Question

A student took five papers in an examination where the full marks were the same for each paper and these marks in these papers were in the proportion $6:7:8:9:10$. The candidate obtained $60\%$ of the total marks in all the papers together. Find the number of papers in which he got more than $50\%$ marks?
(a) 1
(b) 3
(c) 4
(d) 5

Answer 1

Hint: We start solving the problem by assigning the variable for the full marks in each subject. We then take the variables that represent the marks in all the papers following the given proportion. We then use the given condition that the candidate obtained $60\%$ of the total marks in all the papers together to get the relation between two assumed variables. We then make the necessary calculations and check the given condition to get the required answer.

Complete step-by-step solution:
According to the problem, we have a student who took five papers and scored $60\%$ together in all the papers together. We need to find the number of papers in which he got more than $50\%$ of the marks if full marks are the same for every paper and marks is in the ratio of $6:7:8:9:10$.
Let us assume the full marks in each paper be y. So, there are 5 subjects which makes the total marks in all subjects together as $5y$ marks. Let us find the $60\%$ of this $5y$ marks.
We know that $x\%$ of y is defined as $\dfrac{x}{100}\times y$. So, we get $60\%$ of $5y$ as $\dfrac{60}{100}\times 5y$.
$\Rightarrow 60\%$ of $5y$ = $\dfrac{60}{20}\times y$.
 $\Rightarrow 60\%$ of $5y$ = $3y$.
Since the marks in all subjects are in the ratio $6:7:8:9:10$, we take the marks at $6x$, $7x$, $8x$, $9x$, and $10x$.
According to the problem, we have given that the student scored a total of $3y$ marks. So, we get$6x+7x+8x+9x+10x=3y$.
$\Rightarrow 40x=3y$.
$\Rightarrow x=\dfrac{3y}{40}$.
Let us find the values of $6x$, $7x$, $8x$, $9x$, and $10x$.
So, we have $6x=6\times \dfrac{3y}{40}$.
$\Rightarrow 6x=\dfrac{18y}{40}$.
$\Rightarrow 6x=0.45y$ ---(1).
Now, $7x=7\times \dfrac{3y}{40}$.
$\Rightarrow 7x=\dfrac{21y}{40}$.
$\Rightarrow 7x=0.525y$ ---(2).
Now, $8x=8\times \dfrac{3y}{40}$.
$\Rightarrow 8x=\dfrac{24y}{40}$.
$\Rightarrow 8x=0.6y$ ---(3).
Now, $9x=9\times \dfrac{3y}{40}$.
$\Rightarrow 9x=\dfrac{27y}{40}$.
$\Rightarrow 9x=0.675y$---(4).
Now, $10x=10\times \dfrac{3y}{40}$.
$\Rightarrow 10x=\dfrac{30y}{40}$.
$\Rightarrow 10x=0.75y$---(5).
We know that 50% of y is $0.5y$. From equations (1), (2), (3), (4), and (5) we can see that four of the subjects have more than 50% of the full marks.
$\therefore$ The student more than 50% marks in 4 subjects.
The correct option for the given problem is (c).

Note: We can solve this problem by taking 100 marks (any random marks) as full marks of the subject instead of the variable to get a better view. We can also calculate this problem by taking the average of marks and multiplying it with 100. We can also verify this result by making the counter calculation backward. Similarly, we can expect problems to find the marks obtained in each subject.