In the first four examinations,each of $100$ marks,Ravi scored $94, 73, 72,$ and $84$ marks. If a final average greater than or equal to $80$ and less than $90$ is needed to obtain a final grade $'B'$ in a course,find the minimum marks that Ravi must obtain in the fifth examination to get grade $'B'$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(77) Let the marks obtained by Ravi in the fifth examination be $x$.
The total marks for the five examinations are $94 + 73 + 72 + 84 + x = 323 + x$.
The average of the five examinations is $\frac{323 + x}{5}$.
According to the condition for grade $'B'$,the average must be at least $80$ and less than $90$:
$80 \le \frac{323 + x}{5} < 90$.
Multiplying the inequality by $5$:
$400 \le 323 + x < 450$.
Subtracting $323$ from all parts:
$400 - 323 \le x < 450 - 323$.
$77 \le x < 127$.
Since the maximum marks in an examination is $100$,the minimum marks required is $77$.

Explore More

Similar Questions

Find the solution set of the inequality: $4x + 3 \geq 2x + 17$ and $3x - 5 < -2$.

$\left\{x \in R: \frac{14 x}{x+1}-\frac{9 x-30}{x-4} < 0\right\}$ is equal to

The set of all real numbers $x$ for which ${x^2} - |x + 2| + x > 0$ is

Solve for $x$ the following inequality: $\frac{1}{|x|-3} \leq \frac{1}{2}$

Difficult
View Solution

If $\sqrt{9x^2+6x+1} < (2-x)$,then:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo