If 4 students were added to a dance class, would the teacher

Question

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.

Bunuel · Answer

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?

Say # of the students in the class now is x.  The question basically asks whether x+4 is a multiple of 8 , because if it is then the teacher would be able to divide x+4 students evenly into teams of 8.

(1) If 12 students were added, the teacher could divide the students evenly into teams of 8 --> x+12=(x+4)+8 is a multiple of 8, so x+4 is also a multiple of 8. Sufficient.

(2) The number of students in the class is currently not divisible by 8. If x=1 then the answer is NO but if x=12 then the asnwer is YES. Not sufficient.

Answer: A.

vkm · Answer

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.

Bunuel · Answer

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  a-b=-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  a-b=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  a-b=1 , is not divisible by 3;
OR:  a=6  and  b=3 , neither is divisible by 5 --->  a+b=9  and  a-b=3 , neither is divisible by 5;
OR:  a=2  and  b=2 , neither is divisible by 4 --->  a+b=4  and  a-b=0 , both are divisible by 4.
 
Hope it's clear.

Injuin · Answer

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.

Bunuel · Answer

The red part is not correct.

Evenly divisible = divisible.

Injuin · Answer

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.

vkm · Answer

Thanks guys.

I am clear with the explanation.

Sourav199201 · Answer

Consider the no. of students present is X

Now the question is whether (x+4)/8 is integer or not. Y or N?

1. If 12 is added to X it is divided by 8; (x+12)/8 = integer.
 so the values of X = 4,12,20,.... = 4m(m is a odd no.)
 Now using this in basic equation , (4m+4)/8 => (m+1)/2 which is always an integer as m is odd.
 So sufficient.

2. Considering X=1,2,3,4,5,6,7 we cannot find an exact answer as for diff X , (x+4)/8 satisfies and for some it doesn't

Hence A

bumpbot · Answer

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If 4 students were added to a dance class, would the teacher

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