3. In an exhibition hall, there are 24 display boards, each of length 1m 50cm and breadth 1m. There is a 100m long aluminium strip, which is used to frame these boards. a. How many boards will be framed using this strip? b. Also find the length of the aluminium strip required for the remaining boards.
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Answer:
Given total display boards = 24
length of one display boards = 1 m +50 cm
= 1m+ 50/100 m
=1.5m
Breadth of one display board = 1m
:. perimeter of display board = 2× ( l+ b)
= 2× (1.5 +1) m = 2×2.5 m
= 5m
Now ,no. 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
No. of boards left unramed= 24 - 20 = 4
= 4× perimeter of one board
= 4× 2 ( 1.5+ 1)
= 8× 2.5
= 20 m
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