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Question:\(\mathrm{A = \frac{1}{2}(b_1 + b_2)h}\)The formula above gives the area, A, of a trapezoid with bases of length b_1 and...

GMAT Advanced Math : (Adv_Math) Questions

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Advanced Math
Nonlinear equations in 1 variable
MEDIUM
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Question:

\(\mathrm{A = \frac{1}{2}(b_1 + b_2)h}\)


The formula above gives the area, \(\mathrm{A}\), of a trapezoid with bases of length \(\mathrm{b_1}\) and \(\mathrm{b_2}\) and height \(\mathrm{h}\). Which of the following correctly expresses \(\mathrm{b_1}\) in terms of \(\mathrm{A}\), \(\mathrm{b_2}\), and \(\mathrm{h}\)?

A
\(\mathrm{b_1 = \frac{A}{h} - b_2}\)
B
\(\mathrm{b_1 = \frac{2A}{h} - b_2}\)
C
\(\mathrm{b_1 = \frac{2A - b_2}{h}}\)
D
\(\mathrm{b_1 = 2A - h - b_2}\)
Solution

1. TRANSLATE the problem requirement

  • Given: \(\mathrm{A = \frac{1}{2}(b_1 + b_2)h}\)
  • Find: Express \(\mathrm{b_1}\) in terms of the other variables
  • This means: Isolate \(\mathrm{b_1}\) on one side of the equation

2. SIMPLIFY by clearing the fraction first

  • Multiply both sides by 2: \(\mathrm{2A = (b_1 + b_2)h}\)
  • This eliminates the fraction and makes the remaining steps cleaner

3. SIMPLIFY by isolating the binomial term

  • Divide both sides by h: \(\mathrm{\frac{2A}{h} = b_1 + b_2}\)
  • Now we have the sum containing \(\mathrm{b_1}\) isolated

4. SIMPLIFY by isolating b₁

  • Subtract \(\mathrm{b_2}\) from both sides: \(\mathrm{b_1 = \frac{2A}{h} - b_2}\)
  • This gives us \(\mathrm{b_1}\) expressed only in terms of A, \(\mathrm{b_2}\), and h

Answer: B. \(\mathrm{b_1 = \frac{2A}{h} - b_2}\)




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students forget to multiply both sides by 2 when clearing the fraction.

Instead of multiplying both sides by 2, they might try to work with the fraction directly or only multiply one side. Starting from \(\mathrm{A = \frac{1}{2}(b_1 + b_2)h}\), they might divide both sides by h first to get \(\mathrm{\frac{A}{h} = \frac{1}{2}(b_1 + b_2)}\), then subtract \(\mathrm{b_2}\) to get \(\mathrm{\frac{A}{h} - b_2 = \frac{1}{2}b_1}\), and finally multiply by 2. However, this more complicated path increases error likelihood, and if they miss the final multiplication by 2, they end up with \(\mathrm{b_1 = \frac{A}{h} - b_2}\).

This may lead them to select Choice A (\(\mathrm{b_1 = \frac{A}{h} - b_2}\)).

Second Most Common Error:

Poor SIMPLIFY sequencing: Students subtract \(\mathrm{b_2}\) before dividing by h.

After getting \(\mathrm{2A = (b_1 + b_2)h}\), instead of dividing by h first, they might try to subtract \(\mathrm{b_2h}\) from both sides, leading to \(\mathrm{2A - b_2h = b_1h}\), then dividing by h to get \(\mathrm{b_1 = \frac{2A - b_2h}{h}}\). If they then incorrectly simplify this as \(\mathrm{b_1 = \frac{2A - b_2}{h}}\), they get the wrong form.

This may lead them to select Choice C (\(\mathrm{b_1 = \frac{2A - b_2}{h}}\)).

The Bottom Line:

This problem tests systematic algebraic manipulation skills. The key insight is that clearing the fraction first by multiplying both sides by 2 creates the cleanest path to the solution.

Answer Choices Explained
A
\(\mathrm{b_1 = \frac{A}{h} - b_2}\)
B
\(\mathrm{b_1 = \frac{2A}{h} - b_2}\)
C
\(\mathrm{b_1 = \frac{2A - b_2}{h}}\)
D
\(\mathrm{b_1 = 2A - h - b_2}\)
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