actual numbers that are given in the problem but forgot them.

Spade
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1 Answers1

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Make a table of the survey results like this: $$ \begin{array}{cc} &\mathbf{Price}\\ \mathbf{Quality}& \begin{array}{c|c|c|c} &\text{ Satisfied }&\text{ Dissatisfied }&\text{ Total }\\ \hline \text{Satisfied}\rule[-0.8ex]{0ex}{3ex}&B&&E\\ \hline \text{Dissatisfied}\rule[-0.8ex]{0ex}{3ex}&&&F\\ \hline \text{Total}\rule[-0.8ex]{0ex}{3ex}&C&D&A \end{array} \end{array} $$ Now you can see that the number who were satisfied with the quality but dissatisfied with the price must be $E-B$ in order to complete the first row. In the first column, the number satisfied with the price but not with the quality is $C-B$. Now the table looks like this: $$ \begin{array}{cc} &\mathbf{Price}\\ \mathbf{Quality}& \begin{array}{c|c|c|c} &\text{ Satisfied }&\text{ Dissatisfied }&\text{ Total }\\ \hline \text{Satisfied}\rule[-0.8ex]{0ex}{3ex}&B&E-B&E\\ \hline \text{Dissatisfied}\rule[-0.8ex]{0ex}{3ex}&C-B&&F\\ \hline \text{Total}\rule[-0.8ex]{0ex}{3ex}&C&D&A \end{array} \end{array} $$ Finally, the number who are dissatisfied with both the price and the quality is (from the second column) $D - (E - B)$. Or is it (from the second row) $F - (C - B)$? Actually, these are the same: the difference is $(C + D) - (E + F) = A - A = 0$.

FredH
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