Question

In a school examination, a student who gets 36% of the total marks fails by 30 marks and a student who gets 48%
marks gets 60 marks more than the pass marks. What percentage of the total marks is required to pass the
examination?​

Answer 1

Answer:

48-36 =12℅

12% =90 marks

12×30/90

4℅

36+4=40℅ is passing percentage

passing marks =90×40/12=300

passing marks is 300

Answer 2

The required percentage of the total marks to pass the examination is 40%.

Given,

In any school exam, one student fails by 30 marks if he/she gets 36% of the total marks.

Also, a student gets 60 marks more than the pass marks, only if he/she gets 48% marks.

To Find,

The pass marks and required percentage of the total marks to pass the examination.

Solution,

We can solve this mathematical question using the below following method.

First of all, assume the total mark is x and the pass mark is y.

∴The required percentage of the total marks to pass the examination is, $\frac{y}{x}$×100%.

So, 36% of the total marks =  $\frac{36x}{100}$.

Also, as per the first condition, $\frac{36x}{100}$= y-30

⇒36x=100(y-30) . . . . . . . (i)

48% mark of total marks =$\frac{48x}{100}$.

As per the second condition, $\frac{48x}{100}$= y+60.

⇒48x=100(y+60). . . . . (ii)

Now, perform [4×(i)-3×(ii)] we get,

144x-144x=400(y-30)-300(y+60).

⇒400y-12000=300y+18000.

⇒400y-300y=18000+12000.

⇒100y=30000

⇒y=300.

Put the value of y in (i).

We get,

⇒36x=100(300-30).

⇒x=$\frac{27000}{36}$

⇒x=750.

The required percentage is, $\frac{y}{x}$×100%.

=$\frac{300}{750}$×100%.

=40%.

Hence, The required percentage of the total marks to pass the examination is 40%.

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