question_answer

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 aluminum strip, which is used to frame these boards. How many boards will be framed using this strip? Find also the length of the aluminum strip required for the remaining boards.

Answer:

Given, total display boards = 24 length of one display boards \[=1\text{ }m+50\text{ }cm\]       \[=1\text{ }m+\frac{50}{100}m\]        \[=1.5\text{ }m\] Breadth of one display board = 1 m \[\therefore \]  Perimeter of display board \[=2\times (length+Breadth)\] \[=2\times (1.5+1)\text{ }m=2\times 2.5\text{ }m\] \[=5\text{ }m\] length of strip \[=100\text{ }m\] Now, no. of boards will be framed             \[=\frac{Length\,\,of\,\,strip}{Perimeter\,\,of\,\,one\,\,board}\]             \[=\frac{100}{5}=20\] This means that out of 24 only 20 boards will be framed. No. of boards left unframed \[=2420=4\] \[\therefore \] length of the strip required for remaining boards \[=\text{4 }\!\!\times\!\!\text{ perimeter of one board}\] \[=4\times 2(1.5+1)\] \[=4\times 2\times 2.5\] \[=20\text{ }m\].