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:
173 comments Page 8 of 18.

Prakhar Maheshwari said:   7 years ago
Answer is C.

I have directly used option to get the answer.

So, 42/75*100 =56%; (42+33 =75).

Sai said:   7 years ago
We can easily say by seeing option also for time-saving.

Take total Mark's =100
One's mark is 56 % of sum
So here the sum is 100
56% of 100 is 56
And remaking Mark's is 44
So its in the ratio of 44:56
We can take 11*4:14*4
11x:14x
So the Marks are multiple of 11 and 14.

Thank you.

Vel7 said:   7 years ago
Take option c. 42, 33.
Then 42+33=75 so, 75= 100%.
7.5=10%.
0.75=1%.

We calulate 56% 1%= 0.75.
56%=0.75*56= 42
Difference 9. So 42-9=33.

Kalyan said:   8 years ago
Let both of them be A and B.
Given A= 9+X and B = x.
Also A = 56% of their total marks;
So, A = 56/100 * (9+x+x).

Where A= 9+x.
By equating we can get x value 33.

A = 33+9 = 42.
(1)

Sushmitha said:   8 years ago
25(x+9) = 14(2x+9) how its came?

Please explain this step.

Nishant said:   8 years ago
It's actually very simple!! If 1 student got 56%, then other student % must be 100-56%=44%.

Now 56%- 44%=12%.
12%=9(the difference between the marks of both the students).

You can solve it now!

Waqar said:   8 years ago
Thanks @Ashok.

Jajjo said:   8 years ago
How you can take 25 at lhs and 14 rhs?

Vinaya said:   8 years ago
Thanks @Tukuna.

Lingeswaran said:   8 years ago
Thanks @Aman Tomar.


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