Question

In the first four papers each of $100$ marks, Rishi got $95,72,73,83$ marks. If he wants an average of greater than or equal to $75$ marks and less than $80$ marks, find the range of the marks he should score in the fifth paper.
(A) $52 \leqslant x \leqslant 77$
(B) $25 < x < 75$
(C) $75 < x < 80$
(D) $73 < x < 100$

Answer 1

Hint: Start with assuming the unknown marks of the fifth paper as ‘x’. Now use the definition of average that says: $Average = \dfrac{{{\text{Sum of all the observations}}}}{{{\text{Number of observations}}}}$ , to make an equation by substituting the values in the RHS. According to the given condition, the average can also be represented using inequality $75 \leqslant Average \leqslant 80$. Now combine both equations and simplify them to find the required range of unknown ‘x’.

Complete step-by-step answer:
Here in the problem, we are given marks of the first four exams out of $100$. Rishi wants his average marks of five papers to be more than or equal to $75$ and less than $80$ marks. With this information, we need to find the range of values for marks he should have in the fifth exam to satisfy the given condition.
Let us assume that the Rishi’s marks of the fifth paper are $x$ .
Before starting with the solution we must understand the concept of average. The term 'average' refers to the ‘middle’ or ‘central’ point. When used in mathematics, the term refers to a number that is a typical representation of a group of numbers (or data set).
Average can be calculated by the division of the sum of the observations by the total number of observations, i.e. :
$ \Rightarrow Average = \dfrac{{{\text{Sum of all the observations}}}}{{{\text{Number of observations}}}}$
Since we have the marks for four papers as $95,72,73,83$ . The average mark is given as more than or equal to $75$ and less than $80$ marks
$ \Rightarrow 75 \leqslant Average \leqslant 80$ ……….(i)
Now let’s use the formula of average for the given case, we get:
$ \Rightarrow Average = \dfrac{{{\text{Sum of all the observations}}}}{{{\text{Number of observations}}}} = \dfrac{{95 + 72 + 73 + 83 + x}}{5}$
On simplifying it further, we have:
\[ \Rightarrow Average = \dfrac{{95 + 72 + 73 + 83 + x}}{5} = \dfrac{{323 + x}}{5}\]
Using the above inequality (i), we can rewrite the above equation as:
$ \Rightarrow 75 \leqslant \dfrac{{323 + x}}{5} \leqslant 80$
For solving this inequality, we can multiply it by $5$
$ \Rightarrow 75 \times 5 \leqslant \left( {\dfrac{{323 + x}}{5}} \right) \times 5 \leqslant 80 \times 5 \Rightarrow 375 \leqslant 323 + x \leqslant 400$
By subtracting $323$ from both sides, we can have:
 $ \Rightarrow 375 \leqslant 323 + x \leqslant 400 \Rightarrow 375 - 323 \leqslant x \leqslant 400 - 323 \Rightarrow 52 \leqslant x \leqslant 77$
Therefore, we get the range of values for the marks in the fifth paper is given by $52 \leqslant x \leqslant 77$
Hence, the option (A) is the correct answer.

Note: In statistics, the fundamental definition of the terms always helps in the solution of the problem. An alternative approach can be to solve the inequality for more than or equal to and less than separately, i.e. solving $75 \leqslant \dfrac{{323 + x}}{5}$ and $\dfrac{{323 + x}}{5} \leqslant 80$ separately to find a range of required value. You can solve it separately and combine the result to solve the required interval.