Binomial Theorem

Question
CBSEENMA11014377

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 ?

Solution

Let the length of the first piece (or the shortest piece) be x cm
rightwards double arrow   Length of the second piece = (x + 3) cm
Also, the length of the third piece = 2x cm. The maximum length of the board available = 91 cm.
rightwards double arrow straight x plus left parenthesis straight x plus 3 right parenthesis plus 2 straight x less or equal than 91 space rightwards double arrow space 4 straight x plus 3 less or equal than 91 space space rightwards double arrow space 4 straight x less or equal than 88 space rightwards double arrow space straight x less or equal than 22                            ...(i)
Also, the third piece is to be at least 5 cm longer than the second.
 rightwards double arrow                      2 straight x minus left parenthesis straight x plus 3 right parenthesis space greater or equal than space 5 space space rightwards double arrow space space straight x minus 3 greater or equal than 5 space rightwards double arrow space straight x space greater or equal than 8                                    ...(ii)
From (i) and (ii), we have 8 less or equal than straight x less or equal than 22
Hence, the shortest board (or the first board) can have a minimum length of 8 cm and a maximum length of 22 cm.

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