Question

please solve 8 th question......

Answer 1

Let the maximum marks be M & passing marks be P.

Now,
Accounting to 1st Case,

Marks obtained = 20% of M
= $\frac{20}{100} \times m = \frac{1}{5} M = \frac{M}{5}$

P = $\frac{M}{5} + 40 = (\frac{M+ 200}{5} ) ......(1)$

Now,
According to 2nd case,

Marks obtained = 45% of M
=$\frac{45}{100} \times M= \frac{9}{20} M= \frac{9M}{20}$

P = $\frac{9M}{20} - 10 = ( \frac{9M- 200}{20} )......(2)$

Now,
Equalising both equations,

$( \frac{M + 200}{5} ) = ( \frac{9M - 200}{20} ) \\ 20(M + 200) = 5(9M - 200) \\ 20M + 4000 = 45M - 1000 \\ 20M - 45M = - 1000 - 4000 \\ - 25M = - 5000 \\ M = \frac{ - 5000}{ - 25} \\ M = 200$

Maximum marks is 200.

Passing marks is 80.

The percentage required for passing is 40%.