Aptitude - Percentage - Discussion

Discussion Forum : Percentage - General Questions (Q.No. 2)
2.
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:
39, 30
41, 32
42, 33
43, 34
Answer: Option
Explanation:

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

Then, x + 9 = 56 (x + 9 + x)
100

25(x + 9) = 14(2x + 9)

3x = 99

x = 33

So, their marks are 42 and 33.

Discussion:
170 comments Page 5 of 17.

Shrikant sir said:   1 decade ago
Just observe the question, you will find.

We assume total of both marks is 100 %.

1) 100-56 = 44.
2) 56-44 = 12.

3) It means 12% = 9 marks.
Than 44% = ?

You will get 33 marks.
4) Same way.

12% = 9 marks.
56% = ?

You will get 42 marks.

Sailu said:   6 years ago
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).

Arshad said:   9 years ago
Try by Back Solving Method:

Lets try c: 42, 33 so it means total marks are : 42 + 33 = 75 Marks.

Let's try to find out 56 % from marks given in option C.

42 * 100/75 = 56 %

33 * 100/75 = 44 %.

So option C is the correct answer .

Anusha said:   1 decade ago
Solution: Let 1st person marks = x.
2nd person marks = x-9.

From question 1st person marks is 56% of sum of their marks.

x = 56/100(x+x-9).

x = 56/100(2x-9).

x = (56*2x)/100 - (56*9)/100.

x((56*20/100)-1) = (56*9)/100.

x = 42.

Vijay Kumar said:   6 months ago
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}.
(25)

Rahul said:   1 decade ago
This is how I understood :

x = y+9.
x = 0.56(x+y).

x-y = 9.
x-0.56x-0.56y = 0.

0.44x-0.56y = 0.
x = (0.56/0.44)y.
x = (56/44)y.

(56/44)y - (44/44)y = 9.
(12/44)y = 9.
y = 9*44/12.
y = 3/4*44=3*11=33.
y = 33.

x = 33+9.
x = 42.

Hemant singh said:   1 decade ago
By question let first scoted x marks

then 2nd x+9
sum of ther marks {x+x+9}
not by ques the student who ha s scored x+9 marks = su m of marks of both students ie x+9=2x+9 of 56%
solve it now

Vino said:   1 decade ago
No confusion. Listen.
Assume total marks will be 100,
One student=x+9.

Another=x.
So,
For 100 he scored 56, so what marks for them so (x+9+x).
56/100* (x+9+x).

Now x+9 is student 1: So equate,

x+9 = 56/100* (x+9+x).

Fahad khan said:   1 decade ago
@Sarwat.

First assume that the number is x so.
Their 30% is x*30/100 which is equal to 180 i.e. x*30/100 = 180.

So x=60 again 3 times of that number is 180 and their 15% is equal to 180*15/100 = 27 that's the answer.

Soham Garai said:   4 years ago
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.
(4)


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