Skip to main content
Posts
What was the challenge in escaping the tower?
Note that difference between weights in 2 baskets can't exceed 20 pounds. However, chest can withstand the shock even if the difference exceeds that limit
1.
They put 60 pounds chest in one of empty basket & sends it down. (60 pounds vs 0 Pounds)
2. Sosthenes of 80 pounds climbs in other empty basket and rides down to the ground while bringing up the chest. (60 pounds vs 80 pounds)
3. Now chest is taken out and Hilary is sent down while bringing up Sosthenes.(80 pounds vs 100 pounds).
4. Hilary gets out of the basket on ground and Sosthenes get out of the basket at top of the tower.
5. They again send chest of 60 pound to the ground putting it in 1 basket. This brings other empty basket up at the top of the tower.
6. Now, Hilary (who is on ground) climbs in the basket with chest (total weight = 60 + 100 = 160 pounds) while Babylas (170 pounds) enters in the basket at the top of tower. This allows Babylas to ride down to the ground. (160 pounds vs 170 pounds).
Right now, Babylas is on ground and Hilary, Sosthenes are on the top of the tower with chest.
7. Here, Babylas jumps on the ground & Hilary gets out of the basket at the top of tower.
8. Again, because of chest which weighs 60 pounds the basket descends down to the ground bringing up other basket up. (60 pounds vs 0 Pounds).
9. Now, Sosthenes (at top) rides down to the ground bringing up the chest. (60 pounds vs 80 Pounds).
10. Hilary replaces himself with the chest at the top and rides down while bringing Sosthenes up. (100 pounds vs 80 Pounds).
At the moment, Sosthenes is at the top with chest while Hilary and Babylas are on the ground.
11. Now, Sosthenes sends back chest on the ground & then rides to the ground while entering into other empty basket. (60 pounds vs 80 Pounds).
12. And when Sosthenes gets out of the basket on the ground the basket with chest descends on it's own to the ground.
13. Finally, all Sosthenes, Babylas and Hilary take chest out of the basket and safely escape the tower.
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.
================================================================
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.
Assume these matchsticks as fishes. Can you flip the positions of the fishes horizontally i.e. 1 at the leftmost position and 4 at the rightmost position. Condition is that you can move only 3 matchsticks.
That's how it can be done!
