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A car travels from city a to city B, a distance of 240kms. T

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A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post Updated on: 09 Aug 2014, 04:14
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Question Stats:

56% (01:42) correct 44% (01:54) wrong based on 148 sessions

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A car travels from city a to city B, a distance of 240kms. The car travels half the distance at x kmph and the remaining distance at x+20 kmph. If the car has to complete the journey in less than 5 hours what should be the minimum value of x?

a) 20
b) 30
c) 40
d) 50
e) 60

No options were given. This question was given for conceptual understanding and it came with no options. For the sake of the reader, I am making my own optionsl

Source:4gmat

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Originally posted by alphonsa on 08 Aug 2014, 23:30.
Last edited by alphonsa on 09 Aug 2014, 04:14, edited 1 time in total.
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Re: A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post 09 Aug 2014, 03:27
alphonsa wrote:
A car travels from city a to city B, a distance of 240kms. The car travels half the distance at x kmph and the remaining distance at x+20 kmph. If the car has to complete the journey in less than 5 hours what should be the minimum value of x?

No options given

Ans: 40

Please explain


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As for the options x will have to be more than 40 km/hrs since the time taken to cover this distance is less than 5 hrs but if speed is 40km/hr then time required will be equal to 5 hrs...
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Re: A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post 09 Aug 2014, 03:39
1
alphonsa wrote:
A car travels from city a to city B, a distance of 240kms. The car travels half the distance at x kmph and the remaining distance at x+20 kmph. If the car has to complete the journey in less than 5 hours what should be the minimum value of x?

To have the minimum value of speed you need the maximum possible time. So, the time is 5.
Just fill the rate table and use that
rate*time=distance

_______________rate_______time_________distance
first half________x________120/x_________120
second half____x+20____120/(x+20)_____120
Total______________________5_____________240

So, we have the equation:
\(\frac{120}{x}+\frac{120}{x+20}=5\)

On GMAT it is better here to to plug in the answers.

Divide by 120
\(\frac{1}{x}+\frac{1}{x+20}=\frac{5}{120}\)
Common denominator
\(\frac{2x+20}{x(x+20)}=\frac{1}{24}\)
Cross multiply
\(48x+480=x^2+20x\)
\(x^2-28x-480=0\)
Viet's theorem
\(x=40\) or \(x=-12\)
So, the speed is 40.

Don't really know how to solve this equation faster.
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And yes, I like kudos:)

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Re: A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post 09 Aug 2014, 04:33
smyarga wrote:
alphonsa wrote:
A car travels from city a to city B, a distance of 240kms. The car travels half the distance at x kmph and the
remaining distance at x+20 kmph. If the car has to complete the journey in less than 5 hours what should be the minimum value of x?

To have the minimum value of speed you need the maximum possible time. So, the time is 5.
Just fill the rate table and use that
rate*time=distance

_______________rate_______time_________distance
first half________x________120/x_________120
second half____x+20____120/(x+20)_____120
Total______________________5_____________240

So, we have the equation:
\(\frac{120}{x}+\frac{120}{x+20}=5\)

On GMAT it is better here to to plug in the answers.

Divide by 120
\(\frac{1}{x}+\frac{1}{x+20}=\frac{5}{120}\)
Common denominator
\(\frac{2x+20}{x(x+20)}=\frac{1}{24}\)
Cross multiply
\(48x+480=x^2+20x\)
\(x^2-28x-480=0\)
Viet's theorem
\(x=40\) or \(x=-12\)
So, the speed is 40.

Don't really know how to solve this equation faster.


Hi smyarga,

At x=40, the time required will be 5 hrs so speed should be more than that..am I correct ...I did solve the problem like you did and got x=40

Posted from my mobile device
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Re: A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post 09 Aug 2014, 04:36
WoundedTiger wrote:
smyarga wrote:
alphonsa wrote:
A car travels from city a to city B, a distance of 240kms. The car travels half the distance at x kmph and the
remaining distance at x+20 kmph. If the car has to complete the journey in less than 5 hours what should be the minimum value of x?

To have the minimum value of speed you need the maximum possible time. So, the time is 5.
Just fill the rate table and use that
rate*time=distance

_______________rate_______time_________distance
first half________x________120/x_________120
second half____x+20____120/(x+20)_____120
Total______________________5_____________240

So, we have the equation:
\(\frac{120}{x}+\frac{120}{x+20}=5\)

On GMAT it is better here to to plug in the answers.

Divide by 120
\(\frac{1}{x}+\frac{1}{x+20}=\frac{5}{120}\)
Common denominator
\(\frac{2x+20}{x(x+20)}=\frac{1}{24}\)
Cross multiply
\(48x+480=x^2+20x\)
\(x^2-28x-480=0\)
Viet's theorem
\(x=40\) or \(x=-12\)
So, the speed is 40.

Don't really know how to solve this equation faster.


Hi smyarga,

At x=40, the time required will be 5 hrs so speed should be more than that..am I correct ...I did solve the problem like you did and got x=40

Posted from my mobile device


You are right! That's why a wrote that to have minimum speed you need maximum possible time:) because if you go faster, you will come earlier)
_________________

I'm happy, if I make math for you slightly clearer
And yes, I like kudos:)

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Re: A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post 20 Sep 2014, 08:43
smyarga wrote:
alphonsa wrote:
A car travels from city a to city B, a distance of 240kms. The car travels half the distance at x kmph and the remaining distance at x+20 kmph. If the car has to complete the journey in less than 5 hours what should be the minimum value of x?

To have the minimum value of speed you need the maximum possible time. So, the time is 5.
Just fill the rate table and use that
rate*time=distance

_______________rate_______time_________distance
first half________x________120/x_________120
second half____x+20____120/(x+20)_____120
Total______________________5_____________240

So, we have the equation:
\(\frac{120}{x}+\frac{120}{x+20}=5\)

On GMAT it is better here to to plug in the answers.

Divide by 120
\(\frac{1}{x}+\frac{1}{x+20}=\frac{5}{120}\)
Common denominator
\(\frac{2x+20}{x(x+20)}=\frac{1}{24}\)
Cross multiply
\(48x+480=x^2+20x\)
\(x^2-28x-480=0\)
Viet's theorem
\(x=40\) or \(x=-12\)
So, the speed is 40.

Don't really know how to solve this equation faster.




THEY HAVE TOLD THEY HAVE TO COMPLETE THE JOURNEY IN LESS THAN 5 HOURS THEN HOW CAN IT BE EQUAL TO 5/
?
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Re: A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post 14 Dec 2015, 09:39
SunthoshiTejaswi wrote:
smyarga wrote:
alphonsa wrote:
A car travels from city a to city B, a distance of 240kms. The car travels half the distance at x kmph and the remaining distance at x+20 kmph. If the car has to complete the journey in less than 5 hours what should be the minimum value of x?

To have the minimum value of speed you need the maximum possible time. So, the time is 5.
Just fill the rate table and use that
rate*time=distance

_______________rate_______time_________distance
first half________x________120/x_________120
second half____x+20____120/(x+20)_____120
Total______________________5_____________240

So, we have the equation:
\(\frac{120}{x}+\frac{120}{x+20}=5\)

On GMAT it is better here to to plug in the answers.

Divide by 120
\(\frac{1}{x}+\frac{1}{x+20}=\frac{5}{120}\)
Common denominator
\(\frac{2x+20}{x(x+20)}=\frac{1}{24}\)
Cross multiply
\(48x+480=x^2+20x\)
\(x^2-28x-480=0\)
Viet's theorem
\(x=40\) or \(x=-12\)
So, the speed is 40.

Don't really know how to solve this equation faster.




THEY HAVE TOLD THEY HAVE TO COMPLETE THE JOURNEY IN LESS THAN 5 HOURS THEN HOW CAN IT BE EQUAL TO 5/
?



Yes .. which is why the answer should be 50 from the options given .. the actual answer should be a non terminating decimal 40.000000........1
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Re: A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post 18 May 2016, 14:20
I have encountered a problem with solution of this problem. the correct answer is option C - 40 kmph . but if we take the value and solve for
120/x + 120/(x+20) = 5
we get that journey is completed in exactly 5 hrs
But as mentioned in question stem journey is to be completed in less than 5 hours value of correct x should be a bit more than 40.
it can be 40.001 or 41 or 45 or any value>40
So the correct answer appears to be flawed
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Re: A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post 30 Dec 2016, 03:52
less than 5 hours not equal to five hours
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Re: A car travels from city a to city B, a distance of 240kms. T  [#permalink]

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New post 09 Mar 2018, 06:40
alphonsa wrote:
A car travels from city a to city B, a distance of 240kms. The car travels half the distance at x kmph and the remaining distance at x+20 kmph. If the car has to complete the journey in less than 5 hours what should be the minimum value of x?

a) 20
b) 30
c) 40
d) 50
e) 60

No options were given. This question was given for conceptual understanding and it came with no options. For the sake of the reader, I am making my own optionsl

Source:4gmat


Hi Bunuel

This question has an OA which does not match the wording. If I am not wrong,
the answer to this question cannot be 40 km/hr!

At a speed of 40 km/hr, the car would travel 120 km (half the distance) in 3 hours
and the travels the remainder of the distance, traveling at 60 km/hr, in 2 hours.
The total time taken is 5 hours! But as it is explicitly stated in the question, the car
must complete the journey in less than 5 hours. IMO, the OA must be changed!

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.

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Re: A car travels from city a to city B, a distance of 240kms. T &nbs [#permalink] 09 Mar 2018, 06:40
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