GRE Arithmetic & Number Properties Practice & Questions

The question asks for the Least Common Multiple (LCM) of $12$ and $18$. 1. Prime factorize each integer: - $12 = 2^2 \times 3^1$ - $18 = 2^1 \times 3^2$ 2. The LCM takes the highest power of each prime factor: $$\text{LCM}(12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36$$ Therefore, the least possible positive integer divisible by both is $36$.

For any LCM problem, factor into primes and take the maximum exponent for each prime factor.

Let the original price be $P$. 1. An increase of $25\%$ multiplies the price by $(1 + 0.25) = 1.25 = \frac{5}{4}$. 2. A subsequent discount of $20\%$ multiplies that intermediate price by $(1 - 0.20) = 0.80 = \frac{4}{5}$. 3. Set up the cumulative multiplication: $$\text{Final Price} = P \times \frac{5}{4} \times \frac{4}{5} = P \times 1 = P$$ Because the net multiplier is exactly $1$, the final price equals the original price. Hence, $P = \$600$.

Examinees often falsely believe that $+25\%$ followed by $-20\%$ results in a net increase of $5\%$. Because percent changes apply successively to changing bases, $+25\%$ followed by $-20\%$ nets exactly $0\%$.

Always convert successive percentage changes into multiplying factors: $(1 + r_1)(1 - r_2)$.

1. Examine the exponent $k^2 + k$ in Quantity A: $$k^2 + k = k(k + 1)$$ 2. For any integer $k$, either $k$ or $k + 1$ must be an even integer. The product of consecutive integers is always even! 3. Therefore, $k^2 + k$ is guaranteed to be an even integer for all integer values of $k$ (positive, negative, or zero). 4. Since $(-1)^{\text{even}} = 1$, we have $(-1)^m = 1$, which strictly requires that $m$ must also be an even integer. 5. If $m$ is an even integer, the remainder when $m$ is divided by $2$ is always $0$. 6. Thus, Quantity A = $0$, and Quantity B = $0$. The two quantities are equal.

Examinees frequently select (D) assuming that because $k$ can be positive or negative, or because $m$ is not fixed, the parity cannot be determined. Factoring into $k(k+1)$ exposes invariant even parity.

The product of two consecutive integers $k(k+1)$ is ALWAYS even.

To determine the overall percent change, choose a convenient base value such as $100$ employees in 2021. 1. From 2021 to 2022, the count increases by $20\%$: $$100 \times (1 + 0.20) = 120$$ 2. From 2022 to 2023, the count decreases by $10\%$ of the new amount ($120$): $$120 \times (1 - 0.10) = 120 \times 0.90 = 108$$ 3. Calculate the net percent increase from 2021 to 2023: $$\frac{108 - 100}{100} \times 100\% = 8\%$$ Thus, Quantity A is $8\%$, which is equal to Quantity B.

A frequent error is naively subtracting the two percentages ($20\% - 10\% = 10\%$), leading a test taker to mistakenly select Quantity A is greater. Successive percentage changes apply to the updated base, not the original base.

Sequential percentage changes cannot be simply added or subtracted; always apply them multiplicatively to the updated base value.

The relationship between sum, average, and number of terms is given by: $$\text{Sum} = \text{Average} \times \text{Number of terms}$$ 1. Calculate the initial sum of the five numbers: $$\text{Sum}_5 = 24 \times 5 = 120$$ 2. Calculate the sum of the four remaining numbers: $$\text{Sum}_4 = 25 \times 4 = 100$$ 3. The removed number is the difference between the two sums: $$120 - 100 = 20$$ Therefore, the correct choice is $20$.

Test takers who incorrectly assume that removing a number reduces the total sum by the mean ($24$ or $25$) will choose $24$, failing to calculate the total aggregates.

In average problems where items are added or removed, always convert averages to totals (sums) before finding differences.