GRE® Quant Practice | Algebra Q1

The GRE maths sample question given below is a word problem in Linear Algebra. Framing and solving linear equations are the concepts tested in this practice question.

Question 1 : When four is added to six times a number and the result squared, the result obtained is four times the square of the sum of the number and its next multiple. What is the number?

1. $\frac{1}{3}$
2. $\frac{-1}{3}$
3. 3
4. -3
5. -1

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Video Explanation

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Explanatory Answer

Step 1 of solving this GRE Algebra Question: Frame equations from statements

Let 'n' be the number.
Step a: Four is added to six times a number: 4 + 6n
Step b: The result is squared (step a is squared): (4 + 6n)²
Step c: Sum of the number and its next multiple: (n + 2n)
Step d: Four times square of step 3: 4(n + 2n)²

Step 2 of solving this GRE Algebra Question: Equate and solve for "n"

Equating Step b and Step d,
(4 + 6n)² = 4 (n + 2n)²
(2 + 3n)² × 4 = 4 × (3n)²
(2 + 3n)² = 9n²
4 + 12n + 9n² = 9n²
4 + 12n = 0
12n = -4
n = $\frac{-4}{12}$ → n = $\frac{-1}{3}$

Choice B is the correct answer