GRE 2024 Quantitative Reasoning Practice Test Set 8 Question Paper with Solutions PDF

Step 1: Recall the triangle inequality theorem.

In any triangle, the length of one side must be less than the sum of the other two sides and greater than their difference.

So if the given two sides are 8 and 13, then for the unknown side \( x \): \[ |13 - 8| < x < 13 + 8 \]


Step 2: Simplify the inequality.
\[ 5 < x < 21 \]


Step 3: Check the given options.

- (A) \(2\): Not valid, since \(2 < 5\).

- (B) \(4\): Not valid, since \(4 < 5\). Wait correction → \(4\) is also less than 5, so not valid.

- (C) \(6\): Valid, since \(6\) is between 5 and 21.

- (D) \(9\): Valid, since \(9\) is between 5 and 21.

- (E) \(15\): Valid, since \(15\) is between 5 and 21.

- (F) \(22\): Not valid, since \(22 > 21\).


So the correct possible lengths are \(6, 9, 15\).


Final Answer: \[ \boxed{6, \, 9, \, 15} \] Quick Tip: Always apply the triangle inequality: for any triangle with sides \(a, b, c\), each side must satisfy \(|a-b| < c < a+b\).

Step 1: Identify properties of P and Q.

- \( P \) is a negative odd number.
- \( Q \) is a prime other than 2, so \( Q \) is an odd positive integer.

Step 2: Multiply P and Q.

- Odd × Odd = Odd.
- Negative × Positive = Negative.
So \( PQ \) must be a negative odd integer.

Step 3: Check each option.

- (A) True, because PQ is negative odd integer.
- (B) False, because product of two numbers greater than 1 cannot be prime.
- (C) True, every integer (including negative) has a real square root if we consider absolute value (but not in natural numbers). In real numbers, \(\sqrt{|PQ|}\) exists.
- (D) False, PQ cannot be even since both P and Q are odd.

Thus, incorrect options are (B) and (D).


Final Answer: \[ \boxed{B, D} \] Quick Tip: Remember: Odd × Odd = Odd, Odd × Even = Even. Always apply sign rules carefully with negative numbers.

Step 1: Analyze the inequality.

We are given: \[ x < y < x+y \]
From \(y < x+y\), we conclude that \(y > 0\).

Thus, \(y\) is always positive, and \(x\) may be either negative or positive, but smaller than \(y\).



Step 2: Check each option.


- (A) \(-2x\): If \(x < 0\), then \(-2x > 0\) (positive). If \(x > 0\), then \(-2x < 0\) (negative). Therefore, this is not always negative.


- (B) \(-y\): Since \(y > 0\), then \(-y < 0\). This is always negative.


- (C) \(x-y\): Since \(x < y\), subtracting gives \(x-y < 0\). This is always negative.


- (D) \(2x-y\): Consider \(x=5\), \(y=6\). Then \(2x-y = 10-6=4 > 0\), not always negative.


- (E) \(3x\): If \(x < 0\), then \(3x < 0\). If \(x > 0\), then \(3x > 0\). So not always negative.




Step 3: Conclude the must-be-negative expressions.

The only expressions guaranteed to be negative under the given condition are: \[ -y \quad and \quad x-y \]


Final Answer: \[ \boxed{-y and x-y} \] Quick Tip: When solving inequality-based problems, first deduce which variable must always be positive or negative. Then test each option systematically under those conditions.

Step 1: Apply LCM rule.

If a number is divisible by 3 and 14, then it must be divisible by LCM(3,14). \[ LCM(3,14) = 42 \]

So any valid number is a multiple of 42.


Step 2: Check each option.

- (A) Divisible by 6: Since 42 is divisible by 6, the number must be divisible by 6.

- (B) Always divisible by 2: Since 42 is even, multiples of 42 are always even.

- (C) Divisible by 22: Not necessary (42 not divisible by 22).

- (D) Divisible by 3 \& 14 only: Wrong, because multiples of 42 will also be divisible by other numbers like 7, 21 etc.

- (E) Factor of 5: No such requirement.


Thus correct options are (A) and (B).


Final Answer: \[ \boxed{A, B} \] Quick Tip: If a number is divisible by two numbers, check divisibility using their LCM. This avoids missing extra factors.

Step 1: Factorize 6!

We know that: \[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \]
Prime factorization of \( 720 \): \[ 720 = 2^4 \times 3^2 \times 5 \]


Step 2: Condition for \( \frac{6!}{3^m} \) to be an integer.

For the expression to be an integer, \( 3^m \) must divide \( 6! \). Since the highest power of 3 in the factorization is \( 3^2 \), the largest possible value of \( m \) is: \[ m = 2 \]

Step 3: Compare options with the largest \( m \).

Largest possible \( m = 2 \). Options greater than 2 are: 3, 4, and 5.


Step 4: Eliminate carefully.

The question asks for values \emph{greater than the largest possible \( m \). Thus the correct answers are: \[ \boxed{4 and 5} \] Quick Tip: Always use prime factorization to determine the maximum power of a prime dividing a factorial.

Step 1: Understanding the problem.

We are asked about the possible 2-digit numbers that can be formed using 5 distinct digits.

Step 2: Smallest possible two-digit number.

The smallest 2-digit number that can be formed using 5 distinct digits must be at least 12 (since the tens digit cannot be 0).

Step 3: Compare with options.

- Option (1) 10: Less than 12, so valid.

- Option (2) 15: Also less than many possible 2-digit values, valid.

- Option (3) 20: Less than many possible 2-digit numbers, valid.

- Option (4) 40: Less than 2-digit numbers formed later, valid.

- Option (5) 60: Not necessarily less than all, so not correct.

Thus, correct answers are 10, 15, 20, and 40. Quick Tip: When working with problems involving digits, always test the extreme smallest and largest values to narrow down possibilities.

Step 1: Represent relationships.

Let Mark’s age = \( M \). Then Henderson’s age = \( M + 3 \). Alisa’s age = \( M + 9 \).

Step 2: Use given condition.

Alisa is at least 33: \[ M + 9 \geq 33 \quad \Rightarrow \quad M \geq 24 \]
So Henderson = \( M + 3 \geq 27 \).


Step 3: Check options.

- (1) 21: Too small, not possible.

- (2) 24: Too small, not possible.

- (3) 26: Possible if \( M = 23 \), but then Alisa = 32 (<33), so not valid.

- (4) 27: Possible, \( M = 24 \), Alisa = 33, valid.

- (5) 29: Possible, \( M = 26 \), Alisa = 35, valid.

- (6) 35: Possible only if \( M = 32 \), Alisa = 41, valid too.


So valid answers are 27, 29, and 35. Quick Tip: Translate all given conditions into equations and inequalities before checking each option—it avoids careless mistakes.

Step 1: Express sweets given to Elena.

Elena receives \( \frac{4}{7}x \). Thus, remaining sweets after Elena: \[ x - \frac{4}{7}x = \frac{3}{7}x \]

Step 2: Express sweets given to Michael.

Michael receives \( \frac{3}{5} \times \frac{3}{7}x = \frac{9}{35}x \).

Step 3: Conditions for divisibility.

For both Elena and Michael to receive integers, \( x \) must be divisible by 35 (LCM of 7 and 35).

Step 4: Check given options.

- 15: Not divisible by 35.

- 25: Not divisible by 35.

- 35: Divisible by 35.

- 56: Not divisible by 35.

- 70: Divisible by 35.

- 95: Not divisible by 35.


Thus, valid options are 35 and 70. Quick Tip: In problems involving fractions of a total, always check divisibility by the denominators (or their LCM) to ensure integer results.

Step 1: Represent relationships.

Let number of female students = \( F \).

Then number of male students = \( 0.5F \).

Number of faculty = \( 0.6 \times (0.5F) = 0.3F \).

Step 2: Total members.

Total = \( F + 0.5F + 0.3F = 1.8F \).

Step 3: Apply condition.

Since total ≥ 150, \[ 1.8F \geq 150 \quad \Rightarrow \quad F \geq \frac{150}{1.8} = 83.\overline{3} \]
So \( F \) must be a whole number \(≥ \) 84.

Step 4: Faculty = 0.3F must be an integer.

For this, \( F \) must be divisible by 10.

Step 5: Check options.

- If \( F = 70 \), Faculty = 21 (valid).

- If \( F = 90 \), Faculty = 27 (valid).

Other options don’t satisfy the ratio conditions.

Thus, valid answers are 21 and 27. Quick Tip: When multiple categories are expressed as percentages of one another, always express all in terms of a single base variable and then simplify.

Step 1: Define ranges.

Since \( -13 < a < -2 \), possible integer values for \( a \) are: \(-12, -11, -10, -9, -8, -7, -6, -5, -4, -3\).
Since \( 1 < b < 9 \), possible integer values for \( b \) are: \(2, 3, 4, 5, 6, 7, 8\).

Step 2: Product range.

Smallest product = \(-12 \times 8 = -96\).

Largest product = \(-3 \times 2 = -6\).

So possible values of \( ab \) are between -96 and -6.

Step 3: Check options.

- -20: Falls in range, possible.

- -18: Falls in range, possible.

- -15: Falls in range, possible.

- -14: Falls in range, possible.

- -13: Not possible (not divisible by given integers).

- -9: Not possible (since \(a\) is negative and \(b\) is positive, but no integer pair matches).

Thus, correct answers are -20, -18, -15, -14. Quick Tip: For product range questions, calculate extreme values and then verify each option within that interval carefully.