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For a lottery ticket, Emily chose six numbers that average [#permalink]
 17 Oct 2009, 05:44
Question Stats:
71% (02:26) correct
28% (01:35) wrong based on 28 sessions
For a lottery ticket, Emily chose six numbers that average to 10. Did more than half the numbers have 2 digits? (1) One of the numbers was 10. (2) Three of the numbers added up to 40.
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Re: Lottery Ticket [#permalink]
17 Oct 2009, 08:44
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Hmm I don't know a mathematical formula way to figure it out but you can solve it relatively quickly by picking good numbers.
Question stem says (N1+N2....+N6)/6 = 10 => N1+N2....+N6 = 60
Statement 1)
One of the numbers was 10.
This just means N1+N2+N3+N4+N5+10 = 60
N1+N2+N3+N4+N5= 50
Each of the numbers could be 10 or we could have 4 numbers as 1 and the last one as 46.
Insuff.
Statement 2)
Three numbers add to 40. The number of ways to do this could be:
- 1 double digit (eg. 38,1,1)
- 2 double digit (eg. 10,25,5)
- 3 double digit (eg 13,13,14)
The three remaining numbers add to 20. Only two ways to do this:
- 0 double digits (eg. 6,6,8)
- 1 double digit (eg. 10,5,5,)
In short from these combinations you can have more than half in double digits or just 1 double digit (or in fact anything in between). NOT sufficient.
Statements together)
Still not sufficient because the above examples could include a single 10.
ANS = E
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Re: Lottery Ticket [#permalink]
17 Oct 2009, 11:01
yangsta8 wrote: Statement 2)
Three numbers add to 40. The number of ways to do this could be:
- 1 double digit (eg. 38,1,1)
- 2 double digit (eg. 10,25,5)
- 3 double digit (eg 13,13,14)
The three remaining numbers add to 20. Only two ways to do this:
- 0 double digits (eg. 6,6,8)
- 1 double digit (eg. 10,5,5,)
In short from these combinations you can have more than half in double digits or just 1 double digit (or in fact anything in between). NOT sufficient.
Statements together)
Still not sufficient because the above examples could include a single 10.
ANS = E Nice way to break down the Statement 2! Cheers.
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Re: Lottery Ticket [#permalink]
18 Oct 2009, 11:28
There is two built in results.
1)If this is a lottery ticket; I think problem implies that the numbers are different.
2)the sum of numbers is 60.
Problem asks that 3 or more numbers are 2 digited?
Then lets look at to statements.
1)one of the numbers is 10.
So the others sum to 50. The numbers can be 1,2,3,4,40 or 5,9,11,12,13 So insuff.
2)3 of numbers sums to 40. This is same as the other 3 sums to 20. This is insuff too. Because one can choose numbers like that.
1,2,37,5,7,8 or 10, 11, 19 and 12, 5, 3. so insuff.
together
we know that one of the numbers is 10. But there are 2 possibilities. 10 may be in the numbers that sum to 40 or not.
in first possibility, lets assume 10 in the first three, the other two numbers sum to 30. At least one of the other two is two digited. But it is not mandatory that two of them are two digited. But we know that at least 2 of 3 are two digited (10 and one of the other two). Lets look at the 3 other numbers that sum to 20. There need not be a number that is greater than 10. So insuff.
E
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Re: For a lottery ticket, Emily chose six numbers that average [#permalink]
23 Jan 2013, 05:52
badgerboy wrote: For a lottery ticket, Emily chose six numbers that average to 10. DId more than half the numbers have 2 digits? A] One of the numbers was 10. b} Three of the numbers added up to 40. 1. 10 + 1 + 2 + 3 + 4 + 40 = 60, or... (2 two-digits) 10 + 11 + 12 + 13 + 8 + 6 = 60 ... (more than 3 two-digits) INSUFFICIENT! 2. 10 + 11 + 19 + 14 + 6 = 60 (more than 3 two-digits) 10 + 21 + 9 + 9 + 11 = (not more than 3 ) INSUFFICIENT! Together (1) and (2) Samples of statement 2 shows still that it's insufficient Answer: E
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Re: For a lottery ticket, Emily chose six numbers that average
[#permalink]
23 Jan 2013, 05:52
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