Aptitude - Percentage - Discussion

VERY EASY:

Let total marks be = 100.
F =56%.
Therefore,
X = 100-56 = 44,
44-9 = 33,
33+9 = 42.

Total marks = 100%
Obtained marks"×" = 56%
"Y" = 44%.
Here difference is = 12%(it means 9)
So each 1% = 0.75.
10% = 7.5 {100% = 75; 50% = 37.5; 6% = 4.50}
So 56%= 42{ 37.5 + 4.5}.
44% = 33{37.5 - 4.5}.

@All.
I have a solution.
56% = 56\100 = 14\25.
Let's take the total mark = 25.
The higher scorer's mark = 14.
The lower scorer's mark =25 - 14 = 11
The difference given in the question is 9 so,
14parts - 11parts = 9,
value of 3 parts = 9,
so the value of 1 part = 9\3= 3.

Now we have to calculate higher scorer's marks = 14 * 3 = 42.
Then we have to calculate the lower scorer's marks = 11 * 3 = 33.

Still, I can't understand this. Please explain to me.

Let's say two students are X and Y.

For instance, assume that the total mark was 100. Meaning X+Y = 100.

Now, if X gets 56% of the sum of their marks which was 100, then Y must have the remaining percentage of the marks, that's 44%.

Now, compare the difference between the marks that they've obtained and the percentage that they have.

So, difference in percentage = X-Y = 56-44 = 12%
And, the difference in marks = X-Y = 9 (Because in the question it's given that one has 9 marks more than the other.)

Now, we've to find the value of 1 Unit. If you can find that you can easily solve the question.

Comparing their marks and their percentages,

12% = 9 marks.
1 Unit = 9/12.
1 Unit = 0.75.

Now multiply the value of 1 Unit by the percentage they've got.

X = 56*0.75 = 42.
Y = 44*0.75 = 33.
And their difference is also 9.

Let 1st student (A) marks = x.
Therefore,
2nd student (B) marks = x+9.
Now , x+9 = 56% of {(x+9)+(x)}.
=> x+9= 56/100 (x+9+x),
=> x+9 = 0.56(x+9+x),
=> x+9 = 0.56x + 0.56x + 5.04,
=> x+9 = 1.12x + 5.04,
=> x-1.12x = 9 - 5.04,
=> 0.12x = 3.96.
=> x = 3.96/0.12.
=> Marks of A = [x = 33].
Therefore, Marks of B = x+9.
=> 33+9 = 42.

Let the marks obtained by the students be a and b:

Let a = b + 9 -------->1
a = (56/100) * (a+b) ---------> 2.

From equation 1, equation 2 can be written as;

b + 9 = (0.56) * (b + 9 + b),
b + 9 = (0.56)(2b + 9).
b + 9 = 1.12b + 5.04
9 - 5.04 = 1.12b - b.
3.96 = 0.12b,
b = 3.96/0.12,
b = 33.

Then, a=b+9
a=33+9
a=42

Therefore, marks scored by a and b are 42 and 33.

Obtained mark % = 56.
Total mark %= 100.
Remaining % = 100 - 56 = 44%.
Difference % = 56-44 = 12%.
Then the difference mark 9.
So,
12 % = 9.
56% = x.
56*9/12 = 42.
42-9 = 33.

Let total marks be =100.
F =56%.

Therefore,
X = 100-56 = 44,
44-9 = 33,
33+9 = 42.

@All

Here, some people doing the wrong calculation 44-9=35 not 33. Please correct it.