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In a certain parking lot at 10:00 am, the ratio of trucks to
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16 Jan 2020, 01:35
16 Jan 2020, 01:35
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74% (01:03) correct
26% (00:58) wrong based on 50 sessions
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In a certain parking lot at 10:00 am, the ratio of trucks to cars was 1 to 3. Between 10:00 am and 11:00 am, no vehicle left the parking lot, and, for some integer N > 0, N more cars and N more trucks entered the lot and parked there.
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
Quantity A |
Quantity B |
the ratio of trucks to cars in the lot at 11:00 |
\(\frac{1}{3}\) |
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
Re: In a certain parking lot at 10:00 am, the ratio of trucks to
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16 Jan 2020, 06:08
16 Jan 2020, 06:08
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In ratios, when \(numerator < denominator\) that is \(\frac{numerator}{denominator}<1\) and we add a quantity X to both numerator and denominator
it becomes greater than the original fraction.
Consider the number line,
0------------1
Now, we take a fraction \(\frac{1}{3}\) which is between 0 and 1.
We add the same quantity N > 0 to numerator and denominator and the overall fraction moves closer to 1.
Let's see
Adding 1 to both numerator and denominator
\(\frac{1+1}{3+1} = \frac{2}{4} = \frac{1}{2}\)
\(\frac{1}{3} = 0.33\)
\(\frac{1}{2} = 0.5\)
Take another example, add 5 to both numerator and denominator
\(\frac{1+5}{3+5} = \frac{6}{8} = \frac{3}{4}\)
\(\frac{1}{3} = 0.33\)
\(\frac{3}{4} = 0.75\)
From the above examples we see that when \(\frac{A}{B }< 1\) then \(\frac{A+X}{B+X} > \frac{A}{B}\) when X > 0
OA,A
it becomes greater than the original fraction.
Consider the number line,
0------------1
Now, we take a fraction \(\frac{1}{3}\) which is between 0 and 1.
We add the same quantity N > 0 to numerator and denominator and the overall fraction moves closer to 1.
Let's see
Adding 1 to both numerator and denominator
\(\frac{1+1}{3+1} = \frac{2}{4} = \frac{1}{2}\)
\(\frac{1}{3} = 0.33\)
\(\frac{1}{2} = 0.5\)
Take another example, add 5 to both numerator and denominator
\(\frac{1+5}{3+5} = \frac{6}{8} = \frac{3}{4}\)
\(\frac{1}{3} = 0.33\)
\(\frac{3}{4} = 0.75\)
From the above examples we see that when \(\frac{A}{B }< 1\) then \(\frac{A+X}{B+X} > \frac{A}{B}\) when X > 0
OA,A
huda wrote:
In a certain parking lot at 10:00 am, the ratio of trucks to cars was 1 to 3. Between 10:00 am and 11:00 am, no vehicle left the parking lot, and, for some integer N > 0, N more cars and N more trucks entered the lot and parked there.
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
Quantity A |
Quantity B |
the ratio of trucks to cars in the lot at 11:00 |
\(\frac{1}{3}\) |
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
Re: In a certain parking lot at 10:00 am, the ratio of trucks to
[#permalink]
15 Sep 2024, 11:18
15 Sep 2024, 11:18
Hello from the GRE Prep Club BumpBot!
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
