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If 4 students were added to a dance class, would the teacher
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10 May 2012, 05:34
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If 4 students were added to a dance class, would the teacher be able to divide her students evenly into one or more dance teams of 8? (1) If 12 students were added, the teacher could divide the students evenly into teams of 8. (2) The number of students in the class is currently not divisible by 8. Someone please explain this one. I did not got the explanation already present if4studentswereaddedtoadanceclasswouldtheteacher73985.html
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Re: If 4 students were added to a dance class, would the teacher
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10 May 2012, 05:48



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Re: If 4 students were added to a dance class, would the teacher
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23 Aug 2012, 05:36
Hi, I selected E because statement 1 alone is insufficient for all values of X.
From statement 1, if 12 students were added the number of students before lets say = X. i,e 12+x should be divisible by 8. (x = 4, 12,20,..) should be an even number.
Therefore if i add 4 more students as given in question then i cannot divide the dance team in 8 for all values of X.
Please clarify.



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Re: If 4 students were added to a dance class, would the teacher
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23 Aug 2012, 07:49
vkm wrote: Hi, I selected E because statement 1 alone is insufficient for all values of X.
From statement 1, if 12 students were added the number of students before lets say = X. i,e 12+x should be divisible by 8. (x = 4, 12,20,..) should be an even number.
Therefore if i add 4 more students as given in question then i cannot divide the dance team in 8 for all values of X.
Please clarify. First statement says: "if 12 students were added, the teacher could divide the students evenly into teams of 8", which means that x+12 is a multiple of 8. Now, x+12=(x+4)+8=(x+4)+{multiple of 8}. So, we have that the sum of x+4 and some multiple of 8 is a multiple of 8, which means that x+4 must also be a multiple of 8. To elaborate more: If integers \(a\) and \(b\) are both multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\) and \(b=9\), both divisible by 3 > \(a+b=15\) and \(ab=3\), again both divisible by 3. If out of integers \(a\) and \(b\) one is a multiple of some integer \(k>1\) and another is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):Example: \(a=6\), divisible by 3 and \(b=5\), not divisible by 3 > \(a+b=11\) and \(ab=1\), neither is divisible by 3. If integers \(a\) and \(b\) both are NOT multiples of some integer \(k>1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):Example: \(a=5\) and \(b=4\), neither is divisible by 3 > \(a+b=9\), is divisible by 3 and \(ab=1\), is not divisible by 3; OR: \(a=6\) and \(b=3\), neither is divisible by 5 > \(a+b=9\) and \(ab=3\), neither is divisible by 5; OR: \(a=2\) and \(b=2\), neither is divisible by 4 > \(a+b=4\) and \(ab=0\), both are divisible by 4. Hope it's clear.
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Re: If 4 students were added to a dance class, would the teacher
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23 Aug 2012, 08:36
vkm wrote: Hi, I selected E because statement 1 alone is insufficient for all values of X.
From statement 1, if 12 students were added the number of students before lets say = X. i,e 12+x should be divisible by 8. (x = 4, 12,20,..) should be an even number.
Therefore if i add 4 more students as given in question then i cannot divide the dance team in 8 for all values of X.
Please clarify. Evenly divided into =/= divided into even numbers So for the numbers you picked, bare minimum is 4. so is 4+4 divisible by 8? Yes. If 12, 12+4 divisible by 8? Yes. So on and so forth.



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Re: If 4 students were added to a dance class, would the teacher
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23 Aug 2012, 09:15
Injuin wrote: vkm wrote: Hi, I selected E because statement 1 alone is insufficient for all values of X.
From statement 1, if 12 students were added the number of students before lets say = X. i,e 12+x should be divisible by 8. (x = 4, 12,20,..) should be an even number.
Therefore if i add 4 more students as given in question then i cannot divide the dance team in 8 for all values of X.
Please clarify. Evenly divided into =/= divided into even numbersSo for the numbers you picked, bare minimum is 4. so is 4+4 divisible by 8? Yes. If 12, 12+4 divisible by 8? Yes. So on and so forth. The red part is not correct. Evenly divisible = divisible.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If 4 students were added to a dance class, would the teacher
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23 Aug 2012, 09:22
Bunuel wrote: Injuin wrote: vkm wrote: Hi, I selected E because statement 1 alone is insufficient for all values of X.
From statement 1, if 12 students were added the number of students before lets say = X. i,e 12+x should be divisible by 8. (x = 4, 12,20,..) should be an even number.
Therefore if i add 4 more students as given in question then i cannot divide the dance team in 8 for all values of X.
Please clarify. Evenly divided into =/= divided into even numbersSo for the numbers you picked, bare minimum is 4. so is 4+4 divisible by 8? Yes. If 12, 12+4 divisible by 8? Yes. So on and so forth. The red part is not correct. Evenly divisible = divisible. That's what I meant. I can't makes the does not equal sign, but I figured =/= was fine. From what vkm stated, he seemed to be under the impression that x+4/8 would be an even number like 2,4,8 as opposed to being simply divisible.



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Re: If 4 students were added to a dance class, would the teacher
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23 Aug 2012, 11:06
Thanks guys.
I am clear with the explanation.



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