# HELP WITH MATH QUESTION?

How many 2.125 cm square pieces can be cut from the stainless steel sheet that is 24.30 cm long and 45.799 cm wide?

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• Anonymous
1 month ago

I can see three possible answers from the way this is written. This depends on whether your cm square is just the area or the shape. Or if you are describing the length of a side of each square

523 pieces if you are not restricted to the shape. If they have to be literally square, then you get each piece needing to be 1.458 long/wide. so you get 16 x 31 = 496 squares.  If you are meaning 2.125 x 2.125 then you get 11 x 21  = 231. It may be clearer in how it is written in the original.

• Assuming a zero kerf width:

Length 24.30 cm will support INT(24.30/2.125) = 11

Width 45.799 cm will support INT(45.799/2.125) = 21

Ans: 11•21 = 231 pieces can be cut from the stock with zero kerf

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If there is a non-zero kerf width k:

L = 2.125p + (p-1)k

L + k = (2.125+k)p

p = INT( (L+k) )/(2.125+k) )

Note: INT(x) is the largest integer that is not greater than the real number x.

For the given stock size and kerf width k::

n = INT((24.30+k)/(2.125+k))•INT((45.799+k)/(2.125+k))

If the kerf is 1 mm (0.10 cm)

n = INT((24.30+0.10)/(2.125+0.10))*INT((45.799+0.10)/(2.125+0.10))

= INT(24.40/2.225) • INT(45.899/2.225)

= 10 • 20

= 200 pieces with 1 mm kerf

So even as small a kerf as 1 mm will reduce the production from 231 pieces to 200 pieces and increase the waste accordingly.

• Pieces from length

l = 24.3/2.125

l = 11.4352941176

l = 11  (must have whole pieces)

Pieces from width

w = 45.799/2.125

w = 21.5524705882

w = 21  (must have whole pieces)

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Number of pieces from sheet

n = l * w

n = 11 * 21

n = 231

231 pieces <––––––