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 3 of 17.
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.
Mohan said:
9 years ago
To my understanding.
Student 2 score = x.
Student 1 score 9 marks more than student 2 = x + 9 which is 56 percent of the total marks(x + x + 9) right.
x + 9 = 56%(x + x + 9).
Solving the above equation we will get x = 33 which is student 2 mark.
Student 1 has scored 9 marks more than student 2 so 33 + 9 = 42.
Student 2 score = x.
Student 1 score 9 marks more than student 2 = x + 9 which is 56 percent of the total marks(x + x + 9) right.
x + 9 = 56%(x + x + 9).
Solving the above equation we will get x = 33 which is student 2 mark.
Student 1 has scored 9 marks more than student 2 so 33 + 9 = 42.
Sakshi Shinde said:
3 months ago
@All.
Let the marks of the first student be x.
Then, the other student got x + 9 marks.
Total marks = x + (x + 9) = 2x + 9,
Given: x + 9 = 56% of (2x + 9),
=> x + 9 = 0.56(2x + 9),
=> x + 9 = 1.12x + 5.04,
=> -0.12x = -3.96,
=> x = 33,
So, the other student got 33 + 9 = 42.
Answer: 33 and 42 marks.
Let the marks of the first student be x.
Then, the other student got x + 9 marks.
Total marks = x + (x + 9) = 2x + 9,
Given: x + 9 = 56% of (2x + 9),
=> x + 9 = 0.56(2x + 9),
=> x + 9 = 1.12x + 5.04,
=> -0.12x = -3.96,
=> x = 33,
So, the other student got 33 + 9 = 42.
Answer: 33 and 42 marks.
(21)
Jeni said:
2 months ago
Let the marks of the first student be x.
Then, the other student got x + 9 marks.
Total marks = x + (x + 9) = 2x + 9,
Given: x + 9 = 56% of (2x + 9),
=> x + 9 = 0.56(2x + 9),
=> x + 9 = 1.12x + 5.04,
=> -0.12x = -3.96,
=> x = 33,
So, the other student got 33 + 9 = 42.
Answer: 33 and 42 marks.
Then, the other student got x + 9 marks.
Total marks = x + (x + 9) = 2x + 9,
Given: x + 9 = 56% of (2x + 9),
=> x + 9 = 0.56(2x + 9),
=> x + 9 = 1.12x + 5.04,
=> -0.12x = -3.96,
=> x = 33,
So, the other student got 33 + 9 = 42.
Answer: 33 and 42 marks.
(18)
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.
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.
Aparna said:
1 decade ago
Suppose 1 students score is 'x',then others will be 'x+9'
Now the next information is given about the student whose score is 'x+9'
Whenever in the question 'and'/'is' given put '=' sign
56% means 56/100 ,of means put multiplication sign
and sum of the marks means(x+x+9)
Now just solve the equation.
Now the next information is given about the student whose score is 'x+9'
Whenever in the question 'and'/'is' given put '=' sign
56% means 56/100 ,of means put multiplication sign
and sum of the marks means(x+x+9)
Now just solve the equation.
(1)
Being Sunil said:
1 decade ago
Guys, its simple... just equalize 1% as under:
=> (x+9)/56 = x/44.
=> (x+9)/14 = x/11 (denominator divided by 4 to simplify).
=> 14x-11x = 99 (cross multiply & side changed for x's value).
=> x = 99/3 = 33 (one score).
Other score = 33 + 9 = 42.
So answer is (C) 42,33... :-).
=> (x+9)/56 = x/44.
=> (x+9)/14 = x/11 (denominator divided by 4 to simplify).
=> 14x-11x = 99 (cross multiply & side changed for x's value).
=> x = 99/3 = 33 (one score).
Other score = 33 + 9 = 42.
So answer is (C) 42,33... :-).
Naveen Thonpunoori said:
2 years ago
Let's say X, and Y are students.
We can say Y got 44% of the total marks since X got 56%.
Percentage difference b/w X, Y is 56%- 44% = 12%.
X got 9 marks more than Y, So 9 marks are equal to 12%.
So, 1% of marks equal to, 9/12 = 0.75,
X marks are, 0.75 * 56 = 42.
Y marks are, 0.75 * 44 = 33.
We can say Y got 44% of the total marks since X got 56%.
Percentage difference b/w X, Y is 56%- 44% = 12%.
X got 9 marks more than Y, So 9 marks are equal to 12%.
So, 1% of marks equal to, 9/12 = 0.75,
X marks are, 0.75 * 56 = 42.
Y marks are, 0.75 * 44 = 33.
(62)
Nisma said:
5 years ago
2 numbers are (x) and (x+9).
Now,
(x+9) =56% of (x+x+9),
=(2x+9)56%.
Now open brackets
x+9 = 112x/100 + 504/100,
9 - (504/100 ) = (112x-100x)/100,
(900-504)/100= 12x /100,
396/100 = 12x/100.
>> 396 = 12x.
396/12 = x
33=x,
So 2 digits are:
x = 33,
x+9 = 42.
Now,
(x+9) =56% of (x+x+9),
=(2x+9)56%.
Now open brackets
x+9 = 112x/100 + 504/100,
9 - (504/100 ) = (112x-100x)/100,
(900-504)/100= 12x /100,
396/100 = 12x/100.
>> 396 = 12x.
396/12 = x
33=x,
So 2 digits are:
x = 33,
x+9 = 42.
(1)
Anomie said:
5 years ago
x+9=56/100(2x+9),
How did you reach to this step?
How can you equate both student marks although they have a difference of 9 marks
L.H.S = marks of 2nd student
R.H.S = marks in the percentage of 1st student out of the total sum of the marks scored by both of them.
Please anyone explain it.
How did you reach to this step?
How can you equate both student marks although they have a difference of 9 marks
L.H.S = marks of 2nd student
R.H.S = marks in the percentage of 1st student out of the total sum of the marks scored by both of them.
Please anyone explain it.
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