Carcass wrote:
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:
A. 39, 30
B. 41, 32
C. 42, 33
D. 43, 34
E. 41, 36
Let x = the smaller score
So, x + 9 = the larger score
The larger score is 56% of the sum of the two scores.
We can write: x + 9 = 56% of [x + (x + 9)]
In other words: x + 9 = 0.56(2x + 9)
Expand: x + 9 = 1.12x + 5.04
Subtract 5.04 from both sides: x + 3.96 = 1.12x
Subtract x from both sides: 3.96 = 0.12x
Solve: x = 3.96/0.12 = 396/12 = 33
So 33 is the smaller score, and 42 is the larger score.
Answer: C