Aptitude - Percentage - Discussion

Total marks = 100%.
X = 56%.
Y = 100-56.
= 44%.

Now the difference marks between X and Y.
= (56-44)%,
= 12%.

Now,
12% = 9,
1% = 9÷12,
56%= 9÷12 × 56,
= 42.

Next;
Y = 44%.
12% = 9.
1% = 9÷12,
44%= 9÷12 × 44,
= 33.

25(x+9) = 14(2x+9),
25x + 234 = 28x + 126,
25x - 28x = 126 - 234,
3x = 108,
x = 108/3,
x = 33,
y = 33 + 9 = 42.

2 students appeared for the exam.
1 -> 56% and 9 mark more than others,
1 -> x+9.
2 -> x.
x+9 = 56% of (x+9 +x),
x+9 = 56/100 (2x+9),
100(x+9) = 56 (2x + 9),
100x + 900 = 112x + 504,
100x - 112x + 900 - 504 = 0 .
-12x + 396 = 0,
12x = 396.
x = 396/12.
x = 33.

1 -> x+9 = 33+9 = 42.
2-> 33.
So, the answer is 42 and 33.

Let the marks of the first student be x.

Then, the other student got x + 9 marks.
Total marks = x + (x + 9) = 2x + 9,
Given: x + 9 = 56% of (2x + 9),
=> x + 9 = 0.56(2x + 9),
=> x + 9 = 1.12x + 5.04,
=> -0.12x = -3.96,
=> x = 33,

So, the other student got 33 + 9 = 42.
Answer: 33 and 42 marks.

@All.
Let the marks of the first student be x.
Then, the other student got x + 9 marks.
Total marks = x + (x + 9) = 2x + 9,
Given: x + 9 = 56% of (2x + 9),
=> x + 9 = 0.56(2x + 9),
=> x + 9 = 1.12x + 5.04,
=> -0.12x = -3.96,
=> x = 33,
So, the other student got 33 + 9 = 42.
Answer: 33 and 42 marks.

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.