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08 Jun 2018, 10:33
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (02:12) correct
33%
(02:01)
wrong
based on 422
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A rectangular sandbox has a width of W feet, a perimeter of P feet, and an area of A square feet. Which of the following equations must be true?
a. \(W^2 + PW + A = 0\)
b \(W^2 - PW + A = 0\)
c. \(2W^2 + PW + 2A = 0\)
d. \(2W^2 - PW - 2A = 0\)
5. \(2W^2 - PW + 2A = 0\)
a. \(W^2 + PW + A = 0\)
b \(W^2 - PW + A = 0\)
c. \(2W^2 + PW + 2A = 0\)
d. \(2W^2 - PW - 2A = 0\)
5. \(2W^2 - PW + 2A = 0\)
Updated on: 10 Jun 2018, 04:31
Originally posted by GMATGuruNY on 08 Jun 2018, 11:12.
Last edited by GMATGuruNY on 10 Jun 2018, 04:31, edited 1 time in total.
Last edited by GMATGuruNY on 10 Jun 2018, 04:31, edited 1 time in total.
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The GMAT will not use the term perimeter to refer to a rectangular solid.
The posted problem should read as follows:
Let the rectangular base be a SQUARE with a side of 1, implying the following values:
W = s = 1.
P = 4s = 4*1 = 4.
A = s² = 1² = 1.
When we plug W=1, P=4 and A=1 into the answer choices, only E works:
2W² - PW + 2A = 2(1²) - 4(1) + 2(1) = 0.
The posted problem should read as follows:
QZ
Let the rectangular base be a SQUARE with a side of 1, implying the following values:
W = s = 1.
P = 4s = 4*1 = 4.
A = s² = 1² = 1.
When we plug W=1, P=4 and A=1 into the answer choices, only E works:
2W² - PW + 2A = 2(1²) - 4(1) + 2(1) = 0.
General Discussion
08 Jun 2018, 10:52
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QZ
We know we must have W, P, and A in the equation.
We have L, but we cannot use L straightforwardly in the equation itself.
W and L yield P
That fact means L can be defined in terms of two of the "allowed" variables: W and P*
W and L yield A. Area is measured in square units.
Every answer has a W\(^2\) term. That fact suggests in a second way that L is defined in terms of another variable.
1) Change how L is defined
From perimeter equation, find L in terms of W and P
\(P = 2L + 2W=2(L+W)\)
\(\frac{P}{2}=L + W\)
\(L=\frac{P}{2}-W\)
2) Find the squared term from the equation for Area
(A is measured in square units and must be included in the equation - area is the next logical step)
Set up the equation for area, substitute for \(L\) from above
\(A=(W*L)\)
\(A=W*(\frac{P}{2}-W)\)
\(\frac{PW}{2} \\
- W^2=A\)
\(PW -2W^2=2A\)
Move LHS terms to RHS. Signs change
\(2W^2 - PW + 2A = 0\)
Answer E
*There is no L in the equation, but W, P, and A ARE in the equation. Both P and A involve both W and L. Thus L must be defined in terms of \(W\) and \(P\)
