Aptitude - Percentage - Discussion

Total 100%.

A have 56%.

i.e B = 100%-56% = 44%.
A have 9 mark more than B,
i.e A - B=9.
56% - 44%=9,
12% = 9,
4% = 3.

A = 56% / 4% = 14%.
14 * 3 = 42.
B = 44 %/4%
B = 11 * 3 = 33.

so, A = 42, B = 33.

As one got X marks another X+9 so, the sum of both is 2x+9.

Let one student obtain x marks.

So, another obtained x+9 marks.

According to question.

X+9 = 56/100'- (2x+9).

By cross multiplying.

3x = 99.
x = 33.

So, one obtained 33 and another obtained 33 + 9 = 42 as x+9.

Simple way using ratio.

56 : 44.
i.e 14:11.
33 is divisible by 11 so the answer is C.

Thanks @Diksha.

Understand the statement :

One of them secured 9 marks more than the other.
So it is (9 + x) , right ?

Next part is " 56% of the sum of their marks"
So here, sum is (x + 9 + x).
And its 56% is (56/100) * (x + 9 + x).

So now, the important part is " .. 9 marks more than the other AND HIS MARKS WAS 56% of ... "

In this line, it says HIS Marks, so whose marks it is ?? It is the marks of ( 9 + x ).
(9 + x ) is the 56% of their sum.

Because of this line, we have equal to symbol in the equation,

So the equation is : (9 + x) = (56/100) * (x + 9 + x).

@Pawan.

From the QN, Sum of their mark % is 56%.

@Vel.

Two students x and y were appear then one get more than 9marks from the other so we take that one member as y as an assumption so y =x+9.

Here we have x and x+9 as two members. Then that x+9 's mark is 56%of sum of their marks. Hence we take (x+9) = 56% (x+ (x+9) ) = option c.

Answer = C.

1st students marks percentage = 56%,
2nd percentage= 100-56 = 44%,
Difference between marks = 9,
Difference between percentage=56-44 = 12%.
9 = 12% ->1 = 0.75 ->100 = 75(total marks).
56% = 0.75X56 = 42 ,44% = 0.75X44 = 33.

2 numbers are (x) and (x+9).
Now,
(x+9) =56% of (x+x+9),
=(2x+9)56%.

Now open brackets
x+9 = 112x/100 + 504/100,
9 - (504/100 ) = (112x-100x)/100,
(900-504)/100= 12x /100,
396/100 = 12x/100.
>> 396 = 12x.
396/12 = x
33=x,
So 2 digits are:
x = 33,
x+9 = 42.