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 14 of 18.
Hariharan E said:
8 years ago
Please explain how to get the answer?
Vishwa said:
8 years ago
@Hariharan.
Try to understand.
Assume the first person as A second person as B.
A = 56%.
B = 44%.
Now just take the difference between their%.
A-B=12%.
In the problem itself, they have given that A is 9 more than B.
This 9 marks is nothing but that difference 12%.
Hence 12%-9.
56%-?=42=A.
B=A-9=42-9=33.
Try to understand.
Assume the first person as A second person as B.
A = 56%.
B = 44%.
Now just take the difference between their%.
A-B=12%.
In the problem itself, they have given that A is 9 more than B.
This 9 marks is nothing but that difference 12%.
Hence 12%-9.
56%-?=42=A.
B=A-9=42-9=33.
Shivashankar said:
8 years ago
Thank you @Aparna.
Asif husain jafri said:
8 years ago
First, sum all the option and find its 56%.
in option 42+33=75.
75 of 56% =42.
42-9=33.
in option 42+33=75.
75 of 56% =42.
42-9=33.
Bismah hanif said:
8 years ago
How? Please explain in detail. @Asif Husain Jafri.
Aman Tomar said:
8 years ago
Hello Guys
Simple method.
A=x+9,
B=x,
x+9=56%(x+x+9),
x+9=56/100(2x+9),
x=33,
A=33+9=42.
B=33.
Simple method.
A=x+9,
B=x,
x+9=56%(x+x+9),
x+9=56/100(2x+9),
x=33,
A=33+9=42.
B=33.
Gyanajyoti said:
8 years ago
Bdw why (x+9+w) became multiple with 56/100?
Lingeswaran said:
8 years ago
Thanks @Aman Tomar.
Vinaya said:
8 years ago
Thanks @Tukuna.
Jajjo said:
8 years ago
How you can take 25 at lhs and 14 rhs?
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