Question

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.

• A. 68.25%
• B. 66.66%
• C. 56.56%
• D. 63.72%
• E. 62.06

Answer 1

Answer: (E)

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.