Question

In an exhibition hall there are 24 display boards e
each of length 1 m 50 cm and breadth 1 metre. There is a hundred metre long aluminium strip, which is use 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.​

Answer 1

$\huge{\blue{\underline{\purple{\mathbb{AnSwER}}}}}$

step by step explanation

Total display boards = 24

Length of strip = 100 m

Length of one display board = 1.5 m

Breadth of one display board = 1 m

Perimeter of one display board = 2 × (l + b)

= 2 × (1.5 + 1) = 5 m

Number of boards which will be framed = 100 / 5

= 20

Number of boards unframed = 24 – 20 = 4

Length of strip for required for remaining boards = 4 × Perimeter of one board

= 4 × 5

= 20 m

❇️follow me

Answer 2

solution

Length of one display board = 1.5 m

Breadth of one display board = 1 m

therefore ,

Perimeter of one display board = 2 × (l + b)

= 2 × (1.5 + 1) = 5 m

Number of boards which will be framed = 100 / 5

= 20

Number of boards unframed = 24 – 20 = 4

given [ total no of board = 20

and length of strip = 100 m ]

then ,

Length of strip for required for remaining boards = 4 × Perimeter of one board

= 4 × 5

= 20 m