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In how many ways?

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by smackmartine » Sat Aug 27, 2011 3:54 pm
I don't know whats the correct answer, but here is what I thought.

Case 1:When #100 is selected.

1* 99 =99

Case 2: When #100 is not selected.

For example : if #1 is selected first, we are left with only 1 option i.e #100 ..in order to make the sum greater than 100.
Similarly, when #2 , we left with #99 and #100 --> 2 ways.
So get a pattern. Because the # of ways in which second # is selected varies with the selection of each token # , we get (1+2+3+......+99) ways

So, Total # of ways = (1* 99)+(1+2+3+....+99)
Note : we can use the concept of avg to calculate (1+2+3+....+99), if we don't remember the formula to calculate the sum of arithmetic progression.

Sum=(#of terms)(Avg)= (99-1+1)*[(99+1)/2]
=99*50

Finally, Total # of ways=99 + (99*50)=99(1+50)=(100-1)51=5100-51 = 5049 ways.

Experts, please comment if I missed something.
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by Frankenstein » Sat Aug 27, 2011 7:41 pm
Hi Smack,

There is a minor mistake of counting repetitions in your solution.
1 is picked -> 1 way(only 100)
2 is picked -> 2ways(99,100)
.
.
.
50 is picked -> 50 ways(51-100)
51 is picked -> 49 ways(we need to count from 52 to 100 only because if we are counting 51 times then we are essentially counting the pair(50,51) again as it has been counted in the preceding case)
52 is picked -> 48 ways(53-100)
.
.
.
99 is picked -> 1 way(100)
100 is picked -> 0 ways(all of them have already been counted)
So, sum is (1+2+3..+50)+(49+48+...+1) = 2(1+2...+49) + 50 = 49*50 + 50 = 2500.

One quick way of checking answer at the end:
2 tokens from 100 can be picked simultaneously in 100C2 ways right? That is total cases are 4950. The answer of 5049(>4950) should give us a hint that repetitions are counted.
Cheers!

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by smackmartine » Sat Aug 27, 2011 8:12 pm
Thanks Frankenstein. I should have considered the repetitions.
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by GMATGuruNY » Sun Aug 28, 2011 3:53 am
finance 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 greater than 100?
Let B = the bigger integer in each pair.
The largest possible value of B = 100.
The smallest possible value of B = 51.
If B ≤ 50, there will be no smaller integer that could be added to B to yield a sum greater than 100.
Thus, 51 ≤ B ≤ 100.

To count consecutive integers:
Number = biggest - smallest + 1.

B = 100:
Adding 100 to any integer from 99 to 1, inclusive, will yield a sum greater than 100.
Total options = (99-1) + 1 = 99.

B = 99:
Adding 99 to any integer from 98 to 2, inclusive, will yield a sum greater than 100.
Total options = (98-2) + 1 = 97.

B = 98:
Adding 98 to any integer from 97 to 3, inclusive, will yield a sum greater than 100.
Total options = (97-3) + 1 = 95.

B = 51:
Adding 51 to 50 will yield a sum greater than 100.
Total options = 1.

Note the pattern exhibited by the results above.
The number of options is the set of decreasing consecutive odd integers from 99 to 1, inclusive.
Thus, the total number ways to pick the tokens = the sum of the consecutive odd integers from 1 to 99, inclusive.

Sum of evenly spaced integers = (number of integers) * (average).

To count consecutive odd integers:
Number = (biggest-smallest)/2 + 1.
(99-1)/2 + 1 = 50.

Average of evenly spaced integers = (biggest+smallest)/2.
(99+1)/2 = 50.

Sum = number * average = 50*50 = 2500.
Thus, the total number of ways to pick the tokens = 2500.
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by finance » Sun Aug 28, 2011 6:20 am
GMATGuruNY wrote:
finance 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 greater than 100?
Let B = the bigger integer in each pair.
The largest possible value of B = 100.
The smallest possible value of B = 51.
If B ≤ 50, there will be no smaller integer that could be added to B to yield a sum greater than 100.
Thus, 51 ≤ B ≤ 100.

To count consecutive integers:
Number = biggest - smallest + 1.

B = 100:
Adding 100 to any integer from 99 to 1, inclusive, will yield a sum greater than 100.
Total options = (99-1) + 1 = 99.

B = 99:
Adding 99 to any integer from 98 to 2, inclusive, will yield a sum greater than 100.
Total options = (98-2) + 1 = 97.

B = 98:
Adding 98 to any integer from 97 to 3, inclusive, will yield a sum greater than 100.
Total options = (97-3) + 1 = 95.

B = 51:
Adding 51 to 50 will yield a sum greater than 100.
Total options = 1.

Note the pattern exhibited by the results above.
The number of options is the set of decreasing consecutive odd integers from 99 to 1, inclusive.
Thus, the total number ways to pick the tokens = the sum of the consecutive odd integers from 1 to 99, inclusive.

Sum of evenly spaced integers = (number of integers) * (average).

To count consecutive odd integers:
Number = (biggest-smallest)/2 + 1.
(99-1)/2 + 1 = 50.

Average of evenly spaced integers = (biggest+smallest)/2.
(99+1)/2 = 50.

Sum = number * average = 50*50 = 2500.
Thus, the total number of ways to pick the tokens = 2500.
Thank you GMATguruNY,,,you explanations are always concise,wise and understandable.

Even if I do not want to alter the topic, just wanted to learn your advise...I will have my exam in two weeks and really do not know what to expect...my manhattan scores are around 710 and 720 but my veritas result is 670 whereas Kaplan results around 600:((...(I do not like kaplans' questions and expalantions, but I feel safe with manhattan)...Do you think I have any chance to score over 700?
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by Brent@GMATPrepNow » Sun Aug 28, 2011 1:42 pm
Here's another approach.

Of the 2 selected numbers, one of them will be the smaller number and one will be the larger number.
Since the smaller can range from 1 to 99 (the smaller number cannot be 100), there are 99 cases to consider.

Note: for each case, we'll count the number of ways that the sum can be greater than 100.

Case 1: 1 is the smaller number, which means the larger number must be 100 (1 way for the sum to be greater than 100)
Case 2: 2 is the smaller number, which means the larger number must be 99 or 100 (2 ways)
Case 3: 3 is the smaller number, which means the larger number must be 98, 99 or 100 (3 ways)
Case 4: 4 is the smaller number, which means the larger number must be 97, 98, 99 or 100 (4 ways)
.
.
.
Case 50: 50 is the smaller number, which means the larger number must be 51, 52,..., 99 or 100 (50 ways)

Case 51: 51 is the smaller number, which means the larger number must be 52, 53,..., 99 or 100 (49 ways)
Case 52: 52 is the smaller number, which means the larger number must be 53, 54,..., 99 or 100 (48 ways)
Case 53: 53 is the smaller number, which means the larger number must be 54, 55..., 99 or 100 (47 ways)
.
.
.
Case 98: 98 is the smaller number, which means the larger number must be 99 or 100 (2 ways)
Case 99: 99 is the smaller number, which means the larger number must be 100 (1 way)

So, when we add up the all of these cases we get (1+2+3+...+50)+(49+48+47+...+3+2+1)

To find these two sums, we'll use the fact that the sum of the positive integers from 1 to k equals k(k+1)/2

So, when we add all of the cases together we get 50(50+1)/2 + 49(49+1)/2 = 2500

Cheers,
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by jonathan123456 » Sat Jan 28, 2012 6:38 pm
Why is 99 and 100 not considered as part of the list
98+99 > 100, 98+100>100


B = 98:
Adding 98 to any integer from 97 to 3, inclusive, will yield a sum greater than 100.
Total options = (97-3) + 1 = 95.


GMATGuruNY wrote:
finance 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 greater than 100?
Let B = the bigger integer in each pair.
The largest possible value of B = 100.
The smallest possible value of B = 51.
If B ≤ 50, there will be no smaller integer that could be added to B to yield a sum greater than 100.
Thus, 51 ≤ B ≤ 100.

To count consecutive integers:
Number = biggest - smallest + 1.

B = 100:
Adding 100 to any integer from 99 to 1, inclusive, will yield a sum greater than 100.
Total options = (99-1) + 1 = 99.

B = 99:
Adding 99 to any integer from 98 to 2, inclusive, will yield a sum greater than 100.
Total options = (98-2) + 1 = 97.

B = 98:
Adding 98 to any integer from 97 to 3, inclusive, will yield a sum greater than 100.
Total options = (97-3) + 1 = 95.

B = 51:
Adding 51 to 50 will yield a sum greater than 100.
Total options = 1.

Note the pattern exhibited by the results above.
The number of options is the set of decreasing consecutive odd integers from 99 to 1, inclusive.
Thus, the total number ways to pick the tokens = the sum of the consecutive odd integers from 1 to 99, inclusive.

Sum of evenly spaced integers = (number of integers) * (average).

To count consecutive odd integers:
Number = (biggest-smallest)/2 + 1.
(99-1)/2 + 1 = 50.

Average of evenly spaced integers = (biggest+smallest)/2.
(99+1)/2 = 50.

Sum = number * average = 50*50 = 2500.
Thus, the total number of ways to pick the tokens = 2500.
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by GMATGuruNY » Mon Jan 30, 2012 1:45 am
jonathan123456 wrote:Why is 99 and 100 not considered as part of the list
98+99 > 100, 98+100>100


B = 98:
Adding 98 to any integer from 97 to 3, inclusive, will yield a sum greater than 100.
Total options = (97-3) + 1 = 95.
B = the bigger of the two integers.
When we count the number of options for B=100, we include EVERY combination in which B=100. Note the portion in red:
B = 100:
Adding 100 to any integer from 99 to 1, inclusive, will yield a sum greater than 100.
Total options = (99-1) + 1 = 99.
Since the calculation above accounts for ALL the options when B=100, we EXCLUDE B=100 when we count the number of options for B<100.
The same reasoning applies to B=99.

Thus, when we count the number of options for B=98, we exclude B=100 and B=99, since we have already counted all the combinations that include these values.
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by monge1980 » Tue Jan 31, 2012 7:25 am
I approached this problem in the following, quite fast way:

Total number of ways = 100C2 = 50*99
But we need to eliminate the number of ways to choose 2 numbers from 1 to 50 ... 50C2= 25*49
We also need to eliminate the combination of the upper part numbers with the lower part
51... Eliminate 49,48,47......1 ... 49 ways
52.... Eliminate 48, 47, 46.... 48 ways
...
....
...
99 ... Eliminate 1.... 1 way
Which leads to the elimination of 49+48+47..... = 49*50/2

So at the end we have 50*99-(25*49+25*49)= 50*50=2500
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