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 8 of 17.
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.
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.
NJK said:
1 decade ago
Suraj and Radhika's answers are super. But Suraj answer is little confusing.
Aravind said:
1 decade ago
Why we put equate here between x+9 = 56/100* (x+9+x).
Sri Ram said:
1 decade ago
Took 25 minutes to crack the stupid nut. The commonly confused place is (56%o of TOTAL marks so it is written as (x+9) +x where x is marks scored by one person and x+9 is marks scored by the later.
Aniket said:
1 decade ago
I need help on following question:
A candidate who gets 50% of total marks gets 100 marks more than the pass marks. Another candidate who gets 70% of the first candidates marks failed by 50 marks. How much percent a student needs to pass the exam?
A candidate who gets 50% of total marks gets 100 marks more than the pass marks. Another candidate who gets 70% of the first candidates marks failed by 50 marks. How much percent a student needs to pass the exam?
SpT said:
1 decade ago
@Aniket.
Let total mark be x.
50% of x--> (50/100)*x = (1/2)x
Therefore (1/2)x-100 = P --(1). Here P is the pass mark.
70% of 1st candidate marks--> 70% of (1/2)x
(7/20)x.
Therefore (7/20)x+50 = P --(2).
Then equating (1) and (2) we get,
(1/2)x-100 = (7/20)x+50 ==> 3x = 3000.
x = 1000.
Then 50% of x
(50/100)*1000 = 500.
500-100 = 400;
400 is the Answer.
Let total mark be x.
50% of x--> (50/100)*x = (1/2)x
Therefore (1/2)x-100 = P --(1). Here P is the pass mark.
70% of 1st candidate marks--> 70% of (1/2)x
(7/20)x.
Therefore (7/20)x+50 = P --(2).
Then equating (1) and (2) we get,
(1/2)x-100 = (7/20)x+50 ==> 3x = 3000.
x = 1000.
Then 50% of x
(50/100)*1000 = 500.
500-100 = 400;
400 is the Answer.
Jvn said:
1 decade ago
Percentage diff is 12%.
So 12% is equal to 9 marks.
So for 56% whats the marks(x).
We get x = 56*9/12.
= 42.
So 12% is equal to 9 marks.
So for 56% whats the marks(x).
We get x = 56*9/12.
= 42.
Kumudha said:
1 decade ago
I couldn't understand the question. In question it is given. No.of student = 2 and one secured 9 marks more than other. Upto this clear I'm. What's the meaning of, 56% of their sum of their mark.
JENDA FAYAZ said:
1 decade ago
The diff two students marks % is 9.
56%-44% = 9 marks.
Then 12% = 9 marks.
Then 100% = 9*100/12 = 75 marks.
You have to check the answer is 42,33.
56%-44% = 9 marks.
Then 12% = 9 marks.
Then 100% = 9*100/12 = 75 marks.
You have to check the answer is 42,33.
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.
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.
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