Two students appeared for an examination. One of them secured 19 marks more than the other and his marks were 60% of the sum of their marks. The marks obtained by them are:
- A78 and 59
- B57 and 38
- C45 and 26
- D99 and 80
Solution & Step-by-step Explanation
Let the marks of the second student be x.
Then, the marks of the first student who scored more is x+19.
The sum of their marks is:
Sum=x+(x+19)=2x+19
According to the given condition, the first student's marks are 60% of the sum:
x+19=60% of (2x+19)
x+19=
5
3
(2x+19)
Multiply by 5 to clear the fraction:
5(x+19)=3(2x+19)
5x+95=6x+57
95−57=6x−55x
x=38
So, the marks obtained by the two students are:
Second student: x=38
First student: x+19=38+19=57
Thus, the marks are 57 and 38.
Then, the marks of the first student who scored more is x+19.
The sum of their marks is:
Sum=x+(x+19)=2x+19
According to the given condition, the first student's marks are 60% of the sum:
x+19=60% of (2x+19)
x+19=
5
3
(2x+19)
Multiply by 5 to clear the fraction:
5(x+19)=3(2x+19)
5x+95=6x+57
95−57=6x−55x
x=38
So, the marks obtained by the two students are:
Second student: x=38
First student: x+19=38+19=57
Thus, the marks are 57 and 38.