A certain number of candidates appeared for an examination in which one-fifth of the whole plus $16$ secured first division, one-fourth plus $15$ secured second division and one-fourth minus $25$ secured third division. If the remaining $60$ candidates failed, find the total number of candidates appeared.

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Solution

Let, the number of candidates appeared for an examination be $x$ then,

The number of candidates who secured first division $= \frac{x}{5} + 16$

The number of candidates who secured second division $= \frac{x}{4} + 15$

The number of candidates who secured third division $= \frac{x}{4} - 25$

The number of candidates who failed $= 60$

i.e., $x = \frac{x}{5} + 16 + \frac{x}{4} + 15 + \frac{x}{4} - 25 + 60$

$⟹ x = \frac{x}{5} + \frac{x}{4} + \frac{x}{4} + 66$

$⟹ x = \frac{7 x}{10} + 66$

$⟹ \frac{3 x}{10} = 66$

$⟹ x = 66 \times \frac{10}{3}$

$⟹ x = 220$

$∴$ The number of candidates appeared for an examination $= 220$

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A certain number of candidates appeared for an examination in which one-fifth of the whole plus 16 secured first division, one-fourth plus 15 secured second division and one-fourth minus 25 secured third division. If the remaining 60 candidates failed, find the total number of candidates appeared.