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)

62.06

Answer 2

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.