<p>A wood frame for pouring concrete has an interior perimeter of 14 meters. Its length is one meter greater than its width. The frame is to be braced with twelve-gauge steel cross-wires. Assuming an extra half-meter of wire is used at either end of a cross-wire for anchoring, what length of wire should be cut for each brace?</p>
<p>Perimeter= sum of all sides
Length = x + 1
Width = x
x+x+x+1+x+1=14
4x+2=14
4x=12
x=3</p>
<p>Length=4, width=3
Cross wires meaning on the length side I assume? 0.5 on each side = 1
4 - 1 = 3inches?
Random guess, do you know the answer?</p>
<p>This problem would never appear in a SAT booklet – at least not without a drawing that shows the cross-wire pattern and a clearer question. </p>
<p>If it were written by ETS, this problem might test the Pythagorean Theorem and the ability of the tester to remember to add the anchors. </p>
<p>For this problem, one needs to see that the sides are 3 x 4 (14 perimeter) and that a diagonal (cross-wire) measures 5 meters. Add .5 meter twice, and each wire should measure 6 meters. </p>
<p>But again, it is best to ignore such poorly written problem that tend to appear in high school and in the non-TCB books.</p>
<p>I thought the answer would be 5.5 m because the wire used for anchoring is to be used at either end, not both. Did I interpret the problem incorrectly, or is it ambiguous?</p>
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<p>The problem is ambiguous and different interpretations lead to different answers. This is a worthless problem.</p>