Question

A candidate who gets $36\%$ marks in the examination fails by $24$ marks but another candidate, who gets $43\%$ marks, gets $18$ more marks than the minimum pass marks. Find the maximum marks and the percentage of pass marks.

Answer 1

Let the Passing marks be $x$

and the Total marks be $y$

First Canditate$=36\%ofy+24=x$

Second Canditate$=43\%ofy-18=x$

First cand.$=$Second Cand.

$\frac{36y}{100}+24=\frac{43y}{100}-18$

$\frac{9y}{25}+24=\frac{43y}{100}-18$

$24+18=\frac{43y}{100}-\frac{36y}{100}$

$42=\frac{7y}{100}$

$\frac{4200}{7}=y=600$

Total marks$=600$

$\frac{36}{100}y+24=x$

$\frac{36}{10}\times 600+24=x$

$216+24=x$

$240=x$

Passing marks$=240$