# Easy Percentages Solved QuestionAptitude Discussion

 Q. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
 ✖ A. 39, 30 ✖ B. 41, 32 ✔ C. 42, 33 ✖ D. 43, 34

Solution:
Option(C) is correct

Let their marks be $(x + 9)$ and $x$.

Then,

$x + 9 = \left(\dfrac{56}{100}\right) \times (x + 9 + x)$

$\Rightarrow 25(x + 9) = 14(2x + 9)$

$\Rightarrow 25x + 25 \times 9 = 28x + 14 \times 9$

$\Rightarrow 9 \times (25-14) = 28x -25x$

$\Rightarrow 9 \times 11 = 3x$

$\Rightarrow 3x = 99$

$\Rightarrow x = 33$

So, their marks are $\textbf{42 and 33}$.

## (3) Comment(s)

Syed
()

one more solution there could be

one number is 9 more than other so 56%-44%=12% which show 9

so we can move to 100% by =100/12*9 = 75 now you can calculate the 56% of 75 that is 42 and 44% that is 33.

Anu
()

There is no need to write equations for such sums . Since 56% is given , we just to calculate total of each option and multiply by 56.

Only a total of 75 will give 56% as an integer hence directly option 3.

Jagdeep Singh
()

Let total marks 100x

One student got - 56x............ (i)

secondt got - 44x..............(ii)

Now,

56x-44x =9 (given)

x = 9/12

put the value of x in eq. i and ii

we get 42 and 33