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Peter got $30%$ of the maximum marks in an examination and failed by 10 marks. However, Paul who took the same examination got $40%$ of the total marks and got 15 marks more than the passing marks. What were the passing marks in the examination?









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Option(D) is correct

Let $x$ be the maximum marks in the examination.

Therefore, Peter got $30\%$ of $x$

\(=\dfrac{30}{100}\times x\)


And Paul got $40\%$ of $x$

\(=\dfrac{40}{100}\times x\)


In terms of the maximum marks Paul got $0.4x-0.3x = 0.1x$ more than Peter. --------(1) 

The problem however, states that Paul got 15 marks more than the passing mark and Peter got 10 marks less than the passing mark. 

Therefore, Paul has got 15 + 10 = 25 marks more than Peter.-------- (2)

Equating (1) and (2), we get


\(\Rightarrow x=250\)

$x$ is the maximum mark and is equal to 250 marks.

We know that Peter got $30\%$ of the maximum marks.

Therefore, Peter got

\(=\dfrac{30}{100}\times 250\)

\(=75\) marks

We also know that Peter got 10 marks less than the passing mark.

Therefore, the passing mark will be 10 marks more than what Peter got = $75 + 10 = 85$

Edit: For a shortcut alternative method, check comment by Chirag Goyal.

(3) Comment(s)


Peter got 30 percent of max(let max marks be x)--> (30/x)*100,

Paul got 40 percent of total --->(40/x)*100

let the passing marks be P

so, (30/x)*100=P-10----(1)

and (40/x)*100=P+15----(2)

equate and find P


simple short cut

pet Pa

30+10 40-15



100----------- ?( 250)--(total marks)

now pass percentage

30/100*250+10= 85

Chirag Goyal

$\text{% Difference of Max. Marks}$ $= \text{Difference in Obtained Marks Individually}$

$ 10 \text{ % of }x = 15 - (-10)$

$\Rightarrow x=250$

Passing Marks $= 30 \text{ % of } 250 + 10 = 85$