Hill Top A

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Mon Dec 27, 2010 11:02 pm

A person starts from hill top A, goes downhill, then along a plain and finally climbs to hill top B. He takes a total of 4 hours for the jouney travelling downhill @72kmph, on plains @63kmph and uphill @56kmph. However while returning back, from B to A, he takes 4 hours 40 mins with speeds being same. Find distance AB.

Answer is 273 km, Please give the detailed solution.

Saurabh Goyal
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Source: Beat The GMAT — Problem Solving | Mon Dec 27, 2010 11:02 pm

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Mon Dec 27, 2010 11:19 pm

goyalsau wrote:A person starts from hill top A, goes downhill, then along a plain and finally climbs to hill top B. He takes a total of 4 hours for the jouney travelling downhill @72kmph, on plains @63kmph and uphill @56kmph. However while returning back, from B to A, he takes 4 hours 40 mins with speeds being same. Find distance AB.

Referring to the above image, from A to B, the end of downhill and start of plains is C and end of plains and start of uphill is D.

Say, AC = x km, CD = y km and DB = z km
Thus, distance AB = (x + y + z) km

Therefore while going from A to B,

Total time taken = (x/72) + (y/63) + (z/56) = 4
=> (7x + 8y + 9z) = 4*7*8*9 ........................................ (i)

And while going from B to A,

Total time taken = (z/72) + (y/63) + (x/56) = (4 + 2/3)
=> (9x + 8y + 7z) = (4 + 2/3)*7*8*9 ........................................ (ii)

Add (i) and (ii) : (16x + 16y + 16z) = (4 + 4 + 2/3)*7*8*9 = (26/3)*7*8*9
=> (x + y + z) = (26*7*8*9)/(3*16) = (13*7*3) = 273

Therefore the distance AB is 273 km.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Tue Dec 28, 2010 2:17 am

Anurag@Gurome wrote:
goyalsau wrote:A person starts from hill top A, goes downhill, then along a plain and finally climbs to hill top B. He takes a total of 4 hours for the jouney travelling downhill @72kmph, on plains @63kmph and uphill @56kmph. However while returning back, from B to A, he takes 4 hours 40 mins with speeds being same. Find distance AB.

Referring to the above image, from A to B, the end of downhill and start of plains is C and end of plains and start of uphill is D.

Say, AC = x km, CD = y km and DB = z km
Thus, distance AB = (x + y + z) km

Therefore while going from A to B,
Total time taken = (x/72) + (y/63) + (z/56) = 4
=> (7x + 8y + 9z) = 4*7*8*9 ........................................ (i)
And while going from B to A,
Total time taken = (z/72) + (y/63) + (x/56) = (4 + 2/3)
=> (9x + 8y + 7z) = (4 + 2/3)*7*8*9 ........................................ (ii)
Add (i) and (ii) : (16x + 16y + 16z) = (4 + 4 + 2/3)*7*8*9 = (26/3)*7*8*9
=> (x + y + z) = (26*7*8*9)/(3*16) = (13*7*3) = 273

Therefore the distance AB is 273 km.

I was not able to calculate values for x , y and z.

x + z = 168 , and y = 105 , But what about the individual vales of x and z,

Can we calculate that??????????

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Dec 28, 2010 2:18 am

goyalsau wrote:I was not able to calculate values for x , y and z.

x + z = 168 , and y = 105 , But what about the individual vales of x and z,

Can we calculate that??????????

No, as we have two equations in three unknowns.

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goyalsau: Legendary Member; Posts: 866; Joined: Mon Aug 02, 2010 6:46 pm; Location: Gwalior, India; Thanked: 31 times

by goyalsau » Tue Dec 28, 2010 2:34 am

Anurag@Gurome wrote:
goyalsau wrote:I was not able to calculate values for x , y and z.

x + z = 168 , and y = 105 , But what about the individual vales of x and z,

Can we calculate that??????????
No, as we have two equations in three unknowns.

I asked for the value of x and y and z , Because of this problem.

https://www.beatthegmat.com/chocolates-t ... tml#322006

Saurabh Goyal
[email protected]
-------------------------

EveryBody Wants to Win But Nobody wants to prepare for Win.

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Tue Dec 28, 2010 2:45 am

goyalsau wrote:I asked for the value of x and y and z , Because of this problem.

https://www.beatthegmat.com/chocolates-t ... tml#322006

For that problem too you don't need to solve the unknowns individually. Rahul has solved the problem already in the same post. Refer here : https://www.beatthegmat.com/chocolates-t ... tml#322006

My solution will be same as of him.
Just try to distinguish between solving a set of equations and framing a new equation based on the given set of equations.

The first one is the traditional and common one which involves finding the value of each unknowns. In that case, we must have the same number of equations as the number of unknowns. Otherwise we can't solve them uniquely.

The second one does not necessarily requires the solution of each unknown. In this case what we have to do is to frame the required equation by addition/subtraction of the given equation set and/or multiplication/division of the equation by a constant. If necessary we have to repeat this operations.

Both of these two problems are of the second category.
In this "Hill Top" problem, we need to frame the equation (x + y + z) = 273, based upon the given two equation. Similarly for the "chocolate" problem.

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