GRE 2024 Quantitative Reasoning Practice Test Set 12 Question Paper with Solutions PDF

Step 1: Cookies to cupcakes ratio is \( 3:2 \).

Step 2: Pastries to cookies ratio is \( 6:5 \).

Step 3: To compare cupcakes to pastries, take LCM of cookies = 15.

So, \( 2 \times 5 = 10 \) cupcakes, and \( 6 \times 3 = 18 \) pastries.

Step 4: Simplify ratio \( 10:25 = 2:5 \).



Final Answer: \[ \boxed{2:5} \]

\begin{quicktipbox
Always align ratios by taking LCM of the common term for accurate comparison.
\end{quicktipbox Quick Tip: Always align ratios by taking LCM of the common term for accurate comparison.

Step 1: Let number of pairs bought be \( n \). Price per pair = \( \frac{50}{n} \).

Step 2: With 20% discount, price per pair = \( 0.8 \times \frac{50}{n} = \frac{40}{n} \).

Step 3: For same
)50, pairs = \( \frac{50}{40/n} = \frac{50n}{40} = \frac{5n}{4} \).

Step 4: This is 5 more pairs: \( \frac{5n}{4} = n+5 \).

Step 5: Solve: \( 5n = 4n + 20 \Rightarrow n = 20 \).


Final Answer: \[ \boxed{20} \]

\begin{quicktipbox
In discount problems, equating original and discounted purchase conditions is the key step.
\end{quicktipbox Quick Tip: In discount problems, equating original and discounted purchase conditions is the key step.

Step 1: Solve absolute equation: \( |x-3| = 3 \).

Step 2: Two cases:

1. \( x - 3 = 3 \Rightarrow x = 6 \).

2. \( x - 3 = -3 \Rightarrow x = 0 \).

Step 3: Compare with 2. For \( x=6 \), Quantity A > Quantity B. For \( x=0 \), Quantity B > Quantity A.

Thus, no unique relation.


Final Answer: \[ \boxed{Cannot be determined} \]

\begin{quicktipbox
Absolute value equations often give multiple solutions; always test each solution before comparing.
\end{quicktipbox Quick Tip: Absolute value equations often give multiple solutions; always test each solution before comparing.

Step 1: Formula is \( 4x - y^2 \).

Step 2: Check parities:

- If \( y \) is odd, \( y^2 \) odd, \( 4x - y^2 \) = even - odd = odd.

- If \( y \) is even, \( y^2 \) even, \( 4x - y^2 \) = even - even = even.

Step 3: For \( x \, \textcurrency \, 2y \), we substitute \( y' = 2y \). Then term = \( 4x - (2y)^2 = 4x - 4y^2 = 4(x-y^2) \), always even.


Final Answer: \[ \boxed{x \, \textcurrency \, 2y} \]

\begin{quicktipbox
When testing parity problems, check separately for even and odd substitutions.
\end{quicktipbox Quick Tip: When testing parity problems, check separately for even and odd substitutions.

Step 1: Simplify \( p = 4 \times 6 \times 11 \times n = 264n \).

Step 2: Find remainder when divided by 5. \( 264n \div 5 \). Since \( 264 \equiv 4 \pmod{5} \), remainder = \( 4n \pmod{5} \), varies with \( n \). Could be 0,1,2,3,4.

Step 3: Divide by 33. Since \( 264n = 33 \times 8n \), always divisible by 33, remainder = 0.

Step 4: Compare: Quantity B = 0 always. Quantity A varies but is nonnegative and can be \(>0 \). So Quantity \(A ≥0 \), Quantity B =0. But for \( n \) not multiple of 5, \(A > B \).


Final Answer: \[ \boxed{Quantity A is greater (except when \( n \) multiple of 5, then equal)} \]

\begin{quicktipbox
When comparing remainders, check divisibility carefully. Fixed divisibility often gives consistent results.
\end{quicktipbox Quick Tip: When comparing remainders, check divisibility carefully. Fixed divisibility often gives consistent results.

Step 1: Start with inequality.

We are given: \[ 26 \leq 2x < 64 \]

Step 2: Simplify inequality.

Divide the entire inequality by 2: \[ 13 \leq x < 32 \]

Step 3: Interpret graph.

The value of \( x \) is greater than or equal to 13 but strictly less than 32.

This means the number line must include 13 (closed dot) and exclude 32 (open dot).


Final Answer: \[ \boxed{The correct graph is (D).} \] Quick Tip: When solving inequalities, always divide or multiply carefully, especially with negative numbers, as inequality signs may flip.

Step 1: Find total apples needed.

Each pie requires 4 apples, and there are 300 pies: \[ 300 \times 4 = 1200 apples \]

Step 2: Find apples per bushel.

One bushel = 126 apples.

Step 3: Divide total apples by apples per bushel.
\[ \frac{1200}{126} \approx 9.52 \]

Step 4: Round up.

Since partial bushels cannot be purchased, Sam must buy 11 bushels.


Final Answer: \[ \boxed{11} \] Quick Tip: Always round up when dealing with real-world quantities like bushels or boxes—you cannot buy fractions of physical items.

Step 1: Total numbers possible.

We are arranging 4 digits (1, 2, 3, 4) to form 4-digit numbers. The total possible numbers: \[ 4! = 24 \]

Step 2: Contribution of each digit.

Each digit appears equally in each place value (thousands, hundreds, tens, ones).
So, in each place: \[ \frac{24}{4} = 6 times each digit. \]

Step 3: Sum of digits.

The sum of the digits is: \[ 1 + 2 + 3 + 4 = 10 \]

Step 4: Place value contributions.

Each place value sum = \( 6 \times 10 = 60 \).
Thus, the total contribution = \[ 60 \times (1000 + 100 + 10 + 1) = 60 \times 1111 = 66,660 \]

Step 5: Multiply by number of sets.

We already accounted for all 24 numbers, so total sum = \[ 66,660 \times 24 = 711,040 \]


Final Answer: \[ \boxed{711,040} \] Quick Tip: When all digits are used in all positions equally, calculate their frequency in each place and multiply by place values systematically.

Step 1: Formula for nth term.

For an arithmetic sequence: \[ a_n = a_1 + (n-1)d \]
Here \( a_1 = 30 \), \( a_9 = 126 \).

Step 2: Use given values.
\[ 126 = 30 + (9-1)d \quad \Rightarrow \quad 126 = 30 + 8d \] \[ 96 = 8d \quad \Rightarrow \quad d = 12 \]

Step 3: Write recurrence relation.

So, \[ s_n = s_{n-1} + 12 \]


Final Answer: \[ \boxed{s_n = s_{n-1} + 12} \] Quick Tip: Always use the \( a_n = a_1 + (n-1)d \) formula to check consistency of linear (arithmetic) sequences.

Step 1: Find cereal eaten.

If 22.8% is left, then cereal eaten = \[ 100% - 22.8% = 77.2% \]

Step 2: Convert to decimal.
\[ 77.2% = 0.772 \]

Step 3: Approximate.

Among the options, 0.77 is the closest decimal representation.


Final Answer: \[ \boxed{0.77} \] Quick Tip: Always subtract from 100% to find the "used" or "eaten" portion, then convert to decimal by dividing by 100.

Step 1: Calculate the total score of 5 students.

Average = 85, Number of students = 5.

So, Total = \(85 \times 5 = 425\).


Step 2: Calculate the total score of 7 students after new average.

Average = 88, Number of students = 7.

So, Total = \(88 \times 7 = 616\).


Step 3: Calculate the combined contribution of 2 new students.
\(616 - 425 = 191\).


Step 4: Minimize one score by maximizing the other.

If one student scores 100 (maximum), the other scores \(191 - 100 = 91\).


Final Answer: \[ \boxed{91} \] Quick Tip: To find the minimum possible score of one student, maximize the score of the other. This is a common trick in average problems.

Step 1: Total marbles.
\(10 + 15 + 12 = 37\).


Step 2: Total ways to draw 2 marbles.
\(\binom{37}{2} = \frac{37 \times 36}{2} = 666\).


Step 3: Probability of drawing 2 marbles of same color.

- Red: \(\binom{10}{2} = 45\).

- Green: \(\binom{15}{2} = 105\).

- Blue: \(\binom{12}{2} = 66\).

Total same-color = \(45+105+66 = 216\).


Step 4: Probability of same color.
\(\frac{216}{666} \approx 32.43%\).


Step 5: Probability of different colors.
\(100% - 32.43% = 67.57%\).


Final Answer: \[ \boxed{67.57%} \] Quick Tip: Whenever asked for “different colors”, compute the complement of the probability of “same colors”.

Step 1: Choices for 2 distinct digits.

There are 10 digits, so ways = \(\binom{10}{2} = 45\).

Step 2: Choices for 1 uppercase letter.

There are 26 uppercase letters.

Step 3: Remaining 3 characters from 62 total (26 uppercase + 26 lowercase + 10 digits) minus used.

Used = 3 characters. Remaining = 59. So ways = \(P(59,3) = 59 \times 58 \times 57\).

Step 4: Multiply possibilities.

Total = \(45 \times 26 \times (59 \times 58 \times 57)\).
After simplifying, it equals 456426360.

Final Answer: \[ \boxed{456426360} \] Quick Tip: When no repetition is allowed, always reduce the available character set for each subsequent pick.

Step 1: Total of Set A.
\(25 \times 50 = 1250\).

Step 2: Total of Set B.
\(75 \times 100 = 7500\).

Step 3: Combined total and mean.

Total = \(1250 + 7500 = 8750\).

Mean = \(\frac{8750}{100} = 87.5\).

Step 4: Compare with Quantity B.

87.5 vs 80. Quantity A is greater.

Correction: The actual correct option should be (A) not (B).

Final Answer: \[ \boxed{Quantity A is greater.} \] Quick Tip: When combining averages, always use weighted sums, not just the average of averages.

Step 1: Arrange the numbers in ascending order.

1, 3, 3, 4, 6, 8, 11, 12, 15.

Step 2: Calculate range.
\(a = 15 - 1 = 14\).

Step 3: Find mean.

Sum = 63. Mean = \( \frac{63}{9} = 7\). So, \(b = 7\).

Step 4: Find median.

Middle element (5th) = 6. So, \(c = 6\).

Step 5: Find mode.

Most frequent number = 3. So, \(d = 3\).

Step 6: Order them.
\(d = 3 < c = 6 < b = 7 < a = 14\).

Final Answer: \[ \boxed{d < c < b < a} \] Quick Tip: Always compute mode, median, mean, and range carefully, then compare numerically to form inequalities.