Test: Quantitative Comparison- 1 - GRE MCQ

Column A: In Column A, we need to select 2 students from a class of 36 to be chosen as monitors. The number of ways to select 2 students from 36 can be calculated using the combination formula:

Column B: In Column B, we have 18 girls and 18 boys in a class of 36. We need to select one student from the girls and one from the boys, making the number of ways:

18 × 18 = 324

Comparison:

• Column A has 630 ways.
• Column B has 324 ways.

Therefore, Quantity A is greater than Quantity B.

The correct answer is: A: Quantity A is greater

Column A: In Column A, we need to determine the total number of ways in which six objective-type questions can be answered if each question has 5 choices. Since each question is independent of the others and has 5 choices, we calculate the total number of ways as:

5⁶ = 15,625

Column B: In Column B, there are 5 multiple-choice questions with varying numbers of choices:

• The first question has 3 choices.
• The next two questions each have 4 choices.
• The last two questions each have 5 choices.

The total number of possible answer sequences can be calculated by multiplying the choices for each question:

3 × 4 × 4 × 5 × 5 = 3 × 16 × 25 = 1,200

Comparison

- Column A has 15,625 possible ways.
- Column B has 1,200 possible ways.

Therefore, the quantity in Column A is greater than the quantity in Column B.

Answer: A: The quantity on the left is greater

Column A: We have four digits (1, 2, 4, and 9), and we need to form three-digit numbers. Since repetition is not mentioned, we assume each digit can be used only once.

• For the hundreds place, we have 4 options (1, 2, 4, or 9).
• For the tens place, we have 3 remaining choices after selecting the hundreds digit.
• For the units place, we have 2 remaining choices after selecting the hundreds and tens digits.

The total number of ways for Column A is:
4 × 3 × 2 = 24

Column B: The digits available are 1, 2, 4, and 9 (since 0, 3, 5, 6, 7, and 8 cannot be used). We have the same 4 digits and are forming three-digit numbers.

• The process and calculations for Column B are identical to those in Column A.

Therefore, the total number of ways for Column B is also:
4 × 3 × 2 = 24

Both Column A and Column B yield the same number of three-digit numbers, which is 24.

Answer:

C: Both are equal

Column A: The teacher has three duties (A, B, and C) to assign to three students. Since each student can take one unique duty, this is a permutation problem.
The number of ways to assign three duties to three students is:
3! = 3 × 2 × 1 = 6

Column B: We are asked to find the total ways in which three posts can be filled by three candidates, which is also a permutation problem because each post will be assigned to one candidate.
The number of ways to fill three posts with three candidates is:
3! = 3 × 2 × 1 = 6

Both columns have the same quantity, 6.

Answer:

C: Both are equal

Column A: In Column A, we are asked in how many ways 3 prizes can be distributed among 5 boys if no boy can get more than one prize. Here, we need to choose 3 out of 5 boys to receive a prize, and each prize goes to a different boy.

The number of ways to choose 3 boys from 5 is given by the combination formula:

So, Column A has 10 ways.

Column B: In Column B, we need to distribute 5 prizes among the students of a school, but no additional information is given about the number of students or any restrictions on the distribution (e.g., if one student can receive more than one prize). Without knowing the number of students or whether multiple prizes can go to the same student, we cannot determine a specific number of ways.

Comparison: Since we do not have enough information to calculate the number of ways in Column B, we cannot directly compare it with Column A.

Therefore, the correct answer is: D: The relationship cannot be determined without further information

Column A: The word "PROMISE" has 7 letters, with 3 vowels (O, I, E) and 4 consonants (P, R, M, S). We need to arrange these letters so that no two vowels are adjacent.

1. First, arrange the consonants. There are 4 consonants, and they can be arranged in 4! = 24 ways.
2. Placing the consonants creates 5 possible positions for the vowels (before the first consonant, between each pair of consonants, and after the last consonant).
3. We choose 3 of these 5 positions for the vowels. The vowels can be arranged in these 3 positions in 3! = 6 ways.

So, the total number of ways to arrange the letters in "PROMISE" such that no two vowels are adjacent is:

24 × 6 = 144

Thus, Column A has 144 arrangements.

Column B: In Column B, we are asked to form words from the letters I, N, U, R, E, V, L, A, and S. This set of letters has 9 unique letters, so the number of distinct arrangements of these 9 letters is:

9! = 362,880

Comparison:

- Column A has 144 arrangements.
- Column B has 362,880 arrangements.

Therefore, the quantity in Column B is greater than the quantity in Column A.

Answer: B: The quantity on the right is greater

Therefore, the number of ways Sunita can make her purchases is greater than the number of ways Shelly can make her purchases. The correct answer is:

Option A: The quantity on the left is greater.

Column A: In Column A, we have 5 buses running between two cities, A and B. We need to find the number of ways a man can go by one bus and return by a different bus.

1. For the journey from city A to city B, he has 5 choices of buses.
2. For the return journey, he has 4 choices because he must take a different bus.

The total number of ways to go by one bus and return by a different one is:

5 × 4 = 20

Thus, Column A has 20 possible ways.

Column B: In Column B, we need to find the number of ways a man can go by one bus and return by any of the five buses, including the same bus.

1. For the journey from city A to city B, he has 5 choices.
2. For the return journey, he also has 5 choices because he can return by any bus, including the one he used for the journey to city B.

The total number of ways to go by one bus and return by any bus is:

5 × 5 = 25

Thus, Column B has 25 possible ways.

Comparison:

- Column A has 20 ways.
- Column B has 25 ways.

Therefore, the quantity in Column B is greater than the quantity in Column A.

Answer: B: The quantity on the right is greater

Given Information:

We need to create three-digit numbers from the digits 1, 4, 5, 6, 7, and 8, and the digits can be repeated.

Column A: Three-Digit Odd Numbers

To form a three-digit odd number, the last digit (units place) must be an odd digit. From the given digits, the odd digits are 1, 5, and 7.

1. Units place (odd digit): We have 3 choices (1, 5, or 7).
2. Hundreds place: We can choose any of the 6 digits (1, 4, 5, 6, 7, or 8), so we have 6 choices.
3. Tens place: We can again choose any of the 6 digits, so we have 6 choices.

The total number of three-digit odd numbers is:

6 × 6 × 3 = 108

Thus, Column A has 108 possible numbers.

Column B: Three-Digit Even Numbers

To form a three-digit even number, the last digit (units place) must be an even digit. From the given digits, the even digits are 4, 6, and 8.

1. Units place (even digit): We have 3 choices (4, 6, or 8).
2. Hundreds place: We can choose any of the 6 digits, so we have 6 choices.
3. Tens place: We can again choose any of the 6 digits, so we have 6 choices.

The total number of three-digit even numbers is:

6 × 6 × 3 = 108

Thus, Column B also has 108 possible numbers.

Comparison:

- Column A has 108 ways.
- Column B has 108 ways.

Therefore, both quantities are equal.

Answer: C: Both are equal

Given Information:

We need to form two-digit even numbers using the digits 2, 3, 4, and 5.

In Column A, repetition of digits is allowed. In Column B, repetition is not allowed.

Column A: Two-Digit Even Numbers (Repetition Allowed)

To form a two-digit even number, the units place must be an even digit. From the given digits, the even digits are 2 and 4.

1. Units place (even digit): We have 2 choices (2 or 4).
2. Tens place: Since repetition is allowed, we can use any of the 4 digits (2, 3, 4, or 5), so we have 4 choices.

The total number of two-digit even numbers is:

4 × 2 = 8

Thus, Column A has 8 possible numbers.

Column B: Two-Digit Even Numbers (Repetition Not Allowed)

For a two-digit even number without repetition, the units place must again be an even digit.

1. Units place (even digit): We have 2 choices (2 or 4).
2. Tens place: Since repetition is not allowed, we can only choose from the remaining 3 digits for the tens place.

The total number of two-digit even numbers is:

3 × 2 = 6

Thus, Column B has 6 possible numbers.

Comparison:

- Column A has 8 ways.
- Column B has 6 ways.

Therefore, the quantity in Column A is greater than the quantity in Column B.

Answer: A: The quantity on the left is greater

Column A: We are given the letters I, I, H, O, N, and R, and we need to form different words using these letters. Since there are 6 letters and we are using all of them, we need to account for the repetition of "I."

The number of distinct arrangements of these letters can be calculated using the formula for permutations of multiset:

(6! / 2!) = (720 / 2) = 360

Thus, Column A has 360 possible words.

Column B: We are given the letters N, U, M, B, E, and R and need to form 4-letter words using these letters without repetition. We choose 4 letters out of the 6, and then arrange those 4 letters.

1. Choose 4 letters out of 6: This can be done in binom(6, 4) ways.
2. binom(6, 4) = 15
3. Arrange the 4 chosen letters: Each set of 4 letters can be arranged in 4! ways.
4. 4! = 24

The total number of 4-letter words is:

15 × 24 = 360

Thus, Column B also has 360 possible words.

Comparison:

- Column A has 360 possible words.
- Column B has 360 possible words.

Therefore, both quantities are equal.

Answer: C: Both are equal

Column A: We need to form numbers less than 1000 using the digits 0, 2, 3, and 4, with repetition allowed. The digit 0 cannot be considered a one-digit number, so it cannot be used alone.

Calculating the Possible Numbers:

1. One-digit numbers: The only one-digit numbers are 2, 3, and 4.
   Total: 3 ways
2. Two-digit numbers: The tens place can be filled by 2, 3, or 4, and the units place by 0, 2, 3, or 4.
   Total: 3 × 4 = 12 ways
3. Three-digit numbers: The hundreds place can be filled by 2, 3, or 4, and the tens and units places by 0, 2, 3, or 4.
   Total: 3 × 4 × 4 = 48 ways

The total number of numbers in Column A is:

3 + 12 + 48 = 63

Thus, Column A has 63 possible numbers.

Column B: We need to form three-digit numbers using the digits 1, 2, 3, and 4, with repetition allowed.

1. Hundreds place: Can be filled by any of the 4 digits (1, 2, 3, or 4).
2. Tens place: Can also be filled by any of the 4 digits (1, 2, 3, or 4).
3. Units place: Can be filled by any of the 4 digits (1, 2, 3, or 4).

The total number of three-digit numbers in Column B is:

4 × 4 × 4 = 64

Thus, Column B has 64 possible numbers.

Comparison:

- Column A has 63 ways.
- Column B has 64 ways.

Therefore, the quantity in Column B is greater than the quantity in Column A.

Answer: B: The quantity on the right is greater

Step 1 of solving this GRE question : Find the smallest positive integer that has 7 factors

The number of factors of number N is 7. 7 is a prime number.
Note: If a number N can be prime factorized as a^p × b^q, where 'a' and 'b' are prime factors of N, number of factors of N = (p + 1) (q + 1)
So, 7 = 1 × 7 is the only way to express 7 as a product of 2 numbers

Inference: p + 1 = 1 and q + 1 = 7 or p = 0 and q = 6
So, any number that has 7 factors will have p = 0 and q = 6.
i.e., the number will have only one prime factor.
2 is the smallest prime number. So, the smallest number that will have 7 factors, N = 2⁶ = 64.

Step 2 of solving this GRE question : Evaluate Column A

Compute the number of factors of √N
N = 64
√N = √64 = 8
Prime factorize 8, we get 8 = 2³
Number of factors of 8 = (3 + 1) = 4
Value of column A is 4.

Step 3 of solving this GRE question : Evaluate Column B

Compute the number of factors of (N – 2),
N = 64
N - 2 = 64 – 2 = 62
Prime factorize 62, we get 62 = 2 × 31
Number of factors of 62 = (1 + 1) × ( 1 + 1) = 2 × 2 = 4
Value of column B is 4.

Step 4 of solving this GRE question : The Comparison

Column A: Number of factors of √N = 4
Column B: Number of factors of (N – 2) = 4
Both columns are equal.

Choice C is the correct answer

Column A: In Column A, there are 5 items in each of columns A and B, and a student is asked to match the columns. The total number of possible ways to match 5 items with 5 items is given by the number of permutations of 5, which is:

5! = 120

Thus, Column A has 120 possible answers.

Column B: In Column B, there are 6 items in each of columns A and B, and the student is asked to match the columns. The total number of possible ways to match 6 items with 6 items is given by the number of permutations of 6, which is:

6! = 720

Thus, Column B has 720 possible answers.

Comparison:

- Column A has 120 possible answers.
- Column B has 720 possible answers.

Therefore, the quantity in Column B is greater than the quantity in Column A.

Answer: B: The quantity on the right is greater

Column A: In Column A, we need to select 1 boy from a group of 27 boys and 2 girls from a group of 14 girls.

1. Selecting 1 boy from 27: This can be done in binom(27, 1) = 27 ways.
2. Selecting 2 girls from 14: This can be done in binom(14, 2) = 91 ways.

The total number of ways to make the choice in Column A is:

27 × 91 = 2457

Thus, Column A has 2457 possible ways.

Column B: In Column B, we need to select 2 boys from a group of 27 boys and 1 girl from a group of 14 girls.

1. Selecting 2 boys from 27: This can be done in binom(27, 2) = 351 ways.
2. Selecting 1 girl from 14: This can be done in binom(14, 1) = 14 ways.

The total number of ways to make the choice in Column B is:

351 × 14 = 4914

Thus, Column B has 4914 possible ways.

Comparison:

- Column A has 2457 ways.
- Column B has 4914 ways.

Therefore, the quantity in Column B is greater than the quantity in Column A.

Answer: B: The quantity on the right is greater