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.
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.
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)
Vignseh said:
1 decade ago
Let you considered the to students are A and B.
The mark of A is = x.
Then the mark of B is = x+9.
And the B scored 56% of sum of their total marks of both A and B.
So the mark of B can said by,
B = 56% of (A+B).
So,
x+9 = 56/100 (x+x+9).
Now you solve the equation and find the x value.
The x value is equal to A mark.
And add the 9 with x value for the B's mark.
The mark of A is = x.
Then the mark of B is = x+9.
And the B scored 56% of sum of their total marks of both A and B.
So the mark of B can said by,
B = 56% of (A+B).
So,
x+9 = 56/100 (x+x+9).
Now you solve the equation and find the x value.
The x value is equal to A mark.
And add the 9 with x value for the B's mark.
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)
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.
Anju said:
5 years ago
Listen, 56% of SUM OF THEIR MARKS is equal to 2nd person mark.
So, we can simplify 56/100.
By simplifying it we get 14/25.
So we can say that mark of 2nd person is 14 and the first one is 11(we get 11 by subtracting 25-14).
We know that 14-11 is 9 (it's already given in the answer).
That is 3=9.
So, 1=3.
Therefore 14*3 = 42 and 11*3 = 33.
So, we can simplify 56/100.
By simplifying it we get 14/25.
So we can say that mark of 2nd person is 14 and the first one is 11(we get 11 by subtracting 25-14).
We know that 14-11 is 9 (it's already given in the answer).
That is 3=9.
So, 1=3.
Therefore 14*3 = 42 and 11*3 = 33.
Hiren savalia said:
1 decade ago
Go by option u will get answer easily ...
Let me choose 42 & 33
As all answers have difference of 9 so next step is to check 56 % condition for this answer ....
Now 56 % of sum of marks means 75 is 100 % then 42 is how much ?
calculate & u will get 56 % so u will come to know that the answer that u have chosen is d ri8 1 ... enjy :)
Let me choose 42 & 33
As all answers have difference of 9 so next step is to check 56 % condition for this answer ....
Now 56 % of sum of marks means 75 is 100 % then 42 is how much ?
calculate & u will get 56 % so u will come to know that the answer that u have chosen is d ri8 1 ... enjy :)
(2)
Neelu said:
1 decade ago
Let mark of person A = x.
Mark of person B = x+9.
Total marks secured by both A & B = x+(x+9).
Since %mark of person B = 56/100(x+(x+9)).
Therefore x+9 = 56/100(x+(x+9)).
x+9 = 14/25(2x+9).
25(x+9) = 14(2x+9).
25x+225 = 28x + 126.
28x-25x = 225 - 126.
3x = 99.
x = 99/3.
x = 33.
Hence mark of person A = 33.
Mark of person B = 33+9 = 42.
Mark of person B = x+9.
Total marks secured by both A & B = x+(x+9).
Since %mark of person B = 56/100(x+(x+9)).
Therefore x+9 = 56/100(x+(x+9)).
x+9 = 14/25(2x+9).
25(x+9) = 14(2x+9).
25x+225 = 28x + 126.
28x-25x = 225 - 126.
3x = 99.
x = 99/3.
x = 33.
Hence mark of person A = 33.
Mark of person B = 33+9 = 42.
(1)
Shahira said:
1 month ago
2 students appeared for the exam.
1 -> 56% and 9 mark more than others,
1 -> x+9.
2 -> x.
x+9 = 56% of (x+9 +x),
x+9 = 56/100 (2x+9),
100(x+9) = 56 (2x + 9),
100x + 900 = 112x + 504,
100x - 112x + 900 - 504 = 0 .
-12x + 396 = 0,
12x = 396.
x = 396/12.
x = 33.
1 -> x+9 = 33+9 = 42.
2-> 33.
So, the answer is 42 and 33.
1 -> 56% and 9 mark more than others,
1 -> x+9.
2 -> x.
x+9 = 56% of (x+9 +x),
x+9 = 56/100 (2x+9),
100(x+9) = 56 (2x + 9),
100x + 900 = 112x + 504,
100x - 112x + 900 - 504 = 0 .
-12x + 396 = 0,
12x = 396.
x = 396/12.
x = 33.
1 -> x+9 = 33+9 = 42.
2-> 33.
So, the answer is 42 and 33.
(17)
Uto hussain said:
9 years ago
Let's say 'x' is the mark of the first person.
Then from the question, the mark of a 2nd person is 'x + 9'.
Since the percentage of the 2nd person is given i.e. 56%.
We can solve the x value using the below formula.
Per% of 2nd person = mark of 2nd person * 100/ (total marks i.e x + x + 9).
56 = ((x + 9) *100) / (x + x + 9).
Then from the question, the mark of a 2nd person is 'x + 9'.
Since the percentage of the 2nd person is given i.e. 56%.
We can solve the x value using the below formula.
Per% of 2nd person = mark of 2nd person * 100/ (total marks i.e x + x + 9).
56 = ((x + 9) *100) / (x + x + 9).
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