Question

A student get 30% of the maximum marks but fails by 30 marks and another student gets 45% and gets 15 more marks by minimum. Find maximum marks and % necessary for passing

Answer 1

Step-by-step explanation:

A.T.Q,

$\frac{30}{100} x \times 30 = pass$

$\frac{45}{100} x - 15 = fail$

Here

$x \: = total \: marks$

$pass \: mark \: = fail \: mark$

Therefore,

$\frac{30}{100} x + 30 = \frac{45}{100} x - 15$

$\frac{30x + 3000}{100} = \frac{45x - 1500}{100}$

$30x + 3000 = 45x - 1500$

$30x - 45x = - 3000 - 1500$

$- 15x = - 4500$

$x = 300$

Therefore the maximum marks is 300.

$pass \: \% = \frac{pass \: mark \: \times 100}{total \: mark}$

$pass \: \% = 40\%$

$pass \: \% = \frac{120 \times 100}{300}$

Note : $pass \: mark \: = 30\% \times total \: marks \: + 30$ from the given information

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