$A$ man wants to cut three lengths from a single piece of board of length $91 \, cm$. The second length is to be $3 \, cm$ longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least $5 \, cm$ longer than the second?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Let the length of the shortest piece be $x \, cm$. Then,the lengths of the second and third pieces are $(x+3) \, cm$ and $2x \, cm$ respectively.
Since the three lengths are cut from a single board of length $91 \, cm$:
$x + (x+3) + 2x \leq 91$
$4x + 3 \leq 91$
$4x \leq 88$
$x \leq 22$ ...... $(1)$
Also,the third piece is at least $5 \, cm$ longer than the second piece:
$2x \geq (x+3) + 5$
$2x \geq x + 8$
$x \geq 8$ ...... $(2)$
From $(1)$ and $(2)$,we obtain:
$8 \leq x \leq 22$
Thus,the possible length of the shortest board is greater than or equal to $8 \, cm$ but less than or equal to $22 \, cm$.

Explore More

Similar Questions

If $\frac{2x}{2x^2 + 5x + 2} > \frac{1}{x + 1}$,then

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

The longest side of a triangle is twice the shortest side and the third side is $3 \text{ cm}$ more than the shortest side. If the perimeter of the triangle is at least $51 \text{ cm}$, find the minimum length of the shortest side. (in $\text{ cm}$)

The solution set of the constraints $|x-y| \leqslant 1, x, y \geqslant 0$ is

$\left\{x \in R : \frac{\sqrt{6+x-x^2}}{2x+5} \geq \frac{\sqrt{6+x-x^2}}{x+4}\right\}=$

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