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Question
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.
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Solution
Given, total display boards = 24
Length of one display boards = 1 m + 50 cm
= 1 m + `50/100` m ...[∵ 1 m = 100 cm]
= (1 + 0.5) m
= 1.5 m
Breadth of one display board = 1 m
∴ Perimeter of one display board = 2 × (Length + Breadth)
= 2 × (1.5 + 1) m
= 2 × 2.5 m
= 5 m
Length of strip = 100 m ...[Given]
Now, number of boards will be framed
= `"Length of strip"/"Perimeter of one board"`
= `100/5`
= 20
This means that out of 24 only 20 boards will be framed.
Number of board left unframed = 24 – 20 = 4
∴ Length of the strip required for remaining boards
= 4 × Perimeter of one board
= 4 × 2(1.5 + 1)
= 4 × 2 × 2.5
= 20 m
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