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 2 of 17.
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)
Komal said:
1 year ago
Let 1st student (A) marks = x.
Therefore,
2nd student (B) marks = x+9.
Now , x+9 = 56% of {(x+9)+(x)}.
=> x+9= 56/100 (x+9+x),
=> x+9 = 0.56(x+9+x),
=> x+9 = 0.56x + 0.56x + 5.04,
=> x+9 = 1.12x + 5.04,
=> x-1.12x = 9 - 5.04,
=> 0.12x = 3.96.
=> x = 3.96/0.12.
=> Marks of A = [x = 33].
Therefore, Marks of B = x+9.
=> 33+9 = 42.
Therefore,
2nd student (B) marks = x+9.
Now , x+9 = 56% of {(x+9)+(x)}.
=> x+9= 56/100 (x+9+x),
=> x+9 = 0.56(x+9+x),
=> x+9 = 0.56x + 0.56x + 5.04,
=> x+9 = 1.12x + 5.04,
=> x-1.12x = 9 - 5.04,
=> 0.12x = 3.96.
=> x = 3.96/0.12.
=> Marks of A = [x = 33].
Therefore, Marks of B = x+9.
=> 33+9 = 42.
(47)
Dr.Mahaveer said:
3 years ago
@All.
Hello.
Simply, the solution is;
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.
Hello.
Simply, the solution is;
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.
(39)
AVINASH KUMAR said:
2 years ago
A and B is there;
A + B = 100.
A = 56,B = 44,A - B = 12.
12%U = 9
12/100U = 9
3/25U = 9
3U = 9
1U = 3
A = 56% = 56/100,14/25U
A = 14U = 14 * 3 = 42
B = 11U = 11 * 3 = 33.
A + B = 100.
A = 56,B = 44,A - B = 12.
12%U = 9
12/100U = 9
3/25U = 9
3U = 9
1U = 3
A = 56% = 56/100,14/25U
A = 14U = 14 * 3 = 42
B = 11U = 11 * 3 = 33.
(37)
Shakthi said:
2 years ago
Let one student obtain x marks.
So, another obtained x+9 marks.
According to the question;
X+9 = 56/100'- (2x+9).
By cross multiplying.
3x = 99.
x = 33.
So, one obtained 33 and another obtained 33 + 9 = 42 as x+9.
So, another obtained x+9 marks.
According to the question;
X+9 = 56/100'- (2x+9).
By cross multiplying.
3x = 99.
x = 33.
So, one obtained 33 and another obtained 33 + 9 = 42 as x+9.
(31)
Deepikaravikumar said:
8 months ago
Still, I can't understand this. Please explain to me.
(25)
Vijay Kumar said:
6 months ago
Total marks = 100%
Obtained marks"×" = 56%
"Y" = 44%.
Here difference is = 12%(it means 9)
So each 1% = 0.75.
10% = 7.5 {100% = 75; 50% = 37.5; 6% = 4.50}
So 56%= 42{ 37.5 + 4.5}.
44% = 33{37.5 - 4.5}.
Obtained marks"×" = 56%
"Y" = 44%.
Here difference is = 12%(it means 9)
So each 1% = 0.75.
10% = 7.5 {100% = 75; 50% = 37.5; 6% = 4.50}
So 56%= 42{ 37.5 + 4.5}.
44% = 33{37.5 - 4.5}.
(25)
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)
Ritesh Rawal said:
3 years ago
Your explanation is simple & clear. Thank you very much @Neelu.
(20)
Anke Lakshmi said:
1 year ago
Let the marks obtained by the students be a and b:
Let a = b + 9 -------->1
a = (56/100) * (a+b) ---------> 2.
From equation 1, equation 2 can be written as;
b + 9 = (0.56) * (b + 9 + b),
b + 9 = (0.56)(2b + 9).
b + 9 = 1.12b + 5.04
9 - 5.04 = 1.12b - b.
3.96 = 0.12b,
b = 3.96/0.12,
b = 33.
Then, a=b+9
a=33+9
a=42
Therefore, marks scored by a and b are 42 and 33.
Let a = b + 9 -------->1
a = (56/100) * (a+b) ---------> 2.
From equation 1, equation 2 can be written as;
b + 9 = (0.56) * (b + 9 + b),
b + 9 = (0.56)(2b + 9).
b + 9 = 1.12b + 5.04
9 - 5.04 = 1.12b - b.
3.96 = 0.12b,
b = 3.96/0.12,
b = 33.
Then, a=b+9
a=33+9
a=42
Therefore, marks scored by a and b are 42 and 33.
(19)
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