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James made 126 ornaments and put them all in boxes. If each

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James made 126 ornaments and put them all in boxes. If each

by II » Thu Jul 03, 2008 3:12 pm
James made 126 ornaments and put them all in boxes. If each box contained either 6 ornaments or 24 ornaments, how many of the boxes contained 24 ornaments ?

(1) Fewer than 4 boxes contained 6 ornaments

(2) More than 3 of the boxes contained 24 ornaments

I am very interested getting views and feedback on the following:
- primary and secondary approaches to solving this. Please illustrate your logic.
- guessing strategies you would use on this type of question, if you were running out of time, or if you simply had to guess on this question
- the traps and tricks which are built into this question (if you can see any)
- any key take-aways, strategies to keep in mind for identiying future similar problems etc.

Thanks in advance.
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Re: James made 126 ornaments and put them all in boxes. If e

by Stuart@KaplanGMAT » Thu Jul 03, 2008 5:39 pm
II wrote:James made 126 ornaments and put them all in boxes. If each box contained either 6 ornaments or 24 ornaments, how many of the boxes contained 24 ornaments ?

(1) Fewer than 4 boxes contained 6 ornaments

(2) More than 3 of the boxes contained 24 ornaments

I am very interested getting views and feedback on the following:
- primary and secondary approaches to solving this. Please illustrate your logic.
- guessing strategies you would use on this type of question, if you were running out of time, or if you simply had to guess on this question
- the traps and tricks which are built into this question (if you can see any)
- any key take-aways, strategies to keep in mind for identiying future similar problems etc.

Thanks in advance.
Almost certainly picking #s is going to be quickest, because there aren't going to be many numbers that work.

1) at MOST 3 6s. Since 3*6 is 18, which is less than 24, there's only going to be one possible arrangement that adds up to 126: sufficient.

2) at LEAST 4 24s. 4*24 = 96, which leaves 30. Well, that 30 could be 1*24 and 1*6 or just 5*6: insufficient.

(1) is suff, (2) isn't: choose (A)
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by II » Tue Jul 08, 2008 5:57 am
Any other approaches here ?
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by lion147 » Tue Jul 08, 2008 12:32 pm
II wrote:Any other approaches here ?
Well, I got the same answer in a similar fashion, I don't know how different you think it is:

1) You can either have 1, 2 or 3 boxes of six ornaments; if it's 1 box of 6, you have 126-6=120 ornaments left for the boxes of 24, and 120/24 = 5 (i.e. it divides evenly with no remainder).
2 boxes of 6 means 126-(2*6)=114 left over, 114/24 = 4 rem. 18
3 boxes of 6 means 126-(3*6)=108 left over, 108/24 = 4 rem. 12

So, only one solution here will include all ornaments - 1 box of 6 and 5 boxes of 24.

2) So now you can have 4, or 5 boxes of 24 (any more than 5 is too many for 126 ornaments). 4*24=96, so you have 30 ornaments left over. The rest of this is the same as Stuart's solution above.

My solution seems to be the long way around.
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by II » Tue Jul 08, 2008 1:07 pm
My thinking/analysis/approach on this question:

This is a word translation problem. We are going to have to translate the words into algebra.
Note: Follow this approach for these types of questions:
(1) assign variables to the unknown values (be sure to make a note of what each variable represents)
(2) Write Equations
(3) Solve

Question stem tells us:
- There are 126 ornaments which are put into boxes
- There are 2 types of boxes:
(i) boxes which hold 6 ornaments – let this be x
(ii) boxes which hold 24 ornaments – let this be y

We can take the above information and create an equation:
6x + 24y = 126

The question is asking for the number of “24 boxes” (boxes which hold 24 ornaments).
Using the above variables … it is asking us to find the value of y. The question can be rephrased as “what is y?”

Now lets look at the statements. Is there an "easy statement" with which we can start ... no not really ... so lets start with statement 1.

Fewer than 4 boxes contain 6 ornaments. This means that x < 4.

So x can be 1, 2, or 3. We can substitute these values of x into the equation “6x + 24y = 126” to find the value of y.

6(1) + 24y = 126 --> 24y = 120 --> y = 5 when x = 1. So x=1 is a possible value.
6(2) + 24y = 126 --> 24y = 114 --> y = 4.75 when x = 2. Number of boxes cannot be a fraction so this cannot be a possible value of x.
6(3) + 24y = 126 --> 24y = 108 --> y = 4.5 when x = 3. As above, number of boxes cannot be a fraction.

So there was 1 box with 6 ornaments, and therefore there had to be 5 boxes with 24 ornaments. SUFFICIENT
Cross out BCE from answer choices. Left with A or D.

Statement 2:
More than 3 boxes contain 24 ornaments. This means y > 3.
y can only be 4 or 5 … if y was 6, then 6*24=144, which is greater than 126 (the total number of ornaments).
As with statement 1, substitute the value of y into the equation:
- 6x + 24(4) = 126 --> 6x = 30 --> x = 5, so y = 4 could be a possible value of y, since this fits into the equation.
- 6x + 24(5) = 126 --> 6x = 6 --> x = 1, so y = 5 could be a possible value of y, since this fits into the equation.

Since we have 2 possible values of y (4,5), the information provided in this statement is INSUFFICIENT

Answer is A.
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by Stuart@KaplanGMAT » Tue Jul 08, 2008 2:03 pm
Your solution is certainly correct, it just seems like a LOT more work.
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by II » Tue Jul 08, 2008 2:18 pm
Stuart Kovinsky wrote:Your solution is certainly correct, it just seems like a LOT more work.
Hi Stuart ...

In my explanation, I wrote down my whole thinking process in detail (so hence a lot of text) ... with practice it is becoming and I think it will become more natural and quicker.

Your approach is certainly quicker, since you get straight to the meat.

I am just trying to follow a more structured approach and building my foundation.

Thanks.
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Re: James made 126 ornaments and put them all in boxes. If e

by II » Sun Aug 17, 2008 5:55 am
Stuart Kovinsky wrote:
1) at MOST 3 6s. Since 3*6 is 18, which is less than 24, there's only going to be one possible arrangement that adds up to 126: sufficient.
Can you please expand on your point regarding statement 1 ...

Separate note:
I suppose you could also simplify the original equation (derived from the q-stem) 6x + 24y = 126.
Divide all terms by 6 and you get x + 4y = 21 ... this is much easier to work with, when plugging in the values for x.

Thanks.
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Re: James made 126 ornaments and put them all in boxes. If e

by Stuart@KaplanGMAT » Tue Aug 19, 2008 10:53 am
II wrote:
Stuart Kovinsky wrote:
1) at MOST 3 6s. Since 3*6 is 18, which is less than 24, there's only going to be one possible arrangement that adds up to 126: sufficient.
Can you please expand on your point regarding statement 1 ...

[Thanks.
Think about making change with nothing but nickels (5 cents) and quarters (25 cents).

We are making $1.05 with nothing but nickels and quarters. Here's our question: how many quarters do we have?

Without further information, there are a number of possibilities. However, we know that every combination will have at least 1 nickel, since that's the only way to make the extra 20 cents.

(1) There are at most 4 nickels.

Well, 4 nickels is less than 1 quarter. So, if there are at most 4 nickels, the number of nickels will have no impact on the number of quarters: sufficient.

Let's contrast that with:

(2) There are at most 7 nickels.

With 7 nickels, we could be replacing a quarter (i.e. to make 35 cents, you could use 1Q and 2N or 7N). So, (2) wouldn't not allows to figure out the exact number of quarters: insufficient.
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