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 student taking the exam was $z$ %. Find in terms of $x, y$ and $z$ the percentage of students taking the exam who failed ?

(z - x) / (y - x)

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(x - z) / (y - z)

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(y - x) / (z - y)

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None of these

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Solution

The correct option is A $(z - x) / (y - x)$

Let Pass, fail and total $b, c, a$ respectively
$∴ b \times x + c \times y = a \times z$
and $a = b + c$
$⟹ a x = b x + c x$
solving above two equations
we get, $c (y - x) = a (z - x)$
$⟹ \frac{c}{a} = (\frac{z - x}{y - x})$

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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 student taking the exam was z %. Find in terms of x,y and z the percentage of students taking the exam who failed ?

The correct option is A (z−x)/(y−x)Let Pass, fail and total b,c,a respectively∴b×x+c×y=a×zand a=b+c⟹ax=bx+cxsolving above two equations we get, c(y−x)=a(z−x)⟹ca=(z−xy−x)

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 student taking the exam was $z$ %. Find in terms of $x, y$ and $z$ the percentage of students taking the exam who failed ?

The correct option is A $(z - x) / (y - x)$

Let Pass, fail and total $b, c, a$ respectively
$∴ b \times x + c \times y = a \times z$
and $a = b + c$
$⟹ a x = b x + c x$
solving above two equations
we get, $c (y - x) = a (z - x)$
$⟹ \frac{c}{a} = (\frac{z - x}{y - x})$