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A motorcyclist was sent by the post office to meet a plane at the airport.
The plane landed ahead of schedule, and its mail was taken toward the
post office by horse. After half an hour the horseman met the
motorcyclist on the road and gave him the mail.
The motorcyclist returned to the post office 20 minutes earlier than he was expected.
How many minutes early did the plane land?
Here is calculation of scheduled arrival time.
What is the data given for calculation?
Since, motorcyclist returned to the post office 20 minutes earlier than he was expected, it mean, the horse had saved his 20 minutes of journey. That is, after meeting with horse at some point, the motorcyclist would have needed 20 minutes to go to & come back from airport to the same point. That's how the horse managed to save 20 minutes of motorcyclist.
Let 'T' be the time at which horse met with motorcyclist.
It also means that, the motorcyclist would have taken another 10 minutes to reach at the airport exactly when plane was scheduled for landing. So the scheduled time of arrival of plane is T+10.
However, plane arrived at time T-30 where horse left airport with mail & met motorcyclist exactly half hour later at time T.
In short, plane landed at time T-30 instead of scheduled T+10 shows that plane landed 40 minutes ahead of schedule.
The equation in question was....
First of all X + Y + Z = Z < = 10 in not possible since in that case, X + Y + Z - Z =0 i.e.
X + Y = 0.
Hence, there must be carry 1 forwarded to digit's place. So,
X + Y + Z = 10 + Z
X + Y = 10. ........(1)
Therefore, possible pairs for (X,Y) or for (Y,X) are (1,9), (2,8), (3,7), (4,6) and (5,5) out of which (5,5) is invalid as repeating digits are not allowed.
Now, since carry is forwarded to ten's place addition, it looks like,
1 + X + Y + Z = XY
Putting (1) in above,
1 + 10 + Z = XY
11 + Z = XY ........(2)
The maximum value of Z can be 9 & in that case as per above equation XY = 20. But since 0 is not allowed Z must be less than or equal to 8. Then XY <= 11 + 8 = 19.
So X which seems to be carry to hundred place must be 1 (can't be 0 or 2 as proved above). Hence, X = 1.
If X = 1, then from (1), Y = 9.
And if XY = 19 then from (2) Z = 8.
To conclude, X = 1, Y = 9 and Z = 8.
Final equation, looks like,
11 + 99 + 88 = 198.
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Another Method :
The given equation is nothing but
10X + X + 10Y + Y + 10Z + Z = 100X + 10Y + Z
11(X + Y + Z) = 100X + 10Y + Z
89X = Y + 10Z = 10Z + Y
Since, only whole digits are allowed, maximum value of 10Z + Y can be 98 with Z = 9 & Y = 8. So if X = 2 then 89 x 2 = 198 = 10Z + Y is impossible case. Hence, X = 1.
And only Z = 8 and Y = 9 satisfies 89 x 1 = 10Z + Y = 10(8) + 9 = 89.
In short, X = 1, Y = 9 and Z = 8.
