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A subset B of the set of integers from 1 to 100, inclusive, has the pr

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A subset B of the set of integers from 1 to 100, inclusive, has the pr  [#permalink]

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New post 25 Mar 2019, 00:15
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A
B
C
D
E

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  25% (medium)

Question Stats:

71% (02:04) correct 29% (02:19) wrong based on 28 sessions

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Re: A subset B of the set of integers from 1 to 100, inclusive, has the pr  [#permalink]

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New post 25 Mar 2019, 00:25
Bunuel wrote:
A subset B of the set of integers from 1 to 100, inclusive, has the property that no two elements of B sum to 125. What is the maximum possible number of elements in B?

(A) 50
(B) 51
(C) 62
(D) 65
(E) 68


Primary Set of Integers = {1, 2, 3, 4, 5.....100}

The Unwanted pairs making sum 125 = {25, 26, 27, 28, 29,..........96, 97, 98, 99, 100}

Now, The set making pairs making sum 125 has 100-24 = 76 terms

But we can use half of the terms in the required set and leave the other half

So the requires set will have all terms from 1 to 24 and half of 76 terms i.e. 38 terms

Total Integers in Set B = 24+38 = 62

Answer: Option C
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A subset B of the set of integers from 1 to 100, inclusive, has the pr  [#permalink]

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New post Updated on: 28 Mar 2019, 23:35
Set A = { 1,2,3,4,5...,99,100}

Set B = given when 2 numbers added it will be less than 125

If I divide 125/2 = 62.5 , (If I add 62+63 = 125 )

So we can not consider 63 in the Subset , this gives us the total numbers we can consider are till 62

set B = { 1,2,3,4,5,6,7....60,61,62}.

Originally posted by baru on 28 Mar 2019, 21:02.
Last edited by baru on 28 Mar 2019, 23:35, edited 1 time in total.
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Re: A subset B of the set of integers from 1 to 100, inclusive, has the pr  [#permalink]

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New post 28 Mar 2019, 22:01
Bunuel wrote:
A subset B of the set of integers from 1 to 100, inclusive, has the property that no two elements of B sum to 125. What is the maximum possible number of elements in B?

(A) 50
(B) 51
(C) 62
(D) 65
(E) 68


B = {1, 2, 3, 4, 5, ... 100}

Two elements of B should not sum to 125. 100 is the highest number so we need at least 25 to make 125.
Hence we can include 1 to 24 in our set B. These numbers will not add with any other number to give 125.

Now, we have the following pairs which all add up to 125: {25, 100}, {26, 99}, {27, 98}... and so on till {62, 63}.
These are a total of (62 - 25) + 1 = 38 pairs.

We can pick one and only one number from each pair since if it does not have its partner, it will not make 125.
So we can pick another 38 numbers.

Total numbers in B = 24 + 38 = 62

Answer (C)
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Re: A subset B of the set of integers from 1 to 100, inclusive, has the pr   [#permalink] 28 Mar 2019, 22:01
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