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:
170 comments Page 12 of 17.
ISHITA said:
9 years ago
I can not understand the question. Please someone explain me.
Sujay said:
9 years ago
Let their marks be(x + 9)and x.
Then, x+9 =56/100(x + 9 + x),
= 25(x + 9)= 14(2x + 9),
= 3x = 99,
x = 33.
So,there marks are 42 and 33.
Then, x+9 =56/100(x + 9 + x),
= 25(x + 9)= 14(2x + 9),
= 3x = 99,
x = 33.
So,there marks are 42 and 33.
Sunil said:
9 years ago
What calculation do for this 25(x + 9) = 14(2x + 9)?
Sai said:
9 years ago
Please explain me in a simple way.
Sangay Tenzin said:
9 years ago
25(x + 9) = 14(2x +9).
From where this 25 & 14 come from?
Could you explain me clearly, I am confuse with that.
From where this 25 & 14 come from?
Could you explain me clearly, I am confuse with that.
Sangay Tenzin said:
9 years ago
When I divide 56/100 by 4 I got 14/25. But I did not know from where this 4 come from?
So please help me to get it.
So please help me to get it.
Umang gandhi said:
9 years ago
Let two student a&b.
a scores 9 marks more than b and have 56%.
a = 56%.
b = 100-56 = 44%,
56% = 9 + 44%,
12% = 9.
Let total marks be x.
12%X = 9,
(12/100) * x = 9,
x = 75.
Sum of both a and b = 75.
We check option which satisfies above condition = 42, 33.
a scores 9 marks more than b and have 56%.
a = 56%.
b = 100-56 = 44%,
56% = 9 + 44%,
12% = 9.
Let total marks be x.
12%X = 9,
(12/100) * x = 9,
x = 75.
Sum of both a and b = 75.
We check option which satisfies above condition = 42, 33.
Borra vamsi krishna said:
9 years ago
Please explain in simple way.
Megha said:
9 years ago
In an exam, 5%of the application were found ineligible &85% of the eligible candidates belongs to general category. If 4275 eligible candidates belonged to other categories, then how much candidates applied for the exam?
Please answer this question.
Please answer this question.
Dinesh D said:
9 years ago
@Aniket.
Let the total number of candidates who applied be x.
Ineligible candidates = 0.05x.
Eligible candidates = x-0.05x = 0.95x.
General category candidates who are eligible = (0.95x)*0.85 = 0.8075x.
Other category candidates who are eligible = 4275.
Then x = 0.05x + 4275+0.8075x.
= 0.8575x + 4275.
0.1425x = 4275.
x = 30000.
Let the total number of candidates who applied be x.
Ineligible candidates = 0.05x.
Eligible candidates = x-0.05x = 0.95x.
General category candidates who are eligible = (0.95x)*0.85 = 0.8075x.
Other category candidates who are eligible = 4275.
Then x = 0.05x + 4275+0.8075x.
= 0.8575x + 4275.
0.1425x = 4275.
x = 30000.
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