A student took five papers in an examination, where the full marks were the same for each paper. His marks in this papers were in the proportion of 6:7:8:9:10. In all papers together, the candidate obtain 60% of the total marks. Then the number of papers in which he got more than 50% marks is,

A. 2
B. 3
C. 4
D. 5
E. 1

The OA is C.

I'm confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

Hi LUANDATO,
Lets take a look at your question.

The marks of a student in five papers are in the proportion:
$$6:7:8:9:10$$

Sum of the proportions = 6 + 7 + 8 + 9 + 10 = 40

In all papers together, the candidate obtain 60% of the total marks.
Let the total marks be x, then,
$$x\left(60\%\right)=40$$
$$x\left(\frac{60}{100}\right)=40$$
$$x=40\times\frac{100}{60}\approx66.67$$

So the total marks are 66.67
Since there were five papers, where the full marks were the same for each paper. Hence, we can find the total marks for each of the five papers.
Total marks of each paper can be calculated as:
$$\frac{66.67}{5}\approx13.33$$

We are asked to find the number of papers in which he got more than 50% marks.
So we will first find 50% for each paper.
Total marks for each paper = 13.33
50% marks for each paper = 13.33/2 = 6.67

Now, we can easily see from the given proportion that how many of the marks are greater than 6.67.
$$6:7:8:9:10$$
We can see that four of the marks i.e. 7, 8, 9 and 10 are greater than 6.67 i.e. 50% marks.

Therefore, Option C is correct.

Hope it helps.
I am available if you'd like any follow up.