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.
Hill Top A
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- goyalsau
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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.
![Image](https://s4.postimage.org/1ipkybsf8/land.jpg)
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)
- Total time taken = (z/72) + (y/63) + (x/56) = (4 + 2/3)
=> (9x + 8y + 7z) = (4 + 2/3)*7*8*9 ........................................ (ii)
=> (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
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I was not able to calculate values for x , y and z.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,And while going from B to A,
- Total time taken = (x/72) + (y/63) + (z/56) = 4
=> (7x + 8y + 9z) = 4*7*8*9 ........................................ (i)Add (i) and (ii) : (16x + 16y + 16z) = (4 + 4 + 2/3)*7*8*9 = (26/3)*7*8*9
- Total time taken = (z/72) + (y/63) + (x/56) = (4 + 2/3)
=> (9x + 8y + 7z) = (4 + 2/3)*7*8*9 ........................................ (ii)
=> (x + y + z) = (26*7*8*9)/(3*16) = (13*7*3) = 273
Therefore the distance AB is 273 km.
x + z = 168 , and y = 105 , But what about the individual vales of x and z,
Can we calculate that??????????
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No, as we have two equations in three unknowns.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??????????
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- goyalsau
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I asked for the value of x and y and z , Because of this problem.Anurag@Gurome wrote:No, as we have two equations in three unknowns.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??????????
https://www.beatthegmat.com/chocolates-t ... tml#322006
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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#322006goyalsau wrote:I asked for the value of x and y and z , Because of this problem.
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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