Tokens

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Tokens

by goyalsau » Wed Nov 17, 2010 11:57 pm
There are 100 tokens numbered from 1 to 100. In how many ways can two tokens be drawn simultaneously so that their sum is more than 100?



4950

5050

2500

2550

5000
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by beat_gmat_09 » Thu Nov 18, 2010 12:06 am
For every number selected there are that many number's available for choosing so that the sum is greater than 100.
1 - 1,100. No other number can be chosen, therefore only 1 set.
2 - 2,100; 2,99 - 2 sets.
3 - 3,100; 3,99; 3,98; - 3 sets.
..
..
..
generalizing..
98 - 98 sets are possible. 98,3 ; 98,4; 98,5; 98,6 .... 98,98 ; 98,99; 98,100.
thus total possibilities - 1 + 2 + 3 + .... 100
sum of 1st 100 numbers = (100+1)/2*100 = 5050
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by goyalsau » Thu Nov 18, 2010 12:47 am
beat_gmat_09 wrote:For every number selected there are that many number's available for choosing so that the sum is greater than 100.
1 - 1,100. No other number can be chosen, therefore only 1 set.
2 - 2,100; 2,99 - 2 sets.
3 - 3,100; 3,99; 3,98; - 3 sets.
..
..
..
generalizing..
98 - 98 sets are possible. 98,3 ; 98,4; 98,5; 98,6 .... 98,98 ; 98,99; 98,100.
thus total possibilities - 1 + 2 + 3 + .... 100
sum of 1st 100 numbers = (100+1)/2*100 = 5050
Buddy The official Answer is 2500,

Till 50 till approach is correct,

50 - till 51 to 100 ----- in all 50 nummbers,

but from 51 the problem will start,

51 --- 50 ,{ No 51 } because the digit in 51 , then 52 , 53 .. 100 in all 49 digits

52 ---- 49 , 50 , 51, No 52 this time, 53 , 54 .... 100 in all 51 digits

53 --- 48 , 49, 50 , 51, 52 , ......... 54 , 55 to 100 in all 52 digits

This will be the pattern , But it will very tough to calculate it this way,, so i am looking for a better technique.....
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by Rahul@gurome » Thu Nov 18, 2010 1:39 am
goyalsau wrote:There are 100 tokens numbered from 1 to 100. In how many ways can two tokens be drawn simultaneously so that their sum is more than 100?
  • 4950
    5050
    2500
    2550
    5000
Careful!
This question is full of tricks and traps!

One trick is there are two categories, n ≤ 50 and n > 50 (as goyalsau mentioned). And another one is the tokens are drawn simultaneously! Thus (2, 99) and (99, 2) are basically same.

Now, let's proceed to solve the problem.
As mentioned earlier, let's divide the numbers in two categories.

Category 1: n ≤ 50
For 1 : 1 set -> (1, 100)
For 2 : 2 set -> (2, 99), (2, 100)
For 3 : 3 set -> (3, 98), (3, 99), (3, 100)
� �
For 50 : 50 set -> (50, 51), ... , (50, 99), (50, 100)

Total number of sets = 1 + 2 + 3 + ... + 50

Category 1: n > 50
For 51 : 50 set -> (50, 51), (51, 52), ... ,(51, 100) No (51, 51)
For 52 : 51 set -> (49, 52), (50, 52), ... ,(52, 100) No (52, 52)
For 53 : 52 set -> (48, 53), (49, 53), ... ,(53, 100) No (53, 53)
� �
For 100 : 99 set -> (1, 100), (2, 100) ... , (99, 100) No (100, 100)

Total number of sets = 50 + 51 + 52 + ... + 99


Therefore total number of sets = (1 + 2 + 3 + ... + 50) + (50 + 51 + 52 + ... + 99)
= (1 + 2 + 3 + ... + 99) + 50
= (99)*(99 + 1)/2 + 50
= 99*50 + 50
= 50*100
= 5000

Now for the second trap, all the set are counted twice.
Thus effective number of set = 5000/2 = 2500

The correct answer is C.
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by Deepthi Subbu » Thu Nov 18, 2010 10:28 am
Good and detailed explanation Rahul . But when you say that all sets are counted twice - Can you explain with an example as am unable to decode.

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by Rahul@gurome » Thu Nov 18, 2010 11:15 am
Deepthi Subbu wrote:Good and detailed explanation Rahul . But when you say that all sets are counted twice - Can you explain with an example as am unable to decode.
Observe my solution carefully, for 50 I've taken (50, 51) as valid token set, and for 51 again I've taken (50, 51) as valid token set. Thus the set (50, 51) is counted twice. Same thing happens for any other set too. (2, 99) is counted for 2 and again for 99 etc. Thus all the set are counted twice.
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by diebeatsthegmat » Wed Dec 22, 2010 6:45 pm
Rahul@gurome wrote:
goyalsau wrote:There are 100 tokens numbered from 1 to 100. In how many ways can two tokens be drawn simultaneously so that their sum is more than 100?
  • 4950
    5050
    2500
    2550
    5000
Careful!
This question is full of tricks and traps!

One trick is there are two categories, n ≤ 50 and n > 50 (as goyalsau mentioned). And another one is the tokens are drawn simultaneously! Thus (2, 99) and (99, 2) are basically same.

Now, let's proceed to solve the problem.
As mentioned earlier, let's divide the numbers in two categories.

Category 1: n ≤ 50
For 1 : 1 set -> (1, 100)
For 2 : 2 set -> (2, 99), (2, 100)
For 3 : 3 set -> (3, 98), (3, 99), (3, 100)
� �
For 50 : 50 set -> (50, 51), ... , (50, 99), (50, 100)

Total number of sets = 1 + 2 + 3 + ... + 50

Category 1: n > 50
For 51 : 50 set -> (50, 51), (51, 52), ... ,(51, 100) No (51, 51)
For 52 : 51 set -> (49, 52), (50, 52), ... ,(52, 100) No (52, 52)
For 53 : 52 set -> (48, 53), (49, 53), ... ,(53, 100) No (53, 53)
� �
For 100 : 99 set -> (1, 100), (2, 100) ... , (99, 100) No (100, 100)

Total number of sets = 50 + 51 + 52 + ... + 99


Therefore total number of sets = (1 + 2 + 3 + ... + 50) + (50 + 51 + 52 + ... + 99)
= (1 + 2 + 3 + ... + 99) + 50
= (99)*(99 + 1)/2 + 50
= 99*50 + 50
= 50*100
= 5000

Now for the second trap, all the set are counted twice.
Thus effective number of set = 5000/2 = 2500

The correct answer is C.
nice! :) i got 5000 too

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by thebigkats » Sat Jan 08, 2011 3:52 pm
thanks Rahul. Probably your way was faster. this is how got to this solution (although took me full 3 min :-(

for no 1: 100 (total 1)
2: 100, 99 (total 2)
3: 100, 99, 98 (total 3)
...
50: 100, .. 52, 51 (total 50)

51: 100..52, 50 (but 51, 50 was already counted so leave it) TOTAL = 49
52: 100..53, 51, 50, 49 (but 49, 50 and 51 are already counted so leave them). TOTAL = 48
53: 100..54, 53, 52, 51, 50, 49, 48 (but 53..48 are already counted so leave them) TOTAL = 47
...
99: 100.., 98, 97..2 (but last ones except 100 are already counted so leave them) TOTAL = 1
100: all are already counted so TOTAL=0

So for first 50, total = n(n+1)/2 = 50(50+1)/2 = 25*51=1275
from 52..10, total = n(n+1)/2 = 49 (49+1)/2 = 25*49=1225

Total = 1275 + 1225 = 2500

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by Zerks87 » Wed Jan 19, 2011 10:41 am
I disagree that the answer is 2500.

When I look at the problem I see that there are no combinations that violate that condition that the two randomly selected number must add up to greater than 100.

Therefore there are 100 numbers and 2 slots that can possibly be filled. To me this is a simple combinations question and you should be able to solve using the equation N!/K!(N-K)!. Notice how this is formula that takes into account that order matters since (98,2) and (2,98) is the same combination of numbers.

So in this question N= 100 and K=2.

Thus, the formula is 100!/2!(98!), which reduces to 100*99/2. This can be further reduced to 50*99 = 4950.

Therefore I believe the correct answer is A

Please let me know if you think my logic is wrong.

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by fitzgerald23 » Wed Jan 19, 2011 2:35 pm
I approached this from the other direction, which I thought made this relatively simple. There are 99 possibilities with the number 100. There are 97 different possibilities with 99 (all but 1 and 100), 95 possibilities with 98 (all but, 1, 2, 100, and 99) at which point you recognize the pattern is simply going to be 99, 97, 95, 93....1, which will be for 51 (the only remaining possibility is 51 + 50). All you need to do is add up the evenly spaced set, by finding the mean and amount of numbers in the set:


99+1/2 = 50

99-1/2+1 = 50

50 * 50 = 2500.

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by thebigkats » Wed Jan 19, 2011 4:35 pm
Hi:
Your method has a simple flaw - formula N!/K!(N-K)! returns how many combinations of K elements you can have when you are drawing from a set of N elements. It doesn't talk about the characteristics of any element in set of K. hence we can;t use the formula.
Your method tells us how many different combinations of 2 tokens can we draw but it doesn;t take out combos that don;t add up to 100 or more. for example, this method would also allow a combo (1,2) which is NOT allowed by the problem statement.

HTH,
Kats
Zerks87 wrote:I disagree that the answer is 2500.

When I look at the problem I see that there are no combinations that violate that condition that the two randomly selected number must add up to greater than 100.

Therefore there are 100 numbers and 2 slots that can possibly be filled. To me this is a simple combinations question and you should be able to solve using the equation N!/K!(N-K)!. Notice how this is formula that takes into account that order matters since (98,2) and (2,98) is the same combination of numbers.

So in this question N= 100 and K=2.

Thus, the formula is 100!/2!(98!), which reduces to 100*99/2. This can be further reduced to 50*99 = 4950.

Therefore I believe the correct answer is A

Please let me know if you think my logic is wrong.

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by monge1980 » Fri Apr 22, 2011 12:32 am
Hi guys I approach the problem in another way:

- total number of combo = 100C2 = 4950

- eliminate all the combination of the numbers from 1 to 50 (their sum is never >100) ... 50C2=1225

- subtract 4950-1225=3725 ... these are all the combinations of the numbers from 100 to 51 with all the 100 numbers.

- now,
if we choose 99 we need to eliminate 1
if we choose 98 ....... 2
97 ..... 3
.
.
.
51 ... 49

So we need to eliminatethe sum from 1 to 49 ...

mean=49+1/2=25
number=49-1+1=49
25*49=1225

there we go ... 3725-1225=2500

I think it is a quite fast method.

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by del@btg » Mon Apr 25, 2011 6:22 am
Well, I totally agree for the 1st category (n<=50) part of Mr. Rahul's solution but i do not get why is he making the solution more complex in the second part(n>50)

Category 1. n<=50

No. of comb. = 1+2+3....+50 (as explained very well by Mr. Rahul)

Category 2. n>50

We can simply select any two no.(from 51 to 100) and their sum would always be greater than 100.

No. of comb = 50c2

Adding both the catergories

Total no. of comb = 2500


Please let me know if i am making a mistake somewhere.

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by Ian Stewart » Mon Apr 25, 2011 11:16 am
goyalsau wrote:There are 100 tokens numbered from 1 to 100. In how many ways can two tokens be drawn simultaneously so that their sum is more than 100?



4950

5050

2500

2550

5000
We can take advantage of the fact that the set of numbers is symmetric here to avoid awkward calculation, though the solution is a bit conceptual:

* there are 100C2 = 4950 ways to choose 2 tokens

* the average of our set is 101/2 = 50.5, so when we sum two random numbers from the set, on average that sum will be 101

* when we add the two numbers, three things can happen: the sum will be equal to the average of 101, the sum will be below average, or the sum will be above average. The number of ways to get a below average sum must be equal to the number of ways to get an above average sum because our set is symmetric.

* we can get a sum of 101 in 50 ways (for any number from 1 to 50, there's one pair we can make)

* of the remaining 4950 - 50 = 4900 pairs we could choose, half of them, or 2450 of them, give us an 'above average' sum (so a sum greater than 101)

* so there are 50 + 2450 = 2500 pairs we could choose to give a sum of 101 or more
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by mourinhogmat1 » Mon Jan 30, 2012 12:04 am
A graphical approach to this problem is to visualise the problem like a huge matrix of 100 by 100's.

1 2 3 4 5 6 .......100
1
2
3
4
5
6
.
.
.
100

There are a total of 100^2 options of which half are repeat options of one another. So, unique options are 5000. If you can split the square with a diagonal from 1 on the top left to 100 on the bottom right. (Visualize)

Next, when you look at the likely possibilities which gives >100 total, the table fills up as follows:
I have cut short the table to start from 97 due to graphical limitations while posting this.

.. 97 98 99 100
1 - - - x
2 - - x x
3 - x x x
4 x x x x
5
6
.
.
97
98
99
100

- denotes the the sum of the two values will not > 100.
x denotes that the sum of the two values >100.
Note we dont fill up the x's in the bottom half (on the left) because they are repeats and therefore not unique. Once you fill up till 50 you can expect the value to be half the top square (i.e. 50 by 50) --> 5000/2 = 2500.