Question

A candidate who gets 36% marks in an examination fails by 24 marks but another candidate, who gets 43% marks, gets 18 morr marks than the minimumpass marks. Find the maximum marks and the percentage of pass marks

Answer 1

x = maximum marks that can be gotten.
y = passing marks
first equation is:
.36 * x = y - 24
second equation is:
.43 * x = y + 18
solve for y in the first equation to get:
y = .36 * x + 24
substitute for y in the second equation to get:
.43 * x = .36 * x + 24 + 18
simplify to get:
.43 * x = .36 * x + 42
subtract .36 * x from both sides of this equation to get:
.07 * x = 42
divide both sides of this equation by .07 to get:
x = 600
the most marks that can be gotten is 600.
substitute for x in the first equation to get:
.36 * x = y - 24 becomes:
.36 * 600 = y - 24
simplify to get:
216 = y - 24
add 24 to both sides of this equation to get:
y = 240
that's the passing mark.
the passing percent is 240 / 600 = .4 = 40%
your solutions are:
maximum marks = 600
passing mark percent = 40%

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