Aptitude - Percentage - Discussion

Best explanation, Thanks @Girish P.

How come 25 and 14? Please tell me.

Let marks be of st A=x, and st B = x+9.
Then,

x+9=56/100(x+x+9).
100(x+9)=56(x+x+9).
100x+900x=56x+56+504,
100X+900= 112x+504,
112x-100x=900-504.
12x=396,
x=396/12,
x=33,
x+9=33+9 = 42.

@Vel.

Let x be the student who scores 56% marks.
That is, x = y+9.

X = 56% marks of the sum of their marks.
=> Y + 9 = 56% of sum of their marks --> (i)
And x + y = sum of their marks.

Putting the value of x,
(Y+ 9) + y = sum of their marks
2y + 9 = sum of their marks --> (ii)

Now putting the value of (ii) into (i):
Y + 9 = (56/100) ( 2Y+ 9).
3y = 99,
Y = 33, X = 42.

Can't we assume the mark's of another person :-X & those who have less:-X-9?

Likewise;
(X+X-9)*56/100 = X.

Let student 1 = x.
2nd student secured 9marks more = (x+9).

This marks = 56%(total sum marks).

x+9 = 56/100(x+x+9)
100(x+9) = 56(2x+9)
100x+900 = 112x+504
900-504 = 112x-100x
396 = 12x
x = 396/12=33.
x = 33.

x+9 = 33+9 = 42.
(33, 42).

Why we add x and x+9?

As per my method, the Solution is;
Gap in marks= 9.
Gap in percentage= 12%,
1st student marks = 56% = 56/12 * 9 = 42.
2nd student marks = 44% = 44/12 * 9 = 33.

@Anthea. You do by simply dividing 56/100 by least divisible 2 then you got answer 14/25. Hope it's clear !

A simple Answer.

Here given 2 students (assume A&B)
* One of the student(A) marks is = 56%
* And remaing student (B) marks is =44% (100%-56%)
Then, the gap between them is=12% (56-44).

So we know that one person got more marks than another person = 9mrks.

So,
A.
12%-----------> 9mrks.
56%----------> ?
Then {56*9/12}.
Ans 42.

Similarly for B;
12%-------> 9mrks
44%-------> ?
=> {44*9/12} = 33.