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The time required to travel d miles at s miles per hour

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The time required to travel d miles at s miles per hour [#permalink] New post 17 Mar 2018, 03:11
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73% (00:44) correct 26% (00:28) wrong based on 142 sessions
\(ds \neq 0\)

Quantity A
Quantity B
The time required to travel \(d\) miles at \(s\) miles per hour
The time required to travel \(\frac{d}{2}\) miles at \(2s\) miles per hour


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.
[Reveal] Spoiler: OA

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Re: ds different from zero [#permalink] New post 17 Mar 2018, 23:26
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distance travelled/ time taken = speed
fill up the information in the formula
1st case:
\(\frac{d}{t} =s\)
or, \(\frac{d}{s} = t1\) (time taken for the first case)
2nd case:
\(\frac{d}{2/2s} = \frac{d}{4s} = \frac{d}{s} * \frac{1}{4} = t2\) (time taken for the second case)

1st case is similar to the 2nd case except second case is multiplied by \(0.25\).
If a \(+ve\) number is multiplied by less than \(1\) its value decreases hence B<A
option A
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Re: ds different from zero [#permalink] New post 04 May 2018, 07:33
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Carcass wrote:
\(ds \neq 0\)

Quantity A
Quantity B
The time required to travel \(d\) miles at \(s\) miles per hour
The time required to travel \(\frac{d}{2}\) miles at \(2s\) miles per hour


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: The time required to travel d miles at s miles per hour
time = distance/rate
So, here the time = d/s

Quantity B: The time required to travel d/2 miles at 2s miles per hour
time = distance/rate
So, here the time = (d/2)/2s = d/(4s)

So, we have:
Quantity A: d/s
Quantity B: d/(4s)

From here, we can solve the question using matching operations
Since the speed (s) must be a POSITIVE value, we can safely multiply each quantity by 4s to get:
Quantity A: 4d
Quantity B: d

Subtract d from each quantity to get:
Quantity A: 3d
Quantity B: 0

Divide each quantity by 3 to get:
Quantity A: d
Quantity B: 0

Since the distance (d) must be positive, we can see that Quantity A is greater.

Answer: A

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Re: ds different from zero [#permalink] New post 04 May 2018, 09:57
for this kind of question, is it okay that I assigned random values to d and s to be clear of the answer. For example, i took D= 10miles and S = 20mph. then I calculated time for A and B. I got the rigth answer here but I want to be sure I can do for all these kind of problems and there wouldnt be exceptions.
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Re: ds different from zero [#permalink] New post 04 May 2018, 10:15
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kruttikaaggarwal wrote:
for this kind of question, is it okay that I assigned random values to d and s to be clear of the answer. For example, i took D= 10miles and S = 20mph. then I calculated time for A and B. I got the rigth answer here but I want to be sure I can do for all these kind of problems and there wouldnt be exceptions.


The strategy of testing values does not reliably lead us to the correct answer. The ONLY time we can be certain of the correct answer is when plugging in two sets of values leads to two DIFFERENT outcomes (where the correct answer is D).

Here's what I mean.

Let's say we must compare the following quantities:
Quantity A: 5
Quantity B: x

Let's assign a random value for x and see what happens.
Let's say x = 3
We get:
Quantity A: 5
Quantity B: 3

Quantity A is greater, so can we conclude that the correct answer is A?
No.

If I try a different value of x, like x = 10, then we get:
Quantity A: 5
Quantity B: 10

This time Quantity B is greater, so the correct answer is D.

For more on this watch:

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Re: ds different from zero [#permalink] New post 04 May 2018, 11:26
Expert's post
kruttikaaggarwal wrote:
for this kind of question, is it okay that I assigned random values to d and s to be clear of the answer. For example, i took D= 10miles and S = 20mph. then I calculated time for A and B. I got the rigth answer here but I want to be sure I can do for all these kind of problems and there wouldnt be exceptions.



Whenever you are plugging in values your target is to break the question (finding a loophole) so as to get a quick solution. If it is a question where answrers are given i.e. multiple choice questions you would want to eleminate as many options as possible from the given possible answer choices. When answers choices are given you can simply plug the answer choices.

In Quantitative comparison questions plugging values is one of the most effective tool to get the option D, as already pointed out above.
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Re: The time required to travel d miles at s miles per hour [#permalink] New post 20 Oct 2019, 15:52
Carcass wrote:
\(ds \neq 0\)

Quantity A
Quantity B
The time required to travel \(d\) miles at \(s\) miles per hour
The time required to travel \(\frac{d}{2}\) miles at \(2s\) miles per hour


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.


Hi,

So I got this question right on the first try; My question then isn't so much about that but about how I could have gotten it wrong, and why that's wrong.

When we solve for QB we get (and you can approach this a few different ways, which is a bit about what my q. is concerning):

Quote:
s = d/t
-plug in

(2s) = (d/2)/t
-simplify the right:

2s = d/2t
-divide by d

2s/d = 1/2t


Now here is where I could have made the mistake. If you cancel out the opposite factors of 2, you will get:

Quote:
s / d = 1 / t
-take the reciprocal of both sides to isolate t

d / s = t


This is the same value as QA and obviously does not work given that the answer is A. So my question is, why can't we cross-cancel here?

When we proceed the other way we get:

Quote:
2s / d = 1 / 2t
-take the reciprocal first

d / 2s = 2t
-multiply by 1/2

d / 4s = t


We can also get this by cross-multiplication:

Quote:
2s / d = 1 / 2t
4st = d
t = d / 4s


So why isn't the cross-canceling valid? I understand that it leads to the wrong answer, but shouldn't it be a valid way to evaluate an expression?

Thanks so much!
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Re: The time required to travel d miles at s miles per hour [#permalink] New post 22 Oct 2019, 10:31
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einalemjs wrote:
2s/d = 1/2t
Now here is where I could have made the mistake. If you cancel out the opposite factors of 2, you will get:
s/d = 1/t


Be careful.
"cancelling" isn't a matter of removing two values that are the same.

If we take: 2s/d = 1/(2t)...
...and divide both sides by 2, we get: s/d = 1/(4t)

If we take: 2s/d = 1/(2t)...
...and multiply both sides by 2, we get: 4s/d = 1/t

As you can see, there isn't an approach that allows us to eliminate both 2's

Cheers,
Brent
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Re: The time required to travel d miles at s miles per hour [#permalink] New post 23 Oct 2019, 10:49
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Maybe a simpler way: time = distance / speed. As in any fraction, if you make the numerator smaller and the denominator larger, the value of the fraction (in this case, time) goes down. So answer A is bigger.

HTH, Chak
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Re: The time required to travel d miles at s miles per hour [#permalink] New post 25 Oct 2019, 18:42
Chakolate wrote:
Maybe a simpler way: time = distance / speed. As in any fraction, if you make the numerator smaller and the denominator larger, the value of the fraction (in this case, time) goes down. So answer A is bigger.

HTH, Chak


That's beautifully simple!
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Re: The time required to travel d miles at s miles per hour [#permalink] New post 25 Oct 2019, 18:43
GreenlightTestPrep wrote:
einalemjs wrote:
2s/d = 1/2t
Now here is where I could have made the mistake. If you cancel out the opposite factors of 2, you will get:
s/d = 1/t


Be careful.
"cancelling" isn't a matter of removing two values that are the same.

If we take: 2s/d = 1/(2t)...
...and divide both sides by 2, we get: s/d = 1/(4t)

If we take: 2s/d = 1/(2t)...
...and multiply both sides by 2, we get: 4s/d = 1/t

As you can see, there isn't an approach that allows us to eliminate both 2's

Cheers,
Brent


Thank you--this answers my question!
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Re: The time required to travel d miles at s miles per hour [#permalink] New post 30 Oct 2019, 20:44
Maybe a simpler way: time = distance / speed. As in any fraction, if you make the numerator smaller and the denominator larger, the value of the fraction (in this case, time) goes down. So answer A is bigger.
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Re: The time required to travel d miles at s miles per hour [#permalink] New post 10 Jul 2020, 04:41
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Notice that
Q(A) T=D/R
Q(B) T=(d/2)/(2R).
That means Q(B) has lower value than Q(A). its numerator halved and denominator doubled.
Hence, A is the answer.
Re: The time required to travel d miles at s miles per hour   [#permalink] 10 Jul 2020, 04:41
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