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x/3 is a square of a prime number and 15 < x < 195
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12 Sep 2021, 21:52
12 Sep 2021, 21:52
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\(\frac{x}{3}\) is a square of a prime number and \(15 < x < 195\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
OA edited on 14-Sep 2021
Quantity A |
Quantity B |
Number of possible values of \(x\) |
\(4\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
OA edited on 14-Sep 2021
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: x/3 is a square of a prime number and 15 < x < 195
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13 Sep 2021, 07:08
13 Sep 2021, 07:08
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KarunMendiratta wrote:
\(\frac{x}{3}\) is a square of a prime number and \(15 < x < 195\)
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Number of possible values of \(x\) |
\(4\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Some squares of prime numbers:
2² = 4
3² = 9
5² = 25
7² = 49
11² = 121
.
.
.
Next, we are told that \(15 < x < 195\)
Divide all quantities by 3 to get: \(5 < \frac{x}{3} < 65\)
Since, \(\frac{x}{3}\) is a square of a prime number, and since \(5 < \frac{x}{3} < 65\), we can see that:
\(\frac{x}{3}\) COULD equal 9 or 25 or 49
If \(\frac{x}{3}\) = 9, then x = 27
If \(\frac{x}{3}\) = 25, then x = 75
If \(\frac{x}{3}\) = 49, then x = 147
So, there are 3 possible values of x.
So, we have:
QUANTITY A: 3
QUANTITY B: 4
Answer: A
