A car traveling uphill will take an hour less to cover the

Question

A car traveling uphill will take an hour less to cover the distance if it moved 4km/hr faster. On the other hand it will take 3 more hours to reach the destination if it moved 6km/hr slower. What is the distance that the car travels?

A. 50
B. 80
C. 105
D. 75
E. 65

Source: 4gmat

A. 50
B. 80
C. 105
D. 75
E. 65

Source: 4gmat

Bunuel · Accepted Answer

ALGEBRAIC WAY:

Say it takes the car  t  hours to cover the distance at  r  kilometers per hour, so  d=tr . We are told that:

tr=(t-1)(r+4)  -->  tr=tr+4t-r-4  -->  4t-r-4=0 ;
 tr=(t+3)(r-6)  -->  tr=tr-6t+3r-18  -->  r-2t-6=0 ;

Solving gives:  t=5  and  r=16  -->  d=tr=80 .

Answer: B.

rajathpanta · Answer

Hi bunuel, Is there a simpler way to solve this problem. I am stuck. There are 3 unknowns and only 2 equations can be formed. Unable to comprehend the above method.

Bunuel · Answer

We have:
 tr=(t-1)(r+4)  -->  tr=tr+4t-r-4  -->  4t-r-4=0 ;
 tr=(t+3)(r-6)  -->  tr=tr-6t+3r-18  -->  r-2t-6=0 ;

So, we get two distinct linear equations with two unknowns ( 4t-r-4=0  and  r-2t-6=0 ) --> we can solve for  t  and  r . Now, since  distance=tr , then we can get the distance too.

Hope it's clear.

arichardson26 · Answer

Hi could you explain why you set the two rate/time combinations equal to 0?

tr=(t-1)(r+4) --> tr=tr+4t-r-4 --> 4t-r-4=0;
tr=(t+3)(r-6) --> tr=tr-6t+3r-18 --> r-2t-6=0

rajeevrks27 · Answer

tr=(t-1)(r+4) --> tr=tr+4t-r-4--> Take the tr on the left hand side to the right hand side, the sigh of the moving tr changes and it gets cancelled,whereas the left hand side is left with nothing but zero....Same way for the second part also..

SOURH7WK · Answer

To get 2 simultaneous equation to solve 2 unknown variables.

4t-r=4 Eq 1
r-2t=6 Eq 2

Ousmane · Answer

My approach
d/r - d/(r+4) = 1 ( i) (time lag) 
d/(r-6) + d/r = 3 (ii) (time lag)
From (i) d = r(r+4)/4 (iii)
From (ii) d = r(r-6)/2 (iv)
(iii) = (iv) => r = 16
and r = 16 in (iii) => d = 80
Brother Karamazov

bankerboy30 · Answer

Bunuel - in cases such as this when time and rate are unknown but distance is the same, are these always solvable? I'm thinking more along the lines of data sufficiency

Duncomo · Answer

since all the solutions I see here are algebra ill provide an alternative non-fail safe one that eventually got me this...

x=y hours 
x-4=y+1 hours 
x-10=y+4 hours

of the solutions, the potentially relevant factors of each are as follows

50 - 2,5,10,25
80 - 2,4,5,8,10,16,20,40
105 - 3,5,7,15,21,35
75 - 3,5,15,25
65 - 1,5,13

the logic behind this is that each amount of time must fit into 10,6 and 0 
so you are looking for 3 factors of each solution that are 4 apart and then 6 apart (descending in value)

through this assuming you know factors you can use this to see 80 has 20,16 and 10 that fit this method.

I am aware this likely won't help most people but if you have a good grasp of factors and struggle with algebra this is your best bet!

A car traveling uphill will take an hour less to cover the

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