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A car travelling from city A to B takes an hour lesser to cover the

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A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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A car travelling from city A to B takes an hour lesser to cover the distance if it moved 10kmph faster. On the other hand it will take an hour more to cover the distance if it moved 6kmph slower. What is the distance between the two cities?
A) 150kms
B) 180kms
C) 360kms
D) 120kms

Source: 4gmat


Please explain :(
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A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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alphonsa wrote:
A car travelling from city A to B takes an hour lesser to cover the distance if it moved 10kmph faster. On the other hand it will take an hour more to cover the distance if it moved 6kmph slower. What is the distance between the two cities?
A) 150kms
B) 180kms
C) 360kms
D) 120kms

Source: 4gmat


Please explain :(


We need to setup the equations for this problem and solve for the values.

Let D, T1, and R1 be the values initially.

It is said that, If car travels at rate of R2= R1+ 10 it will reach B by T2=T1-1.

Since the distance is equal in both the rates.

T1R1= T2R2

T1R1= (T1-1)(R1+10)

On simplifying we get,

10T1- R1= 10---------------- (1)

Then, the 2nd line in the problem says,

T2=T1+1 and R2= R1-6

T1R1= (T1+1)(R1-6)

On simplifying we get,

-6T1+R1=6-----------------(2)

Solving 2 equation,

10T1-R1= 10
-6T1+R1= 6
(+)__________
4T1=16,

T1=4. So R1= 30.

Distance D= 4*30= 120Kms.


So answer is D.


But, this is not a GMAT problem. Eventhough it is covered in GMAT. A typical GMAT question should contain 5 answer choices..

Hope it helps.
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A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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New post 02 Sep 2015, 18:03
alphonsa wrote:
A car travelling from city A to B takes an hour lesser to cover the distance if it moved 10kmph faster. On the other hand it will take an hour more to cover the distance if it moved 6kmph slower. What is the distance between the two cities?
A) 150kms
B) 180kms
C) 360kms
D) 120kms

Source: 4gmat


Please explain :(


Hi Everyone,

I am having trouble with this problem. I break it down in to the Manhattan style table as follows with \(x\) = to distance between A and B, \(t\) = the original time and \(r\) = the original rate

\(Rate * Time = Distance\)

1: \(r * t = x\)

2: \((r + 10) * (t - 1) = x\)

3: \((r - 6) * (t + 1) = x\)

I foresee that this should be some kind of simple simultaneous equation problem but in this case we have too many variables??? Sorry if this sounds too simple, but distance/rate problems are my absolute nemesis and just don't seem to be making any progress with it :(
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Last edited by DropBear on 05 Sep 2015, 14:36, edited 2 times in total.

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Re: A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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rt=d
(r+10)(t-1)=d➡10t-r-10=0
(r-6)(t+1)=d ➡-6t+r-6=0
adding, 4t-16=0➡t=4
(10)(4)-r-10=0➡r=30
(30)(4)=120

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Re: A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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New post 05 Sep 2016, 15:00
Can anyone please solve this equation step by step

I sorta get hung in the calculation

Thanks

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Re: A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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Tmoni26 wrote:
Can anyone please solve this equation step by step

I sorta get hung in the calculation

Thanks


Lets consider the Distance be D and Speed as S and Time as T

D = S*T

Now as per first statement, the speed is 10 km/hr faster so time taken is 1 hr less.

\(\frac{D}{(S+10)}\) = T-1 or D = (S+10)*(T-1) -- eq 1.

As per second statement, the speed is 6 km/hr slower so it takes 1 extra hour.

\(\frac{D}{(S-6)}\) = T+1 or D = (S-6)*(T+1) -- eq 2

So we have two equations, substitute D = ST in both the equations and you will get following two equations.

10T - S = 10

-6T + S = 6

Adding both the equations you will get 4T = 16

T =4

Substitue in any one equation and you will get S = 30

So, D = S*T = 30*4 = 120

HTH

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Re: A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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New post 21 Oct 2016, 02:22
Another question, please can you shed some light on the substitute D = ST in both equations?

When I solve, I get
ST - S+ 10T = 10 and ST+S-6T = -6.........how do we get rid of the ST in these 2 equations??

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Re: A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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Tmoni26 wrote:
Another question, please can you shed some light on the substitute D = ST in both equations?

When I solve, I get
ST - S+ 10T = 10 and ST+S-6T = -6.........how do we get rid of the ST in these 2 equations??


This is the first eq:

D = (S+10)*(T-1)

D = ST -S + 10T -10

We know that D = ST, substituting the same in equation:

ST = ST - S + 10T - 10

ST - ST + S - 10T + 10 = 0

So we are left with S - 10T + 10 = 0

HTH

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Re: A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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New post 21 Oct 2016, 03:22
Thank you very much, I am grateful for your time.

I initially equated both equations and now had a spare ST that got me wrapped up

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Re: A car travelling from city A to B takes an hour lesser to cover the [#permalink]

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Re: A car travelling from city A to B takes an hour lesser to cover the   [#permalink] 27 Nov 2017, 07:32
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