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# Set A contains 50 different positive integers, the

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Set A contains 50 different positive integers, the [#permalink]

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12 Jul 2006, 15:31
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Set A contains 50 different positive integers, the arithmetic mean of which is 60. Are any less than 35?

(1) 10 of the digits are above 85.
(2) 8 of the digits are above 90.
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Re: DS: Any number less than 35? [#permalink]

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12 Jul 2006, 16:13
Set A contains 50 different positive integers, the arithmetic mean of which is 60. Are any less than 35?

(1) 10 of the digits are above 85.
(2) 8 of the digits are above 90.

The problem statement indicates that the sum of the numbers is 50 * 60 = 300

Consider (1)
10 are above 85
the lowest that these could be is 86-95, and their sum is (86+95)/2 * 10 = 905.
The rest should make up 2095.
Consider the lowest combination of the rest 40 numbers >= 35, this is 35 â€“ 74
Sum of these is (35+74)/2 * 40 = 2180, which is greater than the available limit.

So there should be numbers below 35 in the sequence.

So (1) alone is sufficient.

Consider (2)
Along the same line
Total of 8 above 90 = (91+98)/2 * 8 = 756
The rest should make up = 3000 â€“ 756 = 2244
Lowest possible combination >= 35 is 35 â€“ 76
Sum is (35+76)/2 * 42 = 2331, which is greater than 2244
So again there should be a no below 35

So (2) alone is sufficient,

SO FINALLY, (1) and (2) both alone are sufficient.

Hope this helps...and hopefully this is right.
Anand

kevincan wrote:
Set A contains 50 different positive integers, the arithmetic mean of which is 60. Are any less than 35?

(1) 10 of the digits are above 85.
(2) 8 of the digits are above 90.
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12 Jul 2006, 16:28
Nice work, but a bit too long! Can we find a way that involves crunching smaller numbers?
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12 Jul 2006, 16:39
Ass.: 50 different integers, no integer is less than 35 and their mean is 60. This set is unique. Its elements are 35, 36, 37, 38, ..57, 58, 59, 61, 62, 63, 83, 84, 85.

... see below

Hence, A and B are each alone sufficient.
=> D.

Last edited by game over on 12 Jul 2006, 18:48, edited 1 time in total.
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12 Jul 2006, 16:49
You made it look simple enough..but can u explain...how is that set unique?

game over wrote:
Ass.: 50 different integers, no integer is less than 35 and their mean is 60. This set is unique. Its elements are 35, 36, 37, 38, ..57, 58, 59, 61, 62, 63, 83, 84, 85.
Hence, A and B are each alone sufficient.
=> D.
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12 Jul 2006, 17:12
sorry, my argument wasn't complete.

[You had probably the following in mind:]

consider a new set, which is similar to the old set but has two different elements: one element >60 is replaced by 60. To get mean=60, we can always replace just one element >60 by a new element >85.
How many elements can we replace at most?
Consider: 85 => 60 (this means that 85 is replaced by 60). There are 25 "free points".
=> 84 => 86, 83=>87, 82=>88, 81=> 89, 80 => 90. At most, we can have 5 elements > 85.

Therefore: If we have more than 5 elements >85 (>90), there must be an element <35.
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13 Jul 2006, 00:24
game over wrote:
sorry, my argument wasn't complete.

[You had probably the following in mind:]

consider a new set, which is similar to the old set but has two different elements: one element >60 is replaced by 60. To get mean=60, we can always replace just one element >60 by a new element >85.
How many elements can we replace at most?
Consider: 85 => 60 (this means that 85 is replaced by 60). There are 25 "free points".
=> 84 => 86, 83=>87, 82=>88, 81=> 89, 80 => 90. At most, we can have 4 elements > 85.

Therefore: If we have more than 4 elements >85 (>90), there must be an element <35.

A trivial correction makes yours a very elegant OE!
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13 Jul 2006, 07:01
makes sense....thanks for the explaination.

kevincan wrote:
game over wrote:
sorry, my argument wasn't complete.

[You had probably the following in mind:]

consider a new set, which is similar to the old set but has two different elements: one element >60 is replaced by 60. To get mean=60, we can always replace just one element >60 by a new element >85.
How many elements can we replace at most?
Consider: 85 => 60 (this means that 85 is replaced by 60). There are 25 "free points".
=> 84 => 86, 83=>87, 82=>88, 81=> 89, 80 => 90. At most, we can have 4 elements > 85.

Therefore: If we have more than 4 elements >85 (>90), there must be an element <35.

A trivial correction makes yours a very elegant OE!
13 Jul 2006, 07:01
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