Discussion
In an examination, 40% marks are required to pass. A obtains 10% marks less than those required to pass. B obtains marks \(11\frac{1}{9}\%\) less than those of A, while C obtains \(41\frac{3}{17}\%\) marks less than sum of A and B together. What percentage of marks has C obtained ?
Answer: B
C obtained 40% of the marks.
The pass mark is 40%, and A gets 10% less than that. B gets \(11\frac{1}{9}\% = \frac{1}{9}\) less than A, and C gets \(41\frac{3}{17}\% = \frac{7}{17}\) less than A and B together:
\[ \begin{aligned} A &= 40 \times 0.9 = 36 \\ B &= 36 \times \tfrac{8}{9} = 32 \\ A + B &= 68 \\ C &= 68 \times \tfrac{10}{17} = 40 \end{aligned} \]
The fractions are chosen to divide exactly: \(11\frac{1}{9}\%\) is one-ninth, and \(41\frac{3}{17}\%\) is \(\frac{700}{17}\%\), or seven-seventeenths. Reading 'A obtains 10% marks less' as 30% would give non-whole answers that match no option.
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