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GRE Math Challenge #50- The operation and is defined for all
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Updated on: 18 Feb 2020, 16:21
Updated on: 18 Feb 2020, 16:21
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The operation \(\S\) is defined for all positive numbers r and t by
\(r \S t = \frac{(r-t)^2+ rt}{t}\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
\(r \S t = \frac{(r-t)^2+ rt}{t}\)
Quantity A |
Quantity B |
\(71\S37\) |
\(37\S71\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Kudos for the right answer and explanation
Originally posted by sandy on 03 May 2015, 00:28.
Last edited by GreenlightTestPrep on 18 Feb 2020, 16:21, edited 7 times in total.
Last edited by GreenlightTestPrep on 18 Feb 2020, 16:21, edited 7 times in total.
Updated
Re: GRE Math Challenge #50- The operation and is defined for all
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11 Mar 2018, 07:13
11 Mar 2018, 07:13
sandy wrote:
The operation & is defined for all positive numbers r and t by
\(r&t= \frac{(r-t)^2+ rt}{t}\)
Quantity A: 71•37
Quantity B: 37•71
• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
\(r&t= \frac{(r-t)^2+ rt}{t}\)
Quantity A: 71•37
Quantity B: 37•71
• Quantity A is greater.
• Quantity B is greater.
• Both Quantities are Equal
• Cannot be determined
First, we ought to know two things.
For qaunt A and B, the numerator part would be the same,
Second, if the numerator part is the same the one with smaller denominator is bigger.
Hence A>B
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Re: GRE Math Challenge #50- The operation and is defined for all
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18 Feb 2020, 16:20
18 Feb 2020, 16:20
sandy wrote:
The operation \(\S\) is defined for all positive numbers r and t by
\(r \S t = \frac{(r-t)^2+ rt}{t}\)
\(r \S t = \frac{(r-t)^2+ rt}{t}\)
Quantity A |
Quantity B |
\(71\S37\) |
\(37\S71\) |
Given: \(r \S t = \frac{(r-t)^2+ rt}{t}\)
So, we get:
QUANTITY A: \(71 \S 37 = \frac{(71-37)^2+ (71)(37)}{37}=\frac{(34)^2+ (71)(37)}{37}\)
QUANTITY B: \(37 \S 71 = \frac{(37-71)^2+ (37)(71)}{71}=\frac{(-34)^2+ (37)(71)}{71}\)
If we recognize that \((34)^2 = (-34)^2\) and \((71)(37)=(37)(71)\), we can see that both quantities have the exact same numerator
Since the denominator in Quantity A is less than the denominator in Quantity B, we can conclude that Quantity A > Quantity B
Answer: A
Cheers,
Brent
