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 16 of 17.
Shweta said:
7 years ago
Best explanation, Thanks @Girish P.
Victory said:
6 years ago
How come 25 and 14? Please tell me.
Joseph said:
6 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.
Akshita said:
6 years ago
@Vel.
Let x be the student who scores 56% marks.
That is, x = y+9.
X = 56% marks of the sum of their marks.
=> Y + 9 = 56% of sum of their marks --> (i)
And x + y = sum of their marks.
Putting the value of x,
(Y+ 9) + y = sum of their marks
2y + 9 = sum of their marks --> (ii)
Now putting the value of (ii) into (i):
Y + 9 = (56/100) ( 2Y+ 9).
3y = 99,
Y = 33, X = 42.
Let x be the student who scores 56% marks.
That is, x = y+9.
X = 56% marks of the sum of their marks.
=> Y + 9 = 56% of sum of their marks --> (i)
And x + y = sum of their marks.
Putting the value of x,
(Y+ 9) + y = sum of their marks
2y + 9 = sum of their marks --> (ii)
Now putting the value of (ii) into (i):
Y + 9 = (56/100) ( 2Y+ 9).
3y = 99,
Y = 33, X = 42.
Amit katnoria said:
6 years ago
Can't we assume the mark's of another person :-X & those who have less:-X-9?
Likewise;
(X+X-9)*56/100 = X.
Likewise;
(X+X-9)*56/100 = X.
Sailu said:
6 years ago
Let student 1 = x.
2nd student secured 9marks more = (x+9).
This marks = 56%(total sum marks).
x+9 = 56/100(x+x+9)
100(x+9) = 56(2x+9)
100x+900 = 112x+504
900-504 = 112x-100x
396 = 12x
x = 396/12=33.
x = 33.
x+9 = 33+9 = 42.
(33, 42).
2nd student secured 9marks more = (x+9).
This marks = 56%(total sum marks).
x+9 = 56/100(x+x+9)
100(x+9) = 56(2x+9)
100x+900 = 112x+504
900-504 = 112x-100x
396 = 12x
x = 396/12=33.
x = 33.
x+9 = 33+9 = 42.
(33, 42).
Ramana said:
6 years ago
Why we add x and x+9?
Ashwin said:
5 years ago
@Sarwat
30%of number 180,
30/100*y=180.
y=600.
15%of3(number) =? .
15/100*3(600) =2700.
30%of number 180,
30/100*y=180.
y=600.
15%of3(number) =? .
15/100*3(600) =2700.
Ruhi said:
5 years ago
Total 100%.
One got 56 then the other got 44.
Diff 12%=9 (one got 9 marks more),
56%=42.
And 44%=33.
One got 56 then the other got 44.
Diff 12%=9 (one got 9 marks more),
56%=42.
And 44%=33.
SANJAY said:
5 years ago
@All.
Simply, the solution is;
56 % and balance is 44, Right!
56/100 = 9,
56 - 44 / 100 = 9,
It will give you 75.
THEN,
56/100 X 75 = 42.
44/100 X 75 = 33.
Simply, the solution is;
56 % and balance is 44, Right!
56/100 = 9,
56 - 44 / 100 = 9,
It will give you 75.
THEN,
56/100 X 75 = 42.
44/100 X 75 = 33.
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