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#42001 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-07-15 22:43:33
Thats right!
This is how it is solved.
40 litres of milk and water contain 10% water.
Therefore, the milk content is 36 litres and water is 4 litres.
Lets assume x litres of water are added to make water 20%.
The ratio of water to the total becomes
4+x/40+x
this is equal to 20/100 or 1/5.
4+x/40+x = 1/5,
5(4+x) = 40+x
20+5x = 40+x
4x=20
x=5 litres
#42002 Re: This is Cool » "A Maths Dictionary for Kids" » 2005-07-14 23:18:10
Don't hate Mathematics, Zoe! Mathematics is fun, believe me! 
#42003 Re: Dark Discussions at Cafe Infinity » Wink Murder » 2005-07-14 23:15:58
I wonder if ganesh would like to play?
I am too old...I am more than 10,000 days old! 
#42004 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-07-14 23:02:55
That is close to a 1,000 gallons!
#42005 Re: Jokes » Limericks » 2005-07-14 19:52:23
who looked much like a lizard
#42006 Re: Puzzles and Games » rabbit farm » 2005-07-14 19:41:49
You mean 144? My friend says it is ___!
#42007 Re: Maths Teaching Resources » Welcome » 2005-07-14 18:59:54
I can live on 100 litres of coke a month, without food 
#42008 Re: Help Me ! » I need your help » 2005-07-14 17:57:42
I discovered that 2^20 ends in 76.
Thereafter, any number of the form 2^20n ends in 76.
Similarly, any number of the form 2^100n ends in 376.
Next, I had to know the last few digits of 2^500 and 2^2500.
When I learnt that they are 9376 and 09376, I was excited.
Because, any number of the form 2^500n would then have to end in 9376
and every 2^2500n would have to end in 09376.
This is what Rora was talking about. You are too young to understand this, Maxine. This is about higher powers of the number 2.
#42009 Re: Puzzles and Games » rabbit farm » 2005-07-14 17:47:58
This is much more complicated than what I thought! 
#42010 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2005-07-14 17:10:52
Outstanding! You are really supersmart!
Try this one....But don't post your reply immediately.
Let others too try.
(2) A mixture of 40 liters of milk and water contains 10% water. How much water must be added to make water 20% in the new mixture?
#42011 Re: Guestbook » guinea-pigs » 2005-07-14 17:00:41
I had one which lived with me for three years. I just love them, they are so cute
#42012 Re: Help Me ! » Hi » 2005-07-14 16:46:57
Hey you guys are pretty smart
Thanks, Jenilia. Visit the 'Games and Puzzles' topic regulalry. I shall post mathematical problems and their solutions which could be of some help to you, for the Olympiad.:)
#42013 Jai Ganesh's Puzzles » Problems and Solutions » 2005-07-14 16:43:22
- Jai Ganesh
- Replies: 468
In this topic, I shall give a problem and give the (proper mathematical) solution after a week.:D
The first one:-
The population of a town increases by 5% annually. If its population in 1995 was 138915, what was it in 1992?
#42014 Re: Puzzles and Games » Work done in time! » 2005-07-14 16:39:02
Excellent! You are correct! Tell us how you did it!
#42015 Re: This is Cool » Mathematic complexity » 2005-07-14 16:33:50
If Quasimodo had stumbled upon hidden treasure, then he may have discovered
#42016 Puzzles and Games » Work done in time! » 2005-07-13 23:15:48
- Jai Ganesh
- Replies: 5
A contractor agrees to complete a work in 100 days. He engaged 140 men for the job and after doing the work for 60 days, he came to know that only half the work was completed. How many more men should he hire to finish the work in the agreed time?
#42017 This is Cool » Which of the two is greater? » 2005-07-13 21:39:52
- Jai Ganesh
- Replies: 6
When there are two numbers x and y, such that both x,y ≥1,
does it follow that y^x is always greater than x^y if x is greater than y?
No.
This is true only if y is greater than a certain CRITICAL Value.
Many years back, I tried to find this critical value of y for certain values of x.
Value of x Approximate value of y
10 1.3712886
100 1.04955
1000 1.0069805
10,000 1.000922309
100,000 1.00011514925
1,000,000 1.0000138158
10,000,000 1.00000161283
100,000,000 1.0000001843
1,000,000,000 1.0000000208
Illustration:- y^100 can be greater than 100^y only if the value of y ≥1.04955
#42018 Re: Puzzles and Games » The light switch » 2005-07-13 21:17:21
23+18+12+22 is already greater than 30! So, it should be 'people' as pointed out by Mathsy!:)
#42019 Re: Help Me ! » Hi » 2005-07-13 21:10:05
That's because, in all the odd numbers up to 99,
there are 10 numbers ending in '5'.
5, 15, 25, 35, 45, 55, 65, 75, 85, and 95.
And, when an odd number is multiplied by a
number ending in '5', the resultant always ends in '5':)
#42020 Re: Puzzles and Games » The light switch » 2005-07-13 20:52:58
1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + .............1/ ∞ = 2
I cannot think beyond that.
#42021 Re: Help Me ! » Hi » 2005-07-13 20:43:03
My best wishes to you, Jenilia.
You can post all your mathematical doubts here.
We'd try to help you.
#42022 Re: Help Me ! » I need your help » 2005-07-13 19:04:11
Is it also necessary to prove it for n=1 ?
First, the formula or equation is checked for values like 1,2,3 etc.Only if this test is pased, we asume it is true for an arbitrary number k. Thereafter, if it is proved true for k+1, it becomes a 'proof beyond doubt'.
#42023 Re: Jokes » Limericks » 2005-07-13 17:17:15
There once was a boy from Montreal
Who loved to play basketball
For a team he tried out
But if he made it, I doubt
For you see, he was three feet tall!:D
#42024 Re: Help Me ! » I need your help » 2005-07-13 17:01:10
Mathematical Induction is a brilliant way of proving.
For example, the sum of the first n Natural Numbers is
n(n+1)/2.
For n=1, we get 1.
For n=2, we get 3.
Therefore, the summation formula works for 1 and 2.
Let's assume it is true for an arbitrary Natural Number k.
Therefore, the sum of the first k Natural Numbers would be k(k+1)/2.
For k+1, the sum would be k(k+1)/2 + (k+1),
that is [k(k+1) + 2(k+1)]/2, or [(k+1)(k+2)]/2
And, according to the formula we had at the beginning of the proof,
the sum of the first k+1 Natural Numbers would be [(k+1)(k+2)]/2
Since the two are equal, this can be said to be true for sum up to any natural number.
Similarly, it can be proved that:-
the sum 1² + 2² + 3² + 4² +... n² = [n(n+1)(2n+1)]/6
and also
the sum 1³ + 2³ + 3³ + 4 ³ + ......... n³ = [n(n+1)/2]²
#42025 Re: This is Cool » Mathematic complexity » 2005-07-13 16:35:24
I think it is extremely perplexing here quod it is filled with members who smell like feet, but somehow they actually manage to smell differently as I have struggled to climb up lorries, which resemble large...