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.
Boards framed = 20; Additional strip required = 20 m
Step 1: Convert the length to metres.
Given length = 1 m 50 cm
50 cm = 0.50 m
So, length = (1, ext{m} + 0.50, ext{m} = 1.5, ext{m})
Step 2: Note the breadth.
breadth = (1, ext{m})
Step 3: Find the perimeter of one rectangular board.
Formula: (P = 2(l + b))
(l + b = 1.5 + 1 = 2.5)
(P = 2 imes 2.5 = 5, ext{m})
So, each board needs (5, ext{m}) of aluminium strip for framing.
Step 4: How many boards can be framed with 100 m?
Total strip available = (100, ext{m})
Boards framed = (dfrac{100}{5} = 20)
Step 5: Remaining boards and extra strip needed.
Total boards = 24
Remaining boards = (24 - 20 = 4)
Strip needed for these 4 boards = (4 imes 5 = 20, ext{m})
Answer: Boards framed = 20; Additional strip required = 20 m