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A=\{1,2,3, \ldots, m\} and B=\{1,2,3, \ldots, n\}
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12 Sep 2026, 08:17
12 Sep 2026, 08:17
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50% (02:17) wrong based on 4 sessions
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\(A=\{1,2,3, \ldots, m\}\) and \(B=\{1,2,3, \ldots, n\}\) , where m is a positive integer that leaves a remainder of 2 when divided by 3 , and n is a positive integer that is a multiple of 3.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Percentage of numbers in A that are multiples of 3 |
Percentage of numbers in B that are multiples of 3 |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
- Part of the project: GRE Daily Quant Challenge ALL 4 CORE QUANT AREAS
- Also replying to the unanswered questions
Re: A=\{1,2,3, \ldots, m\} and B=\{1,2,3, \ldots, n\}
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01 Oct 2026, 02:46
01 Oct 2026, 02:46
Expert Reply
Let:
- m leaves remainder 2 when divided by 3 , so m=3 k+2 .
- n is a multiple of 3 , so n=3 j .
Quantity A: Percentage of numbers in \(A=\{1,2, \ldots, m\}\) that are multiples of 3.
In each full block of 3 numbers, there is exactly 1 multiple of 3 .
Since m=3 k+2 , the set A has k complete blocks of 3, plus two extra numbers 3 k+1,3 k+2 , neither of which is a multiple of 3 .
So the number of multiples of 3 in A is k .
Percentage A:
\(\frac{k}{3 k+2} \times 100\)
Quantity B: Percentage of numbers in \(B=\{1,2, \ldots, n\}\) that are multiples of 3, where n=3 j . There are j multiples of 3 in B , and total 3 j numbers.
Percentage B:
\(\frac{j}{3 j} \times 100=\frac{1}{3} \times 100 \approx 33.33 \) %
Now compare \( \frac{k}{3 k+2}\) with \(\frac{1}{3}\) :
\(\frac{k}{3 k+2}<\frac{k}{3 k}=\frac{1}{3}\)
So Quantity A is less than Quantity B.
B) Quantity B is greater.
Answer: (B)
- m leaves remainder 2 when divided by 3 , so m=3 k+2 .
- n is a multiple of 3 , so n=3 j .
Quantity A: Percentage of numbers in \(A=\{1,2, \ldots, m\}\) that are multiples of 3.
In each full block of 3 numbers, there is exactly 1 multiple of 3 .
Since m=3 k+2 , the set A has k complete blocks of 3, plus two extra numbers 3 k+1,3 k+2 , neither of which is a multiple of 3 .
So the number of multiples of 3 in A is k .
Percentage A:
\(\frac{k}{3 k+2} \times 100\)
Quantity B: Percentage of numbers in \(B=\{1,2, \ldots, n\}\) that are multiples of 3, where n=3 j . There are j multiples of 3 in B , and total 3 j numbers.
Percentage B:
\(\frac{j}{3 j} \times 100=\frac{1}{3} \times 100 \approx 33.33 \) %
Now compare \( \frac{k}{3 k+2}\) with \(\frac{1}{3}\) :
\(\frac{k}{3 k+2}<\frac{k}{3 k}=\frac{1}{3}\)
So Quantity A is less than Quantity B.
B) Quantity B is greater.
Answer: (B)
gmatclubot
Re: A=\{1,2,3, \ldots, m\} and B=\{1,2,3, \ldots, n\} [#permalink]
01 Oct 2026, 02:46
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