Question

Question 37

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 the also length of the aluminium strip required for the remaining boards.

Answer 1

Given, total display boards = 24

Length of one display boards = 1m + 50 cm = 1m+$\frac{50}{100}$m [ $\because$ 1 m=100 cm]

= (1 + 0.5)m

= 1.5 m

Breadth of one display board = 1m

$\therefore$ Perimeter of one display = 2$\times$(Length + Breadth)

= 2$\times$(1.5 + 1)m = 2 $\times$2.5m

= 5 m

Length of strip = 100 m

Now, number of boards will be framed = $\frac{\text{Length of strip}}{\text{perimeter one board}}$

=$\frac{100}{5}$ = 20

This means that out of 24 only 20 boards will be framed.

Number of boards left unframed = 24 - 20 = 4

$\therefore$ Length of the strip required for remaining boards = 4$\times$ Perimeter of one board

= 4 $\times 2(1.5+1)=4\times 2\times 2.5$

= 20 m