GRE Quantitative Comparison Practice

A positive integer is said to be 7 greater than a multiple of 13 if it leaves a remainder of $7$ when divided by $13$ — in other words, $x = 13k + 7$ for some non-negative integer $k$. This is a modular arithmetic constraint that limits which integers are candidates for $x$.

Step 1 – List the values of x that satisfy the first condition

Using $x = 13k + 7$:

• $k = 0$: $x = 7$
• $k = 1$: $x = 20$
• $k = 2$: $x = 33$
• $k = 3$: $x = 46$
• $k = 4$: $x = 59$
• $k = 5$: $x = 72$

Step 2 – Apply the second condition: 2512 < x² < 3596

To find the range of $x$, take square roots of all three parts. Since $x$ is a positive integer:

$\sqrt{2512} \approx 50.1 \quad$ and $\quad \sqrt{3596} \approx 59.97$

So $x$ must satisfy $51 \le x \le 59$.

Step 3 – Find x that satisfies both conditions

From the list in Step 1, the only value in the range $51 \le x \le 59$ is $x = 59$.

Verify: $59 = 13 \times 4 + 7$ ✓ and $59^2 = 3481$, and $2512 \lt 3481 \lt 3596$ ✓

Compare Quantity A and Quantity B

Quantity A $= x = 59$

Quantity B $= 55$

Since $59 \gt 55$, Quantity A is greater.

Why the other conclusions don't hold

Quantity A is greater: $x = 59$, which is greater than $55$. Correct.

Quantity B is greater: Would require $x \lt 55$. The only valid candidate, $59$, is greater than $55$. Incorrect.

The two quantities are equal: Would require $x = 55$. But $55 = 13 \times 4 + 3$ (remainder $3$, not $7$), so $55$ does not satisfy the first condition. Incorrect.

The relationship cannot be determined: Both conditions uniquely determine $x = 59$, so the comparison is definitive. Incorrect.

Things to Remember:

• "$r$ greater than a multiple of $n$" means $x = nk + r$, i.e., $x$ leaves remainder $r$ when divided by $n$.
• To bound $x$ from a constraint on $x^2$, take square roots of both sides and restrict to positive integers.
• When both conditions uniquely determine the unknown, the QC answer is always A, B, or C — never D.

In a right triangle, the side opposite the right angle is the hypotenuse. When a point $D$ lies on a leg of a right triangle, angles $\angle BDA$ and $\angle BDC$ form a linear pair — they are supplementary and sum to $180\degree$ because $A$, $D$, and $C$ are collinear.

Step 1 – Find AD

It is given that $AD = \dfrac{2}{5}(AC)$. Since $AC = 30$:

$AD = \dfrac{2}{5} \times 30 = 12$

Therefore $DC = AC - AD = 30 - 12 = 18$.

Step 2 – Analyze triangle ABD

From the figure, $\angle A = 90\degree$, and $D$ lies on $AC$, so $\angle DAB = 90\degree$ as well. In right triangle $ABD$, it is given that $AB = 12$ and $AD = 12$. Since both legs are equal, triangle $ABD$ is an isosceles right triangle, and the two base angles are each $45\degree$:

$\angle ADB = 45\degree$

Step 3 – Find angle BDC

Since $A$, $D$, $C$ are collinear, $\angle ADB$ and $\angle BDC$ are supplementary:

$\angle BDC = 180\degree - \angle ADB = 180\degree - 45\degree = 135\degree$

Compare Quantity A and Quantity B

Quantity A $= 135\degree$, Quantity B $= 120\degree$. Since $135 \gt 120$, Quantity A is greater.

Why the other conclusions don't hold

Quantity A is greater: $\angle BDC = 135\degree \gt 120\degree$. Correct.

Quantity B is greater: Would require $\angle BDC \lt 120\degree$. The only candidate is $135\degree$. Incorrect.

The two quantities are equal: Would require $\angle BDC = 120\degree \neq 135\degree$. Incorrect.

The relationship cannot be determined: All values are uniquely determined. Incorrect.

Things to Remember:

• When $D$ lies between $A$ and $C$ on a line, $\angle BDA + \angle BDC = 180\degree$ (linear pair / supplementary angles).
• An isosceles right triangle (equal legs, right angle between them) has base angles of $45\degree$ each.
• To find an angle using supplementary pairs, subtract the known angle from $180\degree$.

When comparing products of integers, the sign of each integer matters critically: multiplying two negatives gives a positive, and multiplying integers of opposite signs gives a negative. Given only that $w \lt x$ and $y \lt z$ with no sign constraints on any variable, the comparison $wy$ vs. $xz$ can go in any direction depending on the specific values chosen.

Case 1 – Quantity B greater

Let $w = 1, x = 2, y = 1, z = 2$: $w \lt x$ ✓, $y \lt z$ ✓.

Quantity A $= wy = 1$, Quantity B $= xz = 4$. Quantity B is greater.

Case 2 – Quantity A greater

Let $w = -4, x = -1, y = -2, z = -1$: $-4 \lt -1$ ✓, $-2 \lt -1$ ✓.

Quantity A $= wy = (-4)(-2) = 8$, Quantity B $= xz = (-1)(-1) = 1$. Quantity A is greater.

Conclusion

Since valid choices produce different comparison outcomes, the relationship cannot be determined from the given information alone.

Why the other conclusions don’t hold

Quantity A is greater: True in Case 2, false in Case 1. Incorrect.

Quantity B is greater: True in Case 1, false in Case 2. Incorrect.

The two quantities are equal: Not consistently true. Incorrect.

The relationship cannot be determined: Multiple comparison outcomes are achievable. Correct.

Things to Remember:

• An inequality $a \lt b$ does not imply $ac \lt bc$; if $c \lt 0$, multiplying reverses the inequality.
• When signs are unconstrained, test positive-positive and negative-negative cases to probe which quantity is larger.
• Proving “cannot be determined” requires at least two valid cases with different comparison outcomes.