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The time required to travel d miles at s miles per hour [#permalink]
Expert's post 00:00

Question Stats: 78% (00:58) correct 21% (00:35) wrong based on 19 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

_________________ Director  Joined: 07 Jan 2018
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Re: ds different from zero [#permalink]
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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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This is my response to the question and may be incorrect. Feel free to rectify any mistakes GRE Instructor Joined: 10 Apr 2015
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Re: ds different from zero [#permalink]
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Expert's post
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.

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Re: ds different from zero [#permalink]
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. GRE Instructor Joined: 10 Apr 2015
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Re: ds different from zero [#permalink]
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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.

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]
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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Try our free Online GRE Test Re: ds different from zero   [#permalink] 04 May 2018, 11:26
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