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by GmatTakerNo.1 » Sun May 02, 2010 12:52 pm
If Mike left his house at 7am and traveled at 60 mph, he would arrive 10 minutes late to work. If he left his house at the same time and sped to work at 80 mph he would arrive 5 minutes early. What is his distance to work?

The answer must be 60, but I got 50.
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by truplayer256 » Sun May 02, 2010 1:21 pm
Let the original amount of time it takes Mike to get to his work be represented by the variable x and let the distance from Mike's house to his work be D.

D/60 = x + 1/6

D/80 = x - 1/12

60x + 10 = 80x - 20/3

20x = 50/3

x=5/6

D then equals 60 miles.
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by clock60 » Sun May 02, 2010 1:28 pm
GmatTakerNo.1 wrote:If Mike left his house at 7am and traveled at 60 mph, he would arrive 10 minutes late to work. If he left his house at the same time and sped to work at 80 mph he would arrive 5 minutes early. What is his distance to work?

The answer must be 60, but I got 50.
let it be x the number of hours he must be in a way to the work to be on time

at speed 60 m/h his time spend=x+10/60
(x+1/6)*60=distance

at a speed 80 m/h his time x-5/60
(x-1/12)*80=distance

equate both

(x+1/6)*60=(x-1/12)*80, and solve for x
here x=5/6
insert x in any equation
(5/6+1/6)*60=distance
distance=60
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by harshavardhanc » Sun May 02, 2010 11:09 pm
GmatTakerNo.1 wrote:If Mike left his house at 7am and traveled at 60 mph, he would arrive 10 minutes late to work. If he left his house at the same time and sped to work at 80 mph he would arrive 5 minutes early. What is his distance to work?

The answer must be 60, but I got 50.
see that the distance is constant.

therefore, speed is inversely proportional to time. i.e. if one becomes 4/3 of the value (60-->80), the other will become 3/4
(should decrease by 1/4) .

but, this 1/4th is given to be 15 minutes. therefore, actual time taken = 15 * 4 = 60 minutes or 1 hour.

and the distance = 60 mph * 1 hr = 60.
Regards,
Harsha
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by gmatmachoman » Mon May 03, 2010 1:50 am
GmatTakerNo.1 wrote:If Mike left his house at 7am and traveled at 60 mph, he would arrive 10 minutes late to work. If he left his house at the same time and sped to work at 80 mph he would arrive 5 minutes early. What is his distance to work?

The answer must be 60, but I got 50.
Let the distance to travel be x.

@60mph & 10 minutes late, he is short of 10miles from the distance .

@80mph& 5 minutes early, he is 80/12 miles ahead of the distance.

let "t" be the time taken to travel the distance "x".

60 * t = (x-10)

---> t =(x-10)/60

80 *t =(x+(80/12)) ----(equation 2)

substitute the value of t in equation 2

x=60 miles

I agree this is a bit lengthy approach.but it encompasses the fundamental concepts.
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by sumanr84 » Mon May 03, 2010 9:28 am
Let the total distance be X ,

Just calculate the diff in time in both the case i.e. 10 mins late and 5 mins early,
so, the diff in arrival time in both the case will add up to 15 mins


Then this is how the eqn goes :-
[X / 60 ] - [X / 80] = [15 / 60 ] --> convert min to hrs by div by 60,

Solve and you will get X = 60
I am on a break !!
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