Question

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. The possible length of the shortest board, if the third piece is to be at least $5\, cm$ longer than the second, is

• A. less than $8\, cm$
• B. greater than or equal to 8 cm but less than or equal to $22\, cm$
• C. less than $22\, cm$
• D. greater than $22\, cm$

Answer 1

Answer: (B)

Let the shortest side be $x \,cm.$
Then, by given condition, second length $= x + 3\, cm$
Third length $= 2x \,cm$
Also given, total length $= 91$
Hence, sum of all the three lengths should be less than or equal to $91$
$x + x + 3 + 2x \le 91$
$\Rightarrow 4x + 3 \le 91$
Subtracting $(-3)$ to each term, $-3 + 4x + 3 \le 91 - 3$
$\Rightarrow 4x \le 88$
$\Rightarrow \frac{4x}{4} \le\frac{88}{4} \Rightarrow x \le\frac{88}{4}$
$\Rightarrow x \le22\,cm \,...\left(i\right)$
Again, given that
Third length $\ge$ second length $+ 5$
$\Rightarrow 2x \ge (x + 3) + 5$
$\Rightarrow 2x \ge (3 + 5)$
Transferring the term x to $L.H.S.,$
$2x - x \ge 8$
$\Rightarrow x \ge 8\,...(ii)$
From equations (i) and (ii), length of shortest board should be greater than or equal to 8 but less than or equal to 22, i.e., $8 \le x \le 22.$