GRE Quant Study Plan

An equation remains true when the same operation is applied to both sides. Isolating a linear combination of the variables often means rearranging and then dividing by a common coefficient.

It is given that $2x - 3 = 5y + 2$, and the goal is to find the value of $x - \frac{5}{2} y$.

Isolate the target expression

Step 1: Move the constant terms to produce a relation between $2x$ and $5y$. Add $3$ to both sides:

$2x = 5y + 5$.

Step 2: Subtract $5y$ from both sides:

$2x - 5y = 5$.

Step 3: Divide both sides by $2$:

$x - \frac{5}{2} y = \frac{5}{2} = 2.5$.

Check against the choices

Choice D: $2.5$ matches the value of $x - \frac{5}{2} y$.

Choice A: $-0.5$ could come from dividing $5$ by $-10$, or from an arithmetic sign error when moving constants.

Choice B: $0.5$ is $\frac{1}{2}$, which might result from dividing by $10$ instead of by $2$, or confusing $\frac{5}{2}$ with its reciprocal.

Choice C: $1.0$ might come from incorrectly concluding that $2x - 5y = 2$ and then dividing by $2$.

Choice E: $5.0$ is the value of $2x - 5y$ before dividing by $2$; stopping one step early produces this trap.

Things to Remember:

• To evaluate a linear combination, rearrange the given equation until that combination appears, then scale.
• Dividing an equation by $2$ halves every term, including the constant on the right.
• Stopping before the final division often produces a distractor equal to twice the correct answer.

A triangle with side lengths in the ratio $3:4:5$ is a right triangle: $3^2 + 4^2 = 9 + 16 = 25 = 5^2$, so the sides $3$ and $4$ are the legs and $5$ is the hypotenuse. The area of a right triangle is $\frac{1}{2} \times \text{leg}_1 \times \text{leg}_2$. The perimeter is the sum of all three side lengths.

It is given that the side ratio is $3$ to $4$ to $5$ and the enclosed area is $54$. Let the sides be $3k$, $4k$, and $5k$ for some positive scale factor $k$.

Find the scale factor from the area

$\text{Area} = \dfrac{1}{2}(3k)(4k) = 6k^2 = 54$

$k^2 = 9$

$k = 3$ (since length is positive)

Compute the perimeter

$\text{Perimeter} = 3k + 4k + 5k = 12k = 12 \cdot 3 = 36$

Check the choices

Choice C: $36$ is the correct perimeter.

Choice A: $12$ is the sum of the ratio parts $3+4+5$, forgetting to multiply by $k$.

Choice B: $18$ is $6k$ with $k = 3$, which is only the sum of the two legs, not the full perimeter.

Choice D: $48$ might come from treating $4k$ and $5k$ incorrectly as the legs, or from $16k$ with a wrong $k$.

Choice E: $96$ is $12k$ with $k = 8$, which could result from solving $6k = 54$ instead of $6k^2 = 54$.

Things to Remember:

• A $3$-$4$-$5$ triangle is right-angled, so area is half the product of the two shorter sides.
• Scale all three ratio parts by the same positive factor $k$ before computing perimeter.
• Area scales with $k^2$, while perimeter scales with $k$.

For two overlapping sets, the inclusion–exclusion principle says $\lvert B \cup W \rvert = \lvert B \rvert + \lvert W \rvert - \lvert B \cap W \rvert$. The probability of an event, when outcomes are equally likely, is $\dfrac{\text{favorable outcomes}}{\text{total outcomes}}$.

It is given that a herd has $90$ cows; every cow is brown, white, or both. Also, $55$ cows are entirely or partially white, and $75$ are entirely or partially brown. Let $B$ be the set of cows that are entirely or partially brown and $W$ the set that are entirely or partially white. Then $\lvert B \rvert = 75$, $\lvert W \rvert = 55$, and $\lvert B \cup W \rvert = 90$.

Find the both-colored cows

$\lvert B \cap W \rvert = 75 + 55 - 90 = 40$.

Find entirely white cows

Entirely white means in $W$ but not in $B$:

$\lvert W \rvert - \lvert B \cap W \rvert = 55 - 40 = 15$.

Probability

$P(\text{first cow entirely white}) = \dfrac{15}{90} = \dfrac{1}{6}$.

Check the choices

Choice A: $\dfrac{1}{6}$ is correct.

Choice B: $\dfrac{3}{11}$ might come from using $55$ in the denominator or a related fraction with $33$ and $121$.

Choice C: $\dfrac{7}{18} = \dfrac{35}{90}$ could result from using $35$ entirely white cows (for example, $75 + 55 - 95$ or another off-by-count error).

Choice D: $\dfrac{4}{9} = \dfrac{40}{90}$ is the probability of selecting a both-colored cow, not an entirely white one.

Choice E: $\dfrac{11}{18} = \dfrac{55}{90}$ is the probability of selecting a cow that is entirely or partially white.

Things to Remember:

• $\lvert A \cup B \rvert = \lvert A \rvert + \lvert B \rvert - \lvert A \cap B \rvert$.
• “Entirely” one category excludes the intersection with the other category.
• Probability with random selection from a finite herd is favorable count over herd size.