but ur question was whether a 7cm dia ball can pass through the gap. the answer is yes. What is the problem if there's still a gap. :huh:Quote:
Originally Posted by rajraj
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but ur question was whether a 7cm dia ball can pass through the gap. the answer is yes. What is the problem if there's still a gap. :huh:Quote:
Originally Posted by rajraj
Now, you have to think outside the box ! :) Extend the line connecting the corner to the center of the large 40 cm diameter ball to the surface of the ball. The length of that line is 48.28 cm. But the diameter of the ball is 40 cm. If similar length is 8.28 cm what will be diameter of the largest ball? In other words you have to find out the diameter of the largest ball that can pass through that gap. Only then you can make a conclusion. Try 8.25 cm diameter ball. You will have to make a dent in the wall and the floor! :) 8.25 < 8.28 ! :)Quote:
Originally Posted by GP
oops, technical fault.
7cm dia ball cant pass through the gap.
A Ball with dia less than 6.86cm can pass through the gap.
GP: :)
Let us move to triangles:
Two right angled triangles have a common hypotenuse (joined at the hypotenuse of same length). The other two sides of one triangle are 18 and 13 units. The other two sides of the second triangle are 20 and 7 units.
Obviously the units are not decimal (base ten numbers). What is the length of the hypotenuse and what is the base used?
All of you seem to be in IT sector. This should be stress relief for you! :)
i yaam in construction sector, but work yellAm computer la thaan. :)
தசம எண்கள் கிடையாதா? :confused2:
பித்தாகரஸ் தேற்றம் கைகொடுக்கவில்லையே :huh:
Pythogorean theorem will help to find the base(not decimal)! :) First you have to solve for the unknown base ! :)
Quote:
Originally Posted by disk.box
You must be the one who populated Avinasi Road,CBE with all those buildings! :lol:Quote:
Originally Posted by GP
If two triangles have common hypotenuse, it should be a rectangle, right (for any base)? :?Quote:
Originally Posted by rajraj
No! Draw a circle with the hypotenuse as the diameter. Pick any point on the circle. The diameter will subtend a right angle ! :) It is known as Thales' theorem!Quote:
Originally Posted by GP
The base of the number system cannot be lesser than 9 (since there is a number with digit 8) at the same time it cannot be greater than 20, because if so all the numbers decimal equivalent would be the same (the biggest number is 20) and we already observe that the numbers don't meet the requirement in decimal base.Quote:
Originally Posted by rajraj
A tiny piece of Python code to iterate yields the answer to be base 12:
Code:def toDecimal( something,base ):
index = 0
number = 0
for el in something[ : : -1 ]:
number += int( el ) * ( base**index )
index += 1
return number
a1 = "18"
b1 = "13"
a2 = "20"
b2 = "7"
for i in range(9,20):
print "Trying base %i" %i
if toDecimal(a1,i)**2 + toDecimal(b1,i)**2 == toDecimal(a2,i)**2 + toDecimal(b2,i)**2:
print "Success!!! The base is %i" %i
break
Quote:
Trying base 9
Trying base 10
Trying base 11
Trying base 12
Success!!! The base is 12
Rajraj Sir,Quote:
Originally Posted by rajraj
Still I am unable to arrive at the solution :oops: Please post the complete co-ordinates.
Correct! :) You can solve it using paper and pencil. Assume n to be the base ,express the numbers in decimal base and use Pythogorean theorem. You get the equation:Quote:
Originally Posted by Benny Lava
(n+8)^2 + (n+3)^2 = (2n)^2 + 7^2
or
n^2 -11n -12 =0
or
n=12
:)
You get an A- because you used a script ! :lol:
line 1...................... (4,4) to (0,4) Connects the dots in top rowQuote:
Originally Posted by rajraj
line 2...................... (0,4) to (4,0) Connects the dots below the main diagonal in the matrix of dots
line 3....................... (4,0) to (4,3) Connects the bottom three dots in the last column
line 4........................ (4,3) to (1,3) Connects the dots in the row below the top row
line 5......................... (1,3) to (1,1) Connects the bottom three dots in the first column
line 6......................... (1,1) to (3,1) Connects the first three dots in the bottom row
line 7..........................(3,1) to (3,2) Connects the bottom two dots in the third column
Benny Lava: I think you missed something in what I said or misunderstood something! :)
May be, I was not very clear ! :(
Oh! Why didn't I think of that :oops:Quote:
Originally Posted by rajraj
Oh.. I was thinking that you were giving a solution with 6 lines. I was pretty convinced that it couldn't be done with any fewer than 7 lines.Quote:
Originally Posted by rajraj
No Rajraj sir, you were pretty clear.. it was just an assumption on my part that you were posting a solution for connecting with 6 lines.
OTOH, you said even in the original problem (9 dots version) the lines aren't supposed to cross. The solution that I have crosses each other. Could you please post the solution to that puzzle as well? :)
Benny Lava : Here is the solution for 3x3 dot matrix ! :)
.....o.....o.....o <- Start here at (3,3)
.....o.....o.....o
.....o.....o.....o
.................................
line 1...................... (3,3) to (0,3) Connects the dots in top row
line 2...................... (0,3) to (3,0) Connects the dots below the main diagonal in the matrix of dots
line 3....................... (3,0) to (3,3) Connects the three dots in the last column making a triangle
line 4........................ (3,3) to (0,0) Connects the dots in the other diagonal
You should have an arrow like figure! :)
Yes... this is what I have seen but in this solution line 2 and line 4 intersects at (1.5,1.5) hence my original confusion. :)
You are right! I forgot the restriction and gave you the solution for a simpler puzzle. With the added restriction the solution will resemble the letter Z , the letter S or the letter G requiring a minimum of five lines.Quote:
Originally Posted by Benny Lava
(That tells me that I should not do anything serious beyond midnight and three hours beyond my bed time! :lol: )
Ok :)
STRESS RELIEF
When I posted the 4x4 dot matrix puzzle I expected soutions with only horizontal and vertical lines! :) Kirukan gave one. There are others. Can you find them? You will be surprised to see how simple they are!
Benny Lava : Here is the correct solution for 3x3 dot matrix ! :)
.....o.....o.....o <- Start here at (3,3)
.....o.....o.....o
.....o.....o.....o
.................................
line 1...................... (3,3) to (0,3) Connects the dots in top row
line 2...................... (0,3) to (3,0) Connects the dots below the main diagonal in the matrix of dots
line 3....................... (3,0) to (3,2) Connects the bottom two dots in the last column
line 4........................ (3,2) to (1,2) Connects the three dots in the middle row
line 5....................... (1,2) to (1,1) Connects the bottom two dots in the first column
This is similar to the solution for the 4x4 dot matrix.
(There is a pile of small sheets in front of the monitor . I picked the wrong one earlier. I have already given you a few of the other possible solutions with 5 lines! :) )
I hope all of you visiting this thread had enough stress relief! :)
Here is another problem to use what a Greek mathematician discovered:
A five meters long ladder is leaning against a wall. A cubic box ( 1m x 1m x 1m) is against the wall. One edge of the cube touches the ladder and one side of the cube touches the wall with another side on the ground. That is, the one meter cube box fits snugly in the gap between the ladder and the wall sitting on the ground.
What is the distance of the foot(base) of the ladder from the wall?
Close to 400 views! But, no solution yet! :) Time to assemble all information to solve the puzzle!
1. Draw a diagram - cross section
2. The ladder leaning against the wall makes a triangle with the wall and the floor as sides. Hypotenuse has a length of 5 meters.
3. Inside the triangel is a square of side 1 meter. The square divides the large triangle into two small triangles.
4. One triangle sits on the square. That is the length of a side is 1 meter
5. The other small triangle is on the side sharing a side with the square. That is one side of the triangle is 1 meter.
6. The bases ( horizontal sides) of the two small triangles are parallel to each other(and parallel to the ground) .
7. The other sides of the small triangles are also parallel( and parallel to the wall)
8. There must be something special about the two small triangles! :)
9. We learnt a few things in geometry and there is Pythogorean theorem!
Now try! :)
Hint: you have to use one unknown - either the horizontal side of the triangle on the side of the square and relate to the vertical side of the other small triangle or vice versa ! :)
what, this has not been solved even after 10 days?
this doesn't look that much difficult.
i think 2 answers possible for dis.
4.84m and 1.26m?
just a rough calculation.
if wrong, tell me. will once again try it thinking out of box.
GP:
Which one makes sense if you are doing something on the wall? It will help others if you explain how you solved it! :)
I hope it is not by trial and error ! :lol:
let base of the bigger triangle be 'b' and height be 'h'.
Relating the two small triangles,
(h-1)X(b-1)=1X1 => hb=h+b => h=b/(b-1) --> Eqn I
According to Pythogoras,
h^2+b^2=25 --> Eqn II
Solving Eqn I and II,
(b/(b-1))^2+b^2=25 => b^2(1+1/(b-1)^2)=25
Solving the above eqn we can find the value of 'b' and hence 'h'.
of course i used excel spreadsheet for solving the last eqn.
Correct! :)
1.260 is the right answer. With the other length you won't need a ladder! If one is the base the other will be the height. You don't need a ladder for a height of 1.26 m ! :)
In case people are wondering equation one is from the fact that the two small triangles are similar! :)
If nobody posts the next puzzle I will post one soon! :)
How do you solve it without writing a script?
high school maths la indha diagram padicha nyabagam konjam irundhuchu, especially that relationship between smaller triangles.
idhukku script, screenplay yellAm thEvai padala.
it was a nice stress reliever. :wink:
How do you solve it without a spreadsheet/script?Quote:
Originally Posted by GP
Hint: Let (b-1) = x, the distance between the square and the ladder along the floor. Rewrite the last equation with x. Express the 4th degree equation as a square of a quadratic equation. Now, you won't need a spreadsheet/script! :) A simple $4 calculator will do or an approximation! :)
( These puzzles are from practice IQ tests where a laptop is not allowed! :lol: )
A simple time pass puzzle:
Two taps are filling a bath. The first tap, by itself can fill the bath in one hour if the plug is in. The second tap can fill the bath in 30 minutes if the plug is in. With the plug out and both taps turned off, a full bath will empty in 45 minutes. How long does it take to fill an empty bath with both taps running and the plug out?
36 mins?
x/30+x/60-x/45=1
Sorry for the late response. Of course the answer is correct! :clap:
Next question:
A race course owner wants to install photoelectric timing device around his race track. The idea is to be able to conduct races & measure time taken for any race course length between 1000 metres to 31000 metres (in multiples of 1000 metres). Assume that he is free to conduct race in either direction of the race course. However the photoelectric timing device turns out to be pretty expensive. What is the minimum number of sensors required for his purpose and how should they be arranged?
No attempts? Guess I should explain the solution a bit more. Say if the length of the circular track is 3000 metres and the owner wants to conduct time tests for distance 1000 m or 2000 m, he can simply install two photoelectric device 1000 m from each other. For 2000 meter measurement he simply has to conduct race in the opposite direction.Quote:
Originally Posted by Benny Lava
Benny Lava: You have to state the assumptions clearly, unless I misunderstood! I have not been in a race in 60 years! :lol:
Do you need a photo sensor at the starting point?
Is the starting point fixed?
Depending on your answers to these questions the solutions will be different! :)
Hope you had a nice deepaavaLi ! :)
Quote:
Originally Posted by Benny Lava
Rajraj sir, Yes you are right. One sensor each at starting and ending point is needed to measure the time taken for each race. The starting point can be changed. The same photo electric sensor can be used as a starting point or an ending point. :)Quote:
Originally Posted by rajraj
More hints:
Take the 3000 meter circular race track, place one sensor at 0 meters( or 3000 meters if you go around) and another sensor at 2000 meters.
For a 1000 meter race start at the first sensor and end at the second sensor.
For a 2000 meter race start at the second sensor(at 1000 meters) and finish at the first sensor(at 3000 meters or 0 meters).
For a 3000 meter race start at the first sensor and finish at the same sensor going around.
You need only two sensors and you do not have to change directions. But you moved the starting point. That is the key - moving the starting point for different lengths!
Hope this helps in the 31000 meter race course problem! :)
While you are racing to solve the race course puzzle, here is something to relax! :)
On Sunday I watched the movie "A Disappearing Number", video taped version of the drama with the same title produced by the National Theatre(UK). It is about Ramanujan, the mathematician of the millennium. The number 1729 was in the movie. The conversation between Hardy and Ramanujan went somewhat like this when Hardy visited Ramanujan in the hospital.
Hardy: I rode in a taxi with the number 1729. An ordinary number
Ramanujan: No Hardy! It is a special number. It is the smallest number that can be exppressed as.............................
Fill in the blanks!
( Don't search the internet or run to the libray/bookstore for books on Ramanujan! :) )
Some more words of Ramanujan as hint:
Ramanujan: No Hardy! It is the smallest number that can be expressed as the sum of two___________________
Quote:
Originally Posted by rajraj
Yes. For a 31000 meter track, we have 30 different race distances to measure. If a combination of two sensors can measure for distance of N thousand meters, it can also be used to measure for distance of (31 - N) thousand meters. So essentially we need to have 15 combinations. 6C2 is equal to 15 hence 6 sensors are sufficient. Now the only problem is to decide how to place the 6 sensors such that all course lengths are covered :)Quote:
Originally Posted by rajraj