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jogging and walking problem

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by pradeepkaushal9518 » Tue Apr 06, 2010 3:10 am
aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. how many miles from home can aaron jog so that he spends a total of t hours jogging and walking?

a.xt/y
b.(x+y)/xy
c.xyt/x+y
d.x+Y+t/xy
e.y+t/x-t/y
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Source: — Problem Solving |

by neoreaves » Tue Apr 06, 2010 3:27 am
IMO C

Let total distance he has to cover = d

time to walk away from home = t1 = d / x

time to go towards home = t2 = d / y

Total time = t = d/x + d/y
--> t = d(1/x + 1/y)
--> t = d( x +y)/xy
--> txy/(x+y) = d

Answer C
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by srinivasarajui » Tue Apr 06, 2010 3:35 am
In algebraic method
Let
Distance be 'D'
Time he jogs be 'J'
Time he walks be 'W'.

Then it is given that

D= x * J = y *W -- equation 1

and

J + W = t => W = t- J

Substituting the W in equation 1 gives
x*J = y * (t-J)

Then J = (y*t)(x+y)
So D = J * x = x*y*t/(x+y ) SO ans is C

We can also solve easily by Substitution method.

let distance be 10 miles
x be 5 MPH
y be 2 MPH
He takes 2 hours to jog and 5 hours to walk.
So he spends 7 hours in full which is t.

Substituting x, y and t
a. 5*7/2 wrong
b. (5+2)/5*2 wrong
c. 5*2*7/(5+2) correct
d. 5+2+7/(5*2)wrong
e. 2+7/(5-7)/*2 Not clear but lets say its wrong

I think this will help.
Srinu
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by harshavardhanc » Tue Apr 06, 2010 3:49 am
pradeepkaushal9518 wrote:aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. how many miles from home can aaron jog so that he spends a total of t hours jogging and walking?

a.xt/y
b.(x+y)/xy
c.xyt/x+y
d.x+Y+t/xy
e.y+t/x-t/y
Pradeep,

There are already two very good methods explained above, which will guide you to the correct answer.

But, if at all, in the exams you are not able to recall any of the methods, your logic will come in handy. :)

One more method :

You need Miles in the answer. Now, look at the options and just put the units of X, Y, and T. After cancellation, try to figure out which one gives Miles in the end. That's it!

For e.g., let's take option A.

(X * T)/ Y OR

(Miles/hr) * (hr) / (Miles/hr) = hr. Nopes! eliminated!

Similarly, option B :

(X+Y)/XY= (Miles/hr) / (Miles/hr)^2 = hr/Miles. Nopes! eliminated.

You'll find that the expression in C alone has the unit as Miles. :) Options D and E are not even valid.

You've got your answer!
Regards,
Harsha
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by srinivasarajui » Tue Apr 06, 2010 6:44 am
One of the creative ways of finding the answers. Great harshavardhanc
Srinu
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by eaakbari » Tue Apr 06, 2010 6:44 am
The average speed over a distance if there are 2 different speeds is given by the quick formula

2xy/(x+y)
In our case our distance is 2d as we have to find the miles he has to go from home if that is d entire distance is 2d

Since
we need d = s * t

2d = 2txy/(x+y)

answer is txy/x+y
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by kstv » Tue Apr 06, 2010 7:36 am
aaron will jog from home at x miles per hour and then walk back home by the same route at y miles per hour. how many miles from home can aaron jog so that he spends a total of t hours jogging and walking?
a.xt/y b.(x+y)/xy c.xyt/x+y d.x+Y+t/xy e.y+t/x-t/y

After reading all the above posts and being a convert of Stuart Kovinsky method of analysing the answer choices
My approach -------here we have to find the distance
so Dist = Speed * Total time Total time is t, so t has to be in the numerator or D a multiple of t
Option a.x t/y & c.xy t/x+y fits the bill.
we can rule out option a cos' x/y will give the ratio of the speed.

am I going too far ?
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