In an examination, Akhil scored 11.11% less than the pass marks and Akshay scored 12.5% more than the pass marks. Find the percentage of pass marks if the sum of the scores of Akhil and Akshay is 25% more than the total marks.
- 68.25%
- 66.66%
- 56.56%
- 63.72%
- 62.06
The Correct Option is
Solution and Explanation
Let \(‘a’, ‘b’, ‘p’\), and \(‘t’\) be the marks scored by Akhil and Akshay, the pass marks, and the total marks, respectively.
Given that,
\(a = (1-11.11\%)p\) = \((1-\frac{1}9)p = \frac89p\) … (1)
\(b = (1+12.5\%)p = (1+\frac{1}8)p = \frac{9}{8}p\) … (2)
Also, \(a+b = (1+25\%)t = (1+\frac{1}{4})t = \frac{5}4t\) … (3)
Substituting \((1), (2)\) in \((3)\)
\(\frac{8}9p + \frac{9}8p = \frac{5}4t\)
\(\frac{145}{72}p = \frac{5}4t\)
or, \(p = \frac{19}{29}t\)
or, \(p = 62.06\%\) of \(t\)
Hence, option E is the correct answer.
Alternatively:
Let the passing marks be \(720x\).
Akhil scored = \((1-\frac{1}{9})720x = 640x\)
Akshay scored = \((1+\frac{1}{8})720x = 810x\)
So, \(1450x = \frac{125}{100} \times\) Total marks
Total marks = \(1160x\)
So, required percentage = \(\frac{720x}{1160 x}\times 100 = 62.06\%\)
Hence, option E is the correct answer.
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