For all positive values of P, W, L, and A, consider the family of rectangles having perimeter P feet, width...

GMAT Two Part Analysis : (TPA) Questions

Source: Official Guide

Two Part Analysis

Quant - Fitting Values

EASY

...

Notes

Post a Query

For all positive values of \(\mathrm{P}\), \(\mathrm{W}\), \(\mathrm{L}\), and \(\mathrm{A}\), consider the family of rectangles having perimeter \(\mathrm{P}\) feet, width \(\mathrm{W}\) feet, length \(\mathrm{L}\) feet, and area \(\mathrm{A}\) square feet.

In the first column of the table, select the expression in terms of \(\mathrm{P}\) and \(\mathrm{W}\) that is equivalent to \(\mathrm{A}\), and in the second column of the table select the expression in terms of \(\mathrm{P}\) and \(\mathrm{W}\) that is equivalent to \(\mathrm{L}\). Select only two expressions, one in each column.

\(\frac{\mathrm{P}^2}{\mathrm{W}}\)

\(\frac{\mathrm{P}\mathrm{W}}{4}\)

(\(\frac{\mathrm{P}}{2}-\mathrm{W}\))

\(\frac{(\mathrm{P}-\mathrm{W})^2}{4}\)

\((\frac{\mathrm{P}}{2}-\mathrm{W})\mathrm{W}\)

Solution

Phase 1: Owning the Dataset

Visual Representation

Let me draw a rectangle to visualize our relationships:

       W (width)
    ____________
   |            |
L  |    Area    | L (length)
   |     = A    |
   |____________|
        W

Perimeter \(\mathrm{P = 2W + 2L}\)

Using Concrete Numbers

Let's test with \(\mathrm{P = 20}\) feet and \(\mathrm{W = 4}\) feet to understand the relationships:

Since \(\mathrm{P = 2W + 2L}\), we have: \(\mathrm{20 = 2(4) + 2L}\)
Solving: \(\mathrm{20 = 8 + 2L}\), so \(\mathrm{2L = 12}\), thus \(\mathrm{L = 6}\) feet
Area \(\mathrm{A = W \times L = 4 \times 6 = 24}\) square feet

Phase 2: Understanding the Question

We need to find:

Expression for A in terms of P and W
Expression for L in terms of P and W

Finding L in terms of P and W

Starting with \(\mathrm{P = 2W + 2L}\):

\(\mathrm{P = 2W + 2L}\)
\(\mathrm{P - 2W = 2L}\)
\(\mathrm{L = \frac{P - 2W}{2}}\)
\(\mathrm{L = \frac{P}{2} - W}\)

Finding A in terms of P and W

Since \(\mathrm{A = W \times L}\) and we found \(\mathrm{L = \frac{P}{2} - W}\):

\(\mathrm{A = W \times (\frac{P}{2} - W)}\)
\(\mathrm{A = W(\frac{P}{2} - W)}\)

Phase 3: Finding the Answer

Let's match our expressions to the answer choices:

For \(\mathrm{L = \frac{P}{2} - W}\):
Looking at the choices, \(\mathrm{(\frac{P}{2} - W)}\) matches exactly with the third option.

For \(\mathrm{A = W(\frac{P}{2} - W)}\):
This matches exactly with the fifth option: \(\mathrm{(\frac{P}{2} - W)W}\)

Verification with our example

Using \(\mathrm{P = 20}\), \(\mathrm{W = 4}\):

\(\mathrm{L = \frac{P}{2} - W = \frac{20}{2} - 4 = 10 - 4 = 6}\) ✓
\(\mathrm{A = (\frac{P}{2} - W)W = (10 - 4) \times 4 = 6 \times 4 = 24}\) ✓

Phase 4: Solution

Statement 1 (A): Select \(\mathrm{(\frac{P}{2} - W)W}\)
Statement 2 (L): Select \(\mathrm{(\frac{P}{2} - W)}\)

These expressions correctly represent the area and length of a rectangle in terms of its perimeter and width.

Rate this Solution

Tell us what you think about this solution

...

Forum Discussions

Start a new discussion

Post

Previous Attempts

Loading attempts...