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:
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 7 of 18.
Joseph said:
7 years ago
Let marks be of st A=x, and st B = x+9.
Then,
x+9=56/100(x+x+9).
100(x+9)=56(x+x+9).
100x+900x=56x+56+504,
100X+900= 112x+504,
112x-100x=900-504.
12x=396,
x=396/12,
x=33,
x+9=33+9 = 42.
Then,
x+9=56/100(x+x+9).
100(x+9)=56(x+x+9).
100x+900x=56x+56+504,
100X+900= 112x+504,
112x-100x=900-504.
12x=396,
x=396/12,
x=33,
x+9=33+9 = 42.
Victory said:
7 years ago
How come 25 and 14? Please tell me.
Shweta said:
7 years ago
Best explanation, Thanks @Girish P.
Nagesh v said:
7 years ago
Thank you @Shrikant.
Dsr said:
7 years ago
Good explanation, Thanks all.
Roopsingh said:
7 years ago
How come 56/100? Please explain in detail.
MUKUL said:
7 years ago
Let's take x, x+9; of their marks.
x=44/100*(x+x+9) or -+9=56/100*(x+x+9).
Solve; eq x =33. And another x+9.
33+9=42.
x=44/100*(x+x+9) or -+9=56/100*(x+x+9).
Solve; eq x =33. And another x+9.
33+9=42.
Rohan said:
7 years ago
@Hiren Savalia.
Can you elaborate your method?
Can you elaborate your method?
Girish P said:
7 years ago
Easiest way to solve this is as follows:
The total percentage of marks is 56%,
so divide 56 by 2,
You get 33, 1%=6 marks, the student has scored 9 marks more so,
33+9=42, the other student marks will be 9 marks less so simple,
42-9=33.
ANSWER=42 & 33.
The total percentage of marks is 56%,
so divide 56 by 2,
You get 33, 1%=6 marks, the student has scored 9 marks more so,
33+9=42, the other student marks will be 9 marks less so simple,
42-9=33.
ANSWER=42 & 33.
Vasu said:
7 years ago
How it is 56-44/100=12?
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