In an exhibition hall, there are $24$ display boards each of length $1 m 50 c m$ and breadth $1 m$ . There is a $100 m$ long aluminium strip, which is used to frame these boards.
How many boards will be framed using this strip? Find also the length of the aluminium strip required for the remaining boards.

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Solution

Length of the display board $= 1 m 50 c m = 1.5 m$
Breadth of the display board $= 1 m$
$∴$ Perimeter of the one display board
$\Rightarrow 2 \times (l e n g t h + b r e a d t h) = 2 \times (1.5 m \times 1 m) = 2 \times 2.5 m = 5 m$
$∴$ Perimeter of the display board
$\Rightarrow 5 m \times 24 = 120 m$
Number of boards will be framed using this strip $= \frac{100}{5} = 20$ strips
Length of the aluminium strip required for the remaining boards $= (120 - 100) m = 20 m$ .

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In an exhibition hall, there are 24 display boards each of length 1 m 50 cm and breadth 1 m. There is a 100 m long aluminium strip, which is used to frame these boards.How many boards will be framed using this strip? Find also the length of the aluminium strip required for the remaining boards.