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5x + 7y = 1ax + by = 1In the given pair of equations, a and b are constants. The...

GMAT Algebra : (Alg) Questions

Source: Practice Test
Algebra
Linear equations in 2 variables
HARD
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\(5\mathrm{x} + 7\mathrm{y} = 1\)
\(\mathrm{ax} + \mathrm{by} = 1\)

In the given pair of equations, \(\mathrm{a}\) and \(\mathrm{b}\) are constants. The graph of this pair of equations in the \(\mathrm{xy}\)-plane is a pair of perpendicular lines. Which of the following pairs of equations also represents a pair of perpendicular lines?

A

\(10\mathrm{x} + 7\mathrm{y} = 1\)
\(\mathrm{ax} - 2\mathrm{by} = 1\)

B

\(10\mathrm{x} + 7\mathrm{y} = 1\)
\(\mathrm{ax} + 2\mathrm{by} = 1\)

C

\(10\mathrm{x} + 7\mathrm{y} = 1\)
\(2\mathrm{ax} + \mathrm{by} = 1\)

D

\(5\mathrm{x} - 7\mathrm{y} = 1\)
\(\mathrm{ax} + \mathrm{by} = 1\)

Solution

1. INFER the key relationship for perpendicular lines

  • Key concept: Two lines are perpendicular when their slopes are negative reciprocals
  • This means: \(\mathrm{slope_1 \times slope_2 = -1}\)

2. SIMPLIFY to find slopes of the given perpendicular lines

  • First equation \(\mathrm{5x + 7y = 1}\):
    • Subtract \(\mathrm{5x}\): \(\mathrm{7y = -5x + 1}\)
    • Divide by 7: \(\mathrm{y = \frac{-5}{7}x + \frac{1}{7}}\)
    • \(\mathrm{Slope_1 = \frac{-5}{7}}\)
  • Second equation \(\mathrm{ax + by = 1}\):
    • Subtract \(\mathrm{ax}\): \(\mathrm{by = -ax + 1}\)
    • Divide by \(\mathrm{b}\): \(\mathrm{y = \frac{-a}{b}x + \frac{1}{b}}\)
    • \(\mathrm{Slope_2 = \frac{-a}{b}}\)

3. INFER the constraint from perpendicularity

  • Since the given lines are perpendicular:

\(\mathrm{\frac{-5}{7} \times \frac{-a}{b} = -1}\)

  • SIMPLIFY:

\(\mathrm{\frac{5a}{7b} = -1}\)

  • Therefore:

\(\mathrm{\frac{a}{b} = \frac{-7}{5}}\)

4. SIMPLIFY each answer choice to find slopes

Choice B analysis:

  • First equation \(\mathrm{10x + 7y = 1}\):
    • Slope = \(\mathrm{\frac{-10}{7}}\) (using same process as step 2)
  • Second equation \(\mathrm{ax + 2by = 1}\):
    • Rearranging: \(\mathrm{y = \frac{-a}{2b}x + \frac{1}{2b}}\)
    • Slope = \(\mathrm{\frac{-a}{2b}}\)
    • Since \(\mathrm{\frac{a}{b} = \frac{-7}{5}}\):

\(\mathrm{slope = \frac{-(-7/5)}{2} = \frac{7}{10}}\)

5. INFER perpendicularity check

  • Product test:

\(\mathrm{\frac{-10}{7} \times \frac{7}{10} = \frac{-70}{70} = -1}\)

  • The slopes are negative reciprocals!

Answer: B


Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students recognize they need to find slopes but don't establish the constraint \(\mathrm{\frac{a}{b} = \frac{-7}{5}}\) from the given perpendicular relationship.

Without this crucial insight, they attempt to check perpendicularity for each choice without knowing the actual values. This leads to confusion because they can't evaluate expressions like \(\mathrm{\frac{-a}{2b}}\) numerically, causing them to abandon systematic solution and guess.

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when converting standard form to slope-intercept form, particularly with sign handling.

For example, incorrectly writing \(\mathrm{ax + 2by = 1}\) as \(\mathrm{y = \frac{a}{2b}x + \frac{1}{2b}}\) instead of \(\mathrm{y = \frac{-a}{2b}x + \frac{1}{2b}}\). This sign error completely changes the slope and leads to selecting Choice A or Choice C.

The Bottom Line:

This problem requires both recognizing the strategic approach (using the given perpendicular relationship to find a constraint) and executing multiple algebraic conversions accurately. The key insight is that you must use the given information to determine the relationship between a and b before you can meaningfully check the answer choices.

Answer Choices Explained
A

\(10\mathrm{x} + 7\mathrm{y} = 1\)
\(\mathrm{ax} - 2\mathrm{by} = 1\)

B

\(10\mathrm{x} + 7\mathrm{y} = 1\)
\(\mathrm{ax} + 2\mathrm{by} = 1\)

C

\(10\mathrm{x} + 7\mathrm{y} = 1\)
\(2\mathrm{ax} + \mathrm{by} = 1\)

D

\(5\mathrm{x} - 7\mathrm{y} = 1\)
\(\mathrm{ax} + \mathrm{by} = 1\)

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