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A company manufactures cylindrical tins for coffee. Each tin has a radius of r inches, and the height of the...

GMAT Advanced Math : (Adv_Math) Questions

Source: Prism
Advanced Math
Nonlinear functions
HARD
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A company manufactures cylindrical tins for coffee. Each tin has a radius of \(\mathrm{r}\) inches, and the height of the tin is \(\mathrm{2}\) inches less than its diameter. Which function \(\mathrm{A}\) gives the area of the label that covers the lateral surface of the tin, in square inches, in terms of the tin's radius?

A

\(\mathrm{A(r) = 2\pi r(r - 2)}\)

B

\(\mathrm{A(r) = 2\pi r(2r - 2)}\)

C

\(\mathrm{A(r) = 2\pi r(2r + 2)}\)

D

\(\mathrm{A(r) = \pi r^2(2r - 2)}\)

Solution

1. TRANSLATE the problem information

  • Given information:
    • Radius = \(\mathrm{r}\) inches
    • Height = 2 inches less than diameter
    • Need lateral surface area function
  • What this tells us: We have radius but need to express height in terms of radius

2. INFER the approach

  • We need the lateral surface area formula: \(\mathrm{A = 2\pi rh}\)
  • Since we're asked for \(\mathrm{A(r)}\), everything must be in terms of \(\mathrm{r}\)
  • We need to convert the height description into a mathematical expression

3. TRANSLATE the height relationship

  • "Height is 2 inches less than diameter"
  • Diameter = \(\mathrm{2r}\) (since radius = \(\mathrm{r}\))
  • Height = diameter - 2 = \(\mathrm{2r - 2}\)

4. SIMPLIFY by substitution

  • Substitute \(\mathrm{h = 2r - 2}\) into \(\mathrm{A = 2\pi rh}\)
  • \(\mathrm{A(r) = 2\pi r(2r - 2)}\)

Answer: B. \(\mathrm{A(r) = 2\pi r(2r - 2)}\)




Why Students Usually Falter on This Problem


Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "2 inches less than its diameter" and write \(\mathrm{h = d - 2 = r - 2}\) instead of \(\mathrm{h = 2r - 2}\).

They forget that diameter equals \(\mathrm{2r}\), not \(\mathrm{r}\), so they incorrectly think height = \(\mathrm{r - 2}\). When they substitute this into \(\mathrm{A = 2\pi rh}\), they get \(\mathrm{A(r) = 2\pi r(r - 2)}\), leading them to select Choice A.


Second Most Common Error:

Poor INFER reasoning: Students recognize they need \(\mathrm{A = 2\pi rh}\) but don't realize they need to express height in terms of radius first.

They might try to work with the diameter directly or get confused about what to substitute where. This leads to confusion and guessing among the remaining choices.


The Bottom Line:

This problem tests whether you can carefully translate word relationships into mathematical expressions. The key insight is recognizing that "diameter" means \(\mathrm{2r}\), not just \(\mathrm{r}\), when converting the height description.

Answer Choices Explained
A

\(\mathrm{A(r) = 2\pi r(r - 2)}\)

B

\(\mathrm{A(r) = 2\pi r(2r - 2)}\)

C

\(\mathrm{A(r) = 2\pi r(2r + 2)}\)

D

\(\mathrm{A(r) = \pi r^2(2r - 2)}\)

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