d Word Problem 3 | Job Exam Rare Mathematics (JEMrare) - Solved MCQs

Word Problem 3

Problem 3: In an examination a candidate who secures 25% of the maximum marks fails by 60 marks. Another candidate who secures 42 % of the total marks gets 8 marks more than necessary for passing. Find the maximum marks and the percentage for passing.

(word problems on job exam mathematics.)


Let  the maximum marks =’x’

And let there be two candidates ‘A’ and ‘B’

Let pass marks = y% of x

Pass Marks = \frac{y}{100}x

Pass Marks = \frac{yx}{100}----(1)

We consider ‘A’s case

Candidate ‘A’

‘A’ gets 25% marks of total marks and fails by 60 marks. It means had had’A’ got 60 more marks he would has passed the exam ! i-e 25%+60 marks would have made him successful.

so we say mathematically,

Total Marks = x

A Got Marks =\frac{25}{100} x

To Make Him Pass He Needed =\frac{25}{100} x+60----(2)

Eq (1) and (2) both are expressing pass marks , so both should be equal!

\Rightarrow \frac{25}{100} x+60= \frac{yx}{100}

\Rightarrow \frac{25}{100} x+60= \frac{yx}{100}-----(3)

Now we consider ‘B’s case

Candidate ‘B’

“B’ gets 42% marks of total marks ‘x’ which are 8 marks more than needed to pass, i-e if 8 marks were deducted from his score than remaining marks would be equal to passing marks!

We write mathematically;

Total Marks = x

'B' Got Marks = \frac{42}{100}x

'B' Needed Marks To Pass = \frac{42}{100}x-8----(4)

Now eq (1) and (4) both express passing marks, so should be equal!

\Rightarrow \frac{42}{100}x-8=\frac{yx}{100}----(5)

Subtracting eq(3) and (5) from both sides;

\Rightarrow \frac{42}{100}x-8 - \left \{ \frac{25}{100} x+60 \right \}=\frac{yx}{100} - \left \{ \frac{yx}{100} \right \}

\Rightarrow \frac{42}{100}x-8 - \frac{25}{100} x-60 =0

\Rightarrow \frac{42}{100}x- \frac{25}{100} x =68

\Rightarrow \frac{42-25}{100}x =68

\Rightarrow \frac{17}{100}x =68

\Rightarrow x=\frac{6800}{17

\Rightarrow x=400

So, total marks were 400

Put this value of x in eq (3) or (5) {we putting in eq 3}

\Rightarrow \frac{25}{100} \times400+60= \frac{400y}{100}

\Rightarrow 25 \times 4+60= 4y

\Rightarrow 4y=100+60

\Rightarrow 4y=160

\Rightarrow y=\frac{160}{4}

\Rightarrow y=40

So, passing marks were 40%



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