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GRE Math Challenge #127- f(t) = kt for all t, where k
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Updated on: 20 Feb 2020, 15:56
Updated on: 20 Feb 2020, 15:56
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13% (00:58) wrong based on 92 sessions
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\(f(t) = kt\) for all t, where k is a constant, and \(f(3)= \frac{1}{2}\)
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
Quantity A |
Quantity B |
\(k\) |
\(f(1)\) |
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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: GRE Math Challenge #127- f(t) = kt for all t, where k
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13 Jun 2019, 05:02
13 Jun 2019, 05:02
sandy wrote:
f(t) = kt for all t, where k is a constant, and f(3)= \(\frac{1}{2}\)
Quantity A: k
Quantity B: f(1)
Quantity A: k
Quantity B: f(1)
GIVEN: f(t) = kt for all t, where k is a constant
So, f(3) = 3k
Since we're told that f(3) = \(\frac{1}{2}\), we can conclude that 3k = \(\frac{1}{2}\)
Divide both sides by 3 to get: \(k = \frac{1}{6}\)
So, we can now say that f(t) \(= \frac{1}{6}t\)
We get:
QUANTITY A: \(\frac{1}{6}\)
QUANTITY B: f(1) \(= \frac{1}{6}(1) = \frac{1}{6}\)
Answer: C
Cheers,
Brent
Re: GRE Math Challenge #127- f(t) = kt for all t, where k
[#permalink]
03 Jul 2020, 15:34
03 Jul 2020, 15:34
The exercise gives a function f(t)=kt.
And ask you to compare f(1) with k
Replacing t=1 in the function
f(1)=kx1=k
Answer: Quantities are equal (independent of the numeric value of k)
And ask you to compare f(1) with k
Replacing t=1 in the function
f(1)=kx1=k
Answer: Quantities are equal (independent of the numeric value of k)
