Question:
In an examination, the average marks obtained by students who passed was \(x\%\), while the average of those who failed was \(y\%\). The average marks of all the students taking the exam was \(z\%\). Find, in terms of \(x\), \(y\) and \(z\), the percentage of students taking the exam who failed.
In an examination, the average marks obtained by students who passed was \(x\%\), while the average of those who failed was \(y\%\). The average marks of all the students taking the exam was \(z\%\). Find, in terms of \(x\), \(y\) and \(z\), the percentage of students taking the exam who failed.
Show Hint
Write the overall average as a weighted average of the pass and fail group averages, then solve that equation for the fail fraction.
Updated On: Jul 13, 2026
- \(\dfrac{z-x}{y-x}\)
- \(\dfrac{x-z}{y-z}\)
- \(\dfrac{y-x}{z-y}\)
- \(\dfrac{y-z}{x-z}\)
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The Correct Option is A
Solution and Explanation
Step 1: Set up variables for the fractions of students.
Let the total number of students taking the exam be 1, so we work directly in fractions and the answer comes out as a percentage. Let \(f\) be the fraction of students who failed, so \((1-f)\) is the fraction who passed.
Step 2: Write the overall average as a weighted average.
The overall average mark \(z\) is the weighted average of the passed group's average \(x\) and the failed group's average \(y\), weighted by the size of each group.
\[ z = x(1-f) + y f \]
This holds because the total marks scored by all students equal the marks from the passed group plus the marks from the failed group.
Step 3: Expand and collect the terms in \(f\).
\[ z = x - xf + yf \]
\[ z - x = f(y - x) \]
We moved \(x\) to the left side and factored \(f\) out of the two terms that contain it.
Step 4: Solve for \(f\).
\[ f = \frac{z-x}{y-x} \]
Here \(f\) is the fraction of students who failed, so as a percentage it is exactly \(\dfrac{z-x}{y-x}\), since \(x\), \(y\), \(z\) were already given as percentages.
Step 5: Rule out the other options.
Option (B) swaps which quantity is paired with \(z\) in a way that does not follow from this algebra. Options (C) and (D) pair \(z\) with \(y\) instead of \(x\), which does not match the failed fraction derived above.
Final Answer:
The percentage of students who failed is \(\dfrac{z-x}{y-x}\). \[ \boxed{\dfrac{z-x}{y-x}} \]
Let the total number of students taking the exam be 1, so we work directly in fractions and the answer comes out as a percentage. Let \(f\) be the fraction of students who failed, so \((1-f)\) is the fraction who passed.
Step 2: Write the overall average as a weighted average.
The overall average mark \(z\) is the weighted average of the passed group's average \(x\) and the failed group's average \(y\), weighted by the size of each group.
\[ z = x(1-f) + y f \]
This holds because the total marks scored by all students equal the marks from the passed group plus the marks from the failed group.
Step 3: Expand and collect the terms in \(f\).
\[ z = x - xf + yf \]
\[ z - x = f(y - x) \]
We moved \(x\) to the left side and factored \(f\) out of the two terms that contain it.
Step 4: Solve for \(f\).
\[ f = \frac{z-x}{y-x} \]
Here \(f\) is the fraction of students who failed, so as a percentage it is exactly \(\dfrac{z-x}{y-x}\), since \(x\), \(y\), \(z\) were already given as percentages.
Step 5: Rule out the other options.
Option (B) swaps which quantity is paired with \(z\) in a way that does not follow from this algebra. Options (C) and (D) pair \(z\) with \(y\) instead of \(x\), which does not match the failed fraction derived above.
Final Answer:
The percentage of students who failed is \(\dfrac{z-x}{y-x}\). \[ \boxed{\dfrac{z-x}{y-x}} \]
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