it's A !

Steve average speed = x miles and time taken to go from town A to B = y hrs
Distance between town A and B = xy miles - (A)
(1) If Steve had traveled from town A to town B at an average speed that was 10 miles per hour faster,he would have traveled for 5 fewer hours.
so increased speed = x+10 reduced time taken = y-5 Distance travelled between A to B is still the same but can now be expressed as (x+10) (y-5) which is same a (A) = xy
so xy +10y-5x - 50 = xy or 10y - 5x = 50 x 7 y can be 10& 10 , 30&20 etc
Not Suff. we need another eq in x and y
(2) If Steve had traveled from town A to town B at an average speed that was 50 percent greater, the amount of time he traveled would have been the time it actually took reduced by 1/3 of the time it actually took.
50% increase in average speed , so average speed = 3/2 x Time reduces by 1/3 so time taken 2/3y
The distance is 3/2x x 2/3 y = xy 3/2 is the inverse of 2/3 so nothing new is being said. It is obvious that as Distance = Speed X Time if Distance remains the same and speed increase by 3/2 Time will decrease by 1/3.
Not Suff
IMO E

kstv wrote:Steve average speed = x miles and time taken to go from town A to B = y hrs
Distance between town A and B = xy miles - (A)
(1) If Steve had traveled from town A to town B at an average speed that was 10 miles per hour faster,he would have traveled for 5 fewer hours.
so increased speed = x+10 reduced time taken = y-5 Distance travelled between A to B is still the same but can now be expressed as (x+10) (y-5) which is same a (A) = xy
so xy +10y-5x - 50 = xy or 10y - 5x = 50 x 7 y can be 10& 10 , 30&20 etc
Not Suff. we need another eq in x and y
(2) If Steve had traveled from town A to town B at an average speed that was 50 percent greater, the amount of time he traveled would have been the time it actually took reduced by 1/3 of the time it actually took.
50% increase in average speed , so average speed = 3/2 x Time reduces by 1/3 so time taken 2/3y
The distance is 3/2x x 2/3 y = xy 3/2 is the inverse of 2/3 so nothing new is being said. It is obvious that as Distance = Speed X Time if Distance remains the same and speed increase by 3/2 Time will decrease by 1/3.
Not Suff
IMO E

if I combine the two then I can write something like this -

(x+10)(y-5)=distance.........(from stmt A)
(x+0.5x)(2/3)y=distance.....(from stmt B)
equating I get 10=0.5x therefore x=20
and y(1-5/y)=2/3 therefore y=15 now we have x and y and hence calculate the distance
is my way of solving this correct ?
my pick is C
OA is E
Source - kaplan

if I combine the two then I can write something like this -

(x+10)(y-5)=distance.........(from stmt A)
(x+0.5x)(2/3)y=distance.....(from stmt B)
equating I get 10=0.5x therefore x=20
and y(1-5/y)=2/3 therefore y=15 now we have x and y and hence calculate the distance
is my way of solving this correct ?
my pick is C
OA is E
Source - kaplan

Your way of solving is absolutely fine mate but you did a mistake in calculation .


Statement 1 : (x+10)(t-5) = xt [x= distance , t= time ]

10t -5x-50 = 0 ----------- (1)

Statement 2 : (x+x/2)(t-t/3) = xt

(3x/2 )(2t/3) = xt

xt = xt .

So statement 2 dont really give you any thing new and you are only left with statement 1 to solve for 2 unknown variables so can not solve . E.